Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: Suppose that $x$ $y$ , and $z$ are complex numbers such that $xy = -80 - 320i$ $yz = 60$ , and $zx = -96 + 24i$ , where $i$ $=$ $\sqrt{-1}$ . Then there are real numbers $a$ and $b$ such that $x + y + z = a + bi$ . Find $a^2 + b^2$

Solution: The First (pun intended) thing to notice is that $xy$ and $zx$ have a similar structure, but not exactly conjugates, but instead once you take out the magnitudes of both, simply multiples of a root of unity. It turns out that root of unity is $e^{\frac{3\pi i}{2}}$ . Anyway this results in getting that $\left(\frac{-3i}{10}\right)y=z$ . Then substitute this into $yz$ to get, after some calculation, that $y=10e^{\frac{5\pi i}{4}}\sqrt{2}$ and $z=-3e^{\frac{7\pi i}{4}}\sqrt{2}$ . Then plug $z$ into $zx$ , you could do the same thing with $xy$ but $zx$ looks like it's easier due to it being smaller. Anyway you get $x=20+12i$ . Then add all three up, it turns out easier than it seems because for $z$ and $y$ the $\sqrt{2}$ disappears after you expand the root of unity (e raised to a specific power). Long story short, you get $x=20+12i, y=-3+3i, z=-10-10i \implies x+y+z=7+5i \implies a^2+b^2=\boxed{074}$

Answer:  074
Problem: Suppose that $x$ $y$ , and $z$ are complex numbers such that $xy = -80 - 320i$ $yz = 60$ , and $zx = -96 + 24i$ , where $i$ $=$ $\sqrt{-1}$ . Then there are real numbers $a$ and $b$ such that $x + y + z = a + bi$ . Find $a^2 + b^2$

Solution: First we evaluate the magnitudes. $|xy|=80\sqrt{17}$ $|yz|=60$ , and $|zx|=24\sqrt{17}$ . Therefore, $|x^2y^2z^2|=17\cdot80\cdot60\cdot24$ , or $|xyz|=240\sqrt{34}$ . Divide to find that $|z|=3\sqrt{2}$ $|x|=4\sqrt{34}$ , and $|y|=10\sqrt{2}$ [asy] draw((0,0)--(4,0)); dot((4,0),red); draw((0,0)--(-4,0)); draw((0,0)--(0,-4)); draw((0,0)--(-4,1)); dot((-4,1),red); draw((0,0)--(-1,-4)); dot((-1,-4),red); draw((0,0)--(4,4),red); draw((0,0)--(4,-4),red); [/asy] This allows us to see that the argument of $y$ is $\frac{\pi}{4}$ , and the argument of $z$ is $-\frac{\pi}{4}$ . We need to convert the polar form to a standard form. Simple trig identities show $y=10+10i$ and $z=3-3i$ . More division is needed to find what $x$ is. \[x=-20-12i\] \[x+y+z=-7-5i\] \[(-7)^2+(-5)^2=\boxed{74}\] \[QED\blacksquare\] Written by a1b2

Answer:  74
Problem: Suppose that $x$ $y$ , and $z$ are complex numbers such that $xy = -80 - 320i$ $yz = 60$ , and $zx = -96 + 24i$ , where $i$ $=$ $\sqrt{-1}$ . Then there are real numbers $a$ and $b$ such that $x + y + z = a + bi$ . Find $a^2 + b^2$

Solution: Solve this system the way you would if the RHS of all equations were real. Multiply the first and 3rd equations out and then factor out $60$ to find $x^2$ , then use standard techniques that are used to evaluate square roots of irrationals. let \[x = c+di\] , then you get \[c^2 - d^2 = 256\] and \[2cd = 480\] Solve to get $x$ as $20+12i$ and $-20-12i$ . Both will give us the same answer, so use the positive one. Divide $-80-320i$ by $x$ , and you get $10+10i$ as $y$ . This means that $z$ is a multiple of $1-i$ to get a real product, so you find $z$ is $3-3i$ . Now, add the real and imaginary parts separately to get $-7-5i$ , and calculate $a^2 + b^2$ to get $\boxed{74}$ . 
~minor latex improvements done by jske25 and jdong2006

Answer:  74
Problem: Suppose that $x$ $y$ , and $z$ are complex numbers such that $xy = -80 - 320i$ $yz = 60$ , and $zx = -96 + 24i$ , where $i$ $=$ $\sqrt{-1}$ . Then there are real numbers $a$ and $b$ such that $x + y + z = a + bi$ . Find $a^2 + b^2$

Solution: Dividing the first equation by the second equation given, we find that $\frac{xy}{yz}=\frac{x}{z}=\frac{-80-320i}{60}=-\frac{4}{3}-\frac{16}{3}i \implies x=z\left(-\frac{4}{3}-\frac{16}{3}i\right)$ . Substituting this into the third equation, we get $z^2=\frac{-96+24i}{-\frac{4}{3}-\frac{16}{3}i}=3\cdot \frac{-24+6i}{-1-4i}=3\cdot \frac{(-24+6i)(-1+4i)}{1+16}=3\cdot \frac{-102i}{17}=-18i$ . Taking the square root of this is equivalent to halving the argument and taking the square root of the magnitude. Furthermore, the second equation given tells us that the argument of $y$ is the negative of that of $z$ , and their magnitudes multiply to $60$ . Thus, we have $z=\sqrt{-18i}=3-3i$ and $3\sqrt{2}\cdot |y|=60 \implies |y|=10\sqrt{2} \implies y=10+10i$ . To find $x$ , we can use the previous substitution we made to find that $x=z\left(-\frac{4}{3}-\frac{16}{3}i\right)=-\frac{4}{3}\cdot (3-3i)(1+4i)=-4(1-i)(1+4i)=-4(5+3i)=-20-12i$ .
Therefore, $x+y+z=(-20+10+3)+(-12+10-3)i=-7-5i \implies a^2+b^2=(-7)^2+(-5)^2=49+25=\boxed{74}$

Answer:  74
Problem: Suppose that $x$ $y$ , and $z$ are complex numbers such that $xy = -80 - 320i$ $yz = 60$ , and $zx = -96 + 24i$ , where $i$ $=$ $\sqrt{-1}$ . Then there are real numbers $a$ and $b$ such that $x + y + z = a + bi$ . Find $a^2 + b^2$

Solution: We observe that by multiplying $xy,$ $yz,$ and $zx,$ we get $(xyz)^2=(-80-320i)(60)(-96+24i).$ Next, we divide $(xyz)^2$ by $(yz)^2$ to 
get $x^2.$ We have $x^2=\frac{(-80-320i)(60)(-96+24i)}{3600}=256+480i.$ We can write $x$ in the form of $a+bi,$ so we get $(a+bi)^2=256+480i.$ Then, $a^2-b^2+2abi=256+480i,$ $a^2-b^2=256,$ and $2ab=480.$ Solving this system of equations is relatively 
simple. We have two cases, $a=20, b=12,$ and $a=-20, b=-12.$ Case 1: $a=20, b=12,$ so $x=20+12i.$ We solve for $y$ and $z$ by plugging in $x$ to the two equations. We see $y=\frac{-80-320i}{20+12i}=-10-10i$ and $z=\frac{-96+24i}{20+12i}=-3+3i.$ $x+y+z=7+5i,$ so $a=7$ and $b=5.$ Solving, we end up with $7^2+5^2=\boxed{074}$ as our answer. 
Case 2: $a=-20, b=-12,$ so $x=-20-12i.$ Again, we solve for $y$ and $z.$ We find $y=\frac{-80-320i}{-20-12i}=10+10i,$ $z=\frac{-96+24i}{-20-12i}=3-3i,$ so $x+y+z=-7-5i.$ We again have $(-7)^2+(-5)^2=\boxed{074}.$ Solution by Airplane50

Answer:  074
Problem: Suppose that $x$ $y$ , and $z$ are complex numbers such that $xy = -80 - 320i$ $yz = 60$ , and $zx = -96 + 24i$ , where $i$ $=$ $\sqrt{-1}$ . Then there are real numbers $a$ and $b$ such that $x + y + z = a + bi$ . Find $a^2 + b^2$

Solution: According to the Euler's Theory, we can rewrite $x$ $y$ and $z$ as \[x=r_{1}e^{i{\theta}_1}\] \[y=r_{2}e^{i{\theta}_2}\] \[x=r_{3}e^{i{\theta}_3}\] As a result, \[|xy|=r_{1}r_{2}=\sqrt{80^2+320^2}=80\sqrt{17}\] \[|yz|=r_{2}r_{3}=60\] \[|xz|=r_{1}r_{3}=\sqrt{96^2+24^2}=24\sqrt{17}\] Also, it is clear that \[yz=r_{2}e^{i{\theta}_2}r_{3}e^{i{\theta}_3}=|yz|e^{i({\theta}_2+{\theta}_3)}=|yz|=60\] So ${\theta}_2+{\theta}_3=0$ , or \[{\theta}_2=-{\theta}_3\] Also, we have \[xy=-80\sqrt{17}e^{i\arctan{4}}\] \[yz=60\] \[xz=-24\sqrt{17}e^{i\arctan{-\frac{1}{4}}}\] So now we have $r_{1}r_{2}=80\sqrt{17}$ $r_{2}r_{3}=60$ $r_{1}r_{3}=24\sqrt{17}$ ${\theta}_1+{\theta}_2=\arctan{4}$ and ${\theta}_1-{\theta}_2=\arctan {-\frac{1}{4}}$ . Solve these above, we get \[r_{1}=4\sqrt{34}\] \[r_{2}=10\sqrt{2}\] \[r_{3}=3\sqrt{2}\] \[{\theta}_2=\frac{\arctan{4}-\arctan{-\frac{1}{4}}}{2}=\frac{\frac{\pi}{2}}{2}=\frac{\pi}{4}\] So we can get \[y=r_{2}e^{i{\theta}_2}=10\sqrt{2}e^{i\frac{\pi}{4}}=10+10i\] \[z=r_{3}e^{i{\theta}_3}=r_{3}e^{-i{\theta}_2}=3\sqrt{2}e^{-i\frac{\pi}{4}}=3-3i\] Use $xy=-80-320i$ we can find that \[x=-20-12i\] So \[x+y+z=-20-12i+10+10i+3-3i=-7-5i\] So we have $a=-7$ and $b=-5$
As a result, we finally get \[a^2+b^2=(-7)^2+(-5)^2=\boxed{074}\]

Answer:  074
Problem: Suppose that $x$ $y$ , and $z$ are complex numbers such that $xy = -80 - 320i$ $yz = 60$ , and $zx = -96 + 24i$ , where $i$ $=$ $\sqrt{-1}$ . Then there are real numbers $a$ and $b$ such that $x + y + z = a + bi$ . Find $a^2 + b^2$

Solution: We can turn the expression $x+y+z$ into $\sqrt{x^2+y^2+z^2+2xy+2yz+2xz}$ , and this would allow us to plug in the values after some computations.
Based off of the given products, we have
$xy^2z=60(-80-320i)$
$xyz^2=60(-96+24i)$
$x^2yz=(-96+24i)(-80-320i)$
Dividing by the given products, we have
$y^2=\frac{60(-80-320i)}{-96+24i}$
$z^2=\frac{60(-96+24i)}{-80-320i}$
$x^2=\frac{(-96+24i)(-80-320i)}{60}$
Simplifying, we get that this expression becomes $\sqrt{24+70i}$ . This equals $\pm{(7+5i)}$ , so the answer is $7^2+5^2=\boxed{74}$

Answer:  74
Problem: Suppose that $x$ $y$ , and $z$ are complex numbers such that $xy = -80 - 320i$ $yz = 60$ , and $zx = -96 + 24i$ , where $i$ $=$ $\sqrt{-1}$ . Then there are real numbers $a$ and $b$ such that $x + y + z = a + bi$ . Find $a^2 + b^2$

Solution: Multiplying $xy \cdot yz \cdot zx = (xyz)^2$ we obtain $60 \cdot 960(16+30i)$ (too lazy to do $60 \cdot 960$ , you don't need to). Taking the square root, we get $240\sqrt{16+30i}$ . Letting $(a+bi)^2=16+30i,$ we have $a^2+2abi-b^2=16+30i.$ Thus, $(a+b)(a-b)=16,$ and $2ab=30.$ Guessing and checking, we get $a+bi=5+3i$ . Therefore, $xyz=240(5+3i).$ Dividing this by each of the equations provided in the original problem, we get $x=20+12i,y=-10-10i,$ and $z=-3+3i$ $20+12i-10-10i-3+3i=7+5i$ . Finally, $7^2+5^2=\boxed{074}.$

Answer:  074
Problem: A real number $a$ is chosen randomly and uniformly from the interval $[-20, 18]$ . The probability that the roots of the polynomial
$x^4 + 2ax^3 + (2a - 2)x^2 + (-4a + 3)x - 2$
are all real can be written in the form $\dfrac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$

Solution: The polynomial we are given is rather complicated, so we could use Rational Root Theorem to turn the given polynomial into a degree-2 polynomial. With Rational Root Theorem, $x = 1, -1, 2, -2$ are all possible rational roots. Upon plugging these roots into the polynomial, $x = -2$ and $x = 1$ make the polynomial equal 0 and thus, they are roots that we can factor out.
The polynomial becomes:
$(x - 1)(x + 2)(x^2 + (2a - 1)x + 1)$
Since we know $1$ and $-2$ are real numbers, we only need to focus on the quadratic.
We should set the discriminant of the quadratic greater than or equal to 0.
$(2a - 1)^2 - 4 \geq 0$
This simplifies to:
$a \geq \dfrac{3}{2}$
or
$a \leq -\dfrac{1}{2}$
This means that the interval $\left(-\dfrac{1}{2}, \dfrac{3}{2}\right)$ is the "bad" interval. The length of the interval where $a$ can be chosen from is 38 units long, while the bad interval is 2 units long. Therefore, the "good" interval is 36 units long.
$\dfrac{36}{38} = \dfrac{18}{19}$
$18 + 19 = \boxed{037}$

Answer:  037
Problem: Triangle $ABC$ has side lengths $AB = 9$ $BC =$ $5\sqrt{3}$ , and $AC = 12$ . Points $A = P_{0}, P_{1}, P_{2}, ... , P_{2450} = B$ are on segment $\overline{AB}$ with $P_{k}$ between $P_{k-1}$ and $P_{k+1}$ for $k = 1, 2, ..., 2449$ , and points $A = Q_{0}, Q_{1}, Q_{2}, ... , Q_{2450} = C$ are on segment $\overline{AC}$ with $Q_{k}$ between $Q_{k-1}$ and $Q_{k+1}$ for $k = 1, 2, ..., 2449$ . Furthermore, each segment $\overline{P_{k}Q_{k}}$ $k = 1, 2, ..., 2449$ , is parallel to $\overline{BC}$ . The segments cut the triangle into $2450$ regions, consisting of $2449$ trapezoids and $1$ triangle. Each of the $2450$ regions has the same area. Find the number of segments $\overline{P_{k}Q_{k}}$ $k = 1, 2, ..., 2450$ , that have rational length.

Solution: For each $k$ between $2$ and $2450$ , the area of the trapezoid with $\overline{P_kQ_k}$ as its bottom base is the difference between the areas of two triangles, both similar to $\triangle{ABC}$ . Let $d_k$ be the length of segment $\overline{P_kQ_k}$ . The area of the trapezoid with bases $\overline{P_{k-1}Q_{k-1}}$ and $P_kQ_k$ is $\left(\frac{d_k}{5\sqrt{3}}\right)^2 - \left(\frac{d_{k-1}}{5\sqrt{3}}\right)^2 = \frac{d_k^2-d_{k-1}^2}{75}$ times the area of $\triangle{ABC}$ . (This logic also applies to the topmost triangle if we notice that $d_0 = 0$ .) However, we also know that the area of each shape is $\frac{1}{2450}$ times the area of $\triangle{ABC}$ . We then have $\frac{d_k^2-d_{k-1}^2}{75} = \frac{1}{2450}$ . Simplifying, $d_k^2-d_{k-1}^2 = \frac{3}{98}$ . However, we know that $d_0^2 = 0$ , so $d_1^2 = \frac{3}{98}$ , and in general, $d_k^2 = \frac{3k}{98}$ and $d_k = \frac{\sqrt{\frac{3k}{2}}}{7}$ . The smallest $k$ that gives a rational $d_k$ is $6$ , so $d_k$ is rational if and only if $k = 6n^2$ for some integer $n$ .The largest $n$ such that $6n^2$ is less than $2450$ is $20$ , so $k$ has $\boxed{020}$ possible values.

Answer:  020
Problem: Triangle $ABC$ has side lengths $AB = 9$ $BC =$ $5\sqrt{3}$ , and $AC = 12$ . Points $A = P_{0}, P_{1}, P_{2}, ... , P_{2450} = B$ are on segment $\overline{AB}$ with $P_{k}$ between $P_{k-1}$ and $P_{k+1}$ for $k = 1, 2, ..., 2449$ , and points $A = Q_{0}, Q_{1}, Q_{2}, ... , Q_{2450} = C$ are on segment $\overline{AC}$ with $Q_{k}$ between $Q_{k-1}$ and $Q_{k+1}$ for $k = 1, 2, ..., 2449$ . Furthermore, each segment $\overline{P_{k}Q_{k}}$ $k = 1, 2, ..., 2449$ , is parallel to $\overline{BC}$ . The segments cut the triangle into $2450$ regions, consisting of $2449$ trapezoids and $1$ triangle. Each of the $2450$ regions has the same area. Find the number of segments $\overline{P_{k}Q_{k}}$ $k = 1, 2, ..., 2450$ , that have rational length.

Solution: We have that there are $2449$ trapezoids and $1$ triangle of equal area, with that one triangle being $AP_1Q_1$ . Notice, if we "stack" the trapezoids on top of $\bigtriangleup AP_1Q_1$ the way they already are, we'd create a similar triangle, all of which are similar to $\bigtriangleup ABC$ , and since the trapezoids and $\bigtriangleup AP_1Q_1$ have equal area, each of these similar triangles $AP_kQ_k$ have area $\frac{k}{2450}\left[ ABC\right]$ , and so $\frac{\left[ AP_kQ_k\right]}{\left[ABC\right]}=\frac{k}{2450}$ . We want the ratio of the side lengths $P_kQ_k:BC$ . Since area is a 2-dimensional unit of measurement, and side lengths are 1-dimensional, the ratio is simply the square root of the areas, or \[\frac{P_kQ_k}{BC}=\sqrt{\frac{k}{2450}}\] \[\implies P_kQ_k=BC\cdot \sqrt{\frac{k}{2450}}=5\sqrt{3}\cdot\sqrt{\frac{k}{2450}}=\frac{1}{7}\cdot \sqrt{\frac{3k}{2}}=\frac{3}{7}\sqrt{\frac{k}{6}}\] \[\implies k=6n^2<2450\] \[\implies 0<n\leq 20\] so there are $\boxed{020}$ solutions.

Answer:  020
Problem: Triangle $ABC$ has side lengths $AB = 9$ $BC =$ $5\sqrt{3}$ , and $AC = 12$ . Points $A = P_{0}, P_{1}, P_{2}, ... , P_{2450} = B$ are on segment $\overline{AB}$ with $P_{k}$ between $P_{k-1}$ and $P_{k+1}$ for $k = 1, 2, ..., 2449$ , and points $A = Q_{0}, Q_{1}, Q_{2}, ... , Q_{2450} = C$ are on segment $\overline{AC}$ with $Q_{k}$ between $Q_{k-1}$ and $Q_{k+1}$ for $k = 1, 2, ..., 2449$ . Furthermore, each segment $\overline{P_{k}Q_{k}}$ $k = 1, 2, ..., 2449$ , is parallel to $\overline{BC}$ . The segments cut the triangle into $2450$ regions, consisting of $2449$ trapezoids and $1$ triangle. Each of the $2450$ regions has the same area. Find the number of segments $\overline{P_{k}Q_{k}}$ $k = 1, 2, ..., 2450$ , that have rational length.

Solution: Let $T_1$ stand for $AP_1Q_1$ , and $T_k = AP_kQ_k$ . All triangles $T$ are similar by AA. Let the area of $T_1$ be $x$ . The next trapezoid will also have an area of $x$ , as given. Therefore, $T_k$ has an area of $kx$ . The ratio of the areas is equal to the square of the scale factor for any plane figure and its image. Therefore, $P_k Q_k=P_1 Q_1\cdot \sqrt{k}$ , and the same if $Q$ is substituted for $P$ throughout. We want the side $P_k Q_k$ to be rational. Setting up proportions: \[5\sqrt{3} : \sqrt{2450}=35\sqrt{2}\] \[\sqrt{6} : 14\] which shows that $P_1 Q_1=\frac{\sqrt{6}}{14}$ . In order for $\sqrt{k} P_1 Q_1$ to be rational, $\sqrt{k}$ must be some rational multiple of $\sqrt{6}$ . This is achieved at $\sqrt{k}=\sqrt{6}, 2\sqrt{6}, \ldots, 20\sqrt{6}$ . We end there as $21\sqrt{6}=\sqrt{2646}$ . There are 20 numbers from 1 to 20, so there are $\boxed{020}$ solutions.

Answer:  020
Problem: A frog is positioned at the origin of the coordinate plane. From the point $(x, y)$ , the frog can jump to any of the points $(x + 1, y)$ $(x + 2, y)$ $(x, y + 1)$ , or $(x, y + 2)$ . Find the number of distinct sequences of jumps in which the frog begins at $(0, 0)$ and ends at $(4, 4)$

Solution: We solve this problem by working backwards. Notice, the only points the frog can be on to jump to $(4,4)$ in one move are $(2,4),(3,4),(4,2),$ and $(4,3)$ . This applies to any other point, thus we can work our way from $(0,0)$ to $(4,4)$ , recording down the number of ways to get to each point recursively.
$(0,0): 1$
$(1,0)=(0,1)=1$
$(2,0)=(0, 2)=2$
$(3,0)=(0, 3)=3$
$(4,0)=(0, 4)=5$
$(1,1)=2$ $(1,2)=(2,1)=5$ $(1,3)=(3,1)=10$ $(1,4)=(4,1)= 20$
$(2,2)=14, (2,3)=(3,2)=32, (2,4)=(4,2)=71$
$(3,3)=84, (3,4)=(4,3)=207$
$(4,4)=2\cdot \left( (4,2)+(4,3)\right) = 2\cdot \left( 207+71\right)=2\cdot 278=\boxed{556}$

Answer:  556
Problem: A frog is positioned at the origin of the coordinate plane. From the point $(x, y)$ , the frog can jump to any of the points $(x + 1, y)$ $(x + 2, y)$ $(x, y + 1)$ , or $(x, y + 2)$ . Find the number of distinct sequences of jumps in which the frog begins at $(0, 0)$ and ends at $(4, 4)$

Solution: We'll refer to the moves $(x + 1, y)$ $(x + 2, y)$ $(x, y + 1)$ , and $(x, y + 2)$ as $R_1$ $R_2$ $U_1$ , and $U_2$ , respectively. Then the possible sequences of moves that will take the frog from $(0,0)$ to $(4,4)$ are all the permutations of $U_1U_1U_1U_1R_1R_1R_1R_1$ $U_2U_1U_1R_1R_1R_1R_1$ $U_1U_1U_1U_1R_2R_1R_1$ $U_2U_1U_1R_2R_1R_1$ $U_2U_2R_1R_1R_1R_1$ $U_1U_1U_1U_1R_2R_2$ $U_2U_2R_2R_1R_1$ $U_2U_1U_1R_2R_2$ , and $U_2U_2R_2R_2$ . We can reduce the number of cases using symmetry.
Case 1: $U_1U_1U_1U_1R_1R_1R_1R_1$
There are $\frac{8!}{4!4!} = 70$ possibilities for this case.
Case 2: $U_2U_1U_1R_1R_1R_1R_1$ or $U_1U_1U_1U_1R_2R_1R_1$
There are $2 \cdot \frac{7!}{4!2!} = 210$ possibilities for this case.
Case 3: $U_2U_1U_1R_2R_1R_1$
There are $\frac{6!}{2!2!} = 180$ possibilities for this case.
Case 4: $U_2U_2R_1R_1R_1R_1$ or $U_1U_1U_1U_1R_2R_2$
There are $2 \cdot \frac{6!}{2!4!} = 30$ possibilities for this case.
Case 5: $U_2U_2R_2R_1R_1$ or $U_2U_1U_1R_2R_2$
There are $2 \cdot \frac{5!}{2!2!} = 60$ possibilities for this case.
Case 6: $U_2U_2R_2R_2$
There are $\frac{4!}{2!2!} = 6$ possibilities for this case.
Adding up all these cases gives us $70+210+180+30+60+6=\boxed{556}$ ways.

Answer:  556
Problem: A frog is positioned at the origin of the coordinate plane. From the point $(x, y)$ , the frog can jump to any of the points $(x + 1, y)$ $(x + 2, y)$ $(x, y + 1)$ , or $(x, y + 2)$ . Find the number of distinct sequences of jumps in which the frog begins at $(0, 0)$ and ends at $(4, 4)$

Solution: Mark the total number of distinct sequences of jumps for the frog to reach the point $(x,y)$ as $\varphi (x,y)$ . Consider for each point $(x,y)$ in the first quadrant, there are only $4$ possible points in the first quadrant for frog to reach point $(x,y)$ , and these $4$ points are \[(x-1,y); (x-2,y); (x,y-1); (x,y-2)\] . As a result, the way to count $\varphi (x,y)$ is \[\varphi (x,y)=\varphi (x-1,y)+\varphi (x-2,y)+\varphi (x,y-1)+\varphi (x,y-2)\]
Also, for special cases, \[\varphi (0,y)=\varphi (0,y-1)+\varphi (0,y-2)\]
\[\varphi (x,0)=\varphi (x-1,0)+\varphi (x-2,0)\]
\[\varphi (x,1)=\varphi (x-1,1)+\varphi (x-2,1)+\varphi (x,0)\]
\[\varphi (1,y)=\varphi (1,y-1)+\varphi (1,y-2)+\varphi (0,y)\]
\[\varphi (1,1)=\varphi (1,0)+\varphi (0,1)\]
Start with $\varphi (0,0)=1$ , use this method and draw the figure below, we can finally get \[\varphi (4,4)=556\] (In order to make the LaTeX thing more beautiful to look at, I put $0$ to make every number $3$ digits)
\[005-020-071-207-\boxed{556}\] \[003-010-032-084-207\] \[002-005-014-032-071\] \[001-002-005-010-020\] \[001-001-002-003-005\]

Answer:  556
Problem: A frog is positioned at the origin of the coordinate plane. From the point $(x, y)$ , the frog can jump to any of the points $(x + 1, y)$ $(x + 2, y)$ $(x, y + 1)$ , or $(x, y + 2)$ . Find the number of distinct sequences of jumps in which the frog begins at $(0, 0)$ and ends at $(4, 4)$

Solution: Casework Solution:
x-distribution: 1-1-1-1 (1 way to order)
y-distribution: 1-1-1-1 (1 way to order) $\dbinom{8}{4} = 70$ ways total
x-distribution: 1-1-1-1 (1 way to order)
y-distribution: 1-1-2 (3 ways to order) $\dbinom{7}{3} \times 3= 105$ ways total
x-distribution: 1-1-1-1 (1 way to order)
y-distribution: 2-2 (1 way to order) $\dbinom{6}{4} = 15$ ways total
x-distribution: 1-1-2 (3 ways to order)
y-distribution: 1-1-1-1 (1 way to order) $\dbinom{7}{3} \times 3= 105$ ways total
x-distribution: 1-1-2 (3 ways to order)
y-distribution: 1-1-2 (3 ways to order) $\dbinom{6}{3} \times 9= 180$ ways total
x-distribution: 1-1-2 (3 ways to order)
y-distribution: 2-2 (1 way to order) $\dbinom{5}{3} \times 3 = 30$ ways total
x-distribution: 2-2 (1 way to order)
y-distribution: 1-1-1-1 (1 way to order) $\dbinom{6}{4} = 15$ ways total
x-distribution: 2-2 (1 way to order)
y-distribution: 1-1-2 (3 ways to order) $\dbinom{5}{3} \times 3 = 30$ ways total
x-distribution: 2-2 (1 way to order)
y-distribution: 2-2 (1 way to order) $\dbinom{4}{2} = 6$ ways total
$6+30+15+105+180+70+30+15+105=\boxed{556}$ -fidgetboss_4000

Answer:  556
Problem: Octagon $ABCDEFGH$ with side lengths $AB = CD = EF = GH = 10$ and $BC = DE = FG = HA = 11$ is formed by removing 6-8-10 triangles from the corners of a $23$ $\times$ $27$ rectangle with side $\overline{AH}$ on a short side of the rectangle, as shown. Let $J$ be the midpoint of $\overline{AH}$ , and partition the octagon into 7 triangles by drawing segments $\overline{JB}$ $\overline{JC}$ $\overline{JD}$ $\overline{JE}$ $\overline{JF}$ , and $\overline{JG}$ . Find the area of the convex polygon whose vertices are the centroids of these 7 triangles.
[asy] unitsize(6); pair P = (0, 0), Q = (0, 23), R = (27, 23), SS = (27, 0); pair A = (0, 6), B = (8, 0), C = (19, 0), D = (27, 6), EE = (27, 17), F = (19, 23),  G = (8, 23), J = (0, 23/2), H = (0, 17); draw(P--Q--R--SS--cycle); draw(J--B); draw(J--C); draw(J--D); draw(J--EE); draw(J--F); draw(J--G); draw(A--B); draw(H--G); real dark = 0.6; filldraw(A--B--P--cycle, gray(dark)); filldraw(H--G--Q--cycle, gray(dark)); filldraw(F--EE--R--cycle, gray(dark)); filldraw(D--C--SS--cycle, gray(dark)); dot(A); dot(B); dot(C); dot(D); dot(EE); dot(F); dot(G); dot(H); dot(J); dot(H); defaultpen(fontsize(10pt)); real r = 1.3; label("$A$", A, W*r); label("$B$", B, S*r); label("$C$", C, S*r); label("$D$", D, E*r); label("$E$", EE, E*r); label("$F$", F, N*r); label("$G$", G, N*r); label("$H$", H, W*r); label("$J$", J, W*r); [/asy]

Solution: Draw the heptagon whose vertices are the midpoints of octagon $ABCDEFGH$ except $J$ .
We have a homothety since:
1. $J$ passes through corresponding vertices of the two heptagons.
2. By centroid properties, our ratio between the sidelengths is $\frac{2}{3},$ and their area ratio is hence $\frac{4}{9}.$
Compute the area of the large heptagon by dividing into two congruent trapezoids and a triangle. The area of each trapezoid is \[= \frac{1}{2}(17+23)\cdot \frac{19}{2}=190.\] The area of each triangle is \[=  \frac{1}{2}\cdot 17\cdot 4=34.\]
Hence, the area of the large heptagon is \[2\cdot 190+34=414.\] Then, from our homothety, the area of the required heptagon is \[\frac{4}{9}\cdot 414=\boxed{184}.\] ~novus677

Answer:  184
Problem: Find the number of functions $f(x)$ from $\{1, 2, 3, 4, 5\}$ to $\{1, 2, 3, 4, 5\}$ that satisfy $f(f(x)) = f(f(f(x)))$ for all $x$ in $\{1, 2, 3, 4, 5\}$

Solution: We perform casework on the number of fixed points (the number of points where $f(x) = x$ ).  To better visualize this, use the grid from Solution 1.
Case 1: 5 fixed points
Case 2: 4 fixed points
Case 3: 3 fixed points
Case 4: 2 fixed points
Case 5: 1 fixed point
Therefore, the answer is $1+20+150+380+205 = \boxed{756}$

Answer:  756