Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: Point $P$ lies on the diagonal $AC$ of square $ABCD$ with $AP > CP$ . Let $O_{1}$ and $O_{2}$ be the circumcenters of triangles $ABP$ and $CDP$ respectively. Given that $AB = 12$ and $\angle O_{1}PO_{2} = 120^{\circ}$ , then $AP = \sqrt{a} + \sqrt{b}$ , where $a$ and $b$ are positive integers. Find $a + b$

Solution: Reflect $O_1$ across $AP$ to $O_1'$ . By symmetry $O_1’$ is the circumcenter of $\triangle{ADP}$
$\angle{DO_1’P}$ $2*\angle{DAP} = 90^\circ$ , so $\angle{O_1’PD}=45^\circ$
similarly $\angle{DO_2P}$ $2*\angle{DCP} = 90^\circ$ , so $\angle{O_2PD}=45^\circ$
Therefore $\angle{O_1’PO_2}=90^\circ$ , so that $\angle{O_1’PO_1} =120^\circ - 90^\circ = 30^\circ$
By symmetry, $\angle{O_1'PA} = \angle{APO_1} = 0.5*\angle{O_1’PO_1} = 15^\circ$
Therefore, since $O_1’$ is the circumcenter of $\triangle{ADP}$ $\angle{ADP}$ $0.5*(180^\circ - 2*\angle{O_1'PA}) = 75^\circ$
Therefore $\angle{APD} = 180^\circ - 45^\circ - 75^\circ = 60^\circ$
Using sine rule in $\triangle{ADP}$ $AP = (12 * \sin 75^\circ) / \sin 60^\circ =\sqrt{72}+\sqrt{24}$ , and the answer is $72+24=\boxed{096}.$

Answer:  096
Problem: Point $P$ lies on the diagonal $AC$ of square $ABCD$ with $AP > CP$ . Let $O_{1}$ and $O_{2}$ be the circumcenters of triangles $ABP$ and $CDP$ respectively. Given that $AB = 12$ and $\angle O_{1}PO_{2} = 120^{\circ}$ , then $AP = \sqrt{a} + \sqrt{b}$ , where $a$ and $b$ are positive integers. Find $a + b$

Solution: Why not use coordinates? After all, 45 degrees is rather friendly in terms of ordered-pair representation! We can set $A=(0, 12)$ $B=(12,12)$ $C=(12, 0)$ $D=(0, 0)$ . Let this $P=(a, 12-a)$ for some $a$
We also know that the circumcenter is the intersection of all perpendicular bisectors of sides, but two will suffice also due to this property. Therefore, we see that $O_{1}$ is the intersection of $x=6$ and, knowing the midpoint of $AP$ to be $(\frac{a}{2}, \frac{12-a}{2})$ and thus the equation to be $y=x+(12-a)$ , we get $(6, 18-a)$ . Likewise for $O_{2}$ it's $(6, 6-a)$ . Now what do we see? $O_{1}P=O_{2}P$ (just look at the coordinates)! So both of those distances are $4\sqrt{3}$ . Solving for $a$ we get it to be $6+2\sqrt{3}$ , since $AP>CP$ . Multiply by $\sqrt{2}$ because we are looking for $AP$ to get the answer of $\boxed{096}$

Answer:  096
Problem: Point $P$ lies on the diagonal $AC$ of square $ABCD$ with $AP > CP$ . Let $O_{1}$ and $O_{2}$ be the circumcenters of triangles $ABP$ and $CDP$ respectively. Given that $AB = 12$ and $\angle O_{1}PO_{2} = 120^{\circ}$ , then $AP = \sqrt{a} + \sqrt{b}$ , where $a$ and $b$ are positive integers. Find $a + b$

Solution: Let $\angle APD = \theta$ . Then $\angle ADP = 180^{\circ}-45^{\circ}-\theta=135^{\circ}-\theta \implies \angle PDC=\theta-45^{\circ}$ .
Realize that because $O_1$ is a circumcenter, $\angle DO_1P=\angle DCP=45^{\circ} \implies \angle DPO_1=\frac{180^{\circ}-\angle DO_1P}{2}=45^{\circ}$ . Then $\angle O_2PA=75^{\circ}-\theta \implies \angle AOP = 180^{\circ}-2\angle O_2PA = 2\theta+30^{\circ} \implies \angle ABP=\theta + 15^{\circ} \implies \angle PBC=75-\theta$ .
Now, because $P$ lies on diagonal $AC$ $\triangle PDC \cong \triangle PBC \implies \angle PDC = \angle PBC \implies \theta-45^{\circ}=75^{\circ}-\theta \implies \theta = 60^{\circ}$ .
To finish, we look at $\triangle ADP$ . Drop a perpendicular from $D$ to $AP$ at $E$ . Then $\triangle ADE$ is a $45-45-90$ and $\triangle PDE$ is a $30-60-90$ . Therefore, $DE=EA=6\sqrt{2}, EP=2\sqrt{6}$ , so $AP=AE+EP=6\sqrt{2}+2\sqrt{6}=\sqrt{72}+\sqrt{24} \implies \boxed{096}$ $\blacksquare$ ~msc

Answer:  096
Problem: Point $P$ lies on the diagonal $AC$ of square $ABCD$ with $AP > CP$ . Let $O_{1}$ and $O_{2}$ be the circumcenters of triangles $ABP$ and $CDP$ respectively. Given that $AB = 12$ and $\angle O_{1}PO_{2} = 120^{\circ}$ , then $AP = \sqrt{a} + \sqrt{b}$ , where $a$ and $b$ are positive integers. Find $a + b$

Solution: Both $O_1$ and $O_2$ lie on the perpendicular bisector of $AB$
Claim: $O_1O_2=12$ and $O_1P=O_2P$
Proof. Translate $O_1$ and $P$ $12$ units down, and let their images be $O_1'$ and $P'$ , respectively. Note that $\triangle ABP\cong\triangle DCP'$ . Additionally, \[\angle CP'D = \angle BPA = 180^{\circ} - \angle BPC = 180^{\circ} - \angle CPD,\] so $CPDP'$ is cyclic. This means $O_1'$ and $O_2$ coincide, so $O_1O_2=12$ . This also means the circumradii of both triangles are equal, so $O_1P=O_2P$ $\blacksquare$
Let the perpendicular from $P$ intersect $O_1O_2$ at $X$ and $AD$ at $Y$ . Since $\triangle O_1XP$ is 30-60-90, $XP=\frac{6}{\sqrt{3}} = 2\sqrt3$ . Since $YX=6$ $PY=6+2\sqrt3$ , so $AP=6\sqrt2+2\sqrt6 = \sqrt{72}+\sqrt{24} \implies\boxed{96}$

Answer:  96
Problem: Point $P$ lies on the diagonal $AC$ of square $ABCD$ with $AP > CP$ . Let $O_{1}$ and $O_{2}$ be the circumcenters of triangles $ABP$ and $CDP$ respectively. Given that $AB = 12$ and $\angle O_{1}PO_{2} = 120^{\circ}$ , then $AP = \sqrt{a} + \sqrt{b}$ , where $a$ and $b$ are positive integers. Find $a + b$

Solution: \[BP = DP,    \angle PAB =\angle PCD  = 45^\circ \implies  O_{1}P =O_{2}P\] by the Law of Sines.
\[AB= CD, O_{1}A = O_{2}D = O_{1}B = O_{2}C \implies \triangle AO_{1}B = \triangle DO_{2}C.\] They have translational symmetry. The translation vector is \[\vec {AD} \implies  O_{1}O_{2} = AD.\]
\[\triangle PO_{1}O_{2} = \triangle O_{1}AB \implies \angle AO_{1}B = 120^\circ \implies\] \[\angle APB = 60^\circ \implies  \angle ABP = 180^\circ – 60^\circ – 45^\circ = 75^\circ.\] By the Law of Sines \[AP = \frac {AB \cdot \sin 75^\circ}{\sin 60^\circ} = \frac {AB \cdot \sin (30^\circ + 45^\circ)} {\sin 60^\circ}\] \[AP  = AB \cdot (\sin 45^\circ + \cos 45^\circ \cdot \tan 30^\circ),\] \[AP = \frac {AB}{\sqrt{2}} (1 + \frac {1}{\sqrt{3}}) = 6 \sqrt {2} + 2 \sqrt {6} \implies \boxed{096}\] vladimir.shelomovskii@gmail.com, vvsss

Answer:  096
Problem: There are $N$ permutations $(a_{1}, a_{2}, ... , a_{30})$ of $1, 2, \ldots, 30$ such that for $m \in \left\{{2, 3, 5}\right\}$ $m$ divides $a_{n+m} - a_{n}$ for all integers $n$ with $1 \leq n < n+m \leq 30$ . Find the remainder when $N$ is divided by $1000$

Solution: Note that $30=2\cdot 3\cdot 5$ . Since $\gcd(2, 3, 5)=1$ , by CRT, for each value $k=0\ldots 29$ modulo $30$ there exists a unique ordered triple of values $(a, b, c)$ such that $k\equiv a\pmod{2}$ $k\equiv b\pmod{3}$ , and $k\equiv c\pmod{5}$ . Therefore, we can independently assign the residues modulo $2, 3, 5$ , so $N=2!\cdot 3!\cdot 5!=1440$ , and the answer is $\boxed{440}$

Answer:  440
Problem: There are $N$ permutations $(a_{1}, a_{2}, ... , a_{30})$ of $1, 2, \ldots, 30$ such that for $m \in \left\{{2, 3, 5}\right\}$ $m$ divides $a_{n+m} - a_{n}$ for all integers $n$ with $1 \leq n < n+m \leq 30$ . Find the remainder when $N$ is divided by $1000$

Solution: First let's look at the situation when $m$ is equal to 2. It isn't too difficult to see the given conditions are satisfied iff the sequences $S_1 = a_1,a_3,a_5,a_7...$ and $S_2 =a_2,a_4,a_6...$ each are assigned either $\equiv 0 \pmod 2$ or $\equiv 1 \pmod 2$ . Another way to say this is each element in the sequence $a_1,a_3,a_5,a_7...$ would be the same mod 2, and similarly for the other sequence. There are 2! = 2 ways to assign the mods to the sequences.
Now when $m$ is equal to 3, the sequences are $T_1 = a_1, a_4,a_7..$ $T_2 = a_2,a_5,a_8,a_11...$ , and $T_3 = a_3,a_6,a_9...$ . Again, for each sequence, all of its elements are congruent either $0$ $1$ , or $2$ mod $3$ . There are $3! = 6$ ways to assign the mods to the sequences.
Finally do the same thing for $m = 5$ . There are $5!$ ways. In total there are $2 * 6*120 = 1440$ and the answer is $\boxed{440}$

Answer:  440
Problem: Let $P(x) = x^2 - 3x - 9$ . A real number $x$ is chosen at random from the interval $5 \le x \le 15$ . The probability that $\lfloor\sqrt{P(x)}\rfloor = \sqrt{P(\lfloor x \rfloor)}$ is equal to $\frac{\sqrt{a} + \sqrt{b} + \sqrt{c} - d}{e}$ , where $a$ $b$ $c$ $d$ , and $e$ are positive integers. Find $a + b + c + d + e$

Solution: Make the substitution $y=2x-3$ , so $P(x)=\frac{y^2-45}{4}.$ We're looking for solutions to \[\bigg\lfloor{\sqrt{\frac{y^2-45}{4}}\bigg\rfloor}=\sqrt{\frac{\lfloor{y\rfloor}^2-45}{4}}\] with the new bounds $y\in{[7,27]}$ . Since the left side is an integer, it must be that $\frac{\lfloor{y\rfloor}^2-45}{4}$ is a perfect square. For simplicity, write $\lfloor{y\rfloor}=a$ and \[a^2-45=4b^2\implies{(a-2b)(a+2b)=45}.\] Since $a-2b<a+2b$ , it must be that $(a-2b,a+2b)=(1,45),(3,15),(5,9)$ , which gives solutions $(23,11),(9,3),(7,1)$ , respectively. But this gives us three cases to check:
Case 1: $\bigg\lfloor{\sqrt{\frac{y^2-45}{4}}\bigg\rfloor}=11$
In this case, we have \[11\leq{\sqrt{\frac{y^2-45}{4}}}<12\implies{y\in{[23,\sqrt{621})}}.\] Case 2: $\bigg\lfloor{\sqrt{\frac{y^2-45}{4}}\bigg\rfloor}=3$
In this case, we have \[3\leq{\sqrt{\frac{y^2-45}{4}}}<4\implies{y\in{[9,\sqrt{109})}}.\] Case 3: $\bigg\lfloor{\sqrt{\frac{y^2-45}{4}}\bigg\rfloor}=1$
In this case, we have \[1\leq{\sqrt{\frac{y^2-45}{4}}}<2\implies{y\in{[7,\sqrt{61})}}.\] To finish, the total length of the interval from which we choose $y$ is $27-7=20$ . The total length of the success intervals is \[(\sqrt{61}-7)+(\sqrt{109}-9)+(\sqrt{621}-23)=\sqrt{61}+\sqrt{109}+\sqrt{621}-39,\] which means the probability is \[\frac{\sqrt{61}+\sqrt{109}+\sqrt{621}-39}{20}.\] The requested sum is $\boxed{850}$

Answer:  850
Problem: Let $P(x) = x^2 - 3x - 9$ . A real number $x$ is chosen at random from the interval $5 \le x \le 15$ . The probability that $\lfloor\sqrt{P(x)}\rfloor = \sqrt{P(\lfloor x \rfloor)}$ is equal to $\frac{\sqrt{a} + \sqrt{b} + \sqrt{c} - d}{e}$ , where $a$ $b$ $c$ $d$ , and $e$ are positive integers. Find $a + b + c + d + e$

Solution: Note that all the "bounds" have to be less than the number+1, otherwise it wouldn't fit the answer format. Therefore, the answer is $\frac{3*3+\sqrt{9+4(4+9)}-10+\sqrt{9+4(16+9)}-12+\sqrt{9+4(144+9)}}{20} \implies \boxed{850}$

Answer:  850
Problem: Maya lists all the positive divisors of $2010^2$ . She then randomly selects two distinct divisors from this list. Let $p$ be the probability that exactly one of the selected divisors is a perfect square . The probability $p$ can be expressed in the form $\frac {m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$

Solution: $2010^2 = 2^2\cdot3^2\cdot5^2\cdot67^2$ . Thus there are $(2+1)^4$ divisors, $(1+1)^4$ of which are squares (the exponent of each prime factor must either be $0$ or $2$ ). Therefore the probability is \[\frac {2\cdot2^4\cdot(3^4 - 2^4)}{3^4(3^4 - 1)} = \frac {26}{81} \Longrightarrow 26+ 81 = \boxed{107}.\]

Answer:  107
Problem: Maya lists all the positive divisors of $2010^2$ . She then randomly selects two distinct divisors from this list. Let $p$ be the probability that exactly one of the selected divisors is a perfect square . The probability $p$ can be expressed in the form $\frac {m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$

Solution: The prime factorization of $2010^2$ is $67^2\cdot3^2\cdot2^2\cdot5^2$ . Therefore, the number of divisors of $2010^2$ is $3^4$ or $81$ $16$ of which are perfect squares. The number of ways we can choose $1$ perfect square from the two distinct divisors is $\binom{16}{1}\binom{81-16}{1}$ . The total number of ways to pick two divisors is $\binom{81}{2}$
Thus, the probability is \[\frac {\binom{16}{1}\binom{81-16}{1}}{\binom{81}{2}} = \frac {16\cdot65}{81\cdot40} = \frac {26}{81} \Longrightarrow 26+ 81 = \boxed{107}.\]

Answer:  107
Problem: Suppose that $y = \frac34x$ and $x^y = y^x$ . The quantity $x + y$ can be expressed as a rational number $\frac {r}{s}$ , where $r$ and $s$ are relatively prime positive integers. Find $r + s$

Solution: Substitute $y = \frac34x$ into $x^y = y^x$ and solve. \[x^{\frac34x} = \left(\frac34x\right)^x\] \[x^{\frac34x} = \left(\frac34\right)^x \cdot x^x\] \[x^{-\frac14x} = \left(\frac34\right)^x\] \[x^{-\frac14} = \frac34\] \[x = \frac{256}{81}\] \[y = \frac34x = \frac{192}{81}\] \[x + y = \frac{448}{81}\] \[448 + 81 = \boxed{529}\]

Answer:  529
Problem: Suppose that $y = \frac34x$ and $x^y = y^x$ . The quantity $x + y$ can be expressed as a rational number $\frac {r}{s}$ , where $r$ and $s$ are relatively prime positive integers. Find $r + s$

Solution: We solve in general using $c$ instead of $3/4$ . Substituting $y = cx$ , we have:
Dividing by $x^x$ , we get $(x^x)^{c - 1} = c^x$
Taking the $x$ th root, $x^{c - 1} = c$ , or $x = c^{1/(c - 1)}$
In the case $c = \frac34$ $x = \Bigg(\frac34\Bigg)^{ - 4} = \Bigg(\frac43\Bigg)^4 = \frac {256}{81}$ $y = \frac {64}{27}$ $x + y = \frac {256 + 192}{81} = \frac {448}{81}$ , yielding an answer of $448 + 81 = \boxed{529}$

Answer:  529
Problem: Suppose that $y = \frac34x$ and $x^y = y^x$ . The quantity $x + y$ can be expressed as a rational number $\frac {r}{s}$ , where $r$ and $s$ are relatively prime positive integers. Find $r + s$

Solution: Taking the logarithm base $x$ of both sides, we arrive with: \[y = \log_x y^x \Longrightarrow \frac{y}{x} = \log_{x} y = \log_x \left(\frac{3}{4}x\right) = \frac{3}{4}\] Now we proceed by the logarithm rule $\log(ab)=\log a + \log b$ . The equation becomes: \[\log_x \frac{3}{4} + \log_x x = \frac{3}{4}\] \[\Longleftrightarrow \log_x \frac{3}{4} + 1 = \frac{3}{4}\] \[\Longleftrightarrow \log_x \frac{3}{4} = -\frac{1}{4}\] \[\Longleftrightarrow x^{-\frac{1}{4}} = \frac{3}{4}\] \[\Longleftrightarrow \frac{1}{x^{\frac{1}{4}}} = \frac{3}{4}\] \[\Longleftrightarrow x^{\frac{1}{4}} = \frac{4}{3}\] \[\Longleftrightarrow \sqrt[4]{x} = \frac{4}{3}\] \[\Longleftrightarrow x = \left(\frac{4}{3}\right)^4=\frac{256}{81}\] Then find $y$ as in solution 3, and we get $\boxed{529}$

Answer:  529
Problem: Jackie and Phil have two fair coins and a third coin that comes up heads with probability $\frac47$ . Jackie flips the three coins, and then Phil flips the three coins. Let $\frac {m}{n}$ be the probability that Jackie gets the same number of heads as Phil, where $m$ and $n$ are relatively prime positive integers. Find $m + n$

Solution: This can be solved quickly and easily with generating functions
Let $x^n$ represent flipping $n$ heads.
The generating functions for these coins are $(1+x)$ $(1+x)$ ,and $(3+4x)$ in order.
The product is $3+10x+11x^2+4x^3$ . ( $ax^n$ means there are $a$ ways to get $n$ heads, eg there are $10$ ways to get $1$ head, and therefore $2$ tails, here.)
The sum of the coefficients squared (total number of possible outcomes, squared because the event is occurring twice) is $(4 + 11 + 10 + 3)^2 = 28^2 = 784$ and the sum of the squares of each coefficient (the sum of the number of ways that each coefficient can be chosen by the two people) is $4^2 + 11^2 + 10^2 + 3^2=246$ .
The probability is then $\frac{4^2 + 11^2 + 10^2 + 3^2}{28^2}  = \frac{246}{784} = \frac{123}{392}$ .
(Notice the relationship between the addends of the numerator here and the cases in the following solution.)
$123 + 392 = \boxed{515}$

Answer:  515
Problem: Positive integers $a$ $b$ $c$ , and $d$ satisfy $a > b > c > d$ $a + b + c + d = 2010$ , and $a^2 - b^2 + c^2 - d^2 = 2010$ . Find the number of possible values of $a$

Solution: Using the difference of squares $2010 = (a^2 - b^2) + (c^2 - d^2) = (a + b)(a - b) + (c + d)(c - d) \ge a + b + c + d = 2010$ , where equality must hold so $b = a - 1$ and $d = c - 1$ . Then we see $a = 1004$ is maximal and $a = 504$ is minimal, so the answer is $\boxed{501}$

Answer:  501
Problem: Positive integers $a$ $b$ $c$ , and $d$ satisfy $a > b > c > d$ $a + b + c + d = 2010$ , and $a^2 - b^2 + c^2 - d^2 = 2010$ . Find the number of possible values of $a$

Solution: Since $a+b$ must be greater than $1005$ , it follows that the only possible value for $a-b$ is $1$ (otherwise the quantity $a^2 - b^2$ would be greater than $2010$ ). Therefore the only possible ordered pairs for $(a,b)$ are $(504, 503)$ $(505, 504)$ , ... , $(1004, 1003)$ , so $a$ has $\boxed{501}$ possible values.

Answer:  501
Problem: Let $P(x)$ be a quadratic polynomial with real coefficients satisfying $x^2 - 2x + 2 \le P(x) \le 2x^2 - 4x + 3$ for all real numbers $x$ , and suppose $P(11) = 181$ . Find $P(16)$

Solution: Let $Q(x) = x^2 - 2x + 2$ $R(x) = 2x^2 - 4x + 3$ Completing the square , we have $Q(x) = (x-1)^2 + 1$ , and $R(x) = 2(x-1)^2 + 1$ , so it follows that $P(x) \ge Q(x) \ge 1$ for all $x$ (by the Trivial Inequality ).
Also, $1 = Q(1) \le P(1) \le R(1) = 1$ , so $P(1) = 1$ , and $P$ obtains its minimum at the point $(1,1)$ . Then $P(x)$ must be of the form $c(x-1)^2 + 1$ for some constant $c$ ; substituting $P(11) = 181$ yields $c = \frac 95$ . Finally, $P(16) = \frac 95 \cdot (16 - 1)^2 + 1 = \boxed{406}$

Answer:  406
Problem: Let $P(x)$ be a quadratic polynomial with real coefficients satisfying $x^2 - 2x + 2 \le P(x) \le 2x^2 - 4x + 3$ for all real numbers $x$ , and suppose $P(11) = 181$ . Find $P(16)$

Solution: It can be seen that the function $P(x)$ must be in the form $P(x) = ax^2 - 2ax + c$ for some real $a$ and $c$ . This is because the derivative of $P(x)$ is $2ax - 2a$ , and a global minimum occurs only at $x = 1$ (in addition, because of this derivative, the vertex of any quadratic polynomial occurs at $\frac{-b}{2a}$ ). Substituting $(1,1)$ and $(11, 181)$ we obtain two equations:
Solving, we get $a = \frac{9}{5}$ and $c = \frac{14}{5}$ , so $P(x) = \frac{9}{5}x^2 - \frac{18}{5}x + \frac {14}{5}$ . Therefore, $P(16) = \boxed{406}$

Answer:  406
Problem: Let $P(x)$ be a quadratic polynomial with real coefficients satisfying $x^2 - 2x + 2 \le P(x) \le 2x^2 - 4x + 3$ for all real numbers $x$ , and suppose $P(11) = 181$ . Find $P(16)$

Solution: Let $y = x^2 - 2x + 2$ ; note that $2y - 1 = 2x^2 - 4x + 3$ . Setting $y = 2y - 1$ , we find that equality holds when $y = 1$ and therefore when $x^2 - 2x + 2 = 1$ ; this is true iff $x = 1$ , so $P(1) = 1$
Let $Q(x) = P(x) - x$ ; clearly $Q(1) = 0$ , so we can write $Q(x) = (x - 1)Q'(x)$ , where $Q'(x)$ is some linear function. Plug $Q(x)$ into the given inequality:
$x^2 - 3x + 2 \le Q(x) \le 2x^2 - 5x + 3$
$(x - 1)(x - 2) \le (x - 1)Q'(x) \le (x - 1)(2x - 3)$ , and thus
$x - 2 \le Q'(x) \le 2x - 3$
For all $x > 1$ ; note that the inequality signs are flipped if $x < 1$ , and that the division is invalid for $x = 1$ . However,
$\lim_{x \to 1}  x - 2 = \lim_{x \to 1}  2x - 3 = -1$
and thus by the sandwich theorem $\lim_{x \to 1} Q'(x) = -1$ ; by the definition of a continuous function, $Q'(1) = -1$ . Also, $Q(11) = 170$ , so $Q'(11) = 170/(11-1) = 17$ ; plugging in and solving, $Q'(x) = (9/5)(x - 1) - 1$ . Thus $Q(16) = 390$ , and so $P(16) = \boxed{406}$

Answer:  406
Problem: Let $P(x)$ be a quadratic polynomial with real coefficients satisfying $x^2 - 2x + 2 \le P(x) \le 2x^2 - 4x + 3$ for all real numbers $x$ , and suppose $P(11) = 181$ . Find $P(16)$

Solution: Let $Q(x) = P(x) - (x^2-2x+2)$ , then $0\le Q(x) \le (x-1)^2$ (note this is derived from the given inequality chain).  Therefore, $0\le Q(x+1) \le x^2 \Rightarrow Q(x+1) = Ax^2$ for some real value A.
$Q(11) = 10^2A \Rightarrow P(11)-(11^2-22+2)=100A \Rightarrow 80=100A \Rightarrow A=\frac{4}{5}$
$Q(16)=15^2A=180 \Rightarrow P(16)-(16^2-32+2) = 180 \Rightarrow P(16)=180+226= \boxed{406}$

Answer:  406
Problem: Let $P(x)$ be a quadratic polynomial with real coefficients satisfying $x^2 - 2x + 2 \le P(x) \le 2x^2 - 4x + 3$ for all real numbers $x$ , and suppose $P(11) = 181$ . Find $P(16)$

Solution: Let $P(x) = ax^2 + bx + c$ . Plugging in $x = 1$ to the expressions on both sides of the inequality, we see that $a + b + c = 1$ . We see from the problem statement that $121a + 11b + c = 181$ . Since we know the vertex of $P(x)$ lies at $x = 1$ , by symmetry we get $81a -9b + c = 181$ as well. Since we now have three equations, we can solve this trivial system and get our answer of $\boxed{406}$

Answer:  406
Problem: Let $P(x)$ be a quadratic polynomial with real coefficients satisfying $x^2 - 2x + 2 \le P(x) \le 2x^2 - 4x + 3$ for all real numbers $x$ , and suppose $P(11) = 181$ . Find $P(16)$

Solution: Similar to Solution 5, let $P(x) = ax^2 + bx + c$ . Note that $(1,1)$ is a vertex of the polynomial. Additionally, this means that $b = -2a$ (since $\frac{-b}{2a}$ is the minimum $x$ point). Thus, we have $P(x) = ax^2 - 2ax + c$ . Therefore $a - 2a + c = 1$ . Moreover, $99a + c = 181$ . And so our polynomial is $\frac{9}{5}x^2 - \frac{18}{5}x + \frac{14}{5}$ . Plug in $x = 16$ to get $\boxed{406}$

Answer:  406
Problem: Let $P(x)$ be a quadratic polynomial with real coefficients satisfying $x^2 - 2x + 2 \le P(x) \le 2x^2 - 4x + 3$ for all real numbers $x$ , and suppose $P(11) = 181$ . Find $P(16)$

Solution: Very similar to Solution 6, start by noticing that $P(x)$ is between two polynomials, we try to set them equal to find a point where the two polynomials intersect, meaning that $P(x)$ would also have to intersect that point (it must be between the two graphs). Setting $x^2 - 2x + 2 = 2x^2 - 4x + 3$ , we find that $x = 1$ . Note that both of these graphs have the same vertex (at $x = 1$ ), and so $P(x)$ must also have the same vertex $(1, 1)$ . Setting $P(x) = ax^2 - 2ax + a + 1$ (this is where we have a vertex at $(1, 1)$ ), we plug in $11$ and find that $a = 1.8$ . Evaluating $1.8x^2 - 3.6x + 2.8$ when $x = 16$ (our intended goal), we find that $P(16) = \boxed{406}$

Answer:  406
Problem: For a real number $a$ , let $\lfloor a \rfloor$ denote the greatest integer less than or equal to $a$ . Let $\mathcal{R}$ denote the region in the coordinate plane consisting of points $(x,y)$ such that $\lfloor x \rfloor ^2 + \lfloor y \rfloor ^2 = 25$ . The region $\mathcal{R}$ is completely contained in a disk of radius $r$ (a disk is the union of a circle and its interior). The minimum value of $r$ can be written as $\frac {\sqrt {m}}{n}$ , where $m$ and $n$ are integers and $m$ is not divisible by the square of any prime. Find $m + n$

Solution: The desired region consists of 12 boxes, whose lower-left corners are integers solutions of $x^2 + y^2 = 25$ , namely $(\pm5,0), (0,\pm5), (\pm3,\pm4), (\pm4,\pm3).$ Since the points themselves are symmetric about $(0,0)$ , the boxes are symmetric about $\left(\frac12,\frac12\right)$ . The distance from $\left(\frac12,\frac12\right)$ to the furthest point on a box that lays on an axis, for instance $(6,1)$ , is $\sqrt {\frac {11}2^2 + \frac12^2} = \sqrt {\frac {122}4}.$ The distance from $\left(\frac12,\frac12\right)$ to the furthest point on a box in the middle of a quadrant, for instance $(5,4)$ , is $\sqrt {\frac92^2 + \frac72^2} = \sqrt {\frac {130}4}.$ The latter is the larger, and is $\frac {\sqrt {130}}2$ , giving an answer of $130 + 2 = \boxed{132}$

Answer:  132
Problem: Let $(a,b,c)$ be the real solution of the system of equations $x^3 - xyz = 2$ $y^3 - xyz = 6$ $z^3 - xyz = 20$ . The greatest possible value of $a^3 + b^3 + c^3$ can be written in the form $\frac {m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$

Solution: This is almost the same as solution 1. Note $a^3 + b^3 + c^3 = 28 + 3abc$ . Next, let $k = a^3$ . Note that $b = \sqrt [3]{k + 4}$ and $c = \sqrt [3]{k + 18}$ , so we have $28 + 3\sqrt [3]{k(k+4)(k+18)} = 28+3abc=a^3+b^3+c^3=3k+22$ . Move 28 over, divide both sides by 3, then cube to get $k^3-6k^2+12k-8 = k^3+22k^2+18k$ . The $k^3$ terms cancel out, so solve the quadratic to get $k = -2, -\frac{1}{7}$ . We maximize $abc$ by choosing $k = -\frac{1}{7}$ , which gives us $a^3+b^3+c^3 = 3k + 22 = \frac{151}{7}$ . Thus, our answer is $151+7=\boxed{158}$

Answer:  158
Problem: Let $N$ be the number of ways to write $2010$ in the form $2010 = a_3 \cdot 10^3 + a_2 \cdot 10^2 + a_1 \cdot 10 + a_0$ , where the $a_i$ 's are integers, and $0 \le a_i \le 99$ . An example of such a representation is $1\cdot 10^3 + 3\cdot 10^2 + 67\cdot 10^1 + 40\cdot 10^0$ . Find $N$

Solution: Note that $a_2\cdot 10^2 + a_0$ is the base $100$ representation of any number from $0$ to $9999$ , and similarly $10(a_3\cdot 10^2 + a_1)$ is ten times the base $100$ representation of any number from $0$ to $9999$ . Thus, the number of solutions is just the number of solutions to $2010 = 10a+b$ where $0\le a, b\le 9999$ , which is equal to $\boxed{202}$ as $a$ can range from $0$ to $201$

Answer:  202