Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: Let $\overline{AB}$ be a diameter in a circle of radius $5\sqrt2.$ Let $\overline{CD}$ be a chord in the circle that intersects $\overline{AB}$ at a point $E$ such that $BE=2\sqrt5$ and $\angle AEC = 45^{\circ}.$ What is $CE^2+DE^2?$
$\textbf{(A)}\ 96 \qquad\textbf{(B)}\ 98 \qquad\textbf{(C)}\  44\sqrt5 \qquad\textbf{(D)}\ 70\sqrt2 \qquad\textbf{(E)}\ 100$

Solution: [asy] /* Made by sofas103; edited by MRENTHUSIASM */ size(250);  pair O, A, B, C, D, E, P, Q; O = origin; A = (-5*sqrt(2),0); B = (5*sqrt(2),0); E = (5*sqrt(2)-2*sqrt(5),0); P = (5*sqrt(2), -4.5); Q=(5*sqrt(2)-2*sqrt(5)-0.5, 0); path p; p = Circle(O,5*sqrt(2)); C = intersectionpoint(p,E--E+10*dir(135)); D = intersectionpoint(p,E--E+10*dir(-45)); draw(p); dot("$A$",A,1.5*dir(A),linewidth(4)); dot("$B$",B,1.5*dir(B),linewidth(4)); dot("$E$",E,1.5*dir(180+135/2),linewidth(4)); dot("$C$",C,1.5*dir(C),linewidth(4)); dot("$D$",D,1.5*dir(D),linewidth(4)); dot("$P$",P,1.5*dir(P),linewidth(4)); label("$45^{\circ}$",Q,NW); draw(A--B^^C--P--B); draw((5*sqrt(2), 0)--(5*sqrt(2)-0.4, 0)--(5*sqrt(2)-0.4, -0.4)--(5*sqrt(2), -0.4)); [/asy] For ease of notation, let $x = DE$ and $y=EC$ . 
Extend $\overline{CD}$ to point $P$ until $\overline{BP}$ is perpendicular to $AB$ . It's given that $\angle AEC = 45^{\circ}$ , so, by vertical angles, we have $\angle BEP = \angle EPB = 45^{\circ}$
Since $PEB$ is a $45-45-90$ right triangle, we have $BE = BP = 2\sqrt{5}$ and $PE=2\sqrt{10}$ . Hence, $PD = 2\sqrt{10}-x.$
By Power of a Point, we have
\[PB^2 = PD\cdot PC\] \[20 = \left(2\sqrt{10}-x\right)\left(y + 2\sqrt{10}\right).\]
Isolating the variables after expanding gives $x-y = 2\sqrt{10}-10.$
Using Power of a Point again, we have
\[DE\cdot EC = BE\cdot EA\] \[xy = 2\sqrt{5}\left(10\sqrt{2}-2\sqrt{5} \right)\] \[xy = 20\sqrt{10}-20.\] To get $x^2 + y^2$ , we can perform the operation $(x-y)^2 + 2xy$ . Plugging these values in,
\[(x-y)^2 + 2xy = \left(2\sqrt{10}-10\right)^2 + 2\left(20\sqrt{10}-20\right)\] \[= 40 + 100 - 40\sqrt{10} + 40\sqrt{10} - 40 = \boxed{100}\]

Answer: E 100
Problem: Let $\overline{AB}$ be a diameter in a circle of radius $5\sqrt2.$ Let $\overline{CD}$ be a chord in the circle that intersects $\overline{AB}$ at a point $E$ such that $BE=2\sqrt5$ and $\angle AEC = 45^{\circ}.$ What is $CE^2+DE^2?$
$\textbf{(A)}\ 96 \qquad\textbf{(B)}\ 98 \qquad\textbf{(C)}\  44\sqrt5 \qquad\textbf{(D)}\ 70\sqrt2 \qquad\textbf{(E)}\ 100$

Solution: Perhaps not reliable in general, but very useful as a last resort.
The choice of the radius $5\sqrt2$ is strange, and is probably motivated by a nice answer in the end, so we only consider integer options. Notice that a 5 also appears in the condition $BE=2\sqrt5$ , therefore it will likely be present in the answer as well; the only integer containing a factor of 5 amongst the choices is 100, thus the answer is $\boxed{100}$

Answer: E 100
Problem: How many polynomials of the form $x^5 + ax^4 + bx^3 + cx^2 + dx + 2020$ , where $a$ $b$ $c$ , and $d$ are real numbers, have the property that whenever $r$ is a root, so is $\frac{-1+i\sqrt{3}}{2} \cdot r$ ? (Note that $i=\sqrt{-1}$
$\textbf{(A) } 0 \qquad \textbf{(B) }1 \qquad \textbf{(C) } 2 \qquad \textbf{(D) } 3 \qquad \textbf{(E) } 4$

Solution: Let $P(x) = x^5+ax^4+bx^3+cx^2+dx+2020$ . We first notice that $\frac{-1+i\sqrt{3}}{2} = e^{2\pi i / 3}$ .  That is because of Euler's Formula : $e^{ix} = \cos(x) + i \cdot \sin(x)$ $\frac{-1+i\sqrt{3}}{2}$ $-\frac{1}{2} + i \cdot \frac {\sqrt{3}}{2}$ $\cos(120^\circ) + i \cdot \sin(120^\circ) = e^{ 2\pi i / 3}$
In order $r$ to be a root of $P$ $re^{2\pi i / 3}$ must also be a root of P, meaning that 3 of the roots of $P$ must be $r$ $re^{i\frac{2\pi}{3}}$ $re^{i\frac{4\pi}{3}}$ . However, since $P$ is degree 5, there must be two additional roots. Let one of these roots be $w$ , if $w$ is a root, then $we^{2\pi i / 3}$ and $we^{4\pi i / 3}$ must also be roots. However, $P$ is a fifth degree polynomial, and can therefore only have $5$ roots. This implies that $w$ is either $r$ $re^{2\pi i / 3}$ , or $re^{4\pi i / 3}$ . Thus we know that the polynomial $P$ can be written in the form $(x-r)^m(x-re^{2\pi i / 3})^n(x-re^{4\pi i / 3})^p$ . Moreover, by Vieta's, we know that there is only one possible value for the magnitude of $r$ as $||r||^5 = 2020$ , meaning that the amount of possible polynomials $P$ is equivalent to the possible sets $(m,n,p)$ . In order for the coefficients of the polynomial to all be real, $n = p$ due to $re^{2\pi i / 3}$ and $re^{4 \pi i / 3}$ being conjugates and since $m+n+p = 5$ , (as the polynomial is 5th degree) we have two possible solutions for $(m, n, p)$ which are $(1,2,2)$ and $(3,1,1)$ yielding two possible polynomials. The answer is thus $\boxed{2}$

Answer: C 2
Problem: How many polynomials of the form $x^5 + ax^4 + bx^3 + cx^2 + dx + 2020$ , where $a$ $b$ $c$ , and $d$ are real numbers, have the property that whenever $r$ is a root, so is $\frac{-1+i\sqrt{3}}{2} \cdot r$ ? (Note that $i=\sqrt{-1}$
$\textbf{(A) } 0 \qquad \textbf{(B) }1 \qquad \textbf{(C) } 2 \qquad \textbf{(D) } 3 \qquad \textbf{(E) } 4$

Solution: Let $x_1=r$ , then \[x_2=\frac{-1+i\sqrt{3}}{2} r,\] \[x_3=\left( \frac{-1+i\sqrt{3}}{2} \right) ^2 r =\left( \frac{-1-i\sqrt{3}}{2} \right) r,\] \[x_4=\left( \frac{-1+i\sqrt{3}}{2} \right) ^3 r=r,\] which means $x_4$ is the same as $x_1$
Now we have 3 different roots of the polynomial, $x_1$ $x_2$ , and $x_3$ . Next, we will prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there is one root $x_4=p$ which is different from the three roots we already know, then there must be two other roots, \[x_5=\left( \frac{-1+i\sqrt{3}}{2} \right) ^2 p =\left( \frac{-1-i\sqrt{3}}{2} \right) p,\] \[x_6=\left( \frac{-1+i\sqrt{3}}{2} \right) ^3 p=p,\] different from all known roots. We get 6 different roots for the polynomial, which contradicts the limit of 5 roots. Therefore the assumption of a different root is wrong, thus the roots must be chosen from $x_1$ $x_2$ , and $x_3$
The polynomial then can be written like $f(x)=(x-x_1)^m (x-x_2)^n (x-x_3)^q$ , where $m$ $n$ , and $q$ are non-negative integers and $m+n+q=5$ . Since $a$ $b$ $c$ and $d$ are real numbers, then $n$ must be equal to $q$ . Therefore $(m,n,q)$ can only be $(1,2,2)$ or $(3,1,1)$ , so the answer is $\boxed{2}$

Answer: C 2
Problem: How many integers $n \geq 2$ are there such that whenever $z_1, z_2, ..., z_n$ are complex numbers such that
\[|z_1| = |z_2| = ... = |z_n| = 1 \text{    and    } z_1 + z_2 + ... + z_n = 0,\] then the numbers $z_1, z_2, ..., z_n$ are equally spaced on the unit circle in the complex plane?
$\textbf{(A)}\ 1 \qquad\textbf{(B)}\ 2 \qquad\textbf{(C)}\ 3 \qquad\textbf{(D)}\ 4 \qquad\textbf{(E)}\ 5$

Solution: For $n=2$ , we see that if $z_{1}+z_{2}=0$ , then $z_{1}=-z_{2}$ , so they are evenly spaced along the unit circle.
For $n=3$ , WLOG, we can set $z_{1}=1$ . Notice that now Re $(z_{2}+z_{3})=-1$ and Im $\{z_{2}\}$ $-$ Im $\{z_{3}\}$ . This forces $z_{2}$ and $z_{3}$ to be equal to $e^{i\frac{2\pi}{3}}$ and $e^{-i\frac{2\pi}{3}}$ , meaning that all three are equally spaced along the unit circle.
We can now show that we can construct complex numbers when $n\geq 4$ that do not satisfy the conditions in the problem.
Suppose that the condition in the problem holds for some $n=k$ . We can now add two points $z_{k+1}$ and $z_{k+2}$ anywhere on the unit circle such that $z_{k+1}=-z_{k+2}$ , which will break the condition. Now that we have shown that $n=2$ and $n=3$ works, by this construction, any $n\geq 4$ does not work, making the answer $\boxed{2}$

Answer: B 2
Problem: The area of a pizza with radius $4$ is $N$ percent larger than the area of a pizza with radius $3$ inches. What is the integer closest to $N$
$\textbf{(A) } 25 \qquad\textbf{(B) } 33 \qquad\textbf{(C) } 44\qquad\textbf{(D) } 66 \qquad\textbf{(E) } 78$

Solution: The area of the larger pizza is $16\pi$ , while the area of the smaller pizza is $9\pi$ . Therefore, the larger pizza is $\frac{7\pi}{9\pi} \cdot 100\%$ bigger than the smaller pizza. $\frac{7\pi}{9\pi} \cdot 100\% = 77.777....$ , which is closest to $\boxed{78}$

Answer: E 78
Problem: Suppose $a$ is $150\%$ of $b$ . What percent of $a$ is $3b$
$\textbf{(A) } 50 \qquad \textbf{(B) } 66+\frac{2}{3} \qquad \textbf{(C) } 150 \qquad \textbf{(D) } 200 \qquad \textbf{(E) } 450$

Solution: Since $a=1.5b$ , that means $b=\frac{a}{1.5}$ . We multiply by $3$ to get a $3b$ term, yielding $3b=2a$ , and $2a$ is $\boxed{200}$ of $a$

Answer: D 200
Problem: Suppose $a$ is $150\%$ of $b$ . What percent of $a$ is $3b$
$\textbf{(A) } 50 \qquad \textbf{(B) } 66+\frac{2}{3} \qquad \textbf{(C) } 150 \qquad \textbf{(D) } 200 \qquad \textbf{(E) } 450$

Solution: Without loss of generality, let $b=100$ . Then, we have $a=150$ and $3b=300$ . Thus, $\frac{3b}{a}=\frac{300}{150}=2$ , so $3b$ is $200\%$ of $a$ . Hence the answer is $\boxed{200}$

Answer: D 200
Problem: Suppose $a$ is $150\%$ of $b$ . What percent of $a$ is $3b$
$\textbf{(A) } 50 \qquad \textbf{(B) } 66+\frac{2}{3} \qquad \textbf{(C) } 150 \qquad \textbf{(D) } 200 \qquad \textbf{(E) } 450$

Solution: As before, $a = 1.5b$ . Multiply by 2 to obtain $2a = 3b$ . Since $2 = 200\%$ , the answer is $\boxed{200}$

Answer: D 200
Problem: Suppose $a$ is $150\%$ of $b$ . What percent of $a$ is $3b$
$\textbf{(A) } 50 \qquad \textbf{(B) } 66+\frac{2}{3} \qquad \textbf{(C) } 150 \qquad \textbf{(D) } 200 \qquad \textbf{(E) } 450$

Solution: Without loss of generality, let $b=2$ . Then, we have $a=3$ and $3b=6$ . This gives $\frac{3b}{a}=\frac{6}{3}=2$ , so $3b$ is $200\%$ of $a$ , so the answer is $\boxed{200}$

Answer: D 200
Problem: Positive real numbers $x \neq 1$ and $y \neq 1$ satisfy $\log_2{x} = \log_y{16}$ and $xy = 64$ . What is $(\log_2{\tfrac{x}{y}})^2$
$\textbf{(A) } \frac{25}{2} \qquad\textbf{(B) } 20 \qquad\textbf{(C) } \frac{45}{2} \qquad\textbf{(D) } 25 \qquad\textbf{(E) } 32$

Solution: Let $\log_2{x} = \log_y{16}=k$ , so that $2^k=x$ and $y^k=16 \implies y=2^{\frac{4}{k}}$ . Then we have $(2^k)(2^{\frac{4}{k}})=2^{k+\frac{4}{k}}=2^6$
We therefore have $k+\frac{4}{k}=6$ , and deduce $k^2-6k+4=0$ . The solutions to this are $k = 3 \pm \sqrt{5}$
To solve the problem, we now find \begin{align*} (\log_2\tfrac{x}{y})^2&=(\log_2 x - \log_2 y)^2\\ &=(k-\tfrac{4}{k})^2=(3 \pm \sqrt{5} - \tfrac{4}{3 \pm \sqrt{5}})^2 \\ &= (3 \pm \sqrt{5} - [3 \mp \sqrt{5}])^2\\ &= (3 \pm \sqrt{5} - 3 \pm \sqrt{5})^2\\ &=(\pm 2\sqrt{5})^2 \\ &= \boxed{20} ~Edits by BakedPotato66

Answer: B 20
Problem: Positive real numbers $x \neq 1$ and $y \neq 1$ satisfy $\log_2{x} = \log_y{16}$ and $xy = 64$ . What is $(\log_2{\tfrac{x}{y}})^2$
$\textbf{(A) } \frac{25}{2} \qquad\textbf{(B) } 20 \qquad\textbf{(C) } \frac{45}{2} \qquad\textbf{(D) } 25 \qquad\textbf{(E) } 32$

Solution: After obtaining $k + \frac{4}{k} = 6$ , notice that the required answer is $\left(k - \frac{4}{k}\right)^{2} = k^2 - 8 + \frac{16}{k^2} = \left(k^2 + 8 + \frac{16}{k^2}\right) - 16 = \left(k+\frac{4}{k}\right)^2 - 16 = 6^2 - 16 = \boxed{20}$ , as before.

Answer: B 20
Problem: Positive real numbers $x \neq 1$ and $y \neq 1$ satisfy $\log_2{x} = \log_y{16}$ and $xy = 64$ . What is $(\log_2{\tfrac{x}{y}})^2$
$\textbf{(A) } \frac{25}{2} \qquad\textbf{(B) } 20 \qquad\textbf{(C) } \frac{45}{2} \qquad\textbf{(D) } 25 \qquad\textbf{(E) } 32$

Solution: From the given data, $\log_2(x) = \frac{1}{\log_{16}(y)}$ , or $\log_2(x) = \frac{4}{{\log_{2}(y)}}$
We know that $xy=64$ , so $x= \frac{64}{y}$
Thus $\log_2\left(\frac{64}{y}\right) = \frac{4}{{\log_{2}(y)}}$ , so $6-\log_2(y) = \frac{4}{{\log_{2}(y)}}$ , so $6(\log_2(y))-(\log_2(y))^2=4$
Solving for $\log_2(y)$ , we obtain $\log_2(y)=3+\sqrt{5}$
Easy resubstitution further gives $\log_2(x)=\frac{4}{3+\sqrt{5}}$ . Simplifying, we obtain $\log_2(x)= 3-\sqrt{5}$
Looking back at the original problem, we have What is $(\log_2{\tfrac{x}{y}})^2$
Deconstructing this expression using log rules, we get $(\log_2{x}-\log_2{y})^2$
Plugging in our known values, we get $((3-\sqrt{5})-(3+\sqrt{5}))^2$ or $(-2\sqrt{5})^2$
Our answer is $\boxed{20}$

Answer: B 20
Problem: How many ways are there to paint each of the integers $2, 3, \cdots , 9$ either red, green, or blue so that each number has a different color from each of its proper divisors?
$\textbf{(A)}~144\qquad\textbf{(B)}~216\qquad\textbf{(C)}~256\qquad\textbf{(D)}~384\qquad\textbf{(E)}~432$

Solution: The $5$ and $7$ can be painted with no restrictions because the set of integers does not contain a multiple or proper factor of $5$ or $7$ . There are 3 ways to paint each, giving us $\underline{9}$ ways to paint both. The $2$ is the most restrictive number. There are $\underline{3}$ ways to paint $2$ , but without loss of generality, let it be painted red. $4$ cannot be the same color as $2$ or $8$ , so there are $\underline{2}$ ways to paint $4$ , which automatically determines the color for $8$ $6$ cannot be painted red, so there are $\underline{2}$ ways to paint $6$ , but WLOG, let it be painted blue. There are $\underline{2}$ choices for the color for $3$ , which is either red or green in this case. Lastly, there are $\underline{2}$ ways to choose the color for $9$
$9 \cdot 3 \cdot 2 \cdot 2 \cdot 2 \cdot 2 = \boxed{432}$

Answer: E 432
Problem: How many ways are there to paint each of the integers $2, 3, \cdots , 9$ either red, green, or blue so that each number has a different color from each of its proper divisors?
$\textbf{(A)}~144\qquad\textbf{(B)}~216\qquad\textbf{(C)}~256\qquad\textbf{(D)}~384\qquad\textbf{(E)}~432$

Solution: We note that the primes can be colored any of the $3$ colors since they don't have any proper divisors other than $1$ , which is not in the list. Furthermore, $6$ is the only number in the list that has $2$ distinct prime factors (namely, $2$ and $3$ ), so we do casework on $6$
Case 1 $2$ and $3$ are the same color
In this case, we have $3$ primes to choose the color for ( $2$ $5$ , and $7$ ). Afterwards, $4$ $6$ , and $9$ have two possible colors, which will determine the color of $8$ . Thus, there are $3^3\cdot 2^3=216$ possibilities here.
Case 2 $2$ and $3$ are different colors
In this case, we have $4$ primes to color. Without loss of generality, we'll color the $2$ first, then the $3$ . Then there are $3$ color choices for $2,5,7$ , and $2$ color choices for $3$ . This will determine the color of $6$ as well. After that, we only need to choose the color for $4$ and $9$ , which each have $2$ choices. Thus, there are $3^3\cdot 2^3=216$ possibilities here as well.
Adding up gives $216+216=\boxed{432}$

Answer: E 432
Problem: How many ways are there to paint each of the integers $2, 3, \cdots , 9$ either red, green, or blue so that each number has a different color from each of its proper divisors?
$\textbf{(A)}~144\qquad\textbf{(B)}~216\qquad\textbf{(C)}~256\qquad\textbf{(D)}~384\qquad\textbf{(E)}~432$

Solution: $2,4,8$ require different colors each, so there are $6$ ways to color these.
$5$ and $7$ can be any color, so there are $3\times 3$ ways to color these.
$6$ can have $2$ colors once $2$ is colored, and thus $3$ also has $2$ colors following $6$ , which leaves another $2$ for $9$
All together: $6\times 3 \times 3 \times 2 \times 2 \times 2 = 432 \Rightarrow \boxed{432}$

Answer: E 432
Problem: For a certain complex number $c$ , the polynomial \[P(x) = (x^2 - 2x + 2)(x^2 - cx + 4)(x^2 - 4x + 8)\] has exactly 4 distinct roots. What is $|c|$
$\textbf{(A) } 2 \qquad \textbf{(B) } \sqrt{6} \qquad \textbf{(C) } 2\sqrt{2} \qquad \textbf{(D) } 3 \qquad \textbf{(E) } \sqrt{10}$

Solution: The polynomial can be factored further broken down into
$P(x) = (x - [1 - i])(x - [1 + i])(x - [2 - 2i])(x - [2 + 2i])(x^2 - cx + 4)$
by using the quadratic formula on each of the quadratic factors. Since the first four roots are all distinct, the term $(x^2 - cx + 4)$ must be a product of any combination of two (not necessarily distinct) factors from the set: $(x - [1 - i]), (x - [1 + i]), (x - [2 - 2i]),$ and $(x - [2 + 2i])$ . We need the two factors to yield a constant term of $4$ when multiplied together. The only combinations that work are $(x - [1 - i])$ and $(x - [2 + 2i])$ , or $(x - [1+i])$ and $(x - [2-2i])$ . When multiplied together, the polynomial is either $(x^2 + [-3 + i]x + 4)$ or $(x^2+[-3-i]x+4)$ . Therefore, $c = 3 \pm i$ and $|c| = \boxed{10}$

Answer: E 10
Problem: Let $s_k$ denote the sum of the $\textit{k}$ th powers of the roots of the polynomial $x^3-5x^2+8x-13$ . In particular, $s_0=3$ $s_1=5$ , and $s_2=9$ . Let $a$ $b$ , and $c$ be real numbers such that $s_{k+1} = a \, s_k + b \, s_{k-1} + c \, s_{k-2}$ for $k = 2$ $3$ $....$ What is $a+b+c$
$\textbf{(A)} \; -6 \qquad \textbf{(B)} \; 0 \qquad \textbf{(C)} \; 6 \qquad \textbf{(D)} \; 10 \qquad \textbf{(E)} \; 26$

Solution: Let $p, q$ , and $r$ be the roots of the polynomial. Then,
$p^3 - 5p^2 + 8p - 13 = 0$
$q^3 - 5q^2 + 8q - 13 = 0$
$r^3 - 5r^2 + 8r - 13 = 0$
Adding these three equations, we get
$(p^3 + q^3 + r^3) - 5(p^2 + q^2 + r^2) + 8(p + q + r) - 39 = 0$
$s_3 - 5s_2 + 8s_1 = 39$
$39$ can be written as $13s_0$ , giving
$s_3 = 5s_2 - 8s_1 + 13s_0$
We are given that $s_{k+1} = a \, s_k + b \, s_{k-1} + c \, s_{k-2}$ is satisfied for $k = 2$ $3$ $....$ , meaning it must be satisfied when $k = 2$ , giving us $s_3 = a \, s_2 + b \, s_1 + c \, s_0$
Therefore, $a = 5, b = -8$ , and $c = 13$ by matching coefficients.
$5 - 8 + 13 = \boxed{10}$

Answer: D 10
Problem: Let $s_k$ denote the sum of the $\textit{k}$ th powers of the roots of the polynomial $x^3-5x^2+8x-13$ . In particular, $s_0=3$ $s_1=5$ , and $s_2=9$ . Let $a$ $b$ , and $c$ be real numbers such that $s_{k+1} = a \, s_k + b \, s_{k-1} + c \, s_{k-2}$ for $k = 2$ $3$ $....$ What is $a+b+c$
$\textbf{(A)} \; -6 \qquad \textbf{(B)} \; 0 \qquad \textbf{(C)} \; 6 \qquad \textbf{(D)} \; 10 \qquad \textbf{(E)} \; 26$

Solution: Let $p, q$ , and $r$ be the roots of the polynomial. By Vieta's Formulae, we have
$p+q+r = 5$
$pq+qr+rp = 8$
$pqr=13$
We know $s_k = p^k + q^k + r^k$ . Consider $(p+q+r)(s_k) =5s_k$
$5s_k = [p^{k+1} + q^{k+1} + r^{k+1}] + p^k q + p^k r + pq^k + q^k r + pr^k + qr^k$
Using $pqr = 13$ and $s_{k-2} = p^{k-2} + q^{k-2} + r^{k-2}$ , we see $13s_{k-2} = p^{k-1}qr + pq^{k-1}r + pqr^{k-1}$
We have \[\begin{split} 5s_k + 13s_{k-2} &= s_{k+1} + (p^k q + p^k r + p^{k-1}qr) + (pq^k + pq^{k-1}r + q^k r) + (pqr^{k-1} + pr^k + qr^k) \\ &= s_{k+1} + p^{k-1} (pq + pr + qr) + q^{k-1} (pq + pr + qr) + r^{k-1} (pq + pr + qr) \\ &= s_{k+1} + (p^{k-1} + q^{k-1} + r^{k-1})(pq + pr + qr) \\ &= 5s_k + 13s_{k-2} = s_{k+1} + 8s_{k-1}\end{split}\]
Rearrange to get $s_{k+1} = 5s_k - 8s_{k-1} + 13s_{k-2}$
So, $a+ b + c = 5 -8 + 13 = \boxed{10}$

Answer: D 10
Problem: Let $s_k$ denote the sum of the $\textit{k}$ th powers of the roots of the polynomial $x^3-5x^2+8x-13$ . In particular, $s_0=3$ $s_1=5$ , and $s_2=9$ . Let $a$ $b$ , and $c$ be real numbers such that $s_{k+1} = a \, s_k + b \, s_{k-1} + c \, s_{k-2}$ for $k = 2$ $3$ $....$ What is $a+b+c$
$\textbf{(A)} \; -6 \qquad \textbf{(B)} \; 0 \qquad \textbf{(C)} \; 6 \qquad \textbf{(D)} \; 10 \qquad \textbf{(E)} \; 26$

Solution: Let $r,s,t$ be the roots of $x^3-5x^2+8x-13$ . Then:
$r^3=5r^2-8r+13$ \\ $s^3=5s^2-8s+13$ \\ $t^3=5t^2-8t+13$
If we multiply both sides of the equation by $r^k$ , where $k$ is a positive integer, then that won't change the coefficients, but just the degree of the new polynomial and the other term's exponents. We can try multiplying to find $r^4+s^4+t^4$ , but that is just to check. So then with the above information about $r^3,s^3,t^3$ , we see that:
$r^k=5r^{k-1}-8r^{k-2}-13r^{k-3}$ $s^k=5s^{k-1}-8s^{k-2}-13s^{k-3}$ $t^k=5t^{k-1}-8t^{k-2}-13t^{k-3}$
$s_k=r^k+s^k+t^k$
Then: $s_k=5s_{k-1}-8s_{k-2}+13s_{k-3}$
This means that $s_{k+1}=5s_{k}-8s_{k-1}+13s_{k-2}$ , as expected. So we have $a=5, b=-8, c=13$ . So our answer is $5-8+13=\boxed{10}$

Answer: D 10
Problem: In $\triangle ABC$ with integer side lengths, $\cos A = \frac{11}{16}$ $\cos B = \frac{7}{8}$ , and $\cos C = -\frac{1}{4}$ . What is the least possible perimeter for $\triangle ABC$
$\textbf{(A) } 9 \qquad \textbf{(B) } 12 \qquad \textbf{(C) } 23 \qquad \textbf{(D) } 27 \qquad \textbf{(E) } 44$

Solution: Notice that by the Law of Sines, $a:b:c = \sin{A}:\sin{B}:\sin{C}$ , so let's flip all the cosines using $\sin^{2}{x} + \cos^{2}{x} = 1$ $\sin{x}$ is positive for $0^{\circ} < x < 180^{\circ}$ , so we're good there).
$\sin A=\frac{3\sqrt{15}}{16}, \qquad \sin B= \frac{\sqrt{15}}{8}, \qquad \text{and} \qquad\sin C=\frac{\sqrt{15}}{4}$
These are in the ratio $3:2:4$ , so our minimal triangle has side lengths $2$ $3$ , and $4$ $\boxed{9}$ is our answer.

Answer: A 9
Problem: In $\triangle ABC$ with integer side lengths, $\cos A = \frac{11}{16}$ $\cos B = \frac{7}{8}$ , and $\cos C = -\frac{1}{4}$ . What is the least possible perimeter for $\triangle ABC$
$\textbf{(A) } 9 \qquad \textbf{(B) } 12 \qquad \textbf{(C) } 23 \qquad \textbf{(D) } 27 \qquad \textbf{(E) } 44$

Solution: $\angle ACB$ is obtuse since its cosine is negative, so we let the foot of the altitude from $C$ to $AB$ be $H$ . Let $AH=11x$ $AC=16x$ $BH=7y$ , and $BC=8y$ . By the Pythagorean Theorem, $CH=\sqrt{256x^2-121x^2}=3x\sqrt{15}$ and $CH=\sqrt{64y^2-49y^2}=y\sqrt{15}$ . Thus, $y=3x$ . The sides of the triangle are then $16x$ $11x+7(3x)=32x$ , and $24x$ , so for some integers $a,b$ $16x=a$ and $24x=b$ , where $a$ and $b$ are minimal. Hence, $\frac{a}{16}=\frac{b}{24}$ , or $3a=2b$ . Thus the smallest possible positive integers $a$ and $b$ that satisfy this are $a=2$ and $b=3$ , so $x=\frac{1}{8}$ . The sides of the triangle are $2$ $3$ , and $4$ , so $\boxed{9}$ is our answer.

Answer: A 9
Problem: In $\triangle ABC$ with integer side lengths, $\cos A = \frac{11}{16}$ $\cos B = \frac{7}{8}$ , and $\cos C = -\frac{1}{4}$ . What is the least possible perimeter for $\triangle ABC$
$\textbf{(A) } 9 \qquad \textbf{(B) } 12 \qquad \textbf{(C) } 23 \qquad \textbf{(D) } 27 \qquad \textbf{(E) } 44$

Solution: Using the law of cosines, we get the following equations:
\[c^2=a^2+b^2+\frac{ab}{2}\] \[b^2=a^2+c^2-\frac{7ac}{4}\] \[a^2=b^2+c^2-\frac{11bc}{8}\]
Substituting $a^2+c^2-\frac{7ac}{4}$ for $b^2$ in $a^2=b^2+c^2-\frac{11bc}{8}$ and simplifying, we get the following: \[14a+11b=16c\]
Note that since $a, b, c$ are integers, we can solve this for integers. By some trial and error, we get that $(a,b,c) = (3,2,4)$ . Checking to see that this fits the triangle inequality, we find out that this indeed works. Hence, our answer is $3+2+4 = \boxed{9}$

Answer: A 9
Problem: Let \[z=\frac{1+i}{\sqrt{2}}.\] What is \[\left(z^{1^2}+z^{2^2}+z^{3^2}+\dots+z^{{12}^2}\right) \cdot \left(\frac{1}{z^{1^2}}+\frac{1}{z^{2^2}}+\frac{1}{z^{3^2}}+\dots+\frac{1}{z^{{12}^2}}\right)?\]
$\textbf{(A) } 18 \qquad \textbf{(B) } 72-36\sqrt2 \qquad \textbf{(C) } 36 \qquad \textbf{(D) } 72 \qquad \textbf{(E) } 72+36\sqrt2$

Solution: Note that $z = \mathrm{cis }(45^{\circ})$
Also note that $z^{k} = z^{k + 8}$ for all positive integers $k$ because of De Moivre's Theorem. Therefore, we want to look at the exponents of each term modulo $8$
$1^2, 5^2,$ and $9^2$ are all $1 \pmod{8}$
$2^2, 6^2,$ and $10^2$ are all $4 \pmod{8}$
$3^2, 7^2,$ and $11^2$ are all $1 \pmod{8}$
$4^2, 8^2,$ and $12^2$ are all $0 \pmod{8}$
Therefore,
$z^{1^2} = z^{5^2} = z^{9^2} = \mathrm{cis }(45^{\circ})$
$z^{2^2} = z^{6^2} = z^{10^2} = \mathrm{cis }(180^{\circ}) = -1$
$z^{3^2} = z^{7^2} = z^{11^2} = \mathrm{cis }(45^{\circ})$
$z^{4^2} = z^{8^2} = z^{12^2} = \mathrm{cis }(0^{\circ}) = 1$
The term thus $\left(z^{1^2}+z^{2^2}+z^{3^2}+\dots+z^{{12}^2}\right)$ simplifies to $6\mathrm{cis }(45^{\circ})$ , while the term $\left(\frac{1}{z^{1^2}}+\frac{1}{z^{2^2}}+\frac{1}{z^{3^2}}+\dots+\frac{1}{z^{{12}^2}}\right)$ simplifies to $\frac{6}{\mathrm{cis }(45^{\circ})}$ . Upon multiplication, the $\mathrm{cis }(45^{\circ})$ cancels out and leaves us with $\boxed{36}$

Answer: C 36