Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: A circle with integer radius $r$ is centered at $(r, r)$ . Distinct line segments of length $c_i$ connect points $(0, a_i)$ to $(b_i, 0)$ for $1 \le i \le 14$ and are tangent to the circle, where $a_i$ $b_i$ , and $c_i$ are all positive integers and $c_1 \le c_2 \le \cdots \le c_{14}$ . What is the ratio $\frac{c_{14}}{c_1}$ for the least possible value of $r$
$\textbf{(A)} ~\frac{21}{5} \qquad\textbf{(B)} ~\frac{85}{13} \qquad\textbf{(C)} ~7 \qquad\textbf{(D)} ~\frac{39}{5} \qquad\textbf{(E)} ~17$

Solution: Case 1: The tangent and the origin are on the opposite sides of the circle.
In this case, $a, b > 2r$
We can easily prove that \[a + b - 2 r = c . \hspace{1cm} (1)\]
Recall that $c = \sqrt{a^2 + b^2}$
Taking square of (1) and reorganizing all terms, (1) is converted as \[\left( a - 2 r \right) \left( b - 2 r \right) = 2 r^2 .\]
Case 2: The tangent and the origin are on the same sides of the circle.
In this case, $0 < a, b < r$
We can easily prove that \[2 r - a - b  = c . \hspace{1cm} (2)\]
Recall that $c = \sqrt{a^2 + b^2}$
Taking square of (2) and reorganizing all terms, (2) is converted as \[\left( a - 2 r \right) \left( b - 2 r \right) = 2 r^2 .\]
Putting both cases together, for given $r$ , we look for solutions of $a$ and $b$ satisfying \[\left( a - 2 r \right) \left( b - 2 r \right) = 2 r^2 ,\] with either $a, b > 2r$ or $0 < a, b < r$
Now, we need to find the smallest $r$ , such that the number of feasible solutions of $(a, b)$ is at least 14.
For equation \[uv = 2 r^2 ,\] we observe that the R.H.S. is a not a perfect square. Thus, the number of positive $(u, v)$ is equal to the number of positive divisors of $2 r^2$
Second, for each feasible positive solution $(u, v)$ , its opposite $(-u, -v)$ is also a solution. However, $(u,v)$ corresponds to a feasible solution if $(a, b)$ with $a = u + 2r$ and $b = v + 2r$ , but $(-u, -v)$ may not lead to a feasible solution if $(a, b)$ with $a = 2 r - u$ and $b = 2 r - v$
Recall that we are looking for $r$ that leads to at least 14 solutions.
Therefore, the above observations imply that we must have $r$ , such that $2 r^2$ has least 7 positive divisors.
Following this guidance, we find the smallest $r$ is 6. This leads to the following solutions:
$\left( a_1, b_1, c_1 \right) = \left( 3, 4, 5 \right)$ $\left( a_2, b_2, c_2 \right) = \left( 4, 3, 5 \right)$
$\left( a_3, b_3, c_3 \right) = \left( 20, 21, 29 \right)$ $\left( a_4, b_4, c_4 \right) = \left( 21, 20, 29 \right)$
$\left( a_5, b_5, c_5 \right) = \left( 18, 24, 30 \right)$ $\left( a_6, b_6, c_6 \right) = \left( 24, 18, 30 \right)$
$\left( a_7, b_7, c_7 \right) = \left( 16, 30, 34 \right)$ $\left( a_8, b_8, c_8 \right) = \left( 30, 16, 34 \right)$
$\left( a_9, b_9, c_9 \right) = \left( 15, 36, 39 \right)$ $\left( a_{10}, b_{10}, c_{10} \right) = \left( 36, 15, 39 \right)$
$\left( a_{11}, b_{11}, c_{11} \right) = \left( 14, 48, 50 \right)$ $\left( a_{12}, b_{12}, c_{12} \right) = \left( 48, 14, 50 \right)$
$\left( a_{13}, b_{13}, c_{13} \right) = \left( 13, 84, 85 \right)$ $\left( a_{14}, b_{14}, c_{14} \right) = \left( 84, 13, 85 \right)$
Therefore, $\frac{c_{14}}{c_1} = \boxed{17}$

Answer: E 17
Problem: A circle with integer radius $r$ is centered at $(r, r)$ . Distinct line segments of length $c_i$ connect points $(0, a_i)$ to $(b_i, 0)$ for $1 \le i \le 14$ and are tangent to the circle, where $a_i$ $b_i$ , and $c_i$ are all positive integers and $c_1 \le c_2 \le \cdots \le c_{14}$ . What is the ratio $\frac{c_{14}}{c_1}$ for the least possible value of $r$
$\textbf{(A)} ~\frac{21}{5} \qquad\textbf{(B)} ~\frac{85}{13} \qquad\textbf{(C)} ~7 \qquad\textbf{(D)} ~\frac{39}{5} \qquad\textbf{(E)} ~17$

Solution: As $c_i$ is the length of the segment $(0,a_i)$ and $(b_i,0)$ $a_i^2+b_i^2=c_i^2$ . The equation for the line that passes through $(0,a_i)$ and $(b_i,0)$ is $a_ix+b_iy-a_ib_i=0$
By the point-line distance formula from point $(r,r)$ to line $a_ix+b_iy-a_ib_i=0$
\[r = \frac{ |a_ir+b_ir-a_ib_i| }{ \sqrt{a_i^2+b_i^2} }, \quad c_i = \frac{ |a_ir+b_ir-a_ib_i| }{r} = |a_i+b_i- \frac{a_ib_i}{r}|\]
\[\because c_i \quad \text{is an integer}, \quad \therefore r|a_ib_i\]
To find $c_1$ and $c_{14}$ for the smallest $r$ , we will list Pythagorean triples $(a_i,b_i,c_i)$ with relatively prime elements from the smallest. Notice that we only need to find $7$ triples with $a_i<b_i$ , the $7$ other triples will simply be $(b_i,a_i,c_i)$ $a_i$ will not equal $b_i$ because then $c_i = a_i \cdot \sqrt{2}$ , meaning that $a_i$ $b_i$ , and $c_i$ cannot all be integers.
The $7$ smallest triples are: $(3,4,5)$ $(5,12,13)$ $(7,24,25)$ $(8,15,17)$ $(11,60,61)$ $(12,35,37)$ $(13,84,85)$
Therefore, $\frac{c_{14}}{c_1} = \frac{85}{5} = \boxed{17}$

Answer: E 17
Problem: The point $(-1, -2)$ is rotated $270^{\circ}$ counterclockwise about the point $(3, 1)$ . What are the coordinates of its new position?
$\textbf{(A)}\ (-3, -4) \qquad \textbf{(B)}\ (0,5) \qquad \textbf{(C)}\ (2,-1) \qquad \textbf{(D)}\ (4,3) \qquad \textbf{(E)}\ (6,-3)$

Solution: $(-1,-2)$ is $4$ units west and $3$ units south of $(3,1)$ . Performing a counterclockwise rotation of $270^{\circ}$ , which is equivalent to a clockwise rotation of $90^{\circ}$ , the answer is $3$ units west and $4$ units north of $(3,1)$ , or $\boxed{0,5}$

Answer: B 0,5
Problem: The point $(-1, -2)$ is rotated $270^{\circ}$ counterclockwise about the point $(3, 1)$ . What are the coordinates of its new position?
$\textbf{(A)}\ (-3, -4) \qquad \textbf{(B)}\ (0,5) \qquad \textbf{(C)}\ (2,-1) \qquad \textbf{(D)}\ (4,3) \qquad \textbf{(E)}\ (6,-3)$

Solution: We write $(-1, -2)$ as $-1-2i.$ We'd like to rotate about $(3, 1),$ which is $3+i$ in the complex plane, by an angle of $270^{\circ} = \frac{3\pi}{2} \text{ rad}$ counterclockwise.
The formula for rotating the complex number $z$ about the complex number $w$ by an angle of $\theta$ counterclockwise is given as $(z-w)e^{\theta i} + w.$ Plugging in our values $z = -1-2i, w = 3+i, \theta = \frac{3\pi}{2}$ , we evaluate the expression as $((-1-2i) - (3+i))e^{\frac{3\pi i}{2}} + (3+i) = (-4-3i)(-i) + (3+i) = 4i-3i^2+3+i = 4i-3+3+i = 5i,$ which corresponds to $\boxed{0,5}$ on the Cartesian plane.

Answer: B 0,5
Problem: The sequence $a_0,a_1,a_2,\cdots$ is a strictly increasing arithmetic sequence of positive integers such that \[2^{a_7}=2^{27} \cdot a_7.\] What is the minimum possible value of $a_2$
$\textbf{(A)}\ 8 \qquad \textbf{(B)}\ 12 \qquad \textbf{(C)}\ 16 \qquad \textbf{(D)}\ 17 \qquad \textbf{(E)}\ 22$

Solution: We can rewrite the given equation as $2^{a_7-27}=a_7$ . Hence, $a_7$ must be a power of $2$ and larger than $27$ . The first power of 2 that is larger than $27$ , namely $32$ , does satisfy the equation: $2^{32 - 27} = 2^5 = 32$ . In fact, this is the only solution; $2^{a_7-27}$ is exponential whereas $a_7$ is linear, so their graphs will not intersect again.
Now, let the common difference in the sequence be $d$ . Hence, $a_0 = 32 - 7d$ and $a_2 = 32 - 5d$ . To minimize $a_2$ , we maxmimize $d$ . Since the sequence contains only positive integers, $32 - 7d > 0$ and hence $d \leq 4$ . When $d = 4$ $a_2 = \boxed{12}$

Answer: B 12
Problem: Let $f(n) = \left( \frac{-1+i\sqrt{3}}{2} \right)^n + \left( \frac{-1-i\sqrt{3}}{2} \right)^n$ , where $i = \sqrt{-1}$ . What is $f(2022)$
$\textbf{(A)}\ -2 \qquad \textbf{(B)}\ -1 \qquad \textbf{(C)}\ 0 \qquad \textbf{(D)}\ \sqrt{3} \qquad \textbf{(E)}\ 2$

Solution: Converting both summands to exponential form, \begin{align*} -1 + i\sqrt{3} &= 2e^{\frac{2\pi i}{3}}, \\ -1 - i\sqrt{3} &= 2e^{-\frac{2\pi i}{3}} = 2e^{\frac{4\pi i}{3}}. \end{align*} Notice that both are scaled copies of the third roots of unity.
When we replace the summands with their exponential form, we get \[f(n) = \left(e^{\frac{2\pi i}{3}}\right)^n + \left(e^{\frac{4\pi i}{3}}\right)^n.\] When we substitute $n = 2022$ , we get \[f(2022) = \left(e^{\frac{2\pi i}{3}}\right)^{2022} + \left(e^{\frac{4\pi i}{3}}\right)^{2022}.\] We can rewrite $2022$ as $3 \cdot 674$ , how does that help? \[f(2022) = \left(e^{\frac{2\pi i}{3}}\right)^{3 \cdot 674} + \left(e^{\frac{4\pi i}{3}}\right)^{3 \cdot 674} = \left(\left(e^{\frac{2\pi i}{3}}\right)^{3}\right)^{674} + \left(\left(e^{\frac{4\pi i}{3}}\right)^{3}\right)^{674} = 1^{674} + 1^{674} = \boxed{2}.\] Since any third root of unity must cube to $1$

Answer: E 2
Problem: Let $f(n) = \left( \frac{-1+i\sqrt{3}}{2} \right)^n + \left( \frac{-1-i\sqrt{3}}{2} \right)^n$ , where $i = \sqrt{-1}$ . What is $f(2022)$
$\textbf{(A)}\ -2 \qquad \textbf{(B)}\ -1 \qquad \textbf{(C)}\ 0 \qquad \textbf{(D)}\ \sqrt{3} \qquad \textbf{(E)}\ 2$

Solution: The numbers $\frac{-1+i\sqrt{3}}{2}$ and $\frac{-1-i\sqrt{3}}{2}$ are both $\textbf{Eisenstein Units}$ (along with $1$ ), denoted as $\omega$ and $\omega^2$ , respectively. They have the property that when they are cubed, they equal to $1$ . Thus, we can immediately solve: \[\omega^{2022} + \omega^{2 \cdot 2022} = \omega^{3 \cdot 674} + \omega^{3 \cdot 2 \cdot 674} = 1^{674} + 1^{2 \cdot 674} = \boxed{2}.\] ~mathboy100

Answer: E 2
Problem: Let $f(n) = \left( \frac{-1+i\sqrt{3}}{2} \right)^n + \left( \frac{-1-i\sqrt{3}}{2} \right)^n$ , where $i = \sqrt{-1}$ . What is $f(2022)$
$\textbf{(A)}\ -2 \qquad \textbf{(B)}\ -1 \qquad \textbf{(C)}\ 0 \qquad \textbf{(D)}\ \sqrt{3} \qquad \textbf{(E)}\ 2$

Solution: We begin by recognizing this form looks similar to the definition of cosine: \[\cos(x)=\frac{e^{ix}+e^{-ix}}{2}.\] We can convert our two terms into exponential form to find \[f(n) = \left( e^{\frac{2\pi i}{3}} \right ) ^n + \left ( e^{-\frac{2\pi i}{3}} \right ) ^n=e^{\frac{2 \pi i n}{3}} + e^{-\frac{2\pi i n}{3}}.\] This simplifies nicely: \[f(n)=2\cos\left( \frac{2\pi n}{3} \right).\] Thus, \[f(2022)=2\cos \left ( \frac{2\pi (2022) }{3} \right) = 2\cos(1348 \pi) = \boxed{2}.\]

Answer: E 2
Problem: Let $f(n) = \left( \frac{-1+i\sqrt{3}}{2} \right)^n + \left( \frac{-1-i\sqrt{3}}{2} \right)^n$ , where $i = \sqrt{-1}$ . What is $f(2022)$
$\textbf{(A)}\ -2 \qquad \textbf{(B)}\ -1 \qquad \textbf{(C)}\ 0 \qquad \textbf{(D)}\ \sqrt{3} \qquad \textbf{(E)}\ 2$

Solution: Notice how this looks like the closed form of the Fibonacci sequence except different roots. This is motivation to turn this closed formula into a recurrence relation. The base of the exponents are the roots of the characteristic equation $r^3-1=0$ . So we have \begin{align*} a_0&=2\\ a_1&=-1\\ a_2&=-1\\ a_n&=a_{n-3} \end{align*} Every time $n$ is multiple of $3$ as is true when $n=2022$ $a_n= \boxed{2}$ ~lopkiloinm

Answer: E 2
Problem: Let $f(n) = \left( \frac{-1+i\sqrt{3}}{2} \right)^n + \left( \frac{-1-i\sqrt{3}}{2} \right)^n$ , where $i = \sqrt{-1}$ . What is $f(2022)$
$\textbf{(A)}\ -2 \qquad \textbf{(B)}\ -1 \qquad \textbf{(C)}\ 0 \qquad \textbf{(D)}\ \sqrt{3} \qquad \textbf{(E)}\ 2$

Solution: Converting the two terms into rectangular form,
\[f(2022)=\left(\cos{\frac{2\pi}{3}}+i\sin{\frac{2\pi}{3}}\right)^{2022}+\left(\cos{\frac{4\pi}{3}}+i\sin{\frac{4\pi}{3}}\right)^{2022}.\]
By DeMoivre's Theorem,
\[f(2022)=\left(\cos{\left(\frac{2\pi}{3}\cdot{2022}\right)}+i\sin{\left(\frac{2\pi}{3}\cdot{2022}\right)}\right)+\left(\cos{\left(\frac{4\pi}{3}\cdot{2022}\right)}+i\sin{\left(\frac{4\pi}{3}\cdot{2022}\right)}\right).\]
Note that $\cos{\pi\cdot{k}}=1$ if $k$ is even and $-1$ if $k$ is odd, and that $\sin{\pi\cdot{k}}=0$ for all integers $k$
All arguments are even in the second equation for $f(2022)$ , so the two $\cos$ terms are equal to $1$ , and the two $\sin$ terms are equal to $0$
Therefore the answer is $1+1=\boxed{2}.$

Answer: E 2
Problem: How many $4 \times 4$ arrays whose entries are $0$ s and $1$ s are there such that the row sums (the sum of the entries in each row) are $1, 2, 3,$ and $4,$ in some order, and the column sums (the sum of the entries in each column) are also $1, 2, 3,$ and $4,$ in some order? For example, the array \[\left[   \begin{array}{cccc}     1 & 1 & 1 & 0 \\     0 & 1 & 1 & 0 \\     1 & 1 & 1 & 1 \\     0 & 1 & 0 & 0 \\   \end{array} \right]\] satisfies the condition.
$\textbf{(A) }144 \qquad \textbf{(B) }240 \qquad \textbf{(C) }336 \qquad \textbf{(D) }576 \qquad \textbf{(E) }624$

Solution: Note that the arrays and the sum configurations have one-to-one correspondence. Furthermore, the row sum configuration and the column sum configuration are independent of each other. Therefore, the answer is $(4!)^2=\boxed{576}.$

Answer: D 576
Problem: How many $4 \times 4$ arrays whose entries are $0$ s and $1$ s are there such that the row sums (the sum of the entries in each row) are $1, 2, 3,$ and $4,$ in some order, and the column sums (the sum of the entries in each column) are also $1, 2, 3,$ and $4,$ in some order? For example, the array \[\left[   \begin{array}{cccc}     1 & 1 & 1 & 0 \\     0 & 1 & 1 & 0 \\     1 & 1 & 1 & 1 \\     0 & 1 & 0 & 0 \\   \end{array} \right]\] satisfies the condition.
$\textbf{(A) }144 \qquad \textbf{(B) }240 \qquad \textbf{(C) }336 \qquad \textbf{(D) }576 \qquad \textbf{(E) }624$

Solution: In this problem, we call a matrix that satisfies all constraints given in the problem a feasible matrix.
First, we observe that if a matrix is feasible, and we swap two rows or two columns to get a new matrix, then this new matrix is still feasible.
Therefore, any feasible matrix can be obtained through a sequence of such swapping operations from a feasible matrix where for all $i \in \left\{ 1, 2, 3 ,4 \right\}$ , the sum of entries in row $i$ is $i$ and the sum of entries in column $i$ is $i$ , hereafter called as a benchmark matrix.
Second, we observe that there is a unique benchmark matrix, as shown below: \[\left[   \begin{array}{cccc}     0 & 0 & 0 & 1 \\     0 & 0 & 1 & 1 \\     0 & 1 & 1 & 1 \\     1 & 1 & 1 & 1 \\   \end{array} \right]\] With above observations, we now count the number of feasible matrixes.
We construct a feasible matrix in the following steps.
Step 1: We make a permutation of rows of the benchmark matrix.
The number of ways is $4!$
Step 2: We make a permutation of columns of the matrix obtained after Step 1.
The number of ways is $4!$
Following from the rule of product, the total number of feasible matrixes is $4! \cdot 4! = \boxed{576}$

Answer: D 576
Problem: How many $4 \times 4$ arrays whose entries are $0$ s and $1$ s are there such that the row sums (the sum of the entries in each row) are $1, 2, 3,$ and $4,$ in some order, and the column sums (the sum of the entries in each column) are also $1, 2, 3,$ and $4,$ in some order? For example, the array \[\left[   \begin{array}{cccc}     1 & 1 & 1 & 0 \\     0 & 1 & 1 & 0 \\     1 & 1 & 1 & 1 \\     0 & 1 & 0 & 0 \\   \end{array} \right]\] satisfies the condition.
$\textbf{(A) }144 \qquad \textbf{(B) }240 \qquad \textbf{(C) }336 \qquad \textbf{(D) }576 \qquad \textbf{(E) }624$

Solution: Of the sixteen entries in one such array, there are six $0$ s and ten $1$ s. The rows and the columns each have zero, one, two, and three $0$ s, in some order. Once we decide the positions of the $0$ s, we form one such array.
We perform the following steps:
Together, the answer is $4\cdot4\cdot3\cdot3\cdot4=\boxed{576}.$

Answer: D 576
Problem: How many $4 \times 4$ arrays whose entries are $0$ s and $1$ s are there such that the row sums (the sum of the entries in each row) are $1, 2, 3,$ and $4,$ in some order, and the column sums (the sum of the entries in each column) are also $1, 2, 3,$ and $4,$ in some order? For example, the array \[\left[   \begin{array}{cccc}     1 & 1 & 1 & 0 \\     0 & 1 & 1 & 0 \\     1 & 1 & 1 & 1 \\     0 & 1 & 0 & 0 \\   \end{array} \right]\] satisfies the condition.
$\textbf{(A) }144 \qquad \textbf{(B) }240 \qquad \textbf{(C) }336 \qquad \textbf{(D) }576 \qquad \textbf{(E) }624$

Solution: Since exactly $1$ row sum is $4$ and exactly $1$ column sum is $4$ , there is a unique entry in the array such that it, and every other entry in the same row or column, is a $1.$ Since there are $16$ total entries in the array, there are $16$ ways to choose the entry with only $1$ s in its row and column.
Without loss of generality, let that entry be in the top-left corner of the square. Note that there is already $1$ entry numbered $1$ in each unfilled row, and $1$ entry numbered $1$ in each unfilled column. Since exactly $1$ row sum is $1$ and exactly $1$ column sum is $1$ , there is a unique entry in the $3\times3$ array of the empty squares such that it, and every other entry in the same row or column in the $3\times3$ array is a $0.$ Using a process similar to what we used in the first paragraph, we can see that there are $9$ ways to choose the entry with only $0$ in its row and column (in the $3\times3$ array).
Without loss of generality, let that entry be in the bottom-right corner of the square. Then, the remaining empty squares are the $4$ center squares. Of these, one of the columns of the empty $2\times2$ array will have two $1$ s and the other column will have one $1.$ That happens if and only if exactly $1$ of the remaining squares is filled with a $0$ , and there are $4$ ways to choose that square. Filling that square with a $0$ and the other $3$ squares with $1$ s completes the grid.
All in all, there are $4\cdot9\cdot16=\boxed{576}$ ways to complete the grid.

Answer: D 576
Problem: How many $4 \times 4$ arrays whose entries are $0$ s and $1$ s are there such that the row sums (the sum of the entries in each row) are $1, 2, 3,$ and $4,$ in some order, and the column sums (the sum of the entries in each column) are also $1, 2, 3,$ and $4,$ in some order? For example, the array \[\left[   \begin{array}{cccc}     1 & 1 & 1 & 0 \\     0 & 1 & 1 & 0 \\     1 & 1 & 1 & 1 \\     0 & 1 & 0 & 0 \\   \end{array} \right]\] satisfies the condition.
$\textbf{(A) }144 \qquad \textbf{(B) }240 \qquad \textbf{(C) }336 \qquad \textbf{(D) }576 \qquad \textbf{(E) }624$

Solution: Note that swapping any two rows or any two columns or both from the given example array, leads to a new array that satisfies the condition. There are $4$ rows, and you choose $2$ to swap, so $\frac{4!}{2!2!} = 6$ ; likewise for columns. Therefore, $6\cdot6 = 36$ . Clearly, the final answer must be divisible by $36$ , so we can eliminate $\textbf{(B)}$ $\textbf{(C)}$ and $\textbf{(E)}$
Now, you are in exam mode with not much time left. You see that $144 = 4\cdot36$ while $576 = 16\cdot36$ . There are $16$ elements in the array ( $4$ rows and $4$ columns), so you go with $\boxed{576}$ , and this is indeed the correct answer!

Answer: D 576
Problem: In $\triangle{ABC}$ medians $\overline{AD}$ and $\overline{BE}$ intersect at $G$ and $\triangle{AGE}$ is equilateral. Then $\cos(C)$ can be written as $\frac{m\sqrt p}n$ , where $m$ and $n$ are relatively prime positive integers and $p$ is a positive integer not divisible by the square of any prime. What is $m+n+p?$
$\textbf{(A) }44 \qquad \textbf{(B) }48 \qquad \textbf{(C) }52 \qquad \textbf{(D) }56 \qquad \textbf{(E) }60$

Solution: Let $AG = AE = GE = CE = 1$ . Since $G$ is the centroid, $DG = \frac12$ $BG = 2$
\[\angle BGD = \angle AGE = 60^{\circ}\]
By the Law of Cosine in $\triangle BGD$
\[BD^2 = BG^2 + DG^2 - 2 \cdot BG \cdot DG \cdot \cos \angle BGD\]
\[BD = \sqrt {2^2 + \left( \frac{1}{2} \right)^2 - 2 \cdot 2 \cdot \frac12 \cdot \cos \angle BGD} = \frac{\sqrt{13}}{2}, \quad CD = \frac{\sqrt{13}}{2}\]
By the Law of Cosine in $\triangle ACD$
\[AD^2 = AC^2 + CD^2 - 2 \cdot AC \cdot CD \cdot \cos \angle C\]
\[\cos \angle C = \frac{ AC^2 + CD^2 - AD^2 }{ 2 \cdot AC \cdot CD } = \frac{ 2^2 + \left( \frac{\sqrt{13}}{2} \right)^2 - \left( \frac{3}{2} \right)^2 }{ 2 \cdot 2 \cdot \frac{\sqrt{13}}{2} } = \frac{ 5 \sqrt{13} }{26}\]
\[5 + 13 + 26 = \boxed{44}\]

Answer: A 44
Problem: In $\triangle{ABC}$ medians $\overline{AD}$ and $\overline{BE}$ intersect at $G$ and $\triangle{AGE}$ is equilateral. Then $\cos(C)$ can be written as $\frac{m\sqrt p}n$ , where $m$ and $n$ are relatively prime positive integers and $p$ is a positive integer not divisible by the square of any prime. What is $m+n+p?$
$\textbf{(A) }44 \qquad \textbf{(B) }48 \qquad \textbf{(C) }52 \qquad \textbf{(D) }56 \qquad \textbf{(E) }60$

Solution: Using reference triangle $\triangle AGE$ , we can let \[A=(1,0,0),G=(0,1,0),E=(0,0,1),C=(-1,0,2),D=(-\tfrac{1}{2},\tfrac{3}{2},0),B=(0,3,-2).\] If we move $A,B,C$ each over by $(1,0,-2)$ , leaving $\angle C$ unchanged, we have \[A=(2,0,-2),B=(1,3,-4),C=(0,0,0).\] The angle $\theta$ between vectors $\overrightarrow{CA}$ and $\overrightarrow{CB}$ satisfies \[\cos\theta=\frac{(2)(1)+(0)(3)+(-2)(-4)}{\sqrt{\left[2^{2}+0^{2}+(-2)^{2}\right]\left[1^{2}+3^{2}+(-4)^{2}\right]}}=\frac{10}{\sqrt{8\cdot 26}}=\frac{10}{4\sqrt{13}}=\frac{5\sqrt{13}}{26},\] giving the answer, $5+13+26=\boxed{44}$

Answer: A 44
Problem: The figure below depicts a regular $7$ -gon inscribed in a unit circle. [asy]         import geometry; unitsize(3cm); draw(circle((0,0),1),linewidth(1.5)); for (int i = 0; i < 7; ++i) {   for (int j = 0; j < i; ++j) {     draw(dir(i * 360/7) -- dir(j * 360/7),linewidth(1.5));   } } for(int i = 0; i < 7; ++i) {    dot(dir(i * 360/7),5+black); } [/asy] What is the sum of the $4$ th powers of the lengths of all $21$ of its edges and diagonals?
$\textbf{(A) }49 \qquad \textbf{(B) }98 \qquad \textbf{(C) }147 \qquad \textbf{(D) }168 \qquad \textbf{(E) }196$

Solution: There are $7$ segments whose lengths are $2 \sin \frac{\pi}{7}$ $7$ segments whose lengths are $2 \sin \frac{2 \pi}{7}$ $7$ segments whose lengths are $2 \sin \frac{3\pi}{7}$
Therefore, the sum of the $4$ th powers of these lengths is \begin{align*} 7 \cdot 2^4 \sin^4 \frac{\pi}{7} + 7 \cdot 2^4 \sin^4 \frac{2 \pi}{7} + 7 \cdot 2^4 \sin^4 \frac{3 \pi}{7} & = \frac{7 \cdot 2^4}{(2i)^4} \left( e^{i \frac{\pi}{7}} - e^{i \frac{\pi}{7}} \right)^4 + \frac{7 \cdot 2^4}{(2i)^4} \left( e^{i \frac{2 \pi}{7}} - e^{i \frac{2 \pi}{7}} \right)^4 + \frac{7 \cdot 2^4}{(2i)^4} \left( e^{i \frac{3 \pi}{7}} - e^{i \frac{4 \pi}{7}} \right)^4 \\ & = 7 \left( e^{i \frac{4 \pi}{7}} - 4 e^{i \frac{2 \pi}{7}} + 6 - 4 e^{- i \frac{2 \pi}{7}} + e^{- i \frac{4 \pi}{7}} \right) \\ & \quad + 7 \left( e^{i \frac{8 \pi}{7}} - 4 e^{i \frac{4 \pi}{7}} + 6 - 4 e^{- i \frac{4 \pi}{7}} + e^{- i \frac{8 \pi}{7}} \right) \\ & \quad + 7 \left( e^{i \frac{12 \pi}{7}} - 4 e^{i \frac{6 \pi}{7}} + 6 - 4 e^{- i \frac{6 \pi}{7}} + e^{- i \frac{12 \pi}{7}} \right) \\ & = 7 \left( e^{i \frac{4 \pi}{7}} + e^{i \frac{8 \pi}{7}} + e^{i \frac{12 \pi}{7}} + e^{-i \frac{4 \pi}{7}} + e^{-i \frac{8 \pi}{7}} + e^{-i \frac{12 \pi}{7}} \right) \\ & \quad - 7 \cdot 4 \left( e^{i \frac{2 \pi}{7}} + e^{i \frac{4 \pi}{7}} + e^{i \frac{6 \pi}{7}} + e^{-i \frac{2 \pi}{7}} + e^{-i \frac{4 \pi}{7}} + e^{-i \frac{6 \pi}{7}} \right) \\ & \quad + 7 \cdot 6 \cdot 3 \\ & = 7 \left( e^{i \frac{4 \pi}{7}} + e^{-i \frac{6 \pi}{7}} + e^{-i \frac{2 \pi}{7}} + e^{-i \frac{4 \pi}{7}} + e^{i \frac{6 \pi}{7}} + e^{i \frac{2 \pi}{7}} \right) \\ & \quad - 7 \cdot 4 \left( e^{i \frac{2 \pi}{7}} + e^{i \frac{4 \pi}{7}} + e^{i \frac{6 \pi}{7}} + e^{-i \frac{2 \pi}{7}} + e^{-i \frac{4 \pi}{7}} + e^{-i \frac{6 \pi}{7}} \right) \\ & \quad + 7 \cdot 6 \cdot 3 \\ & = -7 + 7 \cdot 4 + 7 \cdot 6 \cdot 3 \\ & = \boxed{147} ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

Answer: C 147