Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: Let $x$ be the least real number greater than $1$ such that $\sin(x)= \sin(x^2)$ , where the arguments are in degrees. What is $x$ rounded up to the closest integer?
$\textbf{(A) } 10 \qquad \textbf{(B) } 13 \qquad \textbf{(C) } 14 \qquad \textbf{(D) } 19 \qquad \textbf{(E) } 20$

Solution: The smallest $x$ to make $\sin(x) = \sin(x^2)$ would require $x=x^2$ , but since $x$ needs to be greater than $1$ , these solutions are not valid.
The next smallest $x$ would require $x=180-x^2$ , or $x^2+x=180$
After a bit of guessing and checking, we find that $12^2+12=156$ , and $13^2+13=182$ , so the solution lies between $12{ }$ and $13$ , making our answer $\boxed{13}.$

Answer: B 13
Problem: Let $x$ be the least real number greater than $1$ such that $\sin(x)= \sin(x^2)$ , where the arguments are in degrees. What is $x$ rounded up to the closest integer?
$\textbf{(A) } 10 \qquad \textbf{(B) } 13 \qquad \textbf{(C) } 14 \qquad \textbf{(D) } 19 \qquad \textbf{(E) } 20$

Solution: For choice $\textbf{(A)},$ we have \begin{align*} \left| \sin x - \sin \left( x^2 \right) \right| & = \left| \sin 10^\circ - \sin \left( \left( 10^2 \right)^\circ \right) \right| \\ & = \left| \sin 10^\circ - \sin 100^\circ \right| \\ & = \left| \sin 10^\circ - \sin 80^\circ \right| \\ & = \sin 80^\circ - \sin 10^\circ . \end{align*} For choice $\textbf{(B)},$ we have \begin{align*} \left| \sin x - \sin \left( x^2 \right) \right| & = \left| \sin 13^\circ - \sin \left( \left( 13^2 \right)^\circ \right) \right| \\ & = \left| \sin 13^\circ - \sin 169^\circ \right| \\ & = \left| \sin 10^\circ - \sin 11^\circ \right| \\ & = \sin 11^\circ - \sin 10^\circ . \end{align*} For choice $\textbf{(C)},$ we have \begin{align*} \left| \sin x - \sin \left( x^2 \right) \right| & = \left| \sin 14^\circ - \sin \left( \left( 14^2 \right)^\circ \right) \right| \\ & = \left| \sin 14^\circ - \sin 196^\circ \right| \\ & = \left| \sin 14^\circ + \sin 16^\circ \right| \\ & = \sin 14^\circ + \sin 16^\circ . \end{align*} For choice $\textbf{(D)},$ we have \begin{align*} \left| \sin x - \sin \left( x^2 \right) \right| & = \left| \sin 19^\circ - \sin \left( \left( 19^2 \right)^\circ \right) \right| \\ & = \left| \sin 19^\circ - \sin 361^\circ \right| \\ & = \left| \sin 19^\circ - \sin 1^\circ \right| \\ & = \sin 19^\circ - \sin 1^\circ . \end{align*} For choice $\textbf{(E)},$ we have \begin{align*} \left| \sin x - \sin \left( x^2 \right) \right| & = \left| \sin 20^\circ - \sin \left( \left( 20^2 \right)^\circ \right) \right| \\ & = \left| \sin 20^\circ - \sin 400^\circ \right| \\ & = \left| \sin 20^\circ - \sin 40^\circ \right| \\ & = \sin 40^\circ - \sin 20^\circ . \end{align*} Therefore, the answer is $\boxed{13},$ as $\sin 11^\circ - \sin 10^\circ$ is the closest to $0.$

Answer: B 13
Problem: Azar and Carl play a game of tic-tac-toe. Azar places an in $X$ one of the boxes in a $3$ -by- $3$ array of boxes, then Carl places an $O$ in one of the remaining boxes. After that, Azar places an $X$ in one of the remaining boxes, and so on until all boxes are filled or one of the players has of their symbols in a row—horizontal, vertical, or diagonal—whichever comes first, in which case that player wins the game. Suppose the players make their moves at random, rather than trying to follow a rational strategy, and that Carl wins the game when he places his third $O$ . How many ways can the board look after the game is over?
$\textbf{(A) } 36 \qquad\textbf{(B) } 112 \qquad\textbf{(C) } 120 \qquad\textbf{(D) } 148 \qquad\textbf{(E) } 160$

Solution: We need to find out the number of configurations with 3 $O$ and 3 $X$ with 3 $O$ in a row, and 3 $X$ not in a row.
$\textbf{Case 1}$ : 3 $O$ are in a horizontal row or a vertical row.
Step 1: We determine the row that 3 $O$ occupy.
The number of ways is 6.
Step 2: We determine the configuration of 3 $X$
The number of ways is $\binom{6}{3} - 2 = 18$
In this case, following from the rule of product, the number of ways is $6 \cdot 18 = 108$
$\textbf{Case 2}$ : 3 $O$ are in a diagonal row.
Step 1: We determine the row that 3 $O$ occupy.
The number of ways is 2.
Step 2: We determine the configuration of 3 $X$
The number of ways is $\binom{6}{3}  = 20$
In this case, following from the rule of product, the number of ways is $2 \cdot 20 = 40$
Putting all cases together, the total number of ways is $108 + 40 = 148$
Therefore, the answer is $\boxed{148}$

Answer: D 148
Problem: Convex quadrilateral $ABCD$ has $AB = 18, \angle{A} = 60^\circ,$ and $\overline{AB} \parallel \overline{CD}.$ In some order, the lengths of the four sides form an arithmetic progression, and side $\overline{AB}$ is a side of maximum length. The length of another side is $a.$ What is the sum of all possible values of $a$
$\textbf{(A) } 24 \qquad \textbf{(B) } 42 \qquad \textbf{(C) } 60 \qquad \textbf{(D) } 66 \qquad \textbf{(E) } 84$

Solution: Let $E$ be a point on $\overline{AB}$ such that $BCDE$ is a parallelogram. Suppose that $BC=ED=b, CD=BE=c,$ and $DA=d,$ so $AE=18-c,$ as shown below. [asy] /* Made by MRENTHUSIASM */ size(250); pair A, B, C, D, E; A = (0,0); B = (18,0); D = A+9*dir(60); C = D+(7,0); E = D+(B-C); dot("$A$",A,1.5*SW,linewidth(4)); dot("$B$",B,1.5*SE,linewidth(4)); dot("$C$",C,1.5*NE,linewidth(4)); dot("$D$",D,1.5*NW,linewidth(4)); dot("$E$",E,1.5*S,linewidth(4)); draw(A--B--C--D--cycle); draw(D--E,dashed); label("$60^\circ$",A,2.5*dir(30),fontsize(10)); label("$18-c$",midpoint(A--E),1.5*S,red); label("$c$",midpoint(E--B),2.25*S,red); label("$b$",midpoint(B--C),scale(1.5)*rotate(90)*dir(midpoint(B--C)--B),red); label("$b$",midpoint(D--E),scale(1.5)*rotate(90)*dir(midpoint(E--D)--E),red); label("$c$",midpoint(C--D),1.5*N,red); label("$d$",midpoint(D--A),scale(1.5)*rotate(90)*dir(midpoint(D--A)--D),red); [/asy] We apply the Law of Cosines to $\triangle ADE:$ \begin{align*} AD^2 + AE^2 - 2\cdot AD\cdot AE\cdot\cos 60^\circ &= DE^2 \\ d^2 + (18-c)^2 - d(18-c) &= b^2 \\ (18-c)^2 - d(18-c) &= b^2 - d^2 \\ (18-c)(18-c-d) &= (b+d)(b-d). \hspace{15mm}(\bigstar) \end{align*} Let $k$ be the common difference of the arithmetic progression of the side-lengths. It follows that $b,c,$ and $d$ are $18-k, 18-2k,$ and $18-3k,$ in some order. It is clear that $0\leq k<6.$
If $k=0,$ then $ABCD$ is a rhombus with side-length $18,$ which is valid.
If $k\neq0,$ then we have six cases:
Together, the sum of all possible values of $a$ is $18+(13+3+8)+(14+12+16)=\boxed{84}.$

Answer: E 84
Problem: Convex quadrilateral $ABCD$ has $AB = 18, \angle{A} = 60^\circ,$ and $\overline{AB} \parallel \overline{CD}.$ In some order, the lengths of the four sides form an arithmetic progression, and side $\overline{AB}$ is a side of maximum length. The length of another side is $a.$ What is the sum of all possible values of $a$
$\textbf{(A) } 24 \qquad \textbf{(B) } 42 \qquad \textbf{(C) } 60 \qquad \textbf{(D) } 66 \qquad \textbf{(E) } 84$

Solution: Let $b, c$ , and $d$ denote the sides $BC, CD$ , and $AD$ respectively. [asy]  size(250); pair A, B, C, D, E; A = (0,0); B = (18,0); D = A+9*dir(60); C = D+(7,0); E = D+(B-C);  pen pdot=linewidth(3)+fontsize(12); dot("$A$",A,SW,pdot); dot("$B$",B,SE,pdot); dot("$C$",C,NE,pdot); dot("$D$",D,NW,pdot);   draw(A--B--C--D--cycle);   label("$60^\circ$",A,5*dir(30),fontsize(10));  label("$\theta$", B, 5*dir(155),fontsize(10)); pen plabel=red+fontsize(12); label("$18$",midpoint(A--B),1.5*S,plabel);  label("$b$", midpoint(B--C), scale(1.5)*rotate(90)* dir((B+C)/2--B), plabel);  label("$c$", (C+D)/2,1.5*N, plabel);  label("$d$",(D+A)/2, scale(1.5)*rotate(90)*dir((D+A)/2--D), plabel);  [/asy] Since $AB\parallel CD$ , we get \[\tfrac{\sqrt 3}{2}\ d = b\sin\theta \quad \textrm{and}\quad \tfrac 12 d + c + b\cos\theta = 18.\] Using $b^2\sin^2\theta + b^2\cos^2\theta = b^2$ , we eliminate $\theta$ from above to get $(36-2c-d)^2+3d^2=4b^2$ , which rearranges to $(36-2c-d)^2-d^2=4(b^2-d^2)$ , and, upon factoring, yields \begin{align}     (18-c)(18-c-d)=(b+d)(b-d). \end{align} We divide into two cases, depending on whether $c$ is the smallest side.
If $c$ is not the smallest side then $18-c=\pm (b-d)$ . If $c=18$ , we get a rhombus of side $18$ , so  one possible value is $a=18$ . Otherwise, we can cancel the common factor from $(1)$ . After rearranging we get \[18-c=-b \quad \textrm{or}\quad  18-c=b+2d.\] The first condition is false because $-b< 0 <18-c$ ; the second condition is false because $b+2d > |b-d| = 18-c$
If $c$ is the smallest side, then $18-c = \pm 3(b-d)$ . Assuming $c<18$ we can cancel common factors in $(1)$ to get \[8b=13d \quad \textrm{or}\quad 8b=7d.\] The first condition yields the solution $(c,d,b)=(3,8,13)$ and the second condition yields the solution $(c,b,d)=(12,14,16)$
Together, the sum of all possible values of $a$ is $18+(3+8+13)+(12+14+16)=\boxed{84}.$

Answer: E 84
Problem: Convex quadrilateral $ABCD$ has $AB = 18, \angle{A} = 60^\circ,$ and $\overline{AB} \parallel \overline{CD}.$ In some order, the lengths of the four sides form an arithmetic progression, and side $\overline{AB}$ is a side of maximum length. The length of another side is $a.$ What is the sum of all possible values of $a$
$\textbf{(A) } 24 \qquad \textbf{(B) } 42 \qquad \textbf{(C) } 60 \qquad \textbf{(D) } 66 \qquad \textbf{(E) } 84$

Solution: Denote $x = AD$ $\theta = \angle B$ .
Hence, $BC = \frac{\sqrt{3}}{2} \cdot \frac{x}{\sin \theta}$ $DC = 18 - \frac{x}{2} - \frac{\sqrt{3}}{2} x \cot \theta$
$\textbf{Case 1}$ $DC = AD = BC = AB$
This is a rhombus. So each side has length $18$
For the following cases, we consider four sides that have distinct lengths.
To make their lengths an arithmetic sequence, we must have $\theta \neq 120^\circ$
Therefore, in the subsequent analysis, we exclude the solution $\theta = 120^\circ$
$\textbf{Case 2}$ $DC < AD < BC < AB$
Because the lengths of these sides form an arithmetic sequence, we have the following system of equations: \[ AB - BC = BC - AD = AD - DC . \]
Hence, \begin{eqnarray*} & 18 - \frac{\sqrt{3}}{2}\cdot\frac{x}{\sin \theta} = \frac{\sqrt{3}}{2}\cdot\frac{x}{\sin \theta} - x = x - \left( 18 - \frac{x}{2} - \frac{\sqrt{3}}{2} x \cot \theta \right) . & \end{eqnarray*}
By solving this system of equations, we get $\left( \cos \theta , \sin \theta , x\right) = \left( \frac{11}{13} , \frac{4 \sqrt{3}}{13} , 8 \right)$
Thus, in this case, $DC = 3$ $AD = 8$ $BC = 13$
$\textbf{Case 3}$ $DC < BC < AD < AB$
Because the lengths of these sides form an arithmetic sequence, we have the following system of equations: \[ AB - AD = AD - BC = BC - DC . \]
Hence, \begin{eqnarray*} & 18 - x = x - \frac{\sqrt{3}}{2}\cdot\frac{x}{\sin \theta} = \frac{\sqrt{3}}{2}\cdot\frac{x}{\sin \theta}  - \left( 18 - \frac{x}{2} - \frac{\sqrt{3}}{2} x \cot \theta \right) . & \end{eqnarray*}
By solving this system of equations, we get $\left( \cos \theta , \sin \theta , x\right) = \left( - \frac{1}{7} , \frac{4 \sqrt{3}}{7} , 16 \right)$
Thus, in this case, $DC = 12$ $AD = 16$ $BC = 14$
$\textbf{Case 4}$ $BC < CD < AD < AB$
By doing the similar analysis, we can show there is no solution in this case.
$\textbf{Case 5}$ $BC < AD < CD < AB$
By doing the similar analysis, we can show there is no solution in this case.
$\textbf{Case 6}$ $AD < CD < BC < AB$
By doing the similar analysis, we can show there is no solution in this case.
$\textbf{Case 7}$ $AD < BC < CD < AB$
By doing the similar analysis, we can show there is no solution in this case.
Therefore, the sum of all possible values of $a$ is \begin{align*} 18 + \left( 3 + 8 + 13 \right) + \left( 12 + 14 + 16 \right) & = 84 . \end{align*}
Therefore, the answer is $\boxed{84}$

Answer: E 84
Problem: Let $m\ge 5$ be an odd integer, and let $D(m)$ denote the number of quadruples $(a_1, a_2, a_3, a_4)$ of distinct integers with $1\le a_i \le m$ for all $i$ such that $m$ divides $a_1+a_2+a_3+a_4$ . There is a polynomial \[q(x) = c_3x^3+c_2x^2+c_1x+c_0\] such that $D(m) = q(m)$ for all odd integers $m\ge 5$ . What is $c_1?$
$\textbf{(A)}\ {-}6\qquad\textbf{(B)}\ {-}1\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ 11$

Solution: For a fixed value of $m,$ there is a total of $m(m-1)(m-2)(m-3)$ possible ordered quadruples $(a_1, a_2, a_3, a_4).$
Let $S=a_1+a_2+a_3+a_4.$ We claim that exactly $\frac1m$ of these $m(m-1)(m-2)(m-3)$ ordered quadruples satisfy that $m$ divides $S:$
Since $\gcd(m,4)=1,$ we conclude that \[\{k+4(0),k+4(1),k+4(2),\ldots,k+4(m-1)\}\] is the complete residue system modulo $m$ for all integers $k.$
Given any ordered quadruple $(a'_1, a'_2, a'_3, a'_4)$ in modulo $m,$ it follows that exactly one of these $m$ ordered quadruples has sum $0$ modulo $m:$ \[\begin{array}{c|c} & \\ [-2.5ex] \textbf{Ordered Quadruple} & \textbf{Sum Modulo }\boldsymbol{m} \\ [0.5ex] \hline & \\ [-2ex] (a'_1, a'_2, a'_3, a'_4) & S'+4(0) \\ [0.5ex] (a'_1+1, a'_2+1, a'_3+1, a'_4+1) & S'+4(1) \\ [0.5ex] (a'_1+2, a'_2+2, a'_3+2, a'_4+2) & S'+4(2) \\ [0.5ex] \cdots & \cdots \\ [0.5ex] (a'_1+m-1, a'_2+m-1, a'_3+m-1, a'_4+m-1) & S'+4(m-1) \\ [0.5ex] \end{array}\] We conclude that $q(m)=\frac1m\cdot[m(m-1)(m-2)(m-3)]=(m-1)(m-2)(m-3),$ so \[q(x)=(x-1)(x-2)(x-3)=c_3x^3+c_2x^2+c_1x+c_0.\] By Vieta's Formulas, we get $c_1=1\cdot2+1\cdot3+2\cdot3=\boxed{11}.$

Answer: E 11
Problem: Let $m\ge 5$ be an odd integer, and let $D(m)$ denote the number of quadruples $(a_1, a_2, a_3, a_4)$ of distinct integers with $1\le a_i \le m$ for all $i$ such that $m$ divides $a_1+a_2+a_3+a_4$ . There is a polynomial \[q(x) = c_3x^3+c_2x^2+c_1x+c_0\] such that $D(m) = q(m)$ for all odd integers $m\ge 5$ . What is $c_1?$
$\textbf{(A)}\ {-}6\qquad\textbf{(B)}\ {-}1\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ 11$

Solution: As before, note that we have $m(m-1)(m-2)$ numbers we can choose as $a,b,c.$ From here, there is exactly one possible value of $1 \leq d \leq m$ that could make $S=a+b+c+d$ divisible by $m.$ However, there is a $\frac{3}{m}$ chance that this value of $d$ has already been chosen as $a,b$ or $c$ . Thus our polynomial is $m(m-1)(m-2)\left(1-\frac{3}{m}\right)=m(m-1)(m-2)\left(\frac{m-3}{m}\right)=(m-1)(m-2)(m-3)$ . By Vieta's, $c_1 = 2+3+6=\boxed{11}$

Answer: E 11
Problem: The product of the lengths of the two congruent sides of an obtuse isosceles triangle is equal to the product of the base and twice the triangle's height to the base. What is the measure, in degrees, of the vertex angle of this triangle?
$\textbf{(A)} \: 105 \qquad\textbf{(B)} \: 120 \qquad\textbf{(C)} \: 135 \qquad\textbf{(D)} \: 150 \qquad\textbf{(E)} \: 165$

Solution: Let the lengths of the two congruent sides of the triangle be $x$ , then the product desired is $x^2$
Notice that the product of the base and twice the height is $4$ times the area of the triangle.
Set the vertex angle to be $a$ , we derive the equation:
$x^2=4\left(\frac{1}{2}x^2\sin(a)\right)$
$\sin(a)=\frac{1}{2}$
As the triangle is obtuse, $a=150^\circ$ only. We get $\boxed{150}.$

Answer: D 150
Problem: The product of the lengths of the two congruent sides of an obtuse isosceles triangle is equal to the product of the base and twice the triangle's height to the base. What is the measure, in degrees, of the vertex angle of this triangle?
$\textbf{(A)} \: 105 \qquad\textbf{(B)} \: 120 \qquad\textbf{(C)} \: 135 \qquad\textbf{(D)} \: 150 \qquad\textbf{(E)} \: 165$

Solution: Denote by $a$ the length of each congruent side. Denote by $\theta$ the degree measure of each acute angle.
Denote by $\phi$ the degree measure of the obtuse angle.
Hence, this problem tells us the following relationship: \[ a^2 = 2 a \cos \theta \cdot 2 a \sin \theta . \]
Hence, \begin{align*} 1 & = 2 \cdot 2 \sin \theta \cos \theta \\ & = 2 \sin 2 \theta \\ & = 2 \sin \left( 180^\circ - 2 \theta \right) \\ & = 2 \sin \phi . \end{align*}
Hence, $\phi = 150^\circ$
Therefore, the answer is $\boxed{150}$

Answer: D 150
Problem: Triangle $ABC$ is equilateral with side length $6$ . Suppose that $O$ is the center of the inscribed
circle of this triangle. What is the area of the circle passing through $A$ $O$ , and $C$
$\textbf{(A)} \: 9\pi \qquad\textbf{(B)} \: 12\pi \qquad\textbf{(C)} \: 18\pi \qquad\textbf{(D)} \: 24\pi \qquad\textbf{(E)} \: 27\pi$

Solution: Construct the circle that passes through $A$ $O$ , and $C$ , centered at $X$
Also notice that $\overline{OA}$ and $\overline{OC}$ are the angle bisectors of angle $\angle BAC$ and $\angle BCA$ respectively. We then deduce $\angle AOC=120^\circ$
Consider another point $M$ on Circle $X$ opposite to point $O$
As $AOCM$ is an inscribed quadrilateral of Circle $X$ $\angle AMC=180^\circ-120^\circ=60^\circ$
Afterward, deduce that $\angle AXC=2·\angle AMC=120^\circ$
By the Cosine Rule, we have the equation: (where $r$ is the radius of circle $X$
$2r^2(1-\cos(120^\circ))=6^2$
$r^2=12$
The area is therefore $\pi r^2 = \boxed{12}$

Answer: B 12
Problem: Triangle $ABC$ is equilateral with side length $6$ . Suppose that $O$ is the center of the inscribed
circle of this triangle. What is the area of the circle passing through $A$ $O$ , and $C$
$\textbf{(A)} \: 9\pi \qquad\textbf{(B)} \: 12\pi \qquad\textbf{(C)} \: 18\pi \qquad\textbf{(D)} \: 24\pi \qquad\textbf{(E)} \: 27\pi$

Solution: We have $\angle AOC = 120^\circ$
Denote by $R$ the circumradius of $\triangle AOC$ .
In $\triangle AOC$ , the law of sines implies \[ 2 R = \frac{AC}{\sin \angle AOC} = 4 \sqrt{3} . \]
Hence, the area of the circumcircle of $\triangle AOC$ is \[ \pi R^2 = 12 \pi . \]
Therefore, the answer is $\boxed{12}$

Answer: B 12
Problem: Triangle $ABC$ is equilateral with side length $6$ . Suppose that $O$ is the center of the inscribed
circle of this triangle. What is the area of the circle passing through $A$ $O$ , and $C$
$\textbf{(A)} \: 9\pi \qquad\textbf{(B)} \: 12\pi \qquad\textbf{(C)} \: 18\pi \qquad\textbf{(D)} \: 24\pi \qquad\textbf{(E)} \: 27\pi$

Solution: As in the previous solution, construct the circle that passes through $A$ $O$ , and $C$ , centered at $X$ . Let $Y$ be the intersection of $\overline{OX}$ and $\overline{AB}$
Note that since $\overline{OA}$ is the angle bisector of $\angle BAC$ that $\angle OAC=30^\circ$ . Also by symmetry, $\overline{OX}$ $\perp$ $\overline{AB}$ and $AY = 3$ . Thus $\tan(30^\circ) = \frac{OY}{3}$ so $OY = \sqrt{3}$
Let $r$ be the radius of circle $X$ , and note that $AX = OX = r$ . So $\triangle AYX$ is a right triangle with legs of length $3$ and $r - \sqrt{3}$ and hypotenuse $r$ . By Pythagoras, $3^2 + (r - \sqrt{3})^2 = r^2$ . So $r = 2\sqrt{3}$
Thus the area is $\pi r^2 = \boxed{12}$

Answer: B 12
Problem: Triangle $ABC$ is equilateral with side length $6$ . Suppose that $O$ is the center of the inscribed
circle of this triangle. What is the area of the circle passing through $A$ $O$ , and $C$
$\textbf{(A)} \: 9\pi \qquad\textbf{(B)} \: 12\pi \qquad\textbf{(C)} \: 18\pi \qquad\textbf{(D)} \: 24\pi \qquad\textbf{(E)} \: 27\pi$

Solution: The semiperimeter is $\frac{6+6+6}{2}=9$ units.
The area of the triangle is $9\sqrt{3}$ units squared.  By the formula that says that the area of the triangle is its semiperimeter times its inradius, the inradius $r=\sqrt{3}$ . As $\angle{AOC}=120^\circ$ , we can form an altitude from point $O$ to side $AC$ at point $M$ , forming two 30-60-90 triangles. As $CM=MA=3$ , we can solve for $OC=2\sqrt{3}$ . Now, the area of the circle is just $\pi*(2*\sqrt{3})^2 = 12\pi$ . Select $\boxed{12}$

Answer: B 12
Problem: Triangle $ABC$ is equilateral with side length $6$ . Suppose that $O$ is the center of the inscribed
circle of this triangle. What is the area of the circle passing through $A$ $O$ , and $C$
$\textbf{(A)} \: 9\pi \qquad\textbf{(B)} \: 12\pi \qquad\textbf{(C)} \: 18\pi \qquad\textbf{(D)} \: 24\pi \qquad\textbf{(E)} \: 27\pi$

Solution: Call the diameter of the circle $d$ . If we extend points $A$ and $C$ to meet at a point on the circle and call it $E$ , then $\bigtriangleup OAE=\bigtriangleup OCE$ . Note that both triangles are right, since their hypotenuse is the diameter of the circle. Therefore, $CE=AE=\sqrt{d^2-12}$ . We know this since $OC=OA=OB$ and $OC$ is the hypotenuse of a $30-60-90$ right triangle, with the longer leg being $\frac{6}{2}=3$ so $OC=2\sqrt{3}$ . Applying Ptolemy's Theorem on cyclic quadrilateral $OCEA$ , we get $2({\sqrt{d^2-12}})\cdot{2\sqrt{3}}=6d$ . Squaring and solving we get $d^2=48 \Longrightarrow (2r)^2=48$ so $r^2=12$ . Therefore, the area of the circle is $\boxed{12}$

Answer:  12
Problem: What is the sum of all possible values of $t$ between $0$ and $360$ such that the triangle in the coordinate plane whose vertices are \[(\cos 40^\circ,\sin 40^\circ), (\cos 60^\circ,\sin 60^\circ), \text{ and } (\cos t^\circ,\sin t^\circ)\] is isosceles?
$\textbf{(A)} \: 100 \qquad\textbf{(B)} \: 150 \qquad\textbf{(C)} \: 330 \qquad\textbf{(D)} \: 360 \qquad\textbf{(E)} \: 380$

Solution: Let $A = (\cos 40^{\circ}, \sin 40^{\circ}), B = (\cos 60^{\circ}, \sin 60^{\circ}),$ and $C = (\cos t^{\circ}, \sin t^{\circ}).$ We apply casework to the legs of isosceles $\triangle ABC:$
Together, the sum of all such possible values of $t$ is $20+80+50+230=\boxed{380}.$

Answer: E 380
Problem: Let $c = \frac{2\pi}{11}.$ What is the value of \[\frac{\sin 3c \cdot \sin 6c \cdot \sin 9c \cdot \sin 12c \cdot \sin 15c}{\sin c \cdot \sin 2c \cdot \sin 3c \cdot \sin 4c \cdot \sin 5c}?\]
$\textbf{(A)}\ {-}1 \qquad\textbf{(B)}\ {-}\frac{\sqrt{11}}{5} \qquad\textbf{(C)}\ \frac{\sqrt{11}}{5} \qquad\textbf{(D)}\ \frac{10}{11} \qquad\textbf{(E)}\ 1$

Solution: Plugging in $c$ , we get \[\frac{\sin 3c \cdot \sin 6c \cdot \sin 9c \cdot \sin 12c \cdot \sin 15c}{\sin c \cdot \sin 2c \cdot \sin 3c \cdot \sin 4c \cdot \sin 5c}=\frac{\sin \frac{6\pi}{11} \cdot \sin \frac{12\pi}{11} \cdot \sin \frac{18\pi}{11} \cdot \sin \frac{24\pi}{11} \cdot \sin \frac{30\pi}{11}}{\sin \frac{2\pi}{11} \cdot \sin \frac{4\pi}{11} \cdot \sin \frac{6\pi}{11} \cdot \sin \frac{8\pi}{11} \cdot \sin \frac{10\pi}{11}}.\] Since $\sin(x+2\pi)=\sin(x),$ $\sin(2\pi-x)=\sin(-x),$ and $\sin(-x)=-\sin(x),$ we get \[\frac{\sin \frac{6\pi}{11} \cdot \sin \frac{12\pi}{11} \cdot \sin \frac{18\pi}{11} \cdot \sin \frac{24\pi}{11} \cdot \sin \frac{30\pi}{11}}{\sin \frac{2\pi}{11} \cdot \sin \frac{4\pi}{11} \cdot \sin \frac{6\pi}{11} \cdot \sin \frac{8\pi}{11} \cdot \sin \frac{10\pi}{11}}=\frac{\sin \frac{6\pi}{11} \cdot \sin \frac{-10\pi}{11} \cdot \sin \frac{-4\pi}{11} \cdot \sin \frac{2\pi}{11} \cdot \sin \frac{8\pi}{11}}{\sin \frac{2\pi}{11} \cdot \sin \frac{4\pi}{11} \cdot \sin \frac{6\pi}{11} \cdot \sin \frac{8\pi}{11} \cdot \sin \frac{10\pi}{11}}=\boxed{1}.\]

Answer: E 1
Problem: Let $c = \frac{2\pi}{11}.$ What is the value of \[\frac{\sin 3c \cdot \sin 6c \cdot \sin 9c \cdot \sin 12c \cdot \sin 15c}{\sin c \cdot \sin 2c \cdot \sin 3c \cdot \sin 4c \cdot \sin 5c}?\]
$\textbf{(A)}\ {-}1 \qquad\textbf{(B)}\ {-}\frac{\sqrt{11}}{5} \qquad\textbf{(C)}\ \frac{\sqrt{11}}{5} \qquad\textbf{(D)}\ \frac{10}{11} \qquad\textbf{(E)}\ 1$

Solution: Eisenstein used such a quotient in his proof of quadratic reciprocity . Let $c=\frac{2\pi}{p}$ where $p$ is an odd prime number and $q$ is any integer.
Then $\dfrac{\sin(qc)\sin(2qc)\cdots\sin(\frac{p-1}{2}qc)}{\sin(c)\sin(2c)\cdots\sin(\frac{p-1}{2}c)}$ is the Legendre symbol $\left(\frac{q}{p}\right)$ . Legendre symbol is calculated using quadratic reciprocity which is $\left(\frac{p}{q}\right)\left(\frac{q}{p}\right)=(-1)^{\frac{p-1}{2}\frac{q-1}{2}}$ . The Legendre symbol $\left(\frac{3}{11}\right)=(-1)\left(\frac{11}{3}\right)=(-1)\left(\frac{-1}{3}\right)=(-1)(-1)=\boxed{1}$

Answer: E 1