Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: The solutions to the equation $\log_{3x} 4 = \log_{2x} 8$ , where $x$ is a positive real number other than $\tfrac{1}{3}$ or $\tfrac{1}{2}$ , can be written as $\tfrac {p}{q}$ where $p$ and $q$ are relatively prime positive integers. What is $p + q$
$\textbf{(A) } 5   \qquad     \textbf{(B) } 13   \qquad     \textbf{(C) } 17   \qquad    \textbf{(D) } 31 \qquad   \textbf{(E) } 35$

Solution: We can convert both $4$ and $8$ into $2^2$ and $2^3,$ respectively: \[2\log_{3x} (2) = 3\log_{2x} (2).\] Converting the bases of the right side, we get \begin{align*} \log_{2x} 2 &= \frac{\ln 2}{\ln (2x)} \\ \frac{2}{3}\cdot\log_{3x} (2) &= \frac{\ln 2}{\ln (2x)} \\ 2^\frac{2}{3} &= (3x)^\frac{\ln 2}{\ln (2x)} \\ \frac{2}{3} \cdot \ln 2 &= \frac{\ln 2}{\ln (2x)} \cdot \ln (3x). \end{align*} Dividing both sides by $\ln 2,$ we get $\frac{2}{3} = \frac{\ln (3x)}{\ln (2x)},$ from which \[2\ln (2x) = 3\ln (3x).\] Expanding this equation gives \begin{align*} 2\ln 2 + 2\ln (x) &= 3\ln 3 + 3\ln (x) \\ \ln (x) &= 2\ln 2 - 3\ln 3. \end{align*} Thus, we have \[x = e^{2\ln 2 - 3\ln 3} = \frac{e^{2\ln 2}}{e^{3\ln 3}} = \frac{2^2}{3^3} = \frac{4}{27},\] from which the answer is $4+27=\boxed{31}.$

Answer: D 31
Problem: The solutions to the equation $\log_{3x} 4 = \log_{2x} 8$ , where $x$ is a positive real number other than $\tfrac{1}{3}$ or $\tfrac{1}{2}$ , can be written as $\tfrac {p}{q}$ where $p$ and $q$ are relatively prime positive integers. What is $p + q$
$\textbf{(A) } 5   \qquad     \textbf{(B) } 13   \qquad     \textbf{(C) } 17   \qquad    \textbf{(D) } 31 \qquad   \textbf{(E) } 35$

Solution: Let $y=\log_{3x} 4 = \log_{2x} 8.$ We convert the equations with $y$ to the exponential form: \begin{align*} (3x)^y&=4, \\ (2x)^y&=8. \end{align*} Cubing the first equation and squaring the second equation, we have \begin{align*} (3x)^{3y}&=64, \\ (2x)^{2y}&=64. \end{align*} Applying the Transitive Property, we get \begin{align*} (3x)^{3y}&=(2x)^{2y} \\ (3x)^3&=(2x)^2 \\ 27x^3&=4x^2 \\ x&=\frac{4}{27}, \end{align*} from which the answer is $4+27=\boxed{31}.$

Answer: D 31
Problem: Let $A$ be the set of positive integers that have no prime factors other than $2$ $3$ , or $5$ . The infinite sum \[\frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{8} + \frac{1}{9} + \frac{1}{10} + \frac{1}{12} + \frac{1}{15} + \frac{1}{16} + \frac{1}{18} + \frac{1}{20} + \cdots\] of the reciprocals of the elements of $A$ can be expressed as $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. What is $m+n$
$\textbf{(A) } 16 \qquad \textbf{(B) } 17 \qquad \textbf{(C) } 19 \qquad \textbf{(D) } 23 \qquad \textbf{(E) } 36$

Solution: Note that the fractions of the form $\frac{1}{2^a3^b5^c},$ where $a,b,$ and $c$ are nonnegative integers, span all terms of the infinite sum.
Therefore, the infinite sum becomes \begin{align*} \sum_{a=0}^{\infty}\sum_{b=0}^{\infty}\sum_{c=0}^{\infty}\frac{1}{2^a3^b5^c} &= \left(\sum_{a=0}^{\infty}\frac{1}{2^a}\right)\cdot\left(\sum_{b=0}^{\infty}\frac{1}{3^b}\right)\cdot\left(\sum_{c=0}^{\infty}\frac{1}{5^c}\right) \\ &= \frac{1}{1-\frac12}\cdot\frac{1}{1-\frac13}\cdot\frac{1}{1-\frac15} \\ &= 2\cdot\frac32\cdot\frac54 \\ &= \frac{15}{4} \end{align*} by a product of geometric series, from which the answer is $15+4=\boxed{19}.$

Answer: C 19
Problem: Let $A$ be the set of positive integers that have no prime factors other than $2$ $3$ , or $5$ . The infinite sum \[\frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{8} + \frac{1}{9} + \frac{1}{10} + \frac{1}{12} + \frac{1}{15} + \frac{1}{16} + \frac{1}{18} + \frac{1}{20} + \cdots\] of the reciprocals of the elements of $A$ can be expressed as $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. What is $m+n$
$\textbf{(A) } 16 \qquad \textbf{(B) } 17 \qquad \textbf{(C) } 19 \qquad \textbf{(D) } 23 \qquad \textbf{(E) } 36$

Solution: This solution is the same as Solution 1 but potentially clearer.
Clearly this is just summing over the reciprocals of the numbers of the form $2^i3^j5^k$ , where $i,j,k\in [0,\infty)$ . SO our desired sum is $\sum_{k=0}^{\infty}\sum_{j=0}^{\infty}\sum_{i=0}^{\infty}\frac{1}{2^i3^j5^k}$ . By the infinite geometric series formula, $\sum_{i=0}^{\infty}\frac{1}{2^i3^j5^k}$ is just $\frac{\frac{1}{3^j5^k}}{1-\frac{1}{2}}=\frac{2}{3^j5^k}$ . Applying the infinite geometric series formula again gives that $\sum_{j=0}^{\infty}\frac{2}{3^j5^k}=\frac{\frac{2}{5^k}}{1-\frac{1}{3}}=\frac{3}{5^k}$ . Applying the infinite geometric series formula again yields $\sum_{k=0}^{\infty}\frac{3}{5^k}=\frac{3}{1-\frac{1}{5}}=\frac{15}{4}$ . Hence our final answer is $15+4=\boxed{19}$

Answer: C 19
Problem: Let $A$ be the set of positive integers that have no prime factors other than $2$ $3$ , or $5$ . The infinite sum \[\frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{8} + \frac{1}{9} + \frac{1}{10} + \frac{1}{12} + \frac{1}{15} + \frac{1}{16} + \frac{1}{18} + \frac{1}{20} + \cdots\] of the reciprocals of the elements of $A$ can be expressed as $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. What is $m+n$
$\textbf{(A) } 16 \qquad \textbf{(B) } 17 \qquad \textbf{(C) } 19 \qquad \textbf{(D) } 23 \qquad \textbf{(E) } 36$

Solution: Separate into $7$ separate infinite series's so we can calculate each and find the original sum:
The first infinite sequence shall be all the reciprocals of the powers of $2$ , the second shall be reciprocals of the powers of $3$ , and the third will consist of reciprocals of the powers of $5$ . We can easily calculate these to be $1, \frac{1}{2}, \frac{1}{4}$ respectively.
The fourth infinite series shall be all real numbers in the form $\frac{1}{2^a3^b}$ , where $a,b\geq1$
The fifth is all real numbers in the form $\frac{1}{2^a5^b}$ , where $a,b\geq1$
The sixth is all real numbers in the form $\frac{1}{3^a5^b}$ , where $a,b\geq1$
The seventh infinite series is all real numbers in the form $\frac{1}{2^a3^b5^c}$ , where $a,b,c\geq1$
Let us denote the first sequence as $a_{1}$ , the second as $a_{2}$ , etc. We know $a_{1}=1$ $a_{2}=\frac{1}{2}$ $a_{3}=\frac{1}{4}$ , let us find $a_{4}$ . factoring out $\frac{1}{6}$ from the terms in this subsequence, we would get $a_{4}=\frac{1}{6}(1+a_{1}+a_{2}+a_{4})$
Knowing $a_{1}$ and $a_{2}$ , we can substitute and solve for $a_{4}$ , and we get $\frac{1}{2}$ . If we do similar procedures for the fifth and sixth sequences, we can solve for them too, and we get after solving them $\frac{1}{4}$ and $\frac{1}{8}$
Finally, for the seventh sequence, we see $a_{7}=\frac{a_{8}}{30}$ , where $a_{8}$ is the infinite series the problem is asking us to solve for. The sum of all seven subsequences will equal the one we are looking for, and we need to add the $\frac11$ term back: $1+1+\frac{1}{2}+\frac{1}{4}+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{a_{8}}{30}=a_{8}$ . We solve this to get $\frac{29}{8}=\frac{29a_{8}}{30}$ , or $\frac{15}{4}=a_{8}$ . So our answer is $\frac{15}{4}$ , but we are asked to add the numerator and denominator, which sums up to $\boxed{19}$

Answer: C 19
Problem: Triangle $ABC$ is an isosceles right triangle with $AB=AC=3$ . Let $M$ be the midpoint of hypotenuse $\overline{BC}$ . Points $I$ and $E$ lie on sides $\overline{AC}$ and $\overline{AB}$ , respectively, so that $AI>AE$ and $AIME$ is a cyclic quadrilateral. Given that triangle $EMI$ has area $2$ , the length $CI$ can be written as $\frac{a-\sqrt{b}}{c}$ , where $a$ $b$ , and $c$ are positive integers and $b$ is not divisible by the square of any prime. What is the value of $a+b+c$
$\textbf{(A) }9 \qquad \textbf{(B) }10 \qquad \textbf{(C) }11 \qquad \textbf{(D) }12 \qquad \textbf{(E) }13 \qquad$

Solution: Observe that $\triangle{EMI}$ is isosceles right ( $M$ is the midpoint of diameter arc $EI$ since $m\angle MEI = m\angle MAI = 45^\circ$ ), so $MI=2,MC=\frac{3}{\sqrt{2}}$ . With $\angle{MCI}=45^\circ$ , we can use Law of Cosines to determine that $CI=\frac{3\pm\sqrt{7}}{2}$ . The same calculations hold for $BE$ also, and since $CI<BE$ , we deduce that $CI$ is the smaller root, giving the answer of $\boxed{12}$

Answer: D 12
Problem: Triangle $ABC$ is an isosceles right triangle with $AB=AC=3$ . Let $M$ be the midpoint of hypotenuse $\overline{BC}$ . Points $I$ and $E$ lie on sides $\overline{AC}$ and $\overline{AB}$ , respectively, so that $AI>AE$ and $AIME$ is a cyclic quadrilateral. Given that triangle $EMI$ has area $2$ , the length $CI$ can be written as $\frac{a-\sqrt{b}}{c}$ , where $a$ $b$ , and $c$ are positive integers and $b$ is not divisible by the square of any prime. What is the value of $a+b+c$
$\textbf{(A) }9 \qquad \textbf{(B) }10 \qquad \textbf{(C) }11 \qquad \textbf{(D) }12 \qquad \textbf{(E) }13 \qquad$

Solution: We first claim that $\triangle{EMI}$ is isosceles and right.
Proof: Construct $\overline{MF}\perp\overline{AB}$ and $\overline{MG}\perp\overline{AC}$ . Since $\overline{AM}$ bisects $\angle{BAC}$ , one can deduce that $MF=MG$ . Then by AAS it is clear that $MI=ME$ and therefore $\triangle{EMI}$ is isosceles. Since quadrilateral $AIME$ is cyclic, one can deduce that $\angle{EMI}=90^\circ$ . Q.E.D.
Since the area of $\triangle{EMI}$ is 2, we can find that $MI=ME=2$ $EI=2\sqrt{2}$
Since $M$ is the mid-point of $\overline{BC}$ , it is clear that $AM=\frac{3\sqrt{2}}{2}$
Now let $AE=a$ and $AI=b$ . By Ptolemy's Theorem, in cyclic quadrilateral $AIME$ , we have $2a+2b=6$ . By Pythagorean Theorem, we have $a^2+b^2=8$ . One can solve the simultaneous system and find $b=\frac{3+\sqrt{7}}{2}$ . Then by deducting the length of $\overline{AI}$ from 3 we get $CI=\frac{3-\sqrt{7}}{2}$ , giving the answer of $\boxed{12}$ . (Surefire2019)

Answer: D 12
Problem: Triangle $ABC$ is an isosceles right triangle with $AB=AC=3$ . Let $M$ be the midpoint of hypotenuse $\overline{BC}$ . Points $I$ and $E$ lie on sides $\overline{AC}$ and $\overline{AB}$ , respectively, so that $AI>AE$ and $AIME$ is a cyclic quadrilateral. Given that triangle $EMI$ has area $2$ , the length $CI$ can be written as $\frac{a-\sqrt{b}}{c}$ , where $a$ $b$ , and $c$ are positive integers and $b$ is not divisible by the square of any prime. What is the value of $a+b+c$
$\textbf{(A) }9 \qquad \textbf{(B) }10 \qquad \textbf{(C) }11 \qquad \textbf{(D) }12 \qquad \textbf{(E) }13 \qquad$

Solution: Like above, notice that $\triangle{EMI}$ is isosceles and right, which means that $\dfrac{ME \cdot MI}{2} = 2$ , so $MI^2=4$ and $MI = 2$ . Then construct $\overline{MF}\perp\overline{AB}$ and $\overline{MG}\perp\overline{AC}$ as well as $\overline{MI}$ . It's clear that $MG^2+GI^2 = MI^2$ by Pythagorean, so knowing that $MG = \dfrac{AB}{2} = \dfrac{3}{2}$ allows one to solve to get $GI = \dfrac{\sqrt{7}}{2}$ . By just looking at the diagram, $CI=AC-MF-GI=\dfrac{3-\sqrt{7}}{2}$ . The answer is thus $3+7+2=\boxed{12}$

Answer: D 12
Problem: Triangle $ABC$ is an isosceles right triangle with $AB=AC=3$ . Let $M$ be the midpoint of hypotenuse $\overline{BC}$ . Points $I$ and $E$ lie on sides $\overline{AC}$ and $\overline{AB}$ , respectively, so that $AI>AE$ and $AIME$ is a cyclic quadrilateral. Given that triangle $EMI$ has area $2$ , the length $CI$ can be written as $\frac{a-\sqrt{b}}{c}$ , where $a$ $b$ , and $c$ are positive integers and $b$ is not divisible by the square of any prime. What is the value of $a+b+c$
$\textbf{(A) }9 \qquad \textbf{(B) }10 \qquad \textbf{(C) }11 \qquad \textbf{(D) }12 \qquad \textbf{(E) }13 \qquad$

Solution: Let $A$ lie on $(0,0)$ $E$ on $(0,y)$ $I$ on $(x,0)$ , and $M$ on $\left(\frac{3}{2},\frac{3}{2}\right)$ . Since ${AIME}$ is cyclic, $\angle EMI$ (which is opposite of another right angle) must be a right angle; therefore, $\overrightarrow{ME} \cdot \overrightarrow{MI} = \left<\frac{-3}{2}, y-\frac{3}{2}\right> \cdot \left<x-\frac{3}{2}, -\frac{3}{2}\right> = 0$ . Compute the dot product to arrive at the relation $y=3-x$ . We can set up another equation involving the area of $\triangle EMI$ using the Shoelace Theorem . This is \[2=\frac{1}{2}\left[\frac{3}{2}\left(y-\frac{3}{2}\right)-xy+\frac{3}{2}\left(x+\frac{3}{2}\right)\right].\] Multiplying, substituting $3-x$ for $y$ , and simplifying, we get $x^2 -3x + \frac{1}{2}=0$ . Thus, $(x,y)=\left(\frac{3 \pm \sqrt{7}}{2},\frac{3 \mp \sqrt{7}}{2}\right)$ . But $AI>AE$ , meaning $x=AI=\frac{3 + \sqrt{7}}{2}$ and $CI = 3-\frac{3 + \sqrt{7}}{2}=\frac{3 - \sqrt{7}}{2}$ , and the final answer is $3+7+2=\boxed{12}$

Answer: D 12
Problem: Triangle $ABC$ is an isosceles right triangle with $AB=AC=3$ . Let $M$ be the midpoint of hypotenuse $\overline{BC}$ . Points $I$ and $E$ lie on sides $\overline{AC}$ and $\overline{AB}$ , respectively, so that $AI>AE$ and $AIME$ is a cyclic quadrilateral. Given that triangle $EMI$ has area $2$ , the length $CI$ can be written as $\frac{a-\sqrt{b}}{c}$ , where $a$ $b$ , and $c$ are positive integers and $b$ is not divisible by the square of any prime. What is the value of $a+b+c$
$\textbf{(A) }9 \qquad \textbf{(B) }10 \qquad \textbf{(C) }11 \qquad \textbf{(D) }12 \qquad \textbf{(E) }13 \qquad$

Solution: From $AIME$ cyclic we get $\angle{MEI} = \angle{MAI} = 45^\circ$ and $\angle{MIE} = \angle{MAE} = 45^\circ$ , so $\triangle{EMI}$ is an isosceles right triangle.
From $[EMI]=2$ we get $EM=MI=2$
Notice $\triangle{AEM} \cong \triangle{CIM}$ , because $\angle{AEM}=180-\angle{AIM}=\angle{CIM}$ $EM=IM$ , and $\angle{EAM}=\angle{ICM}=45^\circ$
Let $CI=AE=x$ , so $AI=3-x$
By Pythagoras on $\triangle{EAI}$ we have $x^2+(3-x)^2=EI^2=8$ , and solve this to get $x=CI=\dfrac{3-\sqrt{7}}{2}$ for a final answer of $3+7+2=\boxed{12}$

Answer: D 12
Problem: Triangle $ABC$ is an isosceles right triangle with $AB=AC=3$ . Let $M$ be the midpoint of hypotenuse $\overline{BC}$ . Points $I$ and $E$ lie on sides $\overline{AC}$ and $\overline{AB}$ , respectively, so that $AI>AE$ and $AIME$ is a cyclic quadrilateral. Given that triangle $EMI$ has area $2$ , the length $CI$ can be written as $\frac{a-\sqrt{b}}{c}$ , where $a$ $b$ , and $c$ are positive integers and $b$ is not divisible by the square of any prime. What is the value of $a+b+c$
$\textbf{(A) }9 \qquad \textbf{(B) }10 \qquad \textbf{(C) }11 \qquad \textbf{(D) }12 \qquad \textbf{(E) }13 \qquad$

Solution: Let $CI=a$ $BE=b$ . Because opposite angles in a cyclic quadrilateral are supplementary, we have $\angle EMI=90^{\circ}$ . By the law of cosines, we have $MI^2=a^2+\frac{9}{4}-\frac{3}{2}a$ , and $ME^2=b^2+\frac{9}{4}-\frac{3}{2}b$ . Notice that $EI=2MO$ , where $O$ is the origin of the circle mentioned in the problem. Thus $\frac{2MO*MO}{2}=2\implies MO=\sqrt{2}, EI=2\sqrt{2}$ . By the Pythagorean Theorem, we have $ME^2+MI^2=EI^2\implies a^2+\frac{9}{4}-\frac{3}{2}a+b^2+\frac{9}{4}-\frac{3}{2}b=(2\sqrt{2})^2=8$ . By the Pythagorean Theorem, we have $AE^2+AI^2=EI^2\implies (3-a)^2+(3-b)^2=(2\sqrt{2})^2=8\implies 18-6a-6b+a^2+b^2=8$ . Thus we have $18-6a-6b+a^2+b^2=a^2+\frac{9}{4}-\frac{3}{2}a+b^2+\frac{9}{4}-\frac{3}{2}b\implies 18-6a-6b=\frac{9}{2}-\frac{3}{2}a-\frac{3}{2}b\implies \frac{27}{2}$ $=\frac{9}{2}a+\frac{9}{2}b\implies a+b=3\implies 3-b=a$ . We know that \begin{align*} (3-a)^2+(3-b)^2&=8 \\ (3-a)^2+a^2&=8 \\ 2a^2-6a+9&=8 \\ 2a^2-6a+1&=0 \\ a&=\frac{6\pm \sqrt{36-8}}{2}=\frac{3\pm\sqrt{7}}{2}. \end{align*} We take the smaller solution because we have $AI>AE\implies 3-AI<3-AE\implies CI<CE$ , and we want $CI$ , not $CE$ , thus $CI=\frac{3-\sqrt{7}}{2}$ . Thus our final answer is $3+7+2=\boxed{12}$

Answer: D 12
Problem: The solutions to the equations $z^2=4+4\sqrt{15}i$ and $z^2=2+2\sqrt 3i,$ where $i=\sqrt{-1},$ form the vertices of a parallelogram in the complex plane. The area of this parallelogram can be written in the form $p\sqrt q-r\sqrt s,$ where $p,$ $q,$ $r,$ and $s$ are positive integers and neither $q$ nor $s$ is divisible by the square of any prime number. What is $p+q+r+s?$
$\textbf{(A) } 20 \qquad  \textbf{(B) } 21 \qquad  \textbf{(C) } 22 \qquad  \textbf{(D) } 23 \qquad  \textbf{(E) } 24$

Solution: We solve each equation separately:
Note that the problem is equivalent to finding the area of a parallelogram with consecutive vertices $(x_1,y_1)=\left(\sqrt{10}, \sqrt{6}\right),(x_2,y_2)=\left(\sqrt{3},1\right),(x_3,y_3)=\left(-\sqrt{10},-\sqrt{6}\right),$ and $(x_4,y_4)=\left(-\sqrt{3}, -1\right)$ in the coordinate plane. By the Shoelace Theorem, the area we seek is \[\frac{1}{2} \left|(x_1y_2 + x_2y_3 + x_3y_4 + x_4y_1) - (y_1x_2 + y_2x_3 + y_3x_4 + y_4x_1)\right| = 6\sqrt2-2\sqrt{10},\] so the answer is $6+2+2+10=\boxed{20}.$

Answer: A 20
Problem: The solutions to the equations $z^2=4+4\sqrt{15}i$ and $z^2=2+2\sqrt 3i,$ where $i=\sqrt{-1},$ form the vertices of a parallelogram in the complex plane. The area of this parallelogram can be written in the form $p\sqrt q-r\sqrt s,$ where $p,$ $q,$ $r,$ and $s$ are positive integers and neither $q$ nor $s$ is divisible by the square of any prime number. What is $p+q+r+s?$
$\textbf{(A) } 20 \qquad  \textbf{(B) } 21 \qquad  \textbf{(C) } 22 \qquad  \textbf{(D) } 23 \qquad  \textbf{(E) } 24$

Solution: We solve each equation separately:
We continue with the last paragraph of Solution 1 to get the answer $\boxed{20}.$

Answer: A 20
Problem: The solutions to the equations $z^2=4+4\sqrt{15}i$ and $z^2=2+2\sqrt 3i,$ where $i=\sqrt{-1},$ form the vertices of a parallelogram in the complex plane. The area of this parallelogram can be written in the form $p\sqrt q-r\sqrt s,$ where $p,$ $q,$ $r,$ and $s$ are positive integers and neither $q$ nor $s$ is divisible by the square of any prime number. What is $p+q+r+s?$
$\textbf{(A) } 20 \qquad  \textbf{(B) } 21 \qquad  \textbf{(C) } 22 \qquad  \textbf{(D) } 23 \qquad  \textbf{(E) } 24$

Solution: Rather than thinking about this with complex numbers, notice that if we take two solutions and think of them as vectors, the area of the parallelogram they form is half the desired area. Also, notice that the area of a parallelogram is $ab\sin \theta$ where $a$ and $b$ are the side lengths.
The side lengths are easily found since we are given the squares of $z$ . Thus, the magnitude of $z$ in the first equation is just $\sqrt{16} = 4$ and in the second equation is just $\sqrt{4} = 2$ . Now, we need $\sin \theta$
To find $\theta$ , think about what squaring is in complex numbers. The angle between the squares of the two solutions is twice the angle between the two solutions themselves. In addition, we can find $\cos$ of this angle by taking the dot product of those two complex numbers and dividing by their magnitudes. The vectors are $\Bigl\langle 4, 4\sqrt{15}\Bigr\rangle$ and $\Bigl\langle 2, 2\sqrt{3}\Bigr\rangle$ , so their dot product is $8 + 24\sqrt{5}$ . Dividing by the magnitudes yields: $\dfrac{8+24\sqrt{5}}{4 \cdot 16} = \dfrac{1 + 3\sqrt{5}}{8}$ . This is $\cos 2\theta$ , and recall the identity $\cos 2\theta = 1 - 2\sin^2 \theta$ . This means that $\sin^2 \theta = \dfrac{7 - 3\sqrt{5}}{16}$ , so $\sin \theta = \dfrac{\sqrt{7-  3\sqrt{5}}}{4}$ . Now, notice that $\sqrt{7-  3\sqrt{5}} = \dfrac{3\sqrt{2}-\sqrt{10}}{2}$ (which is not too hard to discover) so $\sin \theta = \dfrac{3\sqrt{2}-\sqrt{10}}{8}$ . Finally, putting everything together yields: $2\cdot 4 \cdot \dfrac{3\sqrt{2}-\sqrt{10}}{8} = 3\sqrt{2} - \sqrt{10}$ as the area of the parallelogram found by treating two of the solutions as vectors. However, drawing a picture out shows that we actually want twice this (each fourth of the parallelogram from the problem is one half of the parallelogram whose area was found above) so the desired area is actually $6\sqrt{2} - 2\sqrt{10}$ . Then, the answer is $\boxed{20}$

Answer: A 20
Problem: In $\triangle PAT,$ $\angle P=36^{\circ},$ $\angle A=56^{\circ},$ and $PA=10.$ Points $U$ and $G$ lie on sides $\overline{TP}$ and $\overline{TA},$ respectively, so that $PU=AG=1.$ Let $M$ and $N$ be the midpoints of segments $\overline{PA}$ and $\overline{UG},$ respectively. What is the degree measure of the acute angle formed by lines $MN$ and $PA?$
$\textbf{(A) } 76 \qquad  \textbf{(B) } 77 \qquad  \textbf{(C) } 78 \qquad  \textbf{(D) } 79 \qquad \textbf{(E) } 80$

Solution: Let $P$ be the origin, and $PA$ lie on the $x$ -axis.
We can find $U=\left(\cos(36), \sin(36)\right)$ and $G=\left(10-\cos(56), \sin(56)\right)$
Then, we have $M=(5, 0)$ and $N$ is the midpoint of $U$ and $G$ , or $\left(\frac{10+\cos(36)-\cos(56)}{2}, \frac{\sin(36)+\sin(56)}{2}\right)$
Notice that the tangent of our desired points is the the absolute difference between the $y$ -coordinates of the two points divided by the absolute difference between the $x$ -coordinates of the two points.
This evaluates to \[\frac{\sin(36)+\sin(56)}{\cos(36)-\cos(56)}\] Now, using sum to product identities, we have this equal to \[\frac{2\sin(46)\cos(10)}{-2\sin(46)\sin({-10})}=\frac{\sin(80)}{\cos(80)}=\tan(80)\] so the answer is $\boxed{80}.$

Answer: E 80
Problem: In $\triangle PAT,$ $\angle P=36^{\circ},$ $\angle A=56^{\circ},$ and $PA=10.$ Points $U$ and $G$ lie on sides $\overline{TP}$ and $\overline{TA},$ respectively, so that $PU=AG=1.$ Let $M$ and $N$ be the midpoints of segments $\overline{PA}$ and $\overline{UG},$ respectively. What is the degree measure of the acute angle formed by lines $MN$ and $PA?$
$\textbf{(A) } 76 \qquad  \textbf{(B) } 77 \qquad  \textbf{(C) } 78 \qquad  \textbf{(D) } 79 \qquad \textbf{(E) } 80$

Solution: We will refer to the Diagram section. In this solution, all angle measures are in degrees.
We rotate $\triangle PUM$ by $180^\circ$ about $M$ to obtain $\triangle AU'M.$ Let $H$ be the intersection of $\overline{PA}$ and $\overline{GU'},$ as shown below. [asy] /* Made by MRENTHUSIASM */ size(375);  pair P, A, T, U, G, M, N, U1, H; P = origin; A = (10,0); U = intersectionpoint(Circle(P,1),P--P+2*dir(36)); G = intersectionpoint(Circle(A,1),A--A+2*dir(180-56)); T = extension(P,U,A,G); M = midpoint(P--A); N = midpoint(U--G); U1 = rotate(180,M)*U; H = intersectionpoint(P--A,G--U1); fill(U--P--M--cycle^^M--U1--A--cycle,yellow); dot("$P$",P,1.5*SW,linewidth(4)); dot("$A$",A,1.5*SE,linewidth(4)); dot("$U$",U,1.5*(0,1),linewidth(4)); dot("$G$",G,1.5*NE,linewidth(4)); dot("$T$",T,1.5*(0,1),linewidth(4)); dot("$M$",M,1.5*S,linewidth(4)); dot("$N$",N,1.5*(0,1),linewidth(4)); dot("$U'$",U1,1.5*S,linewidth(4)); dot("$H$",H,1.5*NW,linewidth(4)); draw(P--A--T--cycle^^U--G^^M--N^^U--U1--A); draw(G--U1,dashed); label("$1$",midpoint(G--A),1.5*dir(30)); label("$1$",midpoint(A--U1),1.5*dir(-30)); label("$1$",midpoint(U--P),1.5*dir(150)); label("$36^\circ$",P,5*dir(18),fontsize(8)); label("$56^\circ$",A,2.5*dir(180-56/2),fontsize(8)); label("$36^\circ$",A,2.5*dir(180+25),fontsize(8)); Label L = Label("$10$", align=(0,0), position=MidPoint, filltype=Fill(3,0,white)); draw(P-(0,1.5)--A-(0,1.5), L=L, arrow=Arrows(),bar=Bars(15)); add(pathticks(U--N, 2, .5, 4, 8, red)); add(pathticks(N--G, 2, .5, 4, 8, red)); add(pathticks(U--M, 1, .5, 0, 8, red)); add(pathticks(M--U1, 1, .5, 0, 8, red)); [/asy] Note that $\triangle GU'A$ is an isosceles triangle with $GA=U'A=1,$ so $\angle AGU'=\angle AU'G=\frac{180-\angle GAU'}{2}=44.$ In $\triangle GHA,$ it follows that $\angle GHA=180-\angle GAH-\angle AGH=80.$
Since $\frac{UM}{UU'}=\frac{UN}{UG}=\frac12,$ we conclude that $\triangle UMN\sim\triangle UU'G$ by SAS, from which $\angle UMN=\angle UU'G$ and $\angle UNM=\angle UGU'.$ By the Converse of the Corresponding Angles Postulate, we deduce that $\overline{MN}\parallel\overline{U'G}.$
Finally, we have $\angle NMA=\angle GHA=\boxed{80}$ by the Corresponding Angles Postulate.

Answer: E 80