## AMC 12 2018 Test A

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**Instructions**

- This is a 25-question, multiple choice test. Each question is followed by answers marked A, B, C, D and E. Only one of these is correct.
- You will receive 6 points for each correct answer, 2.5 points for each problem left unanswered if the year is before 2006, 1.5 points for each problem left unanswered if the year is after 2006, and 0 points for each incorrect answer.
- No aids are permitted other than scratch paper, graph paper, ruler, compass, protractor and erasers (and calculators that are accepted for use on the test if before 2006. No problems on the test will require the use of a calculator).
- Figures are not necessarily drawn to scale.
- You will have
**75 minutes**working time to complete the test.

A large urn contains $100$ balls, of which $36 \%$ are red and the rest are blue. How many of the blue balls must be removed so that the percentage of red balls in the urn will be $72 \%$? (No red balls are to be removed.)

$\textbf{(A)}\ 28 \qquad\textbf{(B)}\ 32 \qquad\textbf{(C)}\ 36 \qquad\textbf{(D)}\ 50 \qquad\textbf{(E)}\ 64$

$\textbf{D}$

36 red balls take $72\%$ of the rest balls. So the number of the rest balls is $36\div72\%=50$. Hence, we must remove $100-50=50$ balls.

While exploring a cave, Carl comes across a collection of $5$-pound rocks worth $\$14$ each, $4$-pound rocks worth $\$11$ each, and $1$-pound rocks worth $\$2$ each. There are at least $20$ of each size. He can carry at most $18$ pounds. What is the maximum value, in dollars, of the rocks he can carry out of the cave?

$\textbf{(A) } 48 \qquad \textbf{(B) } 49 \qquad \textbf{(C) } 50 \qquad \textbf{(D) } 51 \qquad \textbf{(E) } 52$

$\textbf{C}$

The value of $5$-pound rocks is $\$14\div5=\$2.80$ per pound, and the value of $4$-pound rocks is $\$11\div4=\$2.75$ per pound. Carl should choose 5-pound rocks first, then 4 pound rocks, and finally 1-pound rocks. If he chooses three 5-pound rocks and three 1-pound rocks, the value is $\$48$. If he chooses two 5-pound rocks and two 4-pound rocks, the value is $\$50$. So the answer is C.

How many ways can a student schedule $3$ mathematics courses -- algebra, geometry, and number theory -- in a $6$-period day if no two mathematics courses can be taken in consecutive periods? (What courses the student takes during the other $3$ periods is of no concern here.)

$\textbf{(A) }3\qquad\textbf{(B) }6\qquad\textbf{(C) }12\qquad\textbf{(D) }18\qquad\textbf{(E) }24$

$\textbf{E}$

Ignoring distinguishability of classes, we can list out the ways that three periods can be chosen for the classes when periods cannot be consecutive:

Periods $1, 3, 5$

Periods $1, 3, 6$

Periods $1, 4, 6$

Periods $2, 4, 6$

There are $4$ ways to place $3$ non-distinguishable classes into $6$ periods such that no two classes are in consecutive periods. For each of these ways, there are $3! = 6$ orderings of the classes among themselves.

Therefore, we have $4 \cdot 6 = 24$ ways to choose the classes.

Alice, Bob, and Charlie were on a hike and were wondering how far away the nearest town was. When Alice said, "We are at least $6$ miles away," Bob replied, "We are at most $5$ miles away." Charlie then remarked, "Actually the nearest town is at most $4$ miles away." It turned out that none of the three statements were true. Let $d$ be the distance in miles to the nearest town. Which of the following intervals is the set of all possible values of $d$?

$\textbf{(A) } (0,4) \qquad \textbf{(B) } (4,5) \qquad \textbf{(C) } (4,6) \qquad \textbf{(D) } (5,6) \qquad \textbf{(E) } (5,\infty)$

$\textbf{D}$

Think of the distances as if they are on a number line. Alice claims that $d > 6$, Bob says $d < 5$, while Charlie thinks $d < 4$. This means that all possible numbers less than $5$ and greater than $6$ are included. However, since the three statements are actually false, the distance to the nearest town is one of the numbers not covered. Therefore, the answer is $(5,6)$.

What is the sum of all possible values of $k$ for which the polynomials $x^2 - 3x + 2$ and $x^2 - 5x + k$ have a root in common?

$\textbf{(A) }3 \qquad\textbf{(B) }4 \qquad\textbf{(C) }5 \qquad\textbf{(D) }6 \qquad\textbf{(E) }10 \qquad$

$\textbf{E}$

We factor $x^2-3x+2$ into $(x-1)(x-2)$. Thus, either $1$ or $2$ is a root of $x^2-5x+k$. If $1$ is a root, then $1^2-5\cdot1+k=0$, so $k=4$. If $2$ is a root, then $2^2-5\cdot2+k=0$, so $k=6$. The sum of all possible values of $k$ is $10$.

For positive integers $m$ and $n$ such that $m+10<n+1$, both the mean and the median of the set $\{m, m+4, m+10, n+1, n+2, 2n\}$ are equal to $n$. What is $m+n$?

$\textbf{(A) }20\qquad\textbf{(B) }21\qquad\textbf{(C) }22\qquad\textbf{(D) }23\qquad\textbf{(E) }24$

$\textbf{B}$

The mean and median are\[\frac{3m+4n+17}{6}=\frac{m+n+11}{2}=n,\]Solving, we get $\left(m,n\right)=\left(5,16\right)$. So $m+n=21$.

For how many (not necessarily positive) integer values of $n$ is the value of $4000\cdot \left(\dfrac{2}{5}\right)^n$ an integer?

$\textbf{(A) }3 \qquad \textbf{(B) }4 \qquad \textbf{(C) }6 \qquad \textbf{(D) }8 \qquad \textbf{(E) }9 \qquad$

$\textbf{E}$

Note that\[4000\cdot \left(\frac{2}{5}\right)^n=\left(2^5\cdot5^3\right)\cdot \left(\frac{2}{5}\right)^n=2^{5+n}\cdot5^{3-n}\]Since this expression is an integer, we need:

$5+n\geq0,$ from which $n\geq-5.\newline$

$3-n\geq0,$ from which $n\leq3.$

Taking the intersection gives $-5\leq n\leq3.$ So there are $3-(-5)+1=9$ integer values of $n.$

All of the triangles in the diagram below are similar to isosceles triangle $ABC$, in which $AB=AC$. Each of the $7$ smallest triangles has area $1,$ and $\triangle ABC$ has area $40$. What is the area of trapezoid $DBCE$?

$\textbf{(A) } 16 \qquad \textbf{(B) } 18 \qquad \textbf{(C) } 20 \qquad \textbf{(D) } 22 \qquad \textbf{(E) } 24$

$\textbf{E}$

Let $x$ be the area of $ADE$. Note that $x$ is comprised of the $7$ small isosceles triangles and a triangle similar to $ADE$ with side length ratio $3:4$ (so an area ratio of $9:16$). Thus, we have\[x=7+\dfrac{9}{16}x\]This gives $x=16$, so the area of $DBCE=40-x=24$.

Which of the following describes the largest subset of values of $y$ within the closed interval $[0,\pi]$ for which\[\sin(x+y)\leq \sin(x)+\sin(y)\]for every $x$ between $0$ and $\pi$, inclusive?

$\textbf{(A) } y=0 \qquad \textbf{(B) } 0\leq y\leq \dfrac{\pi}{4} \qquad \textbf{(C) } 0\leq y\leq \dfrac{\pi}{2} \qquad \textbf{(D) } 0\leq y\leq \dfrac{3\pi}{4} \qquad \textbf{(E) } 0\leq y\leq \pi$

$\textbf{E}$

On the interval $[0, \pi]$ sine is non-negative. Thus $\sin(x + y) = \sin x \cos y + \sin y \cos x \le \sin x + \sin y$ for all $x, y \in [0, \pi]$ and equality only occurs when $\cos x = \cos y = 1$, which is cosine's maximum value. The answer is $0\le y\le \pi$.

How many ordered pairs of real numbers $(x,y)$ satisfy the following system of equations?\begin{align*}

x+3y&=3 \\

\big||x|-|y|\big|&=1

\end{align*}

$\textbf{(A) } 1 \qquad \textbf{(B) } 2 \qquad \textbf{(C) } 3 \qquad \textbf{(D) } 4 \qquad \textbf{(E) } 8$

$\textbf{C}$

Note that $||x| - |y||$ can take on one of four values: $x + y$, $x - y$, $-x + y$, $-x -y$. So we have 4 cases:

$\textbf{Case 1:}$ $||x| - |y|| = x+y$\begin{align*}

x+3y&=3\\

x+y&=1

\end{align*} We get $(x,y)=(0,1)$

$\textbf{Case 2:}$ $||x| - |y|| = x-y$\begin{align*}

x+3y&=3\\

x-y&=1

\end{align*} We get $(x,y)=\left(\dfrac{3}{2},\dfrac{1}{2}\right)$

$\textbf{Case 3:}$ $||x| - |y|| = -x+y$\begin{align*}

x+3y&=3\\

-x+y&=1

\end{align*} We get $(x,y)=(0,1)$. This is the same solution as we got in Case 1.

$\textbf{Case 4:}$ $||x| - |y|| = -x-y$\begin{align*}

x+3y&=3\\

-x-y&=1

\end{align*} We get $(x,y)=(-3,2)$.

Finally, we need to test all the results $(0, 1)$, $(-3,2)$, and $\left(\dfrac{3}{2}, \dfrac{1}{2}\right)$ in the original system of equations \begin{align*}

x+3y&=3 \\

\big||x|-|y|\big|&=1

\end{align*}

All of them satisfy the original equations. So the answer is $3$.

A paper triangle with sides of lengths $3,4,$ and $5$ inches, as shown, is folded so that point $A$ falls on point $B$. What is the length in inches of the crease?

$\textbf{(A) } 1+\dfrac12 \sqrt2 \qquad \textbf{(B) } \sqrt3 \qquad \textbf{(C) } \dfrac74 \qquad \textbf{(D) } \dfrac{15}{8} \qquad \textbf{(E) } 2$

$\textbf{D}$

The crease line is the perpendicular bisector of side $AB$. Call the midpoint of $AB$ point $D$. Draw this line and call the intersection point with $AC$ as $E$. Now, $\triangle ACB$ is similar to $\triangle ADE$ by $AA$ similarity. Setting up the ratios, we find that\[\frac{BC}{AC}=\frac{DE}{AD} \Rightarrow \frac{3}{4}=\frac{DE}{\frac{5}{2}} \Rightarrow DE=\frac{15}{8}\]Thus, the answer is $\dfrac{15}{8}$.

Let $S$ be a set of $6$ integers taken from $\{1,2,\dots,12\}$ with the property that if $a$ and $b$ are elements of $S$ with $a<b$, then $b$ is not a multiple of $a$. What is the least possible value of an element in $S$?

$\textbf{(A)}\ 2\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 5\qquad\textbf{(E)}\ 7$

$\textbf{C}$

We start with $2$ because every number is a multiple of $1$. We can include every odd number except $1$ to fill out the set, but then $3$ and $9$ would violate the rule, so that won't work.

Now we try with $3$. Our set becomes $\{3,4,5,7,11\}$ but we can add no more valid numbers.

Starting with $4,$ we find that the sequence $4,5,6,7,9,11$ works. So the answer is 4.

How many nonnegative integers can be written in the form\[a_7\cdot3^7+a_6\cdot3^6+a_5\cdot3^5+a_4\cdot3^4+a_3\cdot3^3+a_2\cdot3^2+a_1\cdot3^1+a_0\cdot3^0,\]where $a_i\in \{-1,0,1\}$ for $0\le i \le 7$?

$\textbf{(A) } 512 \qquad \textbf{(B) } 729 \qquad \textbf{(C) } 1094 \qquad \textbf{(D) } 3281 \qquad \textbf{(E) } 59,048$

$\textbf{D}$

Note that all numbers formed from this sum are either positive, negative or zero. The number of positive numbers formed by this sum is equal to the number of negative numbers formed by this sum, because of symmetry. There is only one way to achieve a sum of zero, if all $a_i=0$. The total number of ways to pick $a_i$ from $i=0, 1, 2, 3, ... 7$ is $3^8=6561$. $\dfrac{6561-1}{2}=3280$ gives the number of possible negative integers. The question asks for the number of non-negative integers, so subtracting from the total gives $6561-3280= 3281$.

The solution to the equation $\log_{3x} 4 = \log_{2x} 8$, where $x$ is a positive real number other than $\dfrac{1}{3}$ or $\dfrac{1}{2}$, can be written as $\dfrac {p}{q}$ where $p$ and $q$ are relatively prime positive integers. What is $p + q$?

$\textbf{(A) } 5 \qquad \textbf{(B) } 13 \qquad \textbf{(C) } 17 \qquad \textbf{(D) } 31 \qquad \textbf{(E) } 35$

$\textbf{D}$

We apply the Change of Base Formula, then rearrange:\begin{align*} \frac{\log_2{4}}{\log_2{(3x)}}&=\frac{\log_2{8}}{\log_2{(2x)}} \\ \frac{2}{\log_2{(3x)}}&=\frac{3}{\log_2{(2x)}} \\ 3\log_2{(3x)}&=2\log_2{(2x)} \end{align*}By the logarithmic identity $n\log_b{a}=\log_b{\left(a^n\right)},$ it follows that\begin{align*} \log_2{\left[(3x)^3\right]}&=\log_2{\left[(2x)^2\right]} \\ (3x)^3&=(2x)^2 \\ 27x^3&=4x^2 \\ x&=\frac{4}{27} \end{align*}from which the answer is $4+27=31.$

A scanning code consists of a $7 \times 7$ grid of squares, with some of its squares colored black and the rest colored white. There must be at least one square of each color in this grid of $49$ squares. A scanning code is called $\textit{symmetric}$ if its look does not change when the entire square is rotated by a multiple of $90 ^{\circ}$ counterclockwise around its center, nor when it is reflected across a line joining opposite corners or a line joining midpoints of opposite sides. What is the total number of possible symmetric scanning codes?

$\textbf{(A)} \text{ 510} \qquad \textbf{(B)} \text{ 1022} \qquad \textbf{(C)} \text{ 8190} \qquad \textbf{(D)} \text{ 8192} \qquad \textbf{(E)} \text{ 65,534}$

$\textbf{B}$

Start from the center and label all protruding cells symmetrically.

In the grid, 10 letters are used: $A$, $B$, $C$, $D$, $E$, $F$, $G$, $H$, $J$, and $K$. Each of the letters must have its own color, either white or black. This means, for example, all $K$'s must have the same color for the grid to be symmetrical.

So there are $2^{10}$ ways to color the grid, including a completely black grid and a completely white grid. Since the grid must contain at least one square with each color, the number of ways is $2^{10}-2=1024-2=$ $1022$.

Which of the following describes the set of values of $a$ for which the curves $x^2+y^2=a^2$ and $y=x^2-a$ in the real $xy$-plane intersect at exactly $3$ points?

$\textbf{(A) }a=\dfrac14 \qquad \textbf{(B) }\dfrac14 < a < \dfrac12 \qquad \textbf{(C) }a>\dfrac14 \qquad \textbf{(D) }a=\dfrac12 \qquad \textbf{(E) }a>\dfrac12 \qquad$

$\textbf{E}$

Substituting $y=x^2-a$ into $x^2+y^2=a^2$, we get\[x^2+(x^2-a)^2=a^2 \implies x^2+x^4-2ax^2=0 \implies x^2(x^2-(2a-1))=0\]Since this is a quartic, there are $4$ total roots (counting multiplicity). We see that $x=0$ always has at least one intersection at $(0,-a)$ (and is in fact a double root).

The other two intersection points have $x$ coordinates $\pm\sqrt{2a-1}$. We must have $2a-1> 0$. Thus, the answer is $a>\dfrac12$.

Farmer Pythagoras has a field in the shape of a right triangle. The right triangle's legs have lengths $3$ and $4$ units. In the corner where those sides meet at a right angle, he leaves a small unplanted square $S$ so that from the air it looks like the right angle symbol. The rest of the field is planted. The shortest distance from $S$ to the hypotenuse is $2$ units. What fraction of the field is planted?

$\textbf{(A) } \dfrac{25}{27} \qquad \textbf{(B) } \dfrac{26}{27} \qquad \textbf{(C) } \dfrac{73}{75} \qquad \textbf{(D) } \dfrac{145}{147} \qquad \textbf{(E) } \dfrac{74}{75}$

$\textbf{D}$

Note that the hypotenuse of the field is $5,$ and the area of the field is $6.$ Let $x$ be the side-length of square $S.$ We partition the field into a red triangle, a yellow triangle, and a green triangle, as shown below:

Let the brackets denote areas. By area addition, we set up an equation for $x:$\begin{align*} [\text{Red Triangle}]+[\text{Yellow Triangle}]+[\text{Green Triangle}]&=[\text{Field}] \\ \frac{3x}{2}+\frac{4x}{2}+\frac{5\cdot2}{2}&=6, \end{align*}from which $x=\dfrac27.$ Therefore, the answer is\[\frac{[\text{Field}]-[S]}{[\text{Field}]}=\frac{6-x^2}{6}=\frac{145}{147}\]

Triangle $ABC$ with $AB=50$ and $AC=10$ has area $120$. Let $D$ be the midpoint of $\overline{AB}$, and let $E$ be the midpoint of $\overline{AC}$. The angle bisector of $\angle BAC$ intersects $\overline{DE}$ and $\overline{BC}$ at $F$ and $G$, respectively. What is the area of quadrilateral $FDBG$?

$\textbf{(A) }60 \qquad \textbf{(B) }65 \qquad \textbf{(C) }70 \qquad \textbf{(D) }75 \qquad \textbf{(E) }80 \qquad$

$\textbf{D}$

Let the area of $\triangle ABG$ be $x$, and the area of $\triangle ACG$ be $y$. Then we have $x+y=120$. Since $AG$ bisects $\angle BAC$, the altitude from $G$ to $AB$ in $\triangle ABG$ is the same as the altitude from $G$ to $AC$ in $\triangle ACG$. Hence, we get $\dfrac{x}{y}=\dfrac{AB}{AC}=5$. Combined with $x+y=120$, we get $x=100$.

We see that $DE\parallel BC$ and $D$ is the middle point of $AB$. So the area of quadrilateral $FDBG$ is $\dfrac34x=75$.

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 $m+n$?

$\textbf{(A) } 16 \qquad \textbf{(B) } 17 \qquad \textbf{(C) } 19 \qquad \textbf{(D) } 23 \qquad \textbf{(E) } 36$

$\textbf{C}$

Note that the fractions of the form $\dfrac{1}{2^a3^b5^c},$ 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 $15+4=19.$

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 $\dfrac{a-\sqrt{b}}{c}$, where $a$, $b$, and $c$ are positive integers and $b$ is not divisible by the square of any prime. What is the value of $a+b+c$?

$\textbf{(A) }9 \qquad \textbf{(B) }10 \qquad \textbf{(C) }11 \qquad \textbf{(D) }12 \qquad \textbf{(E) }13 \qquad$

$\textbf{D}$

Since $AIME$ is a cyclic quadrilateral, we have $\angle EMI=180^\circ-\angle A=90^\circ$. 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. So $MI=2,MC=\dfrac{3}{\sqrt{2}}$. With $\angle{MCI}=45^\circ$, we can use Law of Cosines in $\triangle CMI$ to determine that $CI=\dfrac{3\pm\sqrt{7}}{2}$. The same calculations hold for $BE$ also, and since $CI<BE$, we deduce that $CI$ is the smaller root $\dfrac{3-\sqrt{7}}{2}$. Thus, the answer is $3+7+2=12$.

Which of the following polynomials has the greatest real root?

$\textbf{(A) } x^{19}+2018x^{11}+1 \qquad $

$\textbf{(B) } x^{17}+2018x^{11}+1 \qquad $

$\textbf{(C) } x^{19}+2018x^{13}+1 \qquad $

$\textbf{(D) } x^{17}+2018x^{13}+1 \qquad $

$\textbf{(E) } 2019x+2018$

$\textbf{B}$

Denote the polynomials in the answer choices by $A(x),B(x),C(x),D(x),$ and $E(x),$ respectively.

Note that $A(x),B(x),C(x),D(x),$ and $E(x)$ are strictly increasing functions with range $(-\infty,\infty).$ So, each polynomial has exactly one real root. The real root of $E(x)$ is $x=-\dfrac{2018}{2019}\approx-1.000.$ On the other hand, since $A(-1)=B(-1)=C(-1)=D(-1)=-2018$ and $A(0)=B(0)=C(0)=D(0)=1,$ we conclude that the real root for each of $A(x),B(x),C(x),$ and $D(x)$ must satisfy $x\in(-1,0)$ by the Intermediate Value Theorem (IVT).

We analyze the polynomials for $x\in(-1,0):$

We have\begin{align*} B(x)-A(x)=D(x)-C(x)&=x^{17}-x^{19} \\ &=x^{17}\left(1-x^2\right) \\ &<0\end{align*}As the graph of $y=A(x)$ is always above the graph of $y=B(x),$ we deduce that $B(x)$ has a greater real root than $A(x)$ does. By the same reasoning, $D(x)$ has a greater real root than $C(x)$ does.

We also have\begin{align*} B(x)-D(x)&=2018x^{11}-2018x^{13} \\ &=2018x^{11}\left(1-x^2\right) \\ &<0 \end{align*}from which $B(x)$ has a greater real root than $D(x)$ does.

Now, we are left with comparing the real roots of $B(x)$ and $E(x).$ Since $B\left(-\dfrac{1}{\sqrt2}\right)<0<B\left(-\dfrac{1}{2}\right),$ it follows that the real root of $B(x)$ must satisfy $x\in\left(-\dfrac{1}{\sqrt2},-\dfrac{1}{2}\right)$ by IVT. The answer is B.

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 $p+q+r+s?$

$\textbf{(A) } 20 \qquad \textbf{(B) } 21 \qquad \textbf{(C) } 22 \qquad \textbf{(D) } 23 \qquad \textbf{(E) } 24$

$\textbf{A}$

We solve each equation separately:$$z^2=4+4\sqrt{15}i$$Let $z=a+bi$ for some real numbers $a$ and $b.$

Substituting and expanding, we get\begin{align*} (a+bi)^2&=4+4\sqrt{15}i \\ \left(a^2-b^2\right)+2abi&=4+4\sqrt{15}i. \end{align*}Equating the real parts and the imaginary parts, respectively, we get\begin{align*} a^2-b^2&=4 \\ ab&=2\sqrt{15}\end{align*}Solving, we get $(a,b)=\left(\sqrt{10},\sqrt{6}\right),\left(-\sqrt{10},-\sqrt{6}\right).$

The solutions to $z^2=4+4\sqrt{15}i$ are $\boldsymbol{z=\sqrt{10}+\sqrt{6}i,-\sqrt{10}-\sqrt{6}i}.$

For $z^2=2+2\sqrt{3}i$, by the same process, we have $(a,b)=\left(\sqrt3,1\right),\left(-\sqrt3,-1\right).$ The solutions to $z^2=2+2\sqrt{3}i$ are $\boldsymbol{z=\sqrt3+i,-\sqrt3-i}.$

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 $6+2+2+10=20.$

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 $PA?$

$\textbf{(A) } 76 \qquad \textbf{(B) } 77 \qquad \textbf{(C) } 78 \qquad \textbf{(D) } 79 \qquad \textbf{(E) } 80$

$\textbf{E}$

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 $N=\left(\dfrac{10+\cos(36)-\cos(56)}{2}, \dfrac{\sin(36)+\sin(56)}{2}\right)$.

Notice that the tangent of our desired angle 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\[\tan\theta=\frac{\sin(36)+\sin(56)}{\cos(36)-\cos(56)}=\frac{2\sin(46)\cos(10)}{-2\sin(46)\sin({-10})}=\frac{\sin(80)}{\cos(80)}=\tan(80)\]so the answer is $80.$

Alice, Bob, and Carol play a game in which each of them chooses a real number between $0$ and $1.$ The winner of the game is the one whose number is between the numbers chosen by the other two players. Alice announces that she will choose her number uniformly at random from all the numbers between $0$ and $1,$ and Bob announces that he will choose his number uniformly at random from all the numbers between $\dfrac{1}{2}$ and $\dfrac{2}{3}.$ Armed with this information, what number should Carol choose to maximize her chance of winning?

$\textbf{(A) }\dfrac{1}{2}\qquad \textbf{(B) }\dfrac{13}{24} \qquad \textbf{(C) }\dfrac{7}{12} \qquad \textbf{(D) }\dfrac{5}{8} \qquad \textbf{(E) }\dfrac{2}{3}\qquad$

$\textbf{B}$

Let $a,b,$ and $c$ be the numbers that Alice, Bob, and Carol choose, respectively.

Based on the value of $c,$ we construct the following table:

Let $P(c)$ be Carol's probability of winning when she chooses $c.$ We write $P(c)$ as a piecewise function:\[P(c) = \begin{cases} c & \mathrm{if} \ 0<c<\frac12 \\ -12c^2+13c-3 & \mathrm{if} \ \frac12\leq c\leq\frac23 \\ 1-c & \mathrm{if} \ \frac23<c<1 \end{cases}.\]Note that $P(c)$ is continuous in the interval $(0,1),$ increasing in the interval $\left(0,\dfrac12\right),$ increasing and then decreasing in the interval $\left(\dfrac12,\dfrac23\right),$ and decreasing in the interval $\left(\dfrac23,1\right).$ Therefore, the maximum point of $P(c)$ occurs in the interval $\left[\dfrac12,\dfrac23\right],$ namely at $c=-\dfrac{13}{2\cdot(-12)}=\dfrac{13}{24}.$

For a positive integer $n$ and nonzero digits $a$, $b$, and $c$, let $A_n$ be the $n$-digit integer each of whose digits is equal to $a$; let $B_n$ be the $n$-digit integer each of whose digits is equal to $b$, and let $C_n$ be the $2n$-digit (not $n$-digit) integer each of whose digits is equal to $c$. What is the greatest possible value of $a + b + c$ for which there are at least two values of $n$ such that $C_n - B_n = A_n^2$?

$\textbf{(A) } 12 \qquad \textbf{(B) } 14 \qquad \textbf{(C) } 16 \qquad \textbf{(D) } 18 \qquad \textbf{(E) } 20$

$\textbf{D}$

By geometric series, we have\begin{alignat*}{8} A_n&=a\bigl(\phantom{ }\underbrace{111\cdots1}_{n\text{ digits}}\phantom{ }\bigr)&&=a\left(1+10+10^2+\cdots+10^{n-1}\right)&&=a\cdot\frac{10^n-1}{9}, \\ B_n&=b\bigl(\phantom{ }\underbrace{111\cdots1}_{n\text{ digits}}\phantom{ }\bigr)&&=b\left(1+10+10^2+\cdots+10^{n-1}\right)&&=b\cdot\frac{10^n-1}{9}, \\ C_n&=c\bigl(\phantom{ }\underbrace{111\cdots1}_{2n\text{ digits}}\phantom{ }\bigr)&&=c\left(1+10+10^2+\cdots+10^{2n-1}\right)&&=c\cdot\frac{10^{2n}-1}{9}. \end{alignat*}By substitution, we rewrite the given equation $C_n - B_n = A_n^2$ as\[c\cdot\frac{10^{2n}-1}{9} - b\cdot\frac{10^n-1}{9} = a^2\cdot\left(\frac{10^n-1}{9}\right)^2\]Since $n > 0,$ it follows that $10^n > 1.$ We divide both sides by $\dfrac{10^n-1}{9}$ and then rearrange:\begin{align*} c\left(10^n+1\right) - b &= a^2\cdot\frac{10^n-1}{9} \\ 9c\left(10^n+1\right) - 9b &= a^2\left(10^n-1\right) \\ \left(9c-a^2\right)10^n &= 9b-9c-a^2 \end{align*} Given that there are at least two values of $n$, we get \begin{align*} 9c-a^2&=0 \\ 9b-9c-a^2&=0 \end{align*}The first equation implies that $c=\dfrac{a^2}{9}.$ Substituting this into the second equation gives $b=\dfrac{2a^2}{9}.$

To maximize $a + b + c = a + \dfrac{a^2}{3},$ we need to maximize $a.$ Clearly, $a$ must be divisible by $3.$ The possibilities for $(a,b,c)$ are $(9,18,9),(6,8,4),$ or $(3,2,1),$ but $(9,18,9)$ is invalid. Therefore, the greatest possible value of $a + b + c$ is $6+8+4=18.$