## AMC 12 2018 Test B

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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.

Kate bakes a $20$-inch by $18$-inch pan of cornbread. The cornbread is cut into pieces that measure $2$ inches by $2$ inches. How many pieces of cornbread does the pan contain?

$\textbf{(A) } 90 \qquad \textbf{(B) } 100 \qquad \textbf{(C) } 180 \qquad \textbf{(D) } 200 \qquad \textbf{(E) } 360$

$\textbf{A}$

The 20-inch side can be cut into 10 2-inch sides, and the 18-inch side can be cut into 9 2-inch sides. So the total number of pieces is $10\cdot9=90$.

Sam drove $96$ miles in $90$ minutes. His average speed during the first $30$ minutes was $60$ mph (miles per hour), and his average speed during the second $30$ minutes was $65$ mph. What was his average speed, in mph, during the last $30$ minutes?

$\textbf{(A) } 64 \qquad \textbf{(B) } 65 \qquad \textbf{(C) } 66 \qquad \textbf{(D) } 67 \qquad \textbf{(E) } 68$

$\textbf{D}$

Sam drove $60\cdot\dfrac12=30$ miles in the first 30 minutes, and $65\cdot\dfrac12=32.5$ miles in the second 30 minutes. So he drove $96-30-32.5=33.5$ miles in the last 30 minutes, which means the average speed of the last 30 minutes is $33.5\cdot2=67$ mph.

A line with slope $2$ intersects a line with slope $6$ at the point $(40,30)$. What is the distance between the $x$-intercepts of these two lines?

$\textbf{(A) } 5 \qquad \textbf{(B) } 10 \qquad \textbf{(C) } 20 \qquad \textbf{(D) } 25 \qquad \textbf{(E) } 50$

$\textbf{B}$

Given the slopes and point $(40,30)$, we know that the first line is $y=2x-50$, and the second line is $y=6x-210$. The first line intersects the $x-\text{axis}$ at $x=25$. The second line intersects the $x-\text{axis}$ at $x=35$. So the distance between the two intersection points is 10.

A circle has a chord of length $10$, and the distance from the center of the circle to the chord is $5$. What is the area of the circle?

$\textbf{(A) }25\pi \qquad \textbf{(B) }50\pi \qquad \textbf{(C) }75\pi \qquad \textbf{(D) }100\pi \qquad \textbf{(E) }125\pi \qquad$

$\textbf{B}$

Let $O$ be the center of the circle, $\overline{AB}$ be the chord, and $M$ be the midpoint of $\overline{AB},$ as shown below.

Note that $\overline{OM}\perp\overline{AB}.$ Since $OM=AM=BM=5,$ we conclude that $\triangle OMA$ and $\triangle OMB$ are congruent isosceles right triangles. It follows that $r=5\sqrt2,$ so the area of $\odot O$ is $\pi r^2=50\pi$.

How many subsets of $\{2,3,4,5,6,7,8,9\}$ contain at least one prime number?

$\textbf{(A) } 128 \qquad \textbf{(B) } 192 \qquad \textbf{(C) } 224 \qquad \textbf{(D) } 240 \qquad \textbf{(E) } 256$

$\textbf{D}$

There are 4 prime numbers and 4 composite numbers in the set. Since we need at least 1 prime number, we have $2^4-1$ ways to choose the prime numbers, and $2^4$ ways to choose the composite numbers. The answer is $(2^4-1)\cdot2^4=240$.

Suppose $S$ cans of soda can be purchased from a vending machine for $Q$ quarters. Which of the following expressions describes the number of cans of soda that can be purchased for $D$ dollars, where $1$ dollar is worth $4$ quarters?

$\textbf{(A) } \dfrac{4DQ}{S} \qquad \textbf{(B) } \dfrac{4DS}{Q} \qquad \textbf{(C) } \dfrac{4Q}{DS} \qquad \textbf{(D) } \dfrac{DQ}{4S} \qquad \textbf{(E) } \dfrac{DS}{4Q}$

$\textbf{B}$

Each can of soda costs $\dfrac QS$ quarters, or $\dfrac{Q}{4S}$ dollars. Therefore, $D$ dollars can purchase $\dfrac{D}{\left(\tfrac{Q}{4S}\right)}=\dfrac{4DS}{Q}$ cans of soda.

What is the value of\[\log_37\cdot\log_59\cdot\log_711\cdot\log_913\cdots\log_{21}25\cdot\log_{23}27?\]

$\textbf{(A) } 3 \qquad \textbf{(B) } 3\log_{7}23 \qquad \textbf{(C) } 6 \qquad \textbf{(D) } 9 \qquad \textbf{(E) } 10$

$\textbf{C}$

Using the chain rule of logarithms $\log _{a} b \cdot \log _{b} c = \log _{a} c,$ we get\begin{align*} &\log_37\cdot\log_59\cdot\log_711\cdot\log_913\cdots\log_{21}25\cdot\log_{23}27\\

=\ & (\log _{3} 7 \cdot \log _{7} 11 \cdots \log _{23} 27) \cdot (\log _{5} 9 \cdot \log _{9} 13 \cdots \log _{21} 25) \\

=\ & \log _{3} 27 \cdot \log _{5} 25 \\

=\ & 3 \cdot 2 \\

=\ & 6 \end{align*}

Line segment $\overline{AB}$ is a diameter of a circle with $AB = 24$. Point $C$, not equal to $A$ or $B$, lies on the circle. As point $C$ moves around the circle, the centroid (center of mass) of $\triangle ABC$ traces out a closed curve missing two points. To the nearest positive integer, what is the area of the region bounded by this curve?

$\textbf{(A) } 25 \qquad \textbf{(B) } 38 \qquad \textbf{(C) } 50 \qquad \textbf{(D) } 63 \qquad \textbf{(E) } 75$

$\textbf{C}$

For each $\triangle ABC,$ note that the length of one median is $OC=12.$ Let $G$ be the centroid of $\triangle ABC.$ It follows that $OG=\dfrac13 OC=4.$

As shown below, $\triangle ABC_1$ and $\triangle ABC_2$ are two shapes of $\triangle ABC$ with centroids $G_1$ and $G_2,$ respectively:

Therefore, point $G$ traces out a circle (missing two points) with the center $O$ and the radius $\overline{OG},$ as indicated in red. To the nearest positive integer, the area of the region bounded by the red curve is $\pi\cdot OG^2=16\pi\approx50.$

What is\[\sum^{100}_{i=1} \sum^{100}_{j=1} (i+j) ?\]

$\textbf{(A) }100{,}100 \qquad \textbf{(B) }500{,}500\qquad \textbf{(C) }505{,}000 \qquad \textbf{(D) }1{,}001{,}000 \qquad \textbf{(E) }1{,}010{,}000 \qquad$

$\textbf{E}$

Recall that the sum of the first $100$ positive integers is $\sum^{100}_{k=1} k = \dfrac{101\cdot100}{2}=5050.$ It follows that\begin{align*} \sum^{100}_{i=1} \sum^{100}_{j=1} (i+j) &= \sum^{100}_{i=1} \sum^{100}_{j=1}i + \sum^{100}_{i=1} \sum^{100}_{j=1}j \\ &= \sum^{100}_{i=1} (100i) + 100 \sum^{100}_{j=1}j \\ &= 100 \sum^{100}_{i=1}i + 100 \sum^{100}_{j=1}j \\ &= 100\cdot5050 + 100\cdot5050 \\ &= 1{,}010{,}000 \end{align*}

A list of $2018$ positive integers has a unique mode, which occurs exactly $10$ times. What is the least number of distinct values that can occur in the list?

$\textbf{(A) }202 \qquad \textbf{(B) }223 \qquad \textbf{(C) }224 \qquad \textbf{(D) }225 \qquad \textbf{(E) }234 \qquad$

$\textbf{D}$

To minimize the number of distinct values, we want to maximize the number of times a number appears. So we could have $223$ numbers appear $9$ times, $1$ number appear once, and the mode appear $10$ times, giving us a total of $223 + 1 + 1 =225.$

A closed box with a square base is to be wrapped with a square sheet of wrapping paper. The box is centered on the wrapping paper with the vertices of the base lying on the midlines of the square sheet of paper, as shown in the figure on the left. The four corners of the wrapping paper are to be folded up over the sides and brought together to meet at the center of the top of the box, point $A$ in the figure on the right. The box has base length $w$ and height $h$. What is the area of the sheet of wrapping paper?

$\textbf{(A) } 2(w+h)^2 \qquad \textbf{(B) } \dfrac{(w+h)^2}2 \qquad \textbf{(C) } 2w^2+4wh \qquad \textbf{(D) } 2w^2 \qquad \textbf{(E) } w^2h$

$\textbf{A}$

The sheet of paper is made out of the surface area of the box plus the sum of the four yellow triangles, as shown below.

The surface area of the box is $2w^2 + 4wh + 2wh$. The four triangles each have a height and a base of $h$, so they each have an area of $\dfrac{h^2}{2}$. There are four of them, so multiplied by four is $2h^2$. Together, paper's area is $2w^2 + 4wh + 2h^2=2(w+h)^2$.

Side $\overline{AB}$ of $\triangle ABC$ has length $10$. The bisector of angle $A$ meets $\overline{BC}$ at $D$, and $CD = 3$. The set of all possible values of $AC$ is an open interval $(m,n)$. What is $m+n$?

$\textbf{(A) }16 \qquad \textbf{(B) }17 \qquad \textbf{(C) }18 \qquad \textbf{(D) }19 \qquad \textbf{(E) }20 \qquad$

$\textbf{C}$

Let $AC=x.$ By Angle Bisector Theorem, we have $\dfrac{AB}{AC}=\dfrac{BD}{CD},$ from which $BD=CD\cdot\dfrac{AB}{AC}=\dfrac{30}{x}.$

Recall that $x>0.$ We apply the Triangle Inequality to $\triangle ABC:$

$AC+BC>AB \iff x+\left(\dfrac{30}{x}+3\right)>10$. We have $x+\dfrac{30}{x}+3\geq 2\sqrt{30}+3>2\sqrt{25}+3=13>10$. So we have no restriction on $x$.

$AB+BC>AC \iff 10+\left(\dfrac{30}{x}+3\right)>x$

We simplify and factor to get $(x+2)(x-15)<0,$ from which $0<x<15.$

$AB+AC>BC \iff 10+x>\dfrac{30}{x}+3$

We simplify and factor to get $(x+10)(x-3)>0,$ from which $x>3.$

Taking the intersection of the solutions gives $(m,n)=(3,15),$ so the answer is $m+n=18.$

Square $ABCD$ has side length $30$. Point $P$ lies inside the square so that $AP = 12$ and $BP = 26$. The centroids of $\triangle{ABP}$, $\triangle{BCP}$, $\triangle{CDP}$, and $\triangle{DAP}$ are the vertices of a convex quadrilateral. What is the area of that quadrilateral?

$\textbf{(A) }100\sqrt{2}\qquad\textbf{(B) }100\sqrt{3}\qquad\textbf{(C) }200\qquad\textbf{(D) }200\sqrt{2}\qquad\textbf{(E) }200\sqrt{3}$

$\textbf{C}$

We place the diagram in the coordinate plane. Let $A=(0,30),B=(0,0),C=(30,0),D=(30,30),$ and $P=(3x,3y).$

Recall that for any triangle in the coordinate plane, the coordinates of its centroid are the averages of the coordinates of its vertices. Hence, we get the coordinates of centroid of each triangle $G_1=(x,y+10),G_2=(x+10,y),G_3=(x+20,y+10),$ and $G_4=(x+10,y+20).$

Note that $G_1G_3=G_2G_4=20$ and $\overline{G_1G_3}\perp\overline{G_2G_4}.$ Therefore, the area of quadrilateral $G_1G_2G_3G_4$ is\[\frac12\cdot G_1G_3\cdot G_2G_4=200\]

Joey and Chloe and their daughter Zoe all have the same birthday. Joey is $1$ year older than Chloe, and Zoe is exactly $1$ year old today. Today is the first of the $9$ birthdays on which Chloe's age will be an integral multiple of Zoe's age. What will be the sum of the two digits of Joey's age the next time his age is a multiple of Zoe's age?

$\textbf{(A) }7 \qquad \textbf{(B) }8 \qquad \textbf{(C) }9 \qquad \textbf{(D) }10 \qquad \textbf{(E) }11 \qquad$

$\textbf{E}$

Suppose that Chloe is $c$ years old today, so Joey is $c+1$ years old today. After $n$ years, Chloe and Zoe will be $n+c$ and $n+1$ years old, respectively. We are given that\[\frac{n+c}{n+1}=1+\frac{c-1}{n+1}\]is an integer for $9$ nonnegative integers $n.$ It follows that $c-1$ has $9$ positive divisors. The prime factorization of $c-1$ is either $p^8$ or $p^2q^2.$ Since $c-1<100,$ the only possibility is $c-1=2^2\cdot3^2=36,$ from which $c=37.$ We conclude that Joey is $c+1=38$ years old today.

Suppose that Joey's age is a multiple of Zoe's age after $k$ years, in which Joey and Zoe will be $k+38$ and $k+1$ years old, respectively. We are given that\[\frac{k+38}{k+1}=1+\frac{37}{k+1}\]is an integer for some positive integer $k.$ It follows that $37$ is divisible by $k+1,$ so the only possibility is $k=36.$ We conclude that Joey will be $k+38=74$ years old then, from which the answer is $7+4=11.$

How many odd positive $3$-digit integers are divisible by $3$ but do not contain the digit $3$?

$\textbf{(A) } 96 \qquad \textbf{(B) } 97 \qquad \textbf{(C) } 98 \qquad \textbf{(D) } 102 \qquad \textbf{(E) } 120$

$\textbf{A}$

Let $\underline{ABC}$ be one such odd positive $3$-digit integer with hundreds digit $A,$ tens digit $B,$ and ones digit $C.$ Since $\underline{ABC}\equiv0\pmod3,$ we need $A+B+C\equiv0\pmod3$ by the divisibility rule for $3.$

As $A\in\{1,2,4,5,6,7,8,9\}$ and $C\in\{1,5,7,9\},$ there are $8$ possibilities for $A$ and $4$ possibilities for $C.$ Note that each ordered pair $(A,C)$ determines the value of $B$ modulo $3,$ so $B$ can be any element in one of the sets $\{0,6,9\},\{1,4,7\},$ or $\{2,5,8\}.$ We conclude that there are always $3$ possibilities for $B.$

By the Multiplication Principle, the answer is $8\cdot4\cdot3=96.$

The solutions to the equation $(z+6)^8=81$ are connected in the complex plane to form a convex regular polygon, three of whose vertices are labeled $A,B,$ and $C$. What is the least possible area of $\triangle ABC?$

$\textbf{(A) } \dfrac{1}{6}\sqrt{6} \qquad \textbf{(B) } \dfrac{3}{2}\sqrt{2}-\dfrac{3}{2} \qquad \textbf{(C) } 2\sqrt3-3\sqrt2 \qquad \textbf{(D) } \dfrac{1}{2}\sqrt{2} \qquad \textbf{(E) } \sqrt 3-1$

$\textbf{B}$

Recall that translations preserve the shapes and the sizes for all objects. We translate the solutions to the given equation $6$ units right, so they become the solutions to the equation $z^8=81.$

We rewrite $z$ to the polar form\[z=re^{i\theta}\]where $r$ is the magnitude of $z$ such that $r\geq0,$ and $\theta$ is the argument of $z$ such that $0\leq\theta<2\pi.$

By De Moivre's Theorem, we have\[z^8=r^8e^{i8\theta}=81\] from which $$r=\sqrt4,\ \theta=0,\frac{\pi}{4},\frac{\pi}{2},\frac{3\pi}{4},\pi,\frac{5\pi}{4},\frac{3\pi}{2},\frac{7\pi}{4}$$

In the complex plane, the solutions to the equation $z^8=81$ are the vertices of a regular octagon with center $0$ and radius $\sqrt3.$

The least possible area of $\triangle ABC$ occurs when $A,B,$ and $C$ are the consecutive vertices of the octagon. For simplicity purposes, let $A=\sqrt3\left(\cos\dfrac{\pi}{4}+i\sin\dfrac{\pi}{4}\right)=\dfrac{\sqrt6}{2}+\dfrac{\sqrt6}{2}i, B=\sqrt3\left(\cos\dfrac{\pi}{2}+i\sin\dfrac{\pi}{2}\right)=\sqrt3i,$ and $C=\sqrt3\left(\cos\dfrac{3\pi}{4}+i\sin\dfrac{3\pi}{4}\right)=-\dfrac{\sqrt6}{2}+\dfrac{\sqrt6}{2}i,$ as shown below.

Note that $\triangle ABC$ has base $AC=\sqrt6$ and height $\sqrt3-\frac{\sqrt6}{2},$ so its area is\[\frac12\cdot\sqrt6\cdot\left(\sqrt3-\frac{\sqrt6}{2}\right)= \frac{3}{2}\sqrt{2}-\frac{3}{2}\]

Let $p$ and $q$ be positive integers such that\[\frac{5}{9} < \frac{p}{q} < \frac{4}{7}\]and $q$ is as small as possible. What is $q-p$?

$\textbf{(A) } 7 \qquad \textbf{(B) } 11 \qquad \textbf{(C) } 13 \qquad \textbf{(D) } 17 \qquad \textbf{(E) } 19$

$\textbf{A}$

Inverting the given inequality we get\[\frac{7}{4} < \frac{q}{p} < \frac{9}{5}\]which simplifies to\[35p < 20q < 36p\]We can now substitute $q = p + k.$ Note we need to find $k:$\[35p < 20p + 20k < 36p\]which simplifies to\[15p < 20k < 16p\]Cleary $p>k.$ We will now substitute $p = k + x$ to get\[15k + 15x < 20k < 16k + 16x\]The inequality $15k + 15x < 20k$ simplifies to $3x < k.$

The inequality $20k < 16k + 16x$ simplifies to $k < 4x.$

Combining the two inequalities, we get\[3x < k < 4x\]Since $x$ and $k$ are integers, the smallest values of $x$ and $k$ that satisfy the above equation are $2$ and $7$ respectively. Substituting these back in, we arrive with an answer of $q-p=k=7.$

A function $f$ is defined recursively by $f(1)=f(2)=1$ and\[f(n)=f(n-1)-f(n-2)+n\]for all integers $n \geq 3$. What is $f(2018)$?

$\textbf{(A) } 2016 \qquad \textbf{(B) } 2017 \qquad \textbf{(C) } 2018 \qquad \textbf{(D) } 2019 \qquad \textbf{(E) } 2020$

$\textbf{B}$

For all integers $n \geq 7,$ note that\begin{align*} f(n)&=f(n-1)-f(n-2)+n \\ &=[f(n-2)-f(n-3)+n-1]-f(n-2)+n \\ &=-f(n-3)+2n-1 \\ &=-[f(n-4)-f(n-5)+n-3]+2n-1 \\ &=-f(n-4)+f(n-5)+n+2 \\ &=-[f(n-5)-f(n-6)+n-4]+f(n-5)+n+2 \\ &=f(n-6)+6 \end{align*}It follows that\begin{align*} f(2018)&=f(2012)+6 \\ &=f(2006)+12 \\ &=f(2000)+18 \\ & \ \vdots \\ &=f(2)+2016 \\ &=2017 \end{align*}

Mary chose an even $4$-digit number $n$. She wrote down all the divisors of $n$ in increasing order from left to right: $1,2,\ldots,\dfrac{n}{2},n$. At some moment Mary wrote $323$ as a divisor of $n$. What is the smallest possible value of the next divisor written to the right of $323$?

$\textbf{(A) } 324 \qquad \textbf{(B) } 330 \qquad \textbf{(C) } 340 \qquad \textbf{(D) } 361 \qquad \textbf{(E) } 646$

$\textbf{C}$

Let $d$ be the next divisor written to the right of $323.$

Since $n$ is even and $323=17\cdot19,$ we have $n=2\cdot17\cdot19\cdot k=646k$ for some positive integer $k.$ Moreover, since $1000\leq n\leq9998,$ we get $2\leq k\leq15.$ As $d>323,$ it is clear that $d$ must be divisible by $17$ or $19$ or both.

Therefore, the smallest possible value of $d$ is $17\cdot20=340,$ from which $n=\operatorname{lcm}(323,d)=17\cdot19\cdot20=6460.$

Let $ABCDEF$ be a regular hexagon with side length $1$. Denote by $X$, $Y$, and $Z$ the midpoints of sides $\overline {AB}$, $\overline{CD}$, and $\overline{EF}$, respectively. What is the area of the convex hexagon whose interior is the intersection of the interiors of $\triangle ACE$ and $\triangle XYZ$?

$\textbf{(A)}\ \dfrac {3}{8}\sqrt{3} \qquad \textbf{(B)}\ \dfrac {7}{16}\sqrt{3} \qquad \textbf{(C)}\ \dfrac {15}{32}\sqrt{3} \qquad \textbf{(D)}\ \dfrac {1}{2}\sqrt{3} \qquad \textbf{(E)}\ \dfrac {9}{16}\sqrt{3}$

$\textbf{C}$

The shaded area is the area of $\triangle XYZ$ subtracts the areas of three congruent triangles $\triangle XPN$, $\triangle YQO$ and $\triangle ZRM$.

The side length of regular hexagon $ABCDEF$ is $DE=1$. Furthermore, we have $CF=2DE=2$. Hence, the side length of equilateral triangle $XYZ$ is $YZ=\dfrac{1+2}{2}=\dfrac32$. Its area is $\dfrac{\sqrt3}{4}\cdot\left(\dfrac32\right)^2=\dfrac9{16}\sqrt3$.

Notice that $\angle AEF=30^\circ$. We also notice that $\triangle ZRM\sim FME$ and $ZM=\dfrac12 FE=\dfrac12$. So the area of right triangle $ZRM$ is $\dfrac12ZR\cdot RM=\dfrac12\cdot\dfrac12\sin30^\circ\cdot\dfrac12\cos30^\circ=\dfrac{\sqrt3}{32}$.

The shaded area is the area of $\triangle XYZ$ subtracts the areas of three congruent triangles $\triangle XPN$, $\triangle YQO$ and $\triangle ZRM$. So the answer is $\dfrac9{16}\sqrt3-3\cdot\dfrac{\sqrt3}{32}=\dfrac {15}{32}\sqrt{3}$.

In $\triangle{ABC}$ with side lengths $AB = 13$, $AC = 12$, and $BC = 5$, let $O$ and $I$ denote the circumcenter and incenter, respectively. A circle with center $M$ is tangent to the legs $AC$ and $BC$ and to the circumcircle of $\triangle{ABC}$. What is the area of $\triangle{MOI}$?

$\textbf{(A)}\ \dfrac52\qquad\textbf{(B)}\ \dfrac{11}{4}\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ \dfrac{13}{4}\qquad\textbf{(E)}\ \dfrac72$

$\textbf{E}$

In this solution, let the brackets denote areas.

We place the diagram in the coordinate plane: Let $A=(12,0),B=(0,5),$ and $C=(0,0).$

Since $\triangle ABC$ is a right triangle with $\angle ACB=90^\circ,$ its circumcenter is the midpoint of $\overline{AB},$ from which $O=\left(6,\dfrac52\right).$ Note that the circumradius of $\triangle ABC$ is $\dfrac{13}{2}.$

Let $s$ denote the semiperimeter of $\triangle ABC.$ The inradius of $\triangle ABC$ is $\dfrac{[ABC]}{s}=\dfrac{30}{15}=2,$ from which $I=(2,2).$

Since $\odot M$ is also tangent to both coordinate axes, its center is at $M=(a,a)$ and its radius is $a$ for some positive number $a.$ Let $P$ be the point of tangency of $\odot O$ and $\odot M.$ As $\overline{OP}$ and $\overline{MP}$ are both perpendicular to the common tangent line at $P,$ we conclude that $O,M,$ and $P$ are collinear. It follows that $OM=OP-MP,$ or\[\sqrt{(a-6)^2+\left(a-\frac52\right)^2}=\frac{13}{2}-a\]Solving this equation, we have $a=4,$ from which $M=(4,4).$

Finally, we apply the Shoelace Theorem to $\triangle MOI:$\[[MOI]=\frac12\left|\left(4\cdot\frac52+6\cdot2+2\cdot4\right)-\left(4\cdot6+\frac52\cdot2+2\cdot4\right)\right|=\frac72\]

Consider polynomials $P(x)$ of degree at most $3$, each of whose coefficients is an element of $\{0, 1, 2, 3, 4, 5, 6, 7, 8, 9\}$. How many such polynomials satisfy $P(-1) = -9$?

$\textbf{(A) } 110 \qquad \textbf{(B) } 143 \qquad \textbf{(C) } 165 \qquad \textbf{(D) } 220 \qquad \textbf{(E) } 286$

$\textbf{D}$

Suppose that $P(x)=ax^3+bx^2+cx+d.$ This problem is equivalent to counting the ordered quadruples $(a,b,c,d),$ where all of $a,b,c,$ and $d$ are integers from $0$ through $9$ such that\[P(-1)=-a+b-c+d=-9\]Let $a'=9-a$ and $c'=9-c.$ Note that both of $a'$ and $c'$ are integers from $0$ through $9.$ Moreover, the ordered quadruples $(a,b,c,d)$ and the ordered quadruples $(a',b,c',d)$ have one-to-one correspondence.

We rewrite the given equation as $(9-a)+b+(9-c)+d=9,$ or\[a'+b+c'+d=9\]By the stars and bars argument, there are $\dbinom{9+4-1}{4-1}=220$ ordered quadruples $(a',b,c',d).$

Ajay is standing at point $A$ near Pontianak, Indonesia, $0^\circ$ latitude and $110^\circ \textrm{ E}$ longitude. Billy is standing at point $B$ near Big Baldy Mountain, Idaho, USA, $45^\circ \textrm{ N}$ latitude and $115^\circ \textrm{ W}$ longitude. Assume that Earth is a perfect sphere with center $C.$ What is the degree measure of $\angle ACB?$

$\textbf{(A) }105 \qquad \textbf{(B) }112\dfrac{1}{2} \qquad \textbf{(C) }120 \qquad \textbf{(D) }135 \qquad \textbf{(E) }150 \qquad$

$\textbf{C}$

Let $D$ be the orthogonal projection of $B$ onto the equator. Note that $\angle BDA = \angle BDC = 90^\circ$ and $\angle BCD = 45^\circ.$ Recall that $115^\circ \text{ W}$ longitude is the same as $245^\circ \text{ E}$ longitude, so $\angle ACD=135^\circ.$

We obtain the following diagram:

Without the loss of generality, let $AC=BC=1.$ Since $\triangle BCD$ is an isosceles right triangle, we have $BD=CD=\dfrac{\sqrt2}{2}.$

In $\triangle ACD,$ we apply the Law of Cosines to get $AD=\sqrt{AC^2+CD^2-2\cdot AC\cdot CD\cdot\cos\angle ACD}=\dfrac{\sqrt{10}}{2}.$ In right triangle $\triangle ABD,$ we apply the Pythagorean Theorem to get $AB=\sqrt{AD^2+BD^2}=\sqrt{3}.$ In $\triangle ABC,$ we apply the Law of Cosines to get $\cos\angle ACB=\dfrac{AC^2+BC^2-AB^2}{2\cdot AC\cdot BC}=-\dfrac12,$ so $\angle ACB=120^\circ$.

Let $\lfloor x \rfloor$ denote the greatest integer less than or equal to $x$. How many real numbers $x$ satisfy the equation $x^2 + 10,000\lfloor x \rfloor = 10,000x$?

$\textbf{(A) } 197 \qquad \textbf{(B) } 198 \qquad \textbf{(C) } 199 \qquad \textbf{(D) } 200 \qquad \textbf{(E) } 201$

$\textbf{C}$

The equation can be rewrite as $x^2=10,000\{x\}$, where $\{x\}$ is the fractional part of $x$.

Graphing $y=10,000\{x\}$ and $y=x^2$ we see that the former is a set of line segments with slope $10,000$ from $0$ to $1$ with a hole at $x=1$, then $1$ to $2$ with a hole at $x=2$ etc. Here is a graph of $y=x^2$ and $y=16\{x\}$ for visualization.

Notice that when $x=\pm 100$ the graph has a hole at $(\pm 100,10,000)$ which the equation $y=x^2$ passes through and then continues upwards. Thus our set of possible solutions is bounded by $(-100,100)$. We can see that $y=x^2$ intersects each of the lines once and there are $99+99+1=199$ lines between $(-100,100)$. So the answer is 199.

Circles $\omega_1$, $\omega_2$, and $\omega_3$ each have radius $4$ and are placed in the plane so that each circle is externally tangent to the other two. Points $P_1$, $P_2$, and $P_3$ lie on $\omega_1$, $\omega_2$, and $\omega_3$ respectively such that $P_1P_2=P_2P_3=P_3P_1$ and line $P_iP_{i+1}$ is tangent to $\omega_i$ for each $i=1,2,3$, where $P_4 = P_1$. See the figure below. The area of $\triangle P_1P_2P_3$ can be written in the form $\sqrt{a}+\sqrt{b}$ for positive integers $a$ and $b$. What is $a+b$?

$\textbf{(A) }546\qquad\textbf{(B) }548\qquad\textbf{(C) }550\qquad\textbf{(D) }552\qquad\textbf{(E) }554$

$\textbf{D}$

Let $O_i$ be the center of circle $\omega_i$ for $i=1,2,3$. $O_1X$ is perpendicular to line $P_1P_2$. Since $P_1P_2$ is tangent to $\omega_1$, we see that $\triangle O_1P_1X$ is a 30-60-90 triangle. So $O_1X=\dfrac12O_1P_1=2$, $XP_1=\dfrac{\sqrt3}{2}O_1P_1=2\sqrt3$.

Notice that $P_1P_3$ is perpendicular to both $O_1X$ and $O_3P_3$, we have $$XP_3=\sqrt{O_1O_3^2-\left(O_1X+O_3P_3\right)^2}=\sqrt{8^2-(2+4)^2}=2\sqrt7$$ Hence, the side length of equilateral triangle $\triangle P_1P_2P_3$ is $P_1P_3=2\sqrt3+2\sqrt7$. So its area is \[\frac{\sqrt 3}4\cdot\left(2\sqrt 3 + 2\sqrt 7\right)^2 = 10\sqrt 3 + 6\sqrt 7 = \sqrt{300} + \sqrt{252}\] The answer is $300+252=552$.