## AMC 10 2018 Test B

**If you like our free learning materials, please click the advertisement below or anywhere. No paying, no donation, just a simple click. The advertising revenues will be used to provide more and better learning materials. Thank you!**

**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.

In the expression $\left(\underline{\qquad}\times\underline{\qquad}\right)+\left(\underline{\qquad}\times\underline{\qquad}\right)$ each blank is to be filled in with one of the digits $1,2,3,$ or $4,$ with each digit being used once. How many different values can be obtained?

$\textbf{(A) }2 \qquad \textbf{(B) }3 \qquad \textbf{(C) }4 \qquad \textbf{(D) }6 \qquad \textbf{(E) }24 \qquad$

$\textbf{B}$

For digit 1, we have three choices (2, 3, or 4) to pair with it. When the first pair is settled, another pair is also determined. Now we check the values of the three choices. $$1\times2+3\times4=14$$ $$1\times3+2\times4=11$$ $$1\times4+2\times3=10$$ They have different values. So the answer is 3.

A three-dimensional rectangular box with dimensions $X$, $Y$, and $Z$ has faces whose surface areas are $24, 24, 48, 48, 72,$ and $72$ square units. What is $X+Y+Z$?

$\textbf{(A) }18 \qquad \textbf{(B) }22 \qquad \textbf{(C) }24 \qquad \textbf{(D) }30 \qquad \textbf{(E) }36 \qquad$

$\textbf{B}$

Let $X$ be the length of the shortest dimension and $Z$ be the length of the longest dimension. Thus, we have $XY = 24$, $YZ = 72$, and $XZ = 48$. Multiplying them together, we get $X^2Y^2Z^2=24\cdot72\cdot48$, or $XYZ=288$. Dividing it by $XY = 24$, $YZ = 72$, and $XZ = 48$, we get $X=4$, $Y=6$, $Z=12$.

The answer is $4+6+12=22$.

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

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

$\textbf{D}$

There are 4 prime numbers and 4 composite numbers in the set. We can create $2^8-1=255$ nonempty subsets in total. Among the 255 nonempty subsets, $2^4-1=15$ of them contain only composite numbers. Hence, the number of subsets contain at least one prime number is $255-15=240$.

A box contains $5$ chips, numbered $1, 2, 3, 4,$ and $5$. Chips are drawn randomly one at a time without replacement until the sum of the values drawn exceeds $4$. What is the probability that $3$ draws are required?

$\textbf{(A) }\dfrac{1}{15} \qquad \textbf{(B) }\dfrac{1}{10} \qquad \textbf{(C) }\dfrac{1}{6} \qquad \textbf{(D) }\dfrac{1}{5} \qquad \textbf{(E) }\dfrac{1}{4} \qquad$

$\textbf{D}$

There are 4 ways for the first two draws if a third draw is required: (1,2), (1,3), (2,1) and (3,1). Each of the 4 cases has a $\dfrac15\cdot\dfrac14$ chance. So the answer is $\dfrac15\cdot\dfrac14\cdot4=\dfrac15$.

In the figure below, $N$ congruent semicircles are drawn along a diameter of a large semicircle, with their diameters covering the diameter of the large semicircle with no overlap. Let $A$ be the combined area of the small semicircles and $B$ be the area of the region inside the large semicircle but outside the small semicircles. The ratio $A:B$ is $1:18$. What is $N$?

$\textbf{(A) }16 \qquad \textbf{(B) }17 \qquad \textbf{(C) }18 \qquad \textbf{(D) }19 \qquad \textbf{(E) }36 \qquad$

$\textbf{D}$

Let the number of small semicircles be $n$, and the radius of each small semicircle be $r$.

The area of the sum of all small semicircles is $A=n\cdot\dfrac12\pi r^2$.

The area of the large semicircle is $A+B=\dfrac12\pi(nr)^2$

Since $A:B=1:18$, we have $$\dfrac{A}{A+B}=\dfrac{n\cdot\dfrac12\pi r^2}{\dfrac12\pi(nr)^2}=\dfrac{1}{n}=\dfrac1{19}$$ Hence, we get $n=19$.

Sara makes a staircase out of toothpicks as shown:

This is a $3$-step staircase and uses $18$ toothpicks. How many steps would be in a staircase that used $180$ toothpicks?

$\textbf{(A) }10 \qquad \textbf{(B) }11 \qquad \textbf{(C) }12 \qquad \textbf{(D) }24 \qquad \textbf{(E) }30 \qquad$

$\textbf{C}$

There are $1+2+\cdots+n=\dfrac{n(n+1)}{2}$ squares in a $n$-step staircase. Each square has 4 sides. So the total number of sides is $4\cdot\dfrac{n(n+1)}{2}=2n(n+1)$ (including overlapping).

To find out the overlapping, notice that the perimeter of the $n$-step staircase is $4n$. Hence, the number of overlapping is $\dfrac{2n(n+1)-4n}{2}=n^2-n$. Therefore, the number of toothpicks is $n^2-n+4n=n^2+3n$.

When $n^2+3n=180$, we get $n=12$. So the answer is 12.

The faces of each of $7$ standard dice are labeled with the integers from $1$ to $6$. Let $p$ be the probability that when all $7$ dice are rolled, the sum of the numbers on the top faces is $10$. What other sum occurs with the same probability $p$?

$\textbf{(A) }13 \qquad \textbf{(B) }26 \qquad \textbf{(C) }32 \qquad \textbf{(D) }39 \qquad \textbf{(E) }42 \qquad$

$\textbf{D}$

By symmetry, the probability of rolling the smallest number possible is the same as the probability of rolling the largest number possible, the probability of rolling the second smallest number possible is the same as the probability of rolling the second largest number possible, and so on.

Hence, we can match up the values to find the sum with the same probability as $10$. We can start by noticing that $7$ is the smallest possible roll and $42$ is the largest possible roll. The pairs with the same probability are as follows: $$(7, 42), (8, 41), (9, 40), (10, 39)\cdots$$ The number matched up with 10 is 39. So the answer is 39.

In the rectangular parallelepiped shown, $AB=3$, $BC=1$, and $CG=2$. Point $M$ is the midpoint of $\overline{FG}$. What is the volume of the rectangular pyramid with base $BCHE$ and apex $M$?

$\textbf{(A) }1 \qquad \textbf{(B) }\dfrac{4}{3} \qquad \textbf{(C) }\dfrac{3}{2} \qquad \textbf{(D) }\dfrac{5}{3} \qquad \textbf{(E) }2 \qquad$

$\textbf{E}$

In right triangle $ABE$, we have $BE=\sqrt{3^2+2^2}=\sqrt{13}$. So the area of cross-sectional plane $BCHE$ is $BE\cdot BC=\sqrt{13}\cdot1=\sqrt{13}$.

The distance from point $M$ to plane $BCHE$ is the same as the distance from point $F$ to line $BE$, or the distance from point $G$ to line $CH$. The distance from point $F$ to line $BE$ in right triangle $BEF$ is $\dfrac{BF\cdot EF}{BE}=\dfrac{6}{\sqrt{13}}$.

Therefore, the volume of the rectangular pyramid with base $BCHE$ and apex $M$ is $\dfrac13\cdot\sqrt{13}\cdot\dfrac{6}{\sqrt{13}}=2$.

Which of the following expressions is never a prime number when $p$ is a prime number?

$\textbf{(A) }p^2+16 \qquad \textbf{(B) }p^2+24 \qquad \textbf{(C) }p^2+26 \qquad \textbf{(D) }p^2+46 \qquad \textbf{(E) }p^2+96 \qquad$

$\textbf{C}$

If $p$ is not a multiple of 3, we have $p^2\equiv1 \pmod3$. So $p^2+26$ is a multiple of 3, which is not a prime.

If $p$ is a multiple of 3, since $p$ is a prime, we have $p=3$. Now $p^2+26=35$ is still composite.

Therefore, the answer is C.

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?

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

$\textbf{C}$

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 $\pi\cdot OG^2=16\pi\approx50.$

How many of the first $2018$ numbers in the sequence $101, 1001, 10001, 100001, \dots$ are divisible by $101$?

$\textbf{(A) }253 \qquad \textbf{(B) }504 \qquad \textbf{(C) }505 \qquad \textbf{(D) }506 \qquad \textbf{(E) }1009 \qquad$

$\textbf{C}$

Note that $101=x^2+1$ and $100...0001=x^n+1$, where $x=10$. We have that $\dfrac{x^n+1}{x^2+1}$ must have a remainder of $0$. By the remainder theorem, the roots of $x^2+1$ must also be roots of $x^n+1$. Plugging in $i,-i$ to $x^n+1=0$ yields that $n\equiv2\pmod{4}$.

In the given list $10^2+1,10^3+1,10^4+1,\dots,10^{2019}+1$, the desired exponents are $2,6,10,\dots,2018$. So there are $\dfrac{2020}{4}=505$ numbers in the list.

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 $2w^2 + 4wh + 2h^2=2(w+h)^2$.

Let $a_1,a_2,\dots,a_{2018}$ be a strictly increasing sequence of positive integers such that$$a_1+a_2+\cdots+a_{2018}=2018^{2018}.$$What is the remainder when $a_1^3+a_2^3+\cdots+a_{2018}^3$ is divided by $6$?

$\textbf{(A) }0 \qquad \textbf{(B) }1 \qquad \textbf{(C) }2 \qquad \textbf{(D) }3 \qquad \textbf{(E) }4 \qquad$

$\textbf{E}$

First, we need to prove that $a\equiv a^3 \pmod{6}$ for any integer $a$. Notice that $a^3-a=a(a^2-1)$. $a$ and $a^2-1$ have opposite parity. If $a$ is a multiple of 3, then $a(a^2-1)$ is a multiple of 6. If $a$ is not a multiple of 3, then $a^2-1$ is a multiple of 3, which lead to the same conclusion. Hence, we get $a^3-a\equiv a^0 \pmod{6}$, or $a\equiv a^3 \pmod{6}$.

By $a\equiv a^3 \pmod{6}$, we get\[a_1+a_2+\cdots+a_{2018} \equiv a_1^3+a_2^3+\cdots+a_{2018}^3 \pmod{6}\] Since $a_1+a_2+\cdots+a_{2018}=2018^{2018}=(2016+2)^{2018}=(336\cdot6+2)^{2018}$, we have $$2018^{2018}\equiv 2^{2018} \pmod{6}$$ Notice that $2^n\equiv2 \pmod{6}$ when $n$ is odd, and $2^n\equiv4 \pmod{6}$ when $n$ is even. Hence, we get $2^{2018}\equiv 4\pmod6$. The answer is 4.

In rectangle $PQRS$, $PQ=8$ and $QR=6$. Points $A$ and $B$ lie on $\overline{PQ}$, points $C$ and $D$ lie on $\overline{QR}$, points $E$ and $F$ lie on $\overline{RS}$, and points $G$ and $H$ lie on $\overline{SP}$ so that $AP=BQ<4$ and the convex octagon $ABCDEFGH$ is equilateral. The length of a side of this octagon can be expressed in the form $k+m\sqrt{n}$, where $k$, $m$, and $n$ are integers and $n$ is not divisible by the square of any prime. What is $k+m+n$?

$\textbf{(A) }1 \qquad \textbf{(B) }7 \qquad \textbf{(C) }21 \qquad \textbf{(D) }92 \qquad \textbf{(E) }106 \qquad$

$\textbf{B}$

Let the side length of the octagon be $x$. In right triangle $\triangle APH$, we have $AP=\dfrac{8-x}{2}$, $PH=\dfrac{6-x}{2}$, and $AH=x$. By the Pythagorean theorem, we get $$\left(\dfrac{8-x}{2}\right)^2+\left(\dfrac{6-x}{2}\right)^2=x^2$$ Solving the equation, we get $x=-7+3\sqrt{11}$. Hence, the answer is $k+m+n=-7+3+11=7$.

Three young brother-sister pairs from different families need to take a trip in a van. These six children will occupy the second and third rows in the van, each of which has three seats. To avoid disruptions, siblings may not sit right next to each other in the same row, and no child may sit directly in front of his or her sibling. How many seating arrangements are possible for this trip?

$\textbf{(A) }60 \qquad \textbf{(B) }72 \qquad \textbf{(C) }92 \qquad \textbf{(D) }96 \qquad \textbf{(E) }120 \qquad$

$\textbf{D}$

Let the siblings be $Aa$, $Bb$, and $Cc$.

We have $6$ choices for the middle seat of the second row. Without loss of generality, let $A$ takes it. $$\circ\quad A\quad\circ$$ $$\circ\quad\circ\quad\circ$$ Now $a$ must take either the left seat of the third row, or the right seat of the third row. By symmetry, let $a$ takes the left. The final answer should be multiplied by 2. $$\circ\quad A\quad\circ$$ $$a\quad\circ\quad\circ$$ Now we have $4$ choices for the middle seat of the third row. Without loss of generality, let $B$ takes it. $$\circ\quad A\quad\circ$$ $$a\quad B\quad\circ$$ Now $b$ must take the right seat of the second row. Then we have 2 arrangements for the $Cc$ pairs.

The total number of arrangements is $6\times2\times4\times2=96$.

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 $7+4=11.$

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 $323$?

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

$\textbf{C}$

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 $n=\operatorname{lcm}(323,d)=17\cdot19\cdot20=6460.$

Real numbers $x$ and $y$ are chosen independently and uniformly at random from the interval $[0,1]$. Which of the following numbers is closest to the probability that $x,y,$ and $1$ are the side lengths of an obtuse triangle?

$\textbf{(A) }0.21 \qquad \textbf{(B) }0.25 \qquad \textbf{(C) }0.29 \qquad \textbf{(D) }0.50 \qquad \textbf{(E) }0.79 \qquad$

$\textbf{C}$

The Pythagorean Inequality tells us that in an obtuse triangle, $a^{2} + b^{2} < c^{2}$. The triangle inequality tells us that $a + b > c$. So, we have two inequalities:\[x^2 + y^2 < 1\]\[x + y > 1\]The first equation is $\dfrac14$ of a circle with radius $1$, and the second equation is a line from $(0, 1)$ to $(1, 0)$. So, the area is $\dfrac{\pi}{4} - \dfrac12$ which is approximately $0.29$.

How many ordered pairs $(a, b)$ of positive integers satisfy the equation$$a\cdot b + 63 = 20\cdot \text{lcm}(a, b) + 12\cdot\text{gcd}(a,b),$$where $\text{gcd}(a,b)$ denotes the greatest common divisor of $a$ and $b$, and $\text{lcm}(a,b)$ denotes their least common multiple?

$\textbf{(A)} \text{ 0} \qquad \textbf{(B)} \text{ 2} \qquad \textbf{(C)} \text{ 4} \qquad \textbf{(D)} \text{ 6} \qquad \textbf{(E)} \text{ 8}$

$\textbf{B}$

Let $x = \text{lcm}(a, b)$, and $y = \text{gcd}(a, b)$. Therefore, $a\cdot b = \text{lcm}(a, b)\cdot \text{gcd}(a, b) = x\cdot y$. Thus, the equation becomes \[x\cdot y + 63 = 20x + 12y\]The equation can be rearranged as\[(x - 12)(y - 20) = 177\]Since $177 = 3\cdot 59$ and $x > y$, we have $x - 12 = 59$ and $y - 20 = 3$, or $x - 12 = 177$ and $y - 20 = 1$. This gives us the solutions $(71, 23)$ and $(189, 21)$. Since the $\text{gcd}$ must be a divisor of the $\text{lcm}$, the first pair does not work.

By $189=21\cdot3^2$, we have $a = 21 \cdot 3^2$ and $b = 21$, or $ b=21\cdot3^2$ and $a=21$. So there are $2$ solutions.

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 $\dfrac9{16}\sqrt3-3\cdot\dfrac{\sqrt3}{32}=\dfrac {15}{32}\sqrt{3}$.

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.