AMC 10 2017 Test A

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.

Question 1

What is the value of $(2(2(2(2(2(2+1)+1)+1)+1)+1)+1)$?

$\textbf{(A)}\ 70\qquad\textbf{(B)}\ 97\qquad\textbf{(C)}\ 127\qquad\textbf{(D)}\ 159\qquad\textbf{(E)}\ 729$

Solution

$\textbf{C}$

Computing from the innermost parenthesis outwards, we have \begin{align*}
2(2(2(2(2(2+1)+1)+1)+1)+1)+1&=2(2(2(2(2\times3+1)+1)+1)+1)+1\\
&=2(2(2(2(6+1)+1)+1)+1)+1\\
&=2(2(2(2\times7+1)+1)+1)+1\\
&=2(2(2(14+1)+1)+1)+1\\
&=2(2(2\times15+1)+1)+1\\
&=2(2(30+1)+1)+1\\
&=2(2\times31+1)+1\\
&=2(62+1)+1\\
&=2\times63+1\\
&=126+1\\
&=127
\end{align*}

Question 2

Pablo buys popsicles for his friends. The store sells single popsicles for $\$1$ each, $3$-popsicle boxes for $\$2$ each, and $5$-popsicle boxes for $\$3$. What is the greatest number of popsicles that Pablo can buy with $\$8$?

$\textbf{(A)}\ 8\qquad\textbf{(B)}\ 11\qquad\textbf{(C)}\ 12\qquad\textbf{(D)}\ 13\qquad\textbf{(E)}\ 15$

Solution

$\textbf{D}$

The price of a popsicle in a 5-popsicle box is lower than the price of a popsicle in a 3-popsicle box, and the price of a popsicle in a 3-popsicle box is lower than the price of a single prosicle. So Pablo prefers buying 5-popsicle boxes. He can afford two 5-popsicle boxes, then left $\$2$ for a 3-popsicle box. So the totol number of popsicle he can buy is $2\times5+3=13$.

Question 3

Tamara has three rows of two $6$-feet by $2$-feet flower beds in her garden. The beds are separated and also surrounded by $1$-foot-wide walkways, as shown on the diagram. What is the total area of the walkways, in square feet?

$\textbf{(A)}\ 72\qquad\textbf{(B)}\ 78\qquad\textbf{(C)}\ 90\qquad\textbf{(D)}\ 120\qquad\textbf{(E)}\ 150$

Solution

$\textbf{B}$

The length of the garden is $1+6+1+6+1=15$ feet, and the width is $1+2+1+2+1+2+1=10$ feet. So the area of the garden is $15\times10=150\ \text{ft}^2$. The total area of the flower beds is $6\times2\times6=72\ \text{ft}^2$. Therefore, the total area of the walkways is $150-72=78\ \text{ft}^2$.

Question 4

Mia is “helping” her mom pick up $30$ toys that are strewn on the floor. Mia’s mom manages to put $3$ toys into the toy box every $30$ seconds, but each time immediately after those $30$ seconds have elapsed, Mia takes $2$ toys out of the box. How much time, in minutes, will it take Mia and her mom to put all $30$ toys into the box for the first time?

$\textbf{(A)}\ 13.5\qquad\textbf{(B)}\ 14\qquad\textbf{(C)}\ 14.5\qquad\textbf{(D)}\ 15\qquad\textbf{(E)}\ 15.5$

Solution

$\textbf{B}$

Every $30$ seconds, $3$ toys are put in the box and $2$ toys are taken out, so the number of toys in the box increases by $3-2=1$ every $30$ seconds. Then after $27 \times 30 = 810$ seconds, there are $27$ toys in the box. Mia's mom will then put the remaining $3$ toys into the box after $30$ more seconds, so the total time taken is $27\times 30+30=840$ seconds, or $14$ minutes.

Question 5

The sum of two nonzero real numbers is $4$ times their product. What is the sum of the reciprocals of the two numbers?

$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 8\qquad\textbf{(E)}\ 12$

Solution

$\textbf{C}$

Let the two real numbers be $x$ and $y$. Given that $x+y=4xy$, then dividing both sides by $xy$, we get $\dfrac1y+\dfrac1x=4.$ So the answer is 4.

Question 6

Ms. Carroll promised that anyone who got all the multiple choice questions right on the upcoming exam would receive an A on the exam. Which of these statements necessarily follows logically?

$\textbf{(A)} \text{If Lewis did not receive an A, then he got all of the multiple choice questions wrong.}\\$
$\textbf{(B)} \text{If Lewis did not receive an A, then he got at least one of the multiple choice questions wrong.}\\$
$\textbf{(C)} \text{If Lewis got at least one of the multiple choice questions wrong, then he did not receive an A. }\\$
$\textbf{(D)} \text{If Lewis received an A, then he got all of the multiple choice questions right.}\\$
$\textbf{(E)} \text{If Lewis received an A, then he got at least one of the multiple choice questions right.}$

Solution

$\textbf{B}$

Rewriting the given statement:“if someone got all the multiple choice questions right on the upcoming exam, then he or she would receive an A on the exam.” If that someone is Lewis, the statement becomes:“if Lewis got all the multiple choice questions right, then he would receive an A on the exam.”

The contrapositive:“If Lewis did not receive an A, then he did not get all of the multiple choice questions right.” In other words, “If Lewis did not get an A, then he got at least one of the multiple choice questions wrong.”

We know that the contrapositive is always true if the given statement is true. So the answer is B.

Question 7

Jerry and Silvia wanted to go from the southwest corner of a square field to the northeast corner. Jerry walked due east and then due north to reach the goal, but Silvia headed northeast and reached the goal walking in a straight line. Which of the following is closest to how much shorter Silvia's trip was, compared to Jerry's trip?

$\textbf{(A)}\ 30\%\qquad\textbf{(B)}\ 40\%\qquad\textbf{(C)}\ 50\%\qquad\textbf{(D)}\ 60\%\qquad\textbf{(E)}\ 70\%$

Solution

$\textbf{A}$

Let the side length of the square be 1. The distance Jerry traveled is $1+1=2$, and the distance Silvia traveled is $\sqrt{1^2+1^2}=\sqrt2\approx1.4$. So the answer is $1-\dfrac{\sqrt2}2\approx1-\dfrac{1.4}2=30\%$.

Question 8

At a gathering of $30$ people, there are $20$ people who all know each other and $10$ people who know no one. People who know each other hug, and people who do not know each other shake hands. How many handshakes occur?

$\textbf{(A)}\ 240\qquad\textbf{(B)}\ 245\qquad\textbf{(C)}\ 290\qquad\textbf{(D)}\ 480\qquad\textbf{(E)}\ 490$

Solution

$\textbf{B}$

Set A contains 20 people who know each other. Set B contains 10 people who know no one. Handshakes occur when a person from set B meets a person from set A, which happens $10\times20=200$ times, or 2 persons from set B meet each other, which happens $\dbinom{10}{2}=45$ times. So the answer is $200+45=245$.

Question 9

Minnie rides on a flat road at $20$ kilometers per hour (kph), downhill at $30$ kph, and uphill at $5$ kph. Penny rides on a flat road at $30$ kph, downhill at $40$ kph, and uphill at $10$ kph. Minnie goes from town $A$ to town $B$, a distance of $10$ km all uphill, then from town $B$ to town $C$, a distance of $15$ km all downhill, and then back to town $A$, a distance of $20$ km on the flat. Penny goes the other way around using the same route. How many more minutes does it take Minnie to complete the $45$-km ride than it takes Penny?

$\textbf{(A)}\ 45\qquad\textbf{(B)}\ 60\qquad\textbf{(C)}\ 65\qquad\textbf{(D)}\ 90\qquad\textbf{(E)}\ 95$

Solution

$\textbf{C}$

It takes Minnie $\dfrac{10}5=2$ hours from A to B uphill, $\dfrac{15}{30}=\dfrac12$ hour from B to C downhill, and $\dfrac{20}{20}=1$ hour from C to A flat. So Minnie completes the 45-km ride in $2+\dfrac12+1=3\dfrac12$ hours. Similarly, it takes Penny $\dfrac{20}{30}=\dfrac23$ hour from A to C flat, $\dfrac{15}{10}=\dfrac32$ hours from C to B uphill, and $\dfrac{10}{40}=\dfrac14$ hour from B to A downhill. So Penny completes the 45-km ride in $\dfrac23+\dfrac32+\dfrac14=2\dfrac{5}{12}$ hours. Minnie spends $3\dfrac12-2\dfrac{5}{12}=1\dfrac{1}{12}\ \text{hours}=65\ \text{more minutes}$ than Penny.

Question 10

Joy has $30$ thin rods, one each of every integer length from $1$ cm through $30$ cm. She places the rods with lengths $3$ cm, $7$ cm, and $15$ cm on a table. She then wants to choose a fourth rod that she can put with these three to form a quadrilateral with positive area. How many of the remaining rods can she choose as the fourth rod?

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

Solution

$\textbf{B}$

The triangle inequality generalizes to all polygons, so \begin{align*}
x &< 3+7+15\\
15 &< x+3+7
\end{align*} Hence, we get $5<x<25$. There are 19 numbers between 5 and 25, exclusive. Considering that the rods with lengths 7 cm and 15 cm are already on the table, the number of choices for the fourth rod is $19-2=17$.

Question 11

The region consisting of all points in three-dimensional space within $3$ units of line segment $\overline{AB}$ has volume $216\pi$. What is the length $\textit{AB}$?

$\textbf{(A)}\ 6\qquad\textbf{(B)}\ 12\qquad\textbf{(C)}\ 18\qquad\textbf{(D)}\ 20\qquad\textbf{(E)}\ 24$

Solution

$\textbf{D}$

The region can be divided into a cylinder and two identical hemispheres. The radius of the hemisphere is 3. So the volume of two hemispheres is $\dfrac43\pi\times3^3=36\pi$. The height of the cylinder is $\overline{AB}$, and the radius of the base of the cylinder is 3. So the volume of the cylinder is $\pi\times3^2\times\overline{AB}=9\pi\overline{AB}$. Hence, we have $$9\pi\overline{AB}+36\pi=216\pi\rightarrow \overline{AB}=20$$

Question 12

Let $S$ be a set of points $(x,y)$ in the coordinate plane such that two of the three quantities $3,~x+2,$ and $y-4$ are equal and the third of the three quantities is no greater than this common value. Which of the following is a correct description for $S?$

$\textbf{(A)}\ \text{a single point} \qquad\\$
$\textbf{(B)}\ \text{two intersecting lines} \qquad\\$
$\textbf{(C)}\ \text{three lines whose pairwise intersections are three distinct points} \qquad\\$
$\textbf{(D)}\ \text{a triangle} \qquad\\$
$\textbf{(E)}\ \text{three rays with a common endpoint}$

Solution

$\textbf{E}$

If the two equal quantities are 3 and $x+2$, then we have \begin{align*}
x+2&=3\\
y-4&\leq3
\end{align*} Hence, we get $x=1$, $y\leq7$. This is a vertical ray with an endpoint of $(1,7)$.

If the two equal quantities are 3 and $y-4$, then we have \begin{align*}
y-4&=3\\
x+2&\leq3
\end{align*} Hence, we get $x\leq1$, $y=7$. This is horizontal ray with the same endpoint of $(1,7)$.

If the two equal quantities are $x+2$ and $y-4$, then we have \begin{align*}
x+2&=y-4\\
3&\leq x+2
\end{align*} Hence, we get $y=x+6$, $x\geq1$. This is a portion of line $y=x+6$ with an endpoint of $(1,7)$.

Therefore, the answer is E.

Question 13

Define a sequence recursively by $F_{0}=0,~F_{1}=1,$ and $F_{n}=$ the remainder when $F_{n-1}+F_{n-2}$ is divided by $3,$ for all $n\geq 2.$ Thus the sequence starts $0,1,1,2,0,2,\ldots$ What is $F_{2017}+F_{2018}+F_{2019}+F_{2020}+F_{2021}+F_{2022}+F_{2023}+F_{2024}?$

$\textbf{(A)}\ 6\qquad\textbf{(B)}\ 7\qquad\textbf{(C)}\ 8\qquad\textbf{(D)}\ 9\qquad\textbf{(E)}\ 10$

Solution

$\textbf{D}$

A pattern starts to emerge as the function is continued. The repeating pattern of the sequence is $0,1,1,2,0,2,2,1\ldots$ The problem asks for the sum of eight consecutive terms in the sequence. Since there are eight numbers in the repeating pattern, we just need to find the sum of the numbers in the repeating pattern, which is $0+1+1+2+0+2+2+1=9$.

Question 14

Every week Roger pays for a movie ticket and a soda out of his allowance. Last week, Roger's allowance was $A$ dollars. The cost of his movie ticket was $20\%$ of the difference between $A$ and the cost of his soda, while the cost of his soda was $5\%$ of the difference between $A$ and the cost of his movie ticket. To the nearest whole percent, what fraction of $A$ did Roger pay for his movie ticket and soda?

$\textbf{(A)}\ 9\%\qquad\textbf{(B)}\ 19\%\qquad\textbf{(C)}\ 22\%\qquad\textbf{(D)}\ 23\%\qquad\textbf{(E)}\ 25\%$

Solution

$\textbf{D}$

Let $B$ and $C$ be the price of a movie ticket and a soda, respectively. Then we have \begin{align*}
B&=20\%(A-C)\\
C&=5\%(A-B)
\end{align*} Hence, we get $B=\dfrac{19}{99}A$, $C=\dfrac{4}{99}A$. The fraction of $A$ Roger paid for the movie ticket and soda is $\dfrac{19}{99}+\dfrac{4}{99}=\dfrac{23}{99}\approx23\%$.

Question 15

Chloé chooses a real number uniformly at random from the interval $[0, 2017]$. Independently, Laurent chooses a real number uniformly at random from the interval $[0, 4034]$. What is the probability that Laurent's number is greater than Chloé's number? (Assume they cannot be equal)

$\textbf{(A)}\ \dfrac{1}{2}\qquad\textbf{(B)}\ \dfrac{2}{3}\qquad\textbf{(C)}\ \dfrac{3}{4}\qquad\textbf{(D)}\ \dfrac{5}{6}\qquad\textbf{(E)}\ \dfrac{7}{8}$

Solution

$\textbf{C}$

Chloé chooses a real number from $[0, 2017]$ on the $x$-axis. Laurent chooses a real number from $[0, 4034]$ on the $y$-axis. Point $(x,y)$ lies in a $2017\times4034$ rectangle, as shown in the figure.

If Laurent's number is less than Chloé's number, point $(x,y)$ lies in the portion of the rectangle below line $y=x$. The area of the portion of the rectangle below line $y=x$ is $\dfrac12\times2017^2$. So the probability that Laurent's number is less than Chloé's number is $\dfrac{\frac12\times2017^2}{2017\times4034}=\dfrac14$. Hence, the probability that Laurent's number is greater than Chloé's number is $1-\dfrac14=\dfrac34$.

Question 16

There are 10 horses, named Horse 1, Horse 2, $\ldots$, Horse 10. They get their names from how many minutes it takes them to run one lap around a circular race track: Horse $k$ runs one lap in exactly $k$ minutes. At time 0 all the horses are together at the starting point on the track. The horses start running in the same direction, and they keep running around the circular track at their constant speeds. The least time $S>0$, in minutes, at which all 10 horses will again simultaneously be at the starting point is $S=2520$. Let $T>0$ be the least time, in minutes, such that at least 5 of the horses are again at the starting point. What is the sum of the digits of $T$?

$\textbf{(A)}\ 2\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 5\qquad\textbf{(E)}\ 6$

Solution

$\textbf{B}$

$T$ has at least 5 divisors in the set $\{1,2,\ldots,10\}$. We start from number 1:

Number 1 has 1 divisor in the set $\{1,2,\ldots,10\}$.$\newline$
Number 2 has 2 divisors in the set $\{1,2,\ldots,10\}$.$\newline$
Number 3 has 2 divisors in the set $\{1,2,\ldots,10\}$.$\newline$
Number 4 has 3 divisors in the set $\{1,2,\ldots,10\}$.$\newline$
Number 5 has 2 divisors in the set $\{1,2,\ldots,10\}$.$\newline$
Number 6 has 4 divisors in the set $\{1,2,\ldots,10\}$.$\newline$
Number 7 has 2 divisors in the set $\{1,2,\ldots,10\}$.$\newline$
Number 8 has 4 divisors in the set $\{1,2,\ldots,10\}$.$\newline$
Number 9 has 3 divisors in the set $\{1,2,\ldots,10\}$.$\newline$
Number 10 has 4 divisors in the set $\{1,2,\ldots,10\}$.$\newline$
Number 11 has 1 divisor in the set $\{1,2,\ldots,10\}$.$\newline$
Number 12 has 5 divisors in the set $\{1,2,\ldots,10\}$.

So $T=12$. The sum of digits of $T$ is $1+2=3$.

Question 17

Distinct points $P$, $Q$, $R$, $S$ lie on the circle $x^2+y^2=25$ and have integer coordinates. The distances $PQ$ and $RS$ are irrational numbers. What is the greatest possible value of the ratio $\frac{PQ}{RS}$?

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

Solution

$\textbf{D}$

Because $P$, $Q$, $R$, and $S$ are lattice points, there are only a few coordinates that actually satisfy the equation. The coordinates are $(\pm 3,\pm 4), (\pm 4, \pm 3), (0,\pm 5),$ and $(\pm 5,0).$

We want to maximize $PQ$ and minimize $RS.$ The greatest value of $PQ$ happens when $P$ and $Q$ are almost located diagonally on the circle. They can't be directly located diagonally because the distance $PQ$ has to be an irrational number. For example, $P$ and $Q$ could be $(-4,3)$ and $(3,-4)$. Another possible pair could be $(-4,3)$ and $(5,0)$. Comparing the distances of the two pairs, we see that $$\sqrt{[(-4)-3]^2+[3-(-4)]^2}=\sqrt{98}$$ $$\sqrt{[(-4)-5]^2+(3-0)^2}=\sqrt{90}$$ Therefore, the distance from $(3,-4)$ to $(-4,3)$ is one of the greatest choices for $PQ$. The other possible coordinates for $P$ and $Q$ are $(4,3)$ and $(-3, -4)$, $(3, 4)$ and $(-4, -3)$, $(-3, 4)$ and $(4, -3)$.

The least value of $RS$ happens when the two endpoints are in the same quadrant and are very close to each other. This can occur when, for example, $R$ is $(3,4)$ and $S$ is $(4,3).$ Using the distance formula, we get the distance $RS=\sqrt{(3-4)^2+(4-3)^2}=\sqrt2.$

Hence, the greatest possible value of the ratio $\dfrac{PQ}{RS}$ is $\dfrac{\sqrt{98}}{\sqrt{2}}=\sqrt{49}=7$.

Question 18

Amelia has a coin that lands heads with probability $\dfrac{1}{3}$, and Blaine has a coin that lands on heads with probability $\dfrac{2}{5}$. Amelia and Blaine alternately toss their coins until someone gets a head; the first one to get a head wins. All coin tosses are independent. Amelia goes first. The probability that Amelia wins is $\dfrac{p}{q}$, where $p$ and $q$ are relatively prime positive integers. What is $q-p$?

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

Solution

$\textbf{D}$

Let $P(n)$ be the probability that Amelia wins in the $n$th turn. Hence, the probability that Amelia wins in the first turn is $$P(1)=\dfrac13$$ The case that Amelia wins in the second turn only happens when Amelia gets a tail in the first turn, then Blaine gets a tail in the first turn, then Amelia gets a head in the second turn. Therefore, the probability that Amelia wins in the second turn is $$P(2)=\left(\dfrac23\times\dfrac35\right)\times\dfrac13$$ We see that the probability that Amelia wins in the $n$th turn is $$P(n)=\left(\dfrac23\times\dfrac35\right)^{n-1}\times\dfrac13$$Hence, the probability that Amelia wins is \begin{align*}
P(1)+P(2)+P(3)+\cdots&=\dfrac13+\left(\dfrac23\times\dfrac35\right)\times\dfrac13+\left(\dfrac23\times\dfrac35\right)^2\times\dfrac13+\cdots\\
&=\dfrac13+\left(\dfrac25\right)^1\times\dfrac13+\left(\dfrac25\right)^2\times\dfrac13+\cdots\\
&=\dfrac13\left[1+\left(\dfrac25\right)+\left(\dfrac25\right)^2+\cdots\right]\\
&=\dfrac13\times\dfrac{1}{1-\frac25}\\
&=\dfrac59
\end{align*} The answer is $9-5=4$.

Question 19

Alice refuses to sit next to either Bob or Carla. Derek refuses to sit next to Eric. How many ways are there for the five of them to sit in a row of 5 chairs under these conditions?

$\textbf{(A)}\ 12\qquad\textbf{(B)}\ 16\qquad\textbf{(C)}\ 28\qquad\textbf{(D)}\ 32\qquad\textbf{(E)}\ 40$

Solution

$\textbf{C}$

$\textbf{Case 1: }$ Alice sits on an edge seat.

In this case Alice may sit on the first seat or the last seat. By symmetry, we assume that Alice sits on the first seat. Since Bob and Carla can't sit next to Alice, the guy sits on the second seat must be either Derek or Eric. After we pick either Derek or Eric, then either Bob or Carla must sit on the third seat. Then, we can arrange the two remaining people in two ways. The total number of arrangements in this case is $2 \times 2 \times 2 \times 2 = 16$.

$\textbf{Case 2: }$ Alice does not sit in an edge seat.

In this case we have 3 arrangements for Alice. The only two people that can sit next to Alice are Derek and Eric, and there are two ways to permute them, and this also handles the restriction that Derek can't sit next to Eric. Then, there are two ways to arrange Bob and Carla, the remaining people. The total number of arrangements in this case is $3\times 2 \times 2 = 12$.

Adding up all the cases, we have $16+12 = 28$.

Question 20

Let $S(n)$ equal the sum of the digits of positive integer $n$. For example, $S(1507) = 13$. For a particular positive integer $n$, $S(n) = 1274$. Which of the following could be the value of $S(n+1)$?

$\textbf{(A)}\ 1 \qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 12\qquad\textbf{(D)}\ 1239\qquad\textbf{(E)}\ 1265$

Solution

$\textbf{D}$

Note that $n \equiv S(n) \pmod{9}$. This can be seen from the fact that $\sum_{k=0}^{n}10^{k}a_k \equiv \sum_{k=0}^{n}a_k \pmod{9}$. Thus, if $S(n) = 1274$, then $n \equiv 5 \pmod{9}$, and thus $n+1 \equiv S(n+1) \equiv 6 \pmod{9}$. The only answer choice that satisfies $n+1 \equiv 6 \pmod{9}$ is $1239$.

Question 21

A square with side length $x$ is inscribed in a right triangle with sides of length $3$, $4$, and $5$ so that one vertex of the square coincides with the right-angle vertex of the triangle. A square with side length $y$ is inscribed in another right triangle with sides of length $3$, $4$, and $5$ so that one side of the square lies on the hypotenuse of the triangle. What is $\tfrac{x}{y}$?

$\textbf{(A) } \dfrac{12}{13} \qquad \textbf{(B) } \dfrac{35}{37} \qquad \textbf{(C) } 1 \qquad \textbf{(D) } \dfrac{37}{35} \qquad \textbf{(E) } \dfrac{13}{12}$

Solution

$\textbf{D}$

In the first triangle we see that $\triangle CDE\sim\triangle EFB$. So $\dfrac{CD}{DE} = \dfrac{EF}{FB}$. This can be written as $\dfrac{3-x}{x}=\dfrac{x}{4-x}$. Solving, we get $x = \dfrac{12}{7}$.

In the second triangle we see that $\triangle A'B'C'\sim\triangle RB'Q\sim\triangle STC'$. So $\dfrac{A'C'}{A'B'}=\dfrac{RQ}{RB'}=\dfrac{SC'}{ST}$. This can be written as $\dfrac34=\dfrac{y}{RB'}=\dfrac{SC'}{y}$. Hence, we get $RB' = \dfrac{4}{3}y$, $SC' = \dfrac{3}{4}y$. Thus, the hypotenuse $C'B' = C'S + SR + RB' = \dfrac{3}{4}y + y + \dfrac{4}{3}y = 5$. Solving, we get $y = \dfrac{60}{37}$.

Therefore, the answer is $\dfrac{x}{y} = \dfrac{37}{35}$.

Question 22

Sides $\overline{AB}$ and $\overline{AC}$ of equilateral triangle $ABC$ are tangent to a circle at points $B$ and $C$ respectively. What fraction of the area of $\triangle ABC$ lies outside the circle?

$\textbf{(A)}\ \dfrac{4\sqrt{3}\pi}{27}-\dfrac{1}{3}\qquad\textbf{(B)}\ \dfrac{\sqrt{3}}{2}-\dfrac{\pi}{8}\qquad\textbf{(C)}\ \dfrac{1}{2}\qquad\textbf{(D)}\ \sqrt{3}-\dfrac{2\sqrt{3}\pi}{9}\qquad\textbf{(E)}\ \dfrac{4}{3}-\dfrac{4\sqrt{3}\pi}{27}$

Solution

$\textbf{E}$

Let the center of the circle be $O$, and its radius be 1. Then we have $OB=OC=1$. Since $OA$ bisects $\angle BAC$, in right triangle $AOB$ we have $\angle OAB=30^\circ$, $\angle AOB=60^\circ$. Therefore, we get $AB=\sqrt3OB=\sqrt3$. The area of the equilateral triangle $ABC$ is $\dfrac{\sqrt3}4\times\left(\sqrt3\right)^2=\dfrac{3\sqrt3}4$. We see that $\angle BOC=120^\circ$. So the area of $\triangle ABC$ lies inside of the circle is one third the area of the circle subtracts the area of $\triangle BOC$: $$\dfrac13\times\pi\times1^2-\dfrac12\times1^2\times\sin120^\circ=\dfrac13\pi-\dfrac{\sqrt3}4$$ Therefore, the fraction of the area of $\triangle ABC$ lies outside of the circle is $$1-\dfrac{\frac13\pi-\frac{\sqrt3}4}{\frac{3\sqrt3}{4}}=\dfrac{4}{3}-\dfrac{4\sqrt{3}\pi}{27}$$

Question 23

How many triangles with positive area have all their vertices at points $(i,j)$ in the coordinate plane, where $i$ and $j$ are integers between $1$ and $5$, inclusive?

$\textbf{(A)}\ 2128 \qquad\textbf{(B)}\ 2148 \qquad\textbf{(C)}\ 2160 \qquad\textbf{(D)}\ 2200 \qquad\textbf{(E)}\ 2300$

Solution

$\textbf{B}$

There are $5 \times 5 = 25$ points in total. So we have $\dbinom{25}3 = 2300$ ways to choose the three vertices for the triangle. However, there are some cases where the three vertices chosen are in a straight line.

$\textbf{Case 1:}$ three vertices in a row/column/diagonal

There are 5 rows, 5 columns, and 2 diagonals. Each straight line contains 5 points. So we have $(5+5+2)\times\dbinom53=120$ arrangements in this case.

$\textbf{Case 2:}$ three vertices in a line of 4 points

There are 4 lines with exactly 4 points on each of them. So we have $4\times\dbinom43=16$ arrangements in this case.

$\textbf{Case 3:}$ three vertices in a line of 3 points

There are 16 lines with exactly 3 points on each of them. So we have $16\times\dbinom33=16$ arrangements in this case

Subtracting all the cases where the three vertices are in a line, the number of triangles is $2300-120-16-16=2148$.

Question 24

For certain real numbers $a$, $b$, and $c$, the polynomial
$$g(x) = x^3 + ax^2 + x + 10$$
has three distinct roots, and each root of $g(x)$ is also a root of the polynomial
$$f(x) = x^4 + x^3 + bx^2 + 100x + c.$$
What is $f(1)$?

$\textbf{(A)}\ -9009\qquad\textbf{(B)}\ -8008\qquad\textbf{(C)}\ -7007\qquad\textbf{(D)}\ -6006\qquad\textbf{(E)}\ -5005$

Solution

$\textbf{C}$

Given that each root of $g(x)$ is also a root of $f(x)$, we have $$f(x)=g(x)(x-r)$$ where $r$ is the fourth root of $f(x)$. Substituting $f(x)$ and $g(x)$, we get \begin{align*}
x^4 + x^3 + bx^2 + 100x + c&=(x^3+ax^2+x+10)(x-r)\\
&=x^4+(a-r)x^3+(1-ar)x^2+(10-r)x-10r
\end{align*} Comparing the coefficients, we have \begin{align*}
a-r&=1\\
1-ar&=b\\
10-r&=100\\
-10r&=c
\end{align*} Solving, we get $a=-89$, $b=-8009$, $c=900$, $r=-90$. Hence, the polynomial \[f(x)=x^4+x^3-8009x^2+100x+900\] The answer is $f(1)=1+1-8009+100+900=-7007$.

Question 25

How many integers between $100$ and $999$, inclusive, have the property that some permutation of its digits is a multiple of $11$ between $100$ and $999?$ For example, both $121$ and $211$ have this property.

$\textbf{(A)}\ 226\qquad\textbf{(B)}\ 243\qquad\textbf{(C)}\ 270\qquad\textbf{(D)}\ 469\qquad\textbf{(E)}\ 486$

Solution

$\textbf{A}$

The smallest multiple of 11 between 100 and 999 is 110, and the largest multiple of 11 between 100 and 999 is 990. So the number of multiples of 11 between 100 and 999 is $\dfrac{990-110}{11}+1=81$. Some have digits repeated twice (for example, 616), making 3 permutations. Others that have no repeated digits have $3\times2=6$ permutations, but switching the hundreds and units digits also yield a multiple of 11. Switching shows we have overcounted by a factor of 2, so assign $6 \div 2 = 3$ permutations to each multiple.

There are now $81\times3 = 243$ permutations, but we have overcounted. Some multiples of 11 have $0$ as a digit. Since $0$ cannot be the digit of the hundreds place, we must subtract a permutation for each.

There are $110, 220, 330,\cdots, 990$, yielding 9 extra permutations.

Also, there are $209, 308, 407,\cdots, 902$, yielding 8 more permutations.

Subtracting these 17 from the total, we get the answer $243-17=226$.