AMC 10 2019 Test B

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

Alicia had two containers. The first was $\dfrac{5}{6}$ full of water and the second was empty. She poured all the water from the first container into the second container, at which point the second container was $\dfrac{3}{4}$ full of water. What is the ratio of the volume of the first container to the volume of the second container?

$\textbf{(A) } \dfrac{5}{8} \qquad \textbf{(B) } \dfrac{4}{5} \qquad \textbf{(C) } \dfrac{7}{8} \qquad \textbf{(D) } \dfrac{9}{10} \qquad \textbf{(E) } \dfrac{11}{12}$

Solution

$\textbf{D}$

Let the volume of the first container be $x$, and the volume of the second container be $y$. We have $\dfrac34x=\dfrac56y$. So $\dfrac{x}{y}=\dfrac{9}{10}$.

Question 2

Consider the statement, "If $n$ is not prime, then $n-2$ is prime." Which of the following values of $n$ is a counterexample to this statement?

$\textbf{(A) } 11 \qquad \textbf{(B) } 15 \qquad \textbf{(C) } 19 \qquad \textbf{(D) } 21 \qquad \textbf{(E) } 27$

Solution

$\textbf{E}$

Since a counterexample must be a value of $n$ which is not prime, $n$ must be composite, so we eliminate $\text{A}$ and $\text{C}$. Now we subtract $2$ from the remaining answer choices, and we see that the only time $n-2$ is not prime is when $n =27$.

Question 3

In a high school with $500$ students, $40\%$ of the seniors play a musical instrument, while $30\%$ of the non-seniors do not play a musical instrument. In all, $46.8\%$ of the students do not play a musical instrument. How many non-seniors play a musical instrument?

$\textbf{(A) } 66 \qquad\textbf{(B) } 154 \qquad\textbf{(C) } 186 \qquad\textbf{(D) } 220 \qquad\textbf{(E) } 266$

Solution

$\textbf{B}$

Let $x$ be the number of seniors. Then the number of non-seniors is $500-x$. The number of students who do not play a musical instrument is $$(1-40\%)\cdot x+30\%\cdot(500-x)=46.8\%\cdot500$$ Hence, we get $x=280$. Therefore, the number of non-seniors who play a musical instrument is $(500-x)\cdot(1-30\%)=154$.

Question 4

All lines with equation $ax+by=c$ such that $a,b,c$ form an arithmetic progression pass through a common point. What are the coordinates of that point?

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

Solution

$\textbf{A}$

Let the common difference of the arithmetic progression be $d$. Now we have $b=a+d$ and $c=a+2d$. The original equation turns to $ax+ (a+d)y = a+2d$. Separating the variables $a$ and $d$ gives $a(x+y-1)+d(y-2) = 0$. Since this must always be true (regardless of the values of $a$ and $d$), we must have $x+y-1 = 0$ and $y-2 = 0$. So the common point is $(x,y)=(-1,2)$.

Question 5

Triangle $ABC$ lies in the first quadrant. Points $A$, $B$, and $C$ are reflected across the line $y=x$ to points $A'$, $B'$, and $C'$, respectively. Assume that none of the vertices of the triangle lie on the line $y=x$. Which of the following statements is not always true?

$\textbf{(A) }$ Triangle $A'B'C'$ lies in the first quadrant.$\\$
$\textbf{(B) }$ Triangles $ABC$ and $A'B'C'$ have the same area.$\\$
$\textbf{(C) }$ The slope of line $AA'$ is $-1$.$\\$
$\textbf{(D) }$ The slopes of lines $AA'$ and $CC'$ are the same.$\\$
$\textbf{(E) }$ Lines $AB$ and $A'B'$ are perpendicular to each other.

Solution

$\textbf{E}$

Let's analyze all of the options separately.

$\textbf{(A)}$: Clearly $\textbf{(A)}$ is true, because a point in the first quadrant will have positive $x$- and $y$-coordinates, and so its reflection, with the coordinates swapped, $(y,x)$, will also have positive coordinates.

$\textbf{(B)}$: The triangles have the same area, since $\triangle ABC$ and $\triangle A'B'C'$ are the same triangle (congruent). More formally, we can say that area is invariant under reflection.

$\textbf{(C)}$: If point $A$ has coordinates $(p,q)$, then $A'$ will have coordinates $(q,p)$. The gradient is thus $\dfrac{p-q}{q-p} = -1$, so this is true. (We know $p \neq q$ since the question states that none of the points $A$, $B$, or $C$ lies on the line $y=x$, so there is no risk of division by zero).

$\textbf{(D)}$: Repeating the argument for $\textbf{(C)}$, we see that both lines have slope $-1$, so this is also true.

$\textbf{(E)}$: By process of elimination, this must now be the answer. Indeed, if point $A$ has coordinates $(p,q)$ and point $B$ has coordinates $(r,s)$, then $A'$ and $B'$ will, respectively, have coordinates $(q,p)$ and $(s,r)$. The product of the gradients of $AB$ and $A'B'$ is $\dfrac{s-q}{r-p} \cdot \frac{r-p}{s-q} = 1 \neq -1$, so in fact these lines are never perpendicular to each other (using the "negative reciprocal" condition for perpendicularity).

Thus the answer is $\textbf{(E)}$.

Question 6

There is a real $n$ such that $(n+1)! + (n+2)! = n! \cdot 440$. What is the sum of the digits of $n$?

$\textbf{(A) }3\qquad\textbf{(B) }8\qquad\textbf{(C) }10\qquad\textbf{(D) }11\qquad\textbf{(E) }12$

Solution

$\textbf{C}$

Dividing both sides by $n!$ gives\[(n+1)+(n+2)(n+1)=440 \Rightarrow n^2+4n-437=0 \Rightarrow (n-19)(n+23)=0.\]Since $n$ is non-negative, $n=19$. The answer is $1 + 9 = 10$.

Question 7

Each piece of candy in a store costs a whole number of cents. Casper has exactly enough money to buy either $12$ pieces of red candy, $14$ pieces of green candy, $15$ pieces of blue candy, or $n$ pieces of purple candy. A piece of purple candy costs $20$ cents. What is the smallest possible value of $n$?

$\textbf{(A) } 18 \qquad \textbf{(B) } 21 \qquad \textbf{(C) } 24\qquad \textbf{(D) } 25 \qquad \textbf{(E) } 28$

Solution

$\textbf{B}$

The amount of money is a multiple of 12, 14, and 15. By prime factorization, we have $12=2^2\cdot3$, $14=2\cdot7$, and $15=3\cdot5$. So Casper has at least $2^2\cdot3\cdot5\cdot7=420$ cents. The smallest value of $n$ is $420/20=21$.

Question 8

The figure below shows a square and four equilateral triangles, with each triangle having a side lying on a side of the square, such that each triangle has side length $2$ and the third vertices of the triangles meet at the center of the square. The region inside the square but outside the triangles is shaded. What is the area of the shaded region?

$\textbf{(A) } 4 \qquad \textbf{(B) } 12 - 4\sqrt{3} \qquad \textbf{(C) } 3\sqrt{3}\qquad \textbf{(D) } 4\sqrt{3} \qquad \textbf{(E) } 16 - 4\sqrt{3}$

Solution

$\textbf{B}$

Since the side length of the equilateral triangle is 2, its altitude is $\sqrt3$, and its area is $\dfrac12\cdot2\cdot\sqrt3=\sqrt3$. The side length of the square is twice the altitude of the equilateral triangle. So we get the area of the square $(2\sqrt3)^2=12$. Therefore, the area of the shaded region is $12-4\sqrt3$.

Question 9

The function $f$ is defined by$$f(x) = \lfloor|x|\rfloor - |\lfloor x \rfloor|$$for all real numbers $x$, where $\lfloor r \rfloor$ denotes the greatest integer less than or equal to the real number $r$. What is the range of $f$?

$\textbf{(A) }$ $\{-1, 0\}\\$
$\textbf{(B) }$ $\text{The set of nonpositive integers}\\$
$\textbf{(C) }$ $\{-1, 0, 1\}\\$
$\textbf{(D) }$ $\{0\}\\$
$\textbf{(E) }$ $\text{The set of nonnegative integers}$

Solution

$\textbf{A}$

There are four cases we need to consider here.

$\textbf{Case 1:}$ $x$ is a positive integer. Without loss of generality, assume $x=1$. Then $f(1) = 1 - 1 = 0$.

$\textbf{Case 2:}$ $x$ is a positive fraction. Without loss of generality, assume $x=\dfrac{1}{2}$. Then $f\left(\dfrac{1}{2}\right) = 0 - 0 = 0$.

$\textbf{Case 3:}$ $x$ is a negative integer. Without loss of generality, assume $x=-1$. Then $f(-1) = 1 - 1 = 0$.

$\textbf{Case 4:}$ $x$ is a negative fraction. Without loss of generality, assume $x=-\dfrac{1}{2}$. Then $f\left(-\dfrac{1}{2}\right) = 0 - 1 = -1$.

Thus the range of the function $f$ is $ \{-1, 0\}$.

Question 10

In a given plane, points $A$ and $B$ are $10$ units apart. How many points $C$ are there in the plane such that the perimeter of $\triangle ABC$ is $50$ units and the area of $\triangle ABC$ is $100$ square units?

$\textbf{(A) }0\qquad\textbf{(B) }2\qquad\textbf{(C) }4\qquad\textbf{(D) }8\qquad\textbf{(E) }\textrm{infinitely many}$

Solution

$\textbf{A}$

Notice that whatever point we pick for $C$, $AB$ will be the base of the triangle. Without loss of generality, let points $A$ and $B$ be $(0,0)$ and $(10,0)$, since for any other combination of points, we can just rotate the plane to make them $(0,0)$ and $(10,0)$ under a new coordinate system. When we pick point $C$, we have to make sure that its $y$-coordinate is $\pm20$, because that's the only way the area of the triangle can be $100$.

Now when the perimeter is minimized, by symmetry, we put $C$ in the middle, at $(5, 20)$. We can easily see that $AC$ and $BC$ will both be $\sqrt{20^2+5^2} = \sqrt{425}$. The perimeter of this minimal triangle is $2\sqrt{425} + 10$, which is larger than $50$. Since the minimum perimeter is greater than $50$, there is no triangle that satisfies the condition. The answer is 0.

Question 11

Two jars each contain the same number of marbles, and every marble is either blue or green. In Jar $1$ the ratio of blue to green marbles is $9:1$, and the ratio of blue to green marbles in Jar $2$ is $8:1$. There are $95$ green marbles in all. How many more blue marbles are in Jar $1$ than in Jar $2$?

$\textbf{(A) } 5 \qquad\textbf{(B) } 10 \qquad\textbf{(C) } 25 \qquad\textbf{(D) } 45 \qquad\textbf{(E) } 50$

Solution

$\textbf{A}$

Let the number of marbles in each jar be $x$. Then total number of green marbles is $\dfrac1{10}x+\dfrac19x=95\rightarrow x=450$. So the number of blue marble in each jar is $\dfrac9{10}x=405$, $\dfrac89x=400$. The answer is $405-400=5$.

Question 12

What is the greatest possible sum of the digits in the base-seven representation of a positive integer less than $2019$?

$\textbf{(A) } 11 \qquad\textbf{(B) } 14 \qquad\textbf{(C) } 22 \qquad\textbf{(D) } 23 \qquad\textbf{(E) } 27$

Solution

$\textbf{C}$

Note that $2019_{10} = 5613_7$. To maximize the sum of the digits, we want as many $6$s as possible (since $6$ is the highest value in base $7$), and this will occur with either of the numbers $4666_7$ or $5566_7$. Thus, the answer is $4+6+6+6 = 5+5+6+6 =22$.

Question 13

What is the sum of all real numbers $x$ for which the median of the numbers $4,6,8,17,$ and $x$ is equal to the mean of those five numbers?

$\textbf{(A) } -5 \qquad\textbf{(B) } 0 \qquad\textbf{(C) } 5 \qquad\textbf{(D) } \dfrac{15}{4} \qquad\textbf{(E) } \dfrac{35}{4}$

Solution

$\textbf{A}$

The mean is $\dfrac{4+6+8+17+x}{5}=\dfrac{35+x}{5}$.

There are three possibilities for the median: it is either $6$, $8$, or $x$.

$\textbf{Case 1:}$ The median is 6.

$\dfrac{35+x}{5}=6$ has solution $x=-5$, and the sequence is $-5, 4, 6, 8, 17$, which does have median $6$. So this is a valid solution.

$\textbf{Case 2:}$ The median is 8.

$\dfrac{35+x}{5}=8$ gives $x=5$, so the sequence is $4, 5, 6, 8, 17$, which has median $6$. This is not valid.

$\textbf{Case 3:}$ The median is $x$.

$\dfrac{35+x}{5}=x\implies x=\dfrac{35}{4}=8.75$, and the sequence is $4, 6, 8, 8.75, 17$, which has median $8$. This case is therefore again not valid.

Hence, the only possible value of $x$ is $-5.$

Question 14

The base-ten representation for $19!$ is $121,6T5,100,40M,832,H00$, where $T$, $M$, and $H$ denote digits that are not given. What is $T+M+H$?

$\textbf{(A) }3 \qquad\textbf{(B) }8 \qquad\textbf{(C) }12 \qquad\textbf{(D) }14 \qquad\textbf{(E) } 17$

Solution

$\textbf{C}$

First, we get $H = 0$ by noticing that $19!$ will end with $3$ zeroes, as there are three factors of $5$ in its prime factorization, so there would be 3 powers of 10 meaning it will end in 3 zeros. Next, we know that $9$ and $11$ are both factors of $19!$. We can simply use the divisibility rules for $9$ and $11$ for this problem to find $T$ and $M$. For $19!$ to be divisible by $9$, the sum of digits must simply be divisible by $9$. Summing the digits, we get that $T + M + 33$ must be divisible by $9$. So $T + M \equiv 3 \;(\bmod\; 9)$. For a number to be divisible by $11$, the alternating sum must be divisible by $11$ (for example, with the number $2728$, $2-7+2-8 = -11$, so $2728$ is divisible by $11$). Applying the alternating sum test to this problem, we see that $T - M - 7$ must be divisible by 11. So $T - M \equiv 7 \;(\bmod\; 11)$. By inspection, we can see that this holds if $T=4$ and $M=8$. So the answer is $8 + 4 + 0 = 12$.

Question 15

Right triangles $T_1$ and $T_2$ have areas 1 and 2, respectively. A side of $T_1$ is congruent to a side of $T_2$, and a different side of $T_1$ is congruent to a different side of $T_2$. What is the square of the product of the other (third) sides of $T_1$ and $T_2$?

$\textbf{(A) } \dfrac{28}{3} \qquad\textbf{(B) }10\qquad\textbf{(C) } \dfrac{32}{3} \qquad\textbf{(D) } \dfrac{34}{3} \qquad\textbf{(E) }12$

Solution

$\textbf{A}$

Let the two sides which are congruent be $x$ and $y$, where $y > x$. The only way that the conditions of the problem can be satisfied is if $x$ is the shorter leg of $T_{2}$ and the longer leg of $T_{1}$, and $y$ is the longer leg of $T_{2}$ and the hypotenuse of $T_{1}$.

Notice that this means the value we are looking for is the square of $\sqrt{x^{2}+y^{2}} \cdot \sqrt{y^{2}-x^{2}} = \sqrt{y^{4}-x^{4}}$, which is just $y^{4}-x^{4}$.

The area conditions give us two equations: $\dfrac{xy}{2} = 2$ and $\dfrac{x\sqrt{y^{2}-x^{2}}}{2} = 1$.

This means that $y = \dfrac{4}{x}$ and that $\dfrac{4}{x^{2}} = y^{2} - x^{2}$.

Taking the second equation, we get $x^{2}y^{2} - x^{4} = 4$. Since $xy = 4$, $x^{4} = 12$.

By $y = \dfrac{4}{x}$, we get $y^{4} = \dfrac{256}{12} = \dfrac{64}{3}$.

The value we are looking for is $y^{4}-x^{4}=\dfrac{28}{3}$.

Question 16

In $\triangle ABC$ with a right angle at $C,$ point $D$ lies in the interior of $\overline{AB}$ and point $E$ lies in the interior of $\overline{BC}$ so that $AC=CD,$ $DE=EB,$ and the ratio $AC:DE=4:3.$ What is the ratio $AD:DB?$

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

Solution

$\textbf{A}$

Without loss of generality, let $AC = CD = 4$ and $DE = EB = 3$. $\angle CDE = 180^{\circ} - \angle ADC - \angle BDE=180^\circ-\angle A-\angle B = 90^{\circ}$, so $\triangle CDE$ is a $3-4-5$ triangle with $CE = 5$. Then $CB = 5+3 = 8$, and $\triangle ABC$ is a $1-2-\sqrt{5}$ triangle. Hence, we get $\cos\angle A=\dfrac{1}{\sqrt5}$, $\cos\angle B=\dfrac{2}{\sqrt5}$. $AD=2AC\cos\angle A=\dfrac{8}{\sqrt5}$, $BD=2BE\cos\angle B=\dfrac{12}{\sqrt5}$. The answer is $AD:DB=\dfrac{8}{\sqrt5}:\dfrac{12}{\sqrt5}=2:3$.

Question 17

A red ball and a green ball are randomly and independently tossed into bins numbered with positive integers so that for each ball, the probability that it is tossed into bin $k$ is $2^{-k}$ for $k=1,2,3,\ldots.$ What is the probability that the red ball is tossed into a higher-numbered bin than the green ball?

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

Solution

$\textbf{C}$

By symmetry, the probability of the red ball landing in a higher-numbered bin is the same as the probability of the green ball landing in a higher-numbered bin. The probability of both landing in the same bin is $$\sum_{k=1}^{\infty}{2^{-k} \cdot 2^{-k}} = \sum_{k=1}^{\infty}2^{-2k} = \frac{1}{3}$$Therefore, since the other two probabilities are the same, they have to be $\dfrac{1-\frac{1}{3}}{2} = \dfrac{1}{3}$.

Question 18

Henry decides one morning to do a workout, and he walks $\dfrac{3}{4}$ of the way from his home to his gym. The gym is $2$ kilometers away from Henry's home. At that point, he changes his mind and walks $\dfrac{3}{4}$ of the way from where he is back toward home. When he reaches that point, he changes his mind again and walks $\dfrac{3}{4}$ of the distance from there back toward the gym. If Henry keeps changing his mind when he has walked $\dfrac{3}{4}$ of the distance toward either the gym or home from the point where he last changed his mind, he will get very close to walking back and forth between a point $A$ kilometers from home and a point $B$ kilometers from home. What is $|A-B|$?

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

Solution

$\textbf{C}$

Assuming that $A$ is closer to home than $B$. Let $A$ have a distance of $x$ from the home. Then, the distance to the gym is $2-x$. This means point $B$ and point $A$ are $\dfrac{3}{4} \cdot (2-x)$ away from one another. It also means that Point $B$ is located at $\dfrac{3}{4} (2-x) + x.$ So, the distance between the home and point $B$ is also $\dfrac{3}{4} (2-x) + x.$

It follows that point $A$ must be at a distance of $\dfrac{3}{4} \left( \dfrac{3}{4} (2-x) + x \right)$ from point $B$. However, we also said that this distance has length $\dfrac{3}{4} (2-x)$. So, we can set those equal, which gives the equation:\[\frac{3}{4} \left( \frac{3}{4} (2-x) + x \right) = \frac{3}{4} (2-x)\]Solving, we get $x = \dfrac{2}{5}$. This means $A$ is at point $\dfrac{2}{5}$ and $B$ is at point $\dfrac{3}{4} \cdot \dfrac{8}{5} + \dfrac{2}{5} = \dfrac{8}{5}.$

So, $|A - B| = \dfrac{6}{5}= 1\dfrac{1}{5}.$

Question 19

Let $S$ be the set of all positive integer divisors of $100,000.$ How many numbers are the product of two distinct elements of $S?$

$\textbf{(A) }98\qquad\textbf{(B) }100\qquad\textbf{(C) }117\qquad\textbf{(D) }119\qquad\textbf{(E) }121$

Solution

$\textbf{C}$

The prime factorization of $100,000$ is $2^5 \cdot 5^5$. Thus, we choose two numbers $2^a5^b$ and $2^c5^d$ where $0 \le a,b,c,d \le 5$ and $(a,b) \neq (c,d)$, whose product is $2^{a+c}5^{b+d}$, where $0 \le a+c \le 10$ and $0 \le b+d \le 10$.

Notice that this is similar to choosing a divisor of $100,000^2 = 2^{10}5^{10}$, which has $(10+1)(10+1) = 121$ divisors. However, some of the divisors of $2^{10}5^{10}$ cannot be written as a product of two $\textbf{distinct}$ divisors of $2^5 \cdot 5^5$, namely: $1 = 2^05^0$, $2^{10}5^{10}$, $2^{10}$, and $5^{10}$. So the answer is $121-4=117$.

Question 20

As shown in the figure, line segment $\overline{AD}$ is trisected by points $B$ and $C$ so that $AB=BC=CD=2.$ Three semicircles of radius $1,$ $\stackrel{\huge\frown}{AEB},\stackrel{\huge\frown}{BFC},$ and $\stackrel{\huge\frown}{CGD},$ have their diameters on $\overline{AD},$ and are tangent to line $EG$ at $E,F,$ and $G,$ respectively. A circle of radius $2$ has its center on $F.$ The area of the region inside the circle but outside the three semicircles, shaded in the figure, can be expressed in the form
$$\frac{a}{b}\cdot\pi-\sqrt{c}+d,$$
where $a,b,c,$ and $d$ are positive integers and $a$ and $b$ are relatively prime. What is $a+b+c+d$?

$\textbf{(A) } 13 \qquad\textbf{(B) } 14 \qquad\textbf{(C) } 15 \qquad\textbf{(D) } 16\qquad\textbf{(E) } 17$

Solution

$\textbf{E}$

Let the point $AB$ intersects the large circle be $X$, and the point $CD$ intersects the large circle be $Y$. We see that $FX=FY=2$. The distance between $F$ and line $AB$ is 1, which is half of the length $FX$ or $FY$. So we can see that $\angle XFY=120^\circ$.

We can divide the shaded region into 3 part. The radius of the top semicircle is 2, so its area is $\dfrac12\pi(2)^2=2\pi$. The bottom arc corresponds to a central angle of $120^\circ$, so the bottom area is $\dfrac13\pi(2)^2-\dfrac12\cdot2\cdot2\cdot\sin\angle 120^\circ=\dfrac43\pi-\sqrt3$. The rest part is composed of 4 corners of a circle inscribed in a square, so its area is $2^2-\pi(1)^2=4-\pi$.

Therefore, the area of the shaded region is $2\pi+\dfrac43\pi-\sqrt3+4-\pi=\dfrac73\pi-\sqrt3+4$. The answer is $a+b+c+d=7+3+3+4=17$.

Question 21

Debra flips a fair coin repeatedly, keeping track of how many heads and how many tails she has seen in total, until she gets either two heads in a row or two tails in a row, at which point she stops flipping. What is the probability that she gets two heads in a row but she sees a second tail before she sees a second head?

$\textbf{(A) } \dfrac{1}{36} \qquad \textbf{(B) } \dfrac{1}{24} \qquad \textbf{(C) } \dfrac{1}{18} \qquad \textbf{(D) } \dfrac{1}{12} \qquad \textbf{(E) } \dfrac{1}{6}$

Solution

$\textbf{B}$

For Debra to see the second tail before the second head, her first flip can't be heads. Therefore, her first flip must be tails. The shortest sequence of flips by which she can get two heads in a row and see the second tail before she sees the second head is $THTHH$, which has a probability of $\dfrac{1}{2^5} = \dfrac{1}{32}$. Furthermore, she can prolong her coin flipping by adding an extra $TH$ before the last $H$, which itself has a probability of $\dfrac{1}{2^2} = \dfrac{1}{4}$. Since she can do this indefinitely, the answer is $$\dfrac{1}{32}[1+\dfrac14+\left(\dfrac14\right)^2+\cdots]=\dfrac{\frac{1}{32}}{1-\frac{1}{4}} = \dfrac{1}{24}$$

Question 22

Raashan, Sylvia, and Ted play the following game. Each starts with $\$1$. A bell rings every $15$ seconds, at which time each of the players who currently have money simultaneously chooses one of the other two players independently and at random and gives $\$1$ to that player. What is the probability that after the bell has rung $2019$ times, each player will have $\$1$? (For example, Raashan and Ted may each decide to give $\$1$ to Sylvia, and Sylvia may decide to give her her dollar to Ted, at which point Raashan will have $\$0$, Sylvia will have $\$2$, and Ted will have $\$1$, and that is the end of the first round of play. In the second round Rashaan has no money to give, but Sylvia and Ted might choose each other to give their $\$1$ to, and the holdings will be the same at the end of the second round.)

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

Solution

$\textbf{B}$

On the first turn, each player starts off with $\$1$. Each turn after that, there are only two possibilities: either everyone stays at $\$1$, which we will write as $(1-1-1)$, or the distribution of money becomes $\$2-\$1-\$0$ in some order, which we write as $(2-1-0)$. ($(3-0-0)$ cannot be achieved). We will consider these two states separately.

In the $(1-1-1)$ state, each person has two choices for whom to give their dollar to, meaning there are $2^3=8$ possible ways that the money can be rearranged. Note that there are only two ways that we can reach $(1-1-1)$ again:

1. Raashan gives his money to Sylvia, who gives her money to Ted, who gives his money to Raashan.

2. Raashan gives his money to Ted, who gives his money to Sylvia, who gives her money to Raashan.

Thus, the probability of staying in the $(1-1-1)$ state is $\dfrac{1}{4}$, while the probability of going to the $(2-1-0)$ state is $\dfrac{3}{4}$.

In the $(2-1-0)$ state, we will label the person with $\$2$ as person A, the person with $\$1$ as person B, and the person with $\$0$ as person C. Person A has two options for whom to give money to, and person B has 2 options for whom to give money to, meaning there are total $2\cdot 2 = 4$ ways the money can be redistributed. The only way that the distribution can return to $(1-1-1)$ is if A gives $\$1$ to B, and B gives $\$1$ to C. We check the other possibilities to find that they all lead back to $(2-1-0)$. Thus, the probability of going to the $(1-1-1)$ state is $\dfrac{1}{4}$, while the probability of staying in the $(2-1-0)$ state is $\dfrac{3}{4}$.

No matter which state we are in, the probability of going to the $(1-1-1)$ state is always $\dfrac{1}{4}$. This means that, after the bell rings 2018 times, regardless of what state the money distribution is in, there is a $\dfrac{1}{4}$ probability of going to the $(1-1-1)$ state after the 2019th bell ring. Thus, our answer is $\dfrac{1}{4}$.

Question 23

Points $A(6,13)$ and $B(12,11)$ lie on circle $\omega$ in the plane. Suppose that the tangent lines to $\omega$ at $A$ and $B$ intersect at a point on the $x$-axis. What is the area of $\omega$?

$\textbf{(A) }\dfrac{83\pi}{8}\qquad\textbf{(B) }\dfrac{21\pi}{2}\qquad\textbf{(C) } \dfrac{85\pi}{8}\qquad\textbf{(D) }\dfrac{43\pi}{4}\qquad\textbf{(E) }\dfrac{87\pi}{8}$

Solution

$\textbf{C}$

Let the intersection point on $x$ axis be $C(x,0)$. We have $$AC=BC\rightarrow\sqrt{(x-6)^2 + 13^2} = \sqrt{(x-12)^2 + 11^2}\rightarrow x=5$$ Let the center of circle $\omega$ be $O$. We have $OA\perp AC$ and $OB\perp BC$. The slope of $AC$ is $\dfrac{13-0}{6-5}=13$. So the slope of $OA$ is $-\dfrac1{13}$. Since $A(6,13)$ is a point on $OA$, we get $OA:y=-\dfrac1{13}x+\dfrac{175}{13}$. Similarly, we get $OB:y=-\dfrac{7}{11}x+\dfrac{205}{11}$.

$OA:y=-\dfrac1{13}x+\dfrac{175}{13}$ intersects $OB:y=-\dfrac{7}{11}x+\dfrac{205}{11}$ at point $O(\dfrac{37}{4},\dfrac{51}{4})$. Thus the radius of circle $\omega$ is $OA=\dfrac{\sqrt{170}}4$. Hence, the area is $\dfrac{85}{8}\pi$.

Question 24

Define a sequence recursively by $x_0=5$ and\[x_{n+1}=\frac{x_n^2+5x_n+4}{x_n+6}\]for all nonnegative integers $n.$ Let $m$ be the least positive integer such that\[x_m\leq 4+\frac{1}{2^{20}}.\]In which of the following intervals does $m$ lie?

$\textbf{(A) } [9,26] \qquad\textbf{(B) } [27,80] \qquad\textbf{(C) } [81,242]\qquad\textbf{(D) } [243,728] \qquad\textbf{(E) } [729,\infty]$

Solution

$\textbf{C}$

We first prove that $x_n > 4$ for all $n \ge 0$, by induction. Observe that\[x_{n+1} - 4 = \frac{x_n^2 + 5x_n + 4 - 4(x_n+6)}{x_n+6} = \frac{(x_n - 4)(x_n+5)}{x_n+6}.\]so $x_{n+1} > 4$ if and only if $x_{n} > 4$.

We similarly prove that $x_n$ is decreasing:\[x_{n+1} - x_n = \frac{x_n^2 + 5x_n + 4 - x_n(x_n+6)}{x_n+6} = \frac{4-x_n}{x_n+6} < 0.\]Now we need to estimate the value of $x_{n+1}-4$, which we can do using the rearranged equation:\[x_{n+1} - 4 = (x_n-4)\cdot\frac{x_n + 5}{x_n+6}.\]Since $x_n$ is decreasing, $\dfrac{x_n + 5}{x_n+6}$ is also decreasing, so we have\[\frac{9}{10} < \frac{x_n + 5}{x_n+6} \le \frac{10}{11}\]and\[\frac{9}{10}(x_n-4) < x_{n+1} - 4 \le \frac{10}{11}(x_n-4)\]This becomes\[\left(\frac{9}{10}\right)^n = \left(\frac{9}{10}\right)^n \left(x_0-4\right) < x_{n} - 4 \le \left(\frac{10}{11}\right)^n \left(x_0-4\right) = \left(\frac{10}{11}\right)^n\]The problem thus reduces to finding the least value of $n$ such that\[\left(\frac{9}{10}\right)^n < x_{n} - 4 \le \frac{1}{2^{20}} \text{ and } \left(\frac{10}{11}\right)^{n-1} > x_{n-1} - 4 > \frac{1}{2^{20}}\] Taking logarithms, we get $n \ln \dfrac{9}{10} < -20 \ln 2$ and $(n-1)\ln \dfrac{10}{11} > -20 \ln 2$, i.e.\[n > \frac{20\ln 2}{\ln\frac{10}{9}} \text{ and } n-1 < \frac{20\ln 2}{\ln\frac{11}{10}}\]As approximations, we can use $\ln\dfrac{10}{9} \approx \dfrac{1}{9}$, $\ln\dfrac{11}{10} \approx \dfrac{1}{10}$, and $\ln 2\approx 0.7$. These approximations allow us to estimate\[126 < n < 141,\]which gives C$[81,242]$.

Question 25

How many sequences of $0$s and $1$s of length $19$ are there that begin with a $0$, end with a $0$, contain no two consecutive $0$s, and contain no three consecutive $1$s?

$\textbf{(A) }55\qquad\textbf{(B) }60\qquad\textbf{(C) }65\qquad\textbf{(D) }70\qquad\textbf{(E) }75$

Solution

$\textbf{C}$

Any valid sequence of length $n$ will start with a $0$ followed by either $10$ or $110$. Since $n=19$, we need to add several $10$s and $110$s after the first 0 to make the rest 18 numbers.

$\textbf{Case 1:}$ nine $10$s - there is only $1$ way to arrange them.

$\textbf{Case 2:}$ two $110$s and six $10$s - there are $\dbinom82 = 28$ ways to arrange them.

$\textbf{Case 3:}$ four $110$s and three $10$s - there are $\dbinom74 = 35$ ways to arrange them.

$\textbf{Case 4:}$ six $110$s - there is only $1$ way to arrange them.

Summing the four cases gives $1+28+35+1= 65$.