## AMC 10 2021 Fall Test A

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**Instructions**

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

What is the value of $\dfrac{(2112-2021)^2}{169}$?

$\textbf{(A) } 7 \qquad\textbf{(B) } 21 \qquad\textbf{(C) } 49 \qquad\textbf{(D) } 64 \qquad\textbf{(E) } 91$

$\textbf{C}$

$$\frac{(2112-2021)^2}{169}=\frac{91^2}{169}=\frac{91^2}{13^2}=\frac{(13\times7)^2}{13^2}=7^2=49$$

Menkara has a $4 \times 6$ index card. If she shortens the length of one side of this card by $1$ inch, the card would have area $18$ square inches. What would the area of the card be in square inches if instead she shortens the length of the other side by $1$ inch?

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

$\textbf{E}$

If Menkara shortens the length of the 4-inch side by 1 inch, she will get a rectangle of $3\times6=18$ square inches.

If Menkara shortens the length of the 6-inch side by 1 inch, she will get a rectangle of $4\times5=20$ square inches.

Obviously Menkara shortens the 4-inch side in the first time. So in the next time she will get a rectangle of 20 square inches instead.

What is the maximum number of balls of clay of radius $2$ that can completely fit inside a cube of side length $6$ assuming the balls can be reshaped but not compressed before they are packed in the cube?

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

$\textbf{D}$

The volume of the cube is $V_{\text{cube}}=6^3=216,$ and the volume of a clay ball is $V_{\text{ball}}=\dfrac43\times\pi\times2^3=\dfrac{32}{3}\pi$.

Since the balls can be reshaped but not compressed, the maximum number of balls that can completely fit inside a cube is \[\left\lfloor\frac{V_{\text{cube}}}{V_{\text{ball}}}\right\rfloor=\left\lfloor\frac{81}{4\pi}\right\rfloor\] Approximating with $\pi\approx3.14,$ we have $12<4\pi<13,$ or $\left\lfloor\dfrac{81}{13}\right\rfloor \leq \left\lfloor\dfrac{81}{4\pi}\right\rfloor \leq \left\lfloor\dfrac{81}{12}\right\rfloor.$ Simplifying, we get \[6 \leq \left\lfloor\frac{81}{4\pi}\right\rfloor \leq 6\] from which $\left\lfloor\dfrac{81}{4\pi}\right\rfloor=6.$

Mr. Lopez has a choice of two routes to get to work. Route A is $6$ miles long, and his average speed along this route is $30$ miles per hour. Route B is $5$ miles long, and his average speed along this route is $40$ miles per hour, except for a $\dfrac{1}{2}$-mile stretch in a school zone where his average speed is $20$ miles per hour. By how many minutes is Route B quicker than Route A?

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

$\textbf{B}$

For Route A, Mr. Lopez will spend $\dfrac{6}{30}=\dfrac{1}{5}$ hour.

For Route B, Mr. Lopez will spend $\dfrac{5-0.5}{40}+\dfrac{0.5}{20}=\dfrac{11}{80}$ hour.

Route B is quicker than Route A by $\dfrac{1}{5}-\dfrac{11}{80}=\dfrac{1}{16}$ hour, or $\dfrac{1}{16}\times60=\dfrac{15}{4}=3\dfrac{3}{4}$ minutes.

The six-digit number $\underline{2}\,\underline{0}\,\underline{2}\,\underline{1}\,\underline{0}\,\underline{A}$ is prime for only one digit $A.$ What is $A?$

$(\textbf{A})\: 1\qquad(\textbf{B}) \: 3\qquad(\textbf{C}) \: 5 \qquad(\textbf{D}) \: 7\qquad(\textbf{E}) \: 9$

$\textbf{E}$

We need to consider all possible choices for $A$ from 0 to 9.

In case the six-digit number is divisible by 2, $A$ could be 0, 2, 4, 6, 8.

In case the six-digit number is divisible by 5, $A$ could be 0 or 5.

In case the six-digit number is divisible by 3, $2+0+2+1+0+A=A+5$ is a multiple of 3. Hence, $A$ could be 1, 4, or 7.

In case the six-digit number is divisible by 11, $A+1+0-0-2-2=A-3$ is a multiple of 11. Hence, $A$ could be 3.

The only choice we does not mention above is 9. So the answer is 9.

Elmer the emu takes $44$ equal strides to walk between consecutive telephone poles on a rural road. Oscar the ostrich can cover the same distance in $12$ equal leaps. The telephone poles are evenly spaced, and the $41$st pole along this road is exactly one mile ($5280$ feet) from the first pole. How much longer, in feet, is Oscar's leap than Elmer's stride?

$\textbf{(A) }6\qquad\textbf{(B) }8\qquad\textbf{(C) }10\qquad\textbf{(D) }11\qquad\textbf{(E) }15$

$\textbf{B}$

There are $41-1=40$ gaps between the $41$ telephone poles, so the distance of each gap is $5280\div40=132$ feet.

Each of Elmer's strides covers $132\div44=3$ feet.

Each of Oscar's leaps covers $132\div12=11$ feet.

Therefore, Oscar's leap is $11-3=8$ feet longer than Elmer's stride.

As shown in the figure below, point $E$ lies on the opposite half-plane determined by line $CD$ from point $A$ so that $\angle CDE = 110^\circ$. Point $F$ lies on $\overline{AD}$ so that $DE=DF$, and $ABCD$ is a square. What is the degree measure of $\angle AFE$?

$\textbf{(A) }160\qquad\textbf{(B) }164\qquad\textbf{(C) }166\qquad\textbf{(D) }170\qquad\textbf{(E) }174$

$\textbf{D}$

First, $\angle ADE = 360^\circ - \angle ADC - \angle CDE = 160^\circ$.

By $DE=DF$ we know $\triangle DEF$ is isosceles, so $\angle DFE = \dfrac{180^\circ - \angle ADE}{2}=10^\circ$.

Finally, we get $\angle AFE = 180^\circ - \angle DFE = 170^\circ$.

A two-digit positive integer is said to be $\textit{cuddly}$ if it is equal to the sum of its nonzero tens digit and the square of its units digit. How many two-digit positive integers are cuddly?

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

$\textbf{B}$

Note that the number $\overline{x}\overline{y} = 10x + y.$ By the problem statement, \[10x + y = x + y^2 \implies 9x = y^2 - y \implies 9x = y(y-1)\] From this we see that $y(y-1)$ must be divisible by $9.$ This only happens when $y=9.$ Hence, we get $x=8.$ So there is only $1$ cuddly number, which is $89.$

When a certain unfair die is rolled, an even number is $3$ times as likely to appear as an odd number. The die is rolled twice. What is the probability that the sum of the numbers rolled is even?

$\textbf{(A) }\dfrac38\qquad\textbf{(B) }\dfrac49\qquad\textbf{(C) }\dfrac59\qquad\textbf{(D) }\dfrac9{16}\qquad\textbf{(E) }\dfrac58$

$\textbf{E}$

Since an even number is $3$ times more likely to appear than an odd number, the probability of an even number appearing is $\dfrac{3}{4}$. If the sum of the numbers rolled is even, the numbers must both be even or both be odd. So the probability is \[\frac{3}{4}\times \frac{3}{4} + \frac{1}{4} \times \frac{1}{4} = \frac{5}{8}\]

A school has $100$ students and $5$ teachers. In the first period, each student is taking one class, and each teacher is teaching one class. The enrollments in the classes are $50, 20, 20, 5,$ and $5$. Let $t$ be the average value obtained if a teacher is picked at random and the number of students in their class is noted. Let $s$ be the average value obtained if a student was picked at random and the number of students in their class, including the student, is noted. What is $t-s$?

$\textbf{(A)}\ {-}18.5 \qquad\textbf{(B)}\ {-}13.5 \qquad\textbf{(C)}\ 0 \qquad\textbf{(D)}\ 13.5 \qquad\textbf{(E)}\ 18.5$

$\textbf{B}$

Based on the formula of expected value $$E(x)=\sum x\cdot p(x)$$ We have $$t= \frac15\times50 + \frac15\times20 + \frac15\times20 + \frac15\times5 + \frac15\times5=20$$ and $$s= \frac{50}{100}\times50 + \frac{20}{100}\times20 + \frac{20}{100}\times20 + \frac{5}{100}\times5 + \frac{5}{100}\times5=33.5$$ So $$t-s=-13.5$$

Emily sees a ship traveling at a constant speed along a straight section of a river. She walks parallel to the riverbank at a uniform rate faster than the ship. She counts $210$ equal steps walking from the back of the ship to the front. Walking in the opposite direction, she counts $42$ steps of the same size from the front of the ship to the back. In terms of Emily's equal steps, what is the length of the ship?

$\textbf{(A) }70\qquad\textbf{(B) }84\qquad\textbf{(C) }98\qquad\textbf{(D) }105\qquad\textbf{(E) }126$

$\textbf{A}$

Let $x$ be the length of the ship in the unit of Emily's step. In the time that Emily walks $210$ steps, the ship moves $210-x$ steps. Also, in the time that Emily walks $42$ steps, the ship moves $x-42$ steps. Since the ship and Emily both travel at some constant speeds, the ratio of their speeds is $$\dfrac{210}{210-x} = \dfrac{42}{x-42}$$ Hence, we get $x=70$.

The base-nine representation of the number $N$ is $27{,}006{,}000{,}052_{nine}.$ What is the remainder when $N$ is divided by $5?$

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

$\textbf{D}$

\[N= 2\times9^{10} + 7\times9^9 + 6\times9^6 + 5\times9 + 2\] Since $9\equiv-1\pmod{5}$, we have \[N\equiv 2\times(-1)^{10} + 7\times(-1)^9 + 6\times(-1)^6 + 5\times(-1) + 2\pmod{5}\] So we get \[N\equiv-2\pmod{5}\] or \[N\equiv 3\pmod{5}\] Another way to solve the problem is to think about the last digit of base-nine $N$. We can find the pattern of the units digit of the power of 9. When the power is odd, the units digit is 9. When the power is even, the units digit is 1. So the expression of $N$ can be simplified as $$N=2\times1+7\times9+6\times1+5\times9+2=118$$ Hence, the reminder when $N$ is divided by 5 is 3.

Each of $6$ balls is randomly and independently painted either black or white with equal probability. What is the probability that every ball is different in color from more than half of the other $5$ balls?

$\textbf{(A) } \dfrac{1}{64}\qquad\textbf{(B) } \dfrac{1}{6}\qquad\textbf{(C) } \dfrac{1}{4}\qquad\textbf{(D) } \dfrac{5}{16}\qquad\textbf{(E) }\dfrac{1}{2}$

$\textbf{D}$

We have $2^6 = 64$ ways to color the balls in total. Note that for this restriction to be true, there must be $3$ balls of each color. There are $\dbinom63 = 20$ ways for three balls chosen to be painted white, and the others are painted black. Thus, the answer is $\dfrac{20}{64}=\dfrac{5}{16}$.

How many ordered pairs $(x,y)$ of real numbers satisfy the following system of equations?

\begin{align*} x^2+3y&=9 \\ (|x|+|y|-4)^2 &= 1 \end{align*}

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

$\textbf{D}$

From the second equation, we get $|x|+|y|=3\ \text{or}\ 5$.

Case 1: $|x|+|y|=3$. Then we get $|x|=3-|y|\rightarrow x^2=y^2-6|y|+9$. Substituting the value of $x^2$ in the first equation, we get $$y^2-6|y|+3y=0$$ Case 1.1: $y\ge0$, which means $|y|=y$. Hence, we get $y^2-3y=0\rightarrow y=0\ \text{or}\ 3$. So the $(x,y)$ pair could be $(3.0)$, $(-3,0)$, $(0,3)$.

Case 1.2: $y<0$. Now we have $y^2+9y=0\rightarrow y=-9$. So $|x|=3-|y|=-6$, which is impossible.

Case 2: $|x|+|y|=5$. Then we get $|x|=5-|y|\rightarrow x^2=y^2-10|y|+25$. Substituting the value of $x^2$ in the first equation, we get $$y^2-10|y|+3y+16=0$$ By the discussion of the value of $y\ge0$ or $y<0$ in a similar way, we find 2 possible $(x,y)$ pairs: $$\left(\dfrac{-3+\sqrt{105}}{2},\dfrac{-13+\sqrt{105}}{2}\right)$$ $$\left(\dfrac{3-\sqrt{105}}{2},\dfrac{-13+\sqrt{105}}{2}\right)$$

In conclusion, there are 5 pairs of $(x,y)$.

Another way to solving the problem is by graphing. The function graphs of $x^2+3y=9$, $|x|+|y|=3$, and $|x|+|y|=5$ are drawn below.

We see from the graph that there are $5$ intersections, so the answer is 5.

Isosceles triangle $ABC$ has $AB = AC = 3\sqrt6$, and a circle with radius $5\sqrt2$ is tangent to line $AB$ at $B$ and to line $AC$ at $C$. What is the area of the circle that passes through vertices $A$, $B$, and $C?$

$\textbf{(A) }24\pi\qquad\textbf{(B) }25\pi\qquad\textbf{(C) }26\pi\qquad\textbf{(D) }27\pi\qquad\textbf{(E) }28\pi$

$\textbf{C}$

Let $O$ be the center of circle with radius $5\sqrt2$ that is tangent to $AB$ at $B$ and $AC$ at $C$. We see that $AO$ bisects $\angle BAC$. Let $\theta$ be half of $\angle BAC$. We have $$\tan\theta=\dfrac{OB}{AB}=\dfrac{5\sqrt2}{3\sqrt6}=\dfrac{5}{3\sqrt3}$$ Let $P$ be the center of circle that passes through vertices $A$, $B$, and $C$. In $\triangle ABP$, both $AP$ and $BP$ are the radius $r$ of circle $P$. Hence, we get $$\cos\theta=\dfrac{AB/2}{AP}=\dfrac{3\sqrt6}{2r}$$ The area of circle $P$ is $$\pi r^2=\pi\left(\dfrac{3\sqrt6}{2\cos\theta}\right)^2=\dfrac{27}{2}\pi\cdot\dfrac{1}{\cos^2\theta}=\dfrac{27}{2}\pi\left(1+\tan^2\theta\right)=\dfrac{27}{2}\pi\left[1+\left(\dfrac{5}{3\sqrt3}\right)^2\right]=26\pi$$

The graph of

$$f(x) = |\lfloor x \rfloor| - |\lfloor 1 - x \rfloor|$$

is symmetric about which of the following? (Here $\lfloor x \rfloor$ is the greatest integer not exceeding $x$.)

$\textbf{(A) }\text{the }y\text{-axis}\qquad \textbf{(B) }\text{the line }x = 1\qquad \textbf{(C) }\text{the origin}\qquad \textbf{(D) }\text{ the point }\left(\dfrac12, 0\right)\qquad \textbf{(E) }\text{the point }(1,0)$

$\textbf{D}$

Note that\[f(1-x)=|\lfloor 1-x\rfloor|-|\lfloor x\rfloor|=-f(x)\]so $$f\left(\dfrac12+x\right)=-f\left(\dfrac12-x\right)$$ This means that the graph is symmetric about $\left(\dfrac12, 0\right)$.

An architect is building a structure that will place vertical pillars at the vertices of regular hexagon $ABCDEF$, which is lying horizontally on the ground. The six pillars will hold up a flat solar panel that will not be parallel to the ground. The heights of pillars at $A$, $B$, and $C$ are $12$, $9$, and $10$ meters, respectively. What is the height, in meters, of the pillar at $E$?

$\textbf{(A) }9 \qquad\textbf{(B) } 6\sqrt{3} \qquad\textbf{(C) } 8\sqrt{3} \qquad\textbf{(D) } 17 \qquad\textbf{(E) }12\sqrt{3}$

$\textbf{D}$

The three-dimensional and two-dimensional diagram are drawn below.

The pillar at $B$ has height $9$ and the pillar at $A$ has height $12.$ Since the solar panel is flat, the inclination from pillar $B$ to pillar $A$ is $3.$ Call the center of the regular hexagon $G.$ Since $\overrightarrow{CG}\parallel\overrightarrow{BA},$ it follows that the solar panel has height $10+3=13$ at $G.$ Since the solar panel is flat, the heights of the solar panel at $B,G,$ and $E$ are collinear. Therefore, the pillar at $E$ has height $9+4+4=17.$

A farmer's rectangular field is partitioned into $2$ by $2$ grid of $4$ rectangular sections as shown in the figure. In each section the farmer will plant one crop: corn, wheat, soybeans, or potatoes. The farmer does not want to grow corn and wheat in any two sections that share a border, and the farmer does not want to grow soybeans and potatoes in any two sections that share a border. Given these restrictions, in how many ways can the farmer choose crops to plant in each of the four sections of the field?

$\textbf{(A)}\ 12 \qquad \textbf{(B)}\ 64 \qquad \textbf{(C)}\ 84 \qquad \textbf{(D)}\ 90 \qquad \textbf{(E)}\ 144$

$\textbf{C}$

There are $4$ possibilities for the top-left section. It follows that the top-right and bottom-left sections each have $3$ possibilities, so they have $3^2=9$ combinations. We have two cases:

Case 1: The top-right and bottom-left sections have the same crop. Note that $3$ of the $9$ combinations of the top-right and bottom-left sections satisfy this case, from which the bottom-right section has $3$ possibilities. Hence, there are $4\times3\times3=36$ ways in this case.

Case 2: The top-right and bottom-left sections have different crops. Note that $6$ of the $9$ combinations of the top-right and bottom-left sections satisfy this case, from which the bottom-right section has $2$ possibilities. Hence, there are $4\times6\times2=48$ ways in this case.

Together, the answer is $36+48=84.$

A disk of radius $1$ rolls all the way around the inside of a square of side length $s>4$ and sweeps out a region of area $A$. A second disk of radius $1$ rolls all the way around the outside of the same square and sweeps out a region of area $2A$. The value of $s$ can be written as $a+\dfrac{b\pi}{c}$, where $a,b$, and $c$ are positive integers and $b$ and $c$ are relatively prime. What is $a+b+c$?

$\textbf{(A)} ~10\qquad\textbf{(B)} ~11\qquad\textbf{(C)} ~12\qquad\textbf{(D)} ~13\qquad\textbf{(E)} ~14$

$\textbf{A}$

According to the given information, the area of the red region is $A$, and the area of the green region is $2A$. The side length of the small white square at the center of the graph is $s-4.$ There is a white piece at each corner (bounded by two sides of a square and one $90^\circ$ arc) where the disk never sweeps out. The combined area of these four white pieces is $(1+1)^2-\pi\times1^2=4-\pi.$ As a result, we have\[A=s^2-(s-4)^2-(4-\pi)=8s-20+\pi\]Now, we consider the second disk. The part it sweeps is comprised of four quarter circles with radius $2$ and four rectangles with side lengths of $2$ and $s.$ When we add it all together, we have $$2A=4\times\dfrac14\pi\times2^2+4\times2s=8s+4\pi$$ or $A=4s+2\pi$. We equate the expressions for $A,$ and then solve for $s:$\[8s-20+\pi=4s+2\pi\]We get $s=5+\dfrac{\pi}{4},$ so the answer is $5+1+4=10.$

How many ordered pairs of positive integers $(b,c)$ exist where both $x^2+bx+c=0$ and $x^2+cx+b=0$ do not have distinct, real solutions?

$\textbf{(A) } 4 \qquad \textbf{(B) } 6 \qquad \textbf{(C) } 8 \qquad \textbf{(D) } 10 \qquad \textbf{(E) } 12 \qquad$

$\textbf{B}$

A quadratic equation does not have distinct real solutions if and only if the discriminant is nonpositive. We conclude that:

- Since $x^2+bx+c=0$ does not have distinct real solutions, we have $b^2\leq 4c.$
- Since $x^2+cx+b=0$ does not have distinct real solutions, we have $c^2\leq 4b.$

Squaring the first inequality, we get $b^4\leq 16c^2.$ Multiplying the second inequality by $16,$ we get $16c^2\leq 64b.$ Combining these results, we get\[b^4\leq 16c^2\leq 64b.\]We apply casework to the value of $b:$

- If $b=1,$ then $1\leq 16c^2\leq 64,$ from which $c=1,2.$
- If $b=2,$ then $16\leq 16c^2\leq 128,$ from which $c=1,2.$
- If $b=3,$ then $81\leq 16c^2\leq 192,$ from which $c=3.$
- If $b=4,$ then $256\leq 16c^2\leq 256,$ from which $c=4.$

Together, there are $6$ ordered pairs $(b,c),$ namely $(1,1),(1,2),(2,1),(2,2),(3,3),$ and $(4,4).$

Each of $20$ balls is tossed independently and at random into one of $5$ bins. Let $p$ be the probability that some bin ends up with $3$ balls, another with $5$ balls, and the other three with $4$ balls each. Let $q$ be the probability that every bin ends up with $4$ balls. What is $\dfrac{p}{q}$?

$\textbf{(A)}\ 1 \qquad\textbf{(B)}\ 4 \qquad\textbf{(C)}\ 8 \qquad\textbf{(D)}\ 12 \qquad\textbf{(E)}\ 16$

$\textbf{E}$

For simplicity purposes, we assume that the balls and the bins are both distinguishable.

For each ball, we must decide which of the 5 bins should it go into. So we has 5 choices. And 20 balls have $5^{20}$ arrangements.

In case $q$, we need to decide which 4 ball go into the first bin. So there are $\dbinom{20}{4}$ choices. For the second bin, we have $\dbinom{16}{4}$ choices, etc. Hence, the probability is $$q=\dfrac{\dbinom{20}{4}\dbinom{16}{4}\dbinom{12}{4}\dbinom{8}{4}\dbinom{4}{4}}{5^{20}}=\dfrac{\dbinom{20}{4,4,4,4,4}}{5^{20}}$$ In case $p$, first we need to decide which of the 5 bins should be the one with 5 balls, and which 5 balls they are. This gives us $5\times\dbinom{20}{5}$ choices. Next, we need to decide which of the rest 4 bins should be the one with 3 balls, and which 3 balls they are. This gives us $4\times\dbinom{15}{3}$ choices. The rest 12 balls are evenly distributed, so we have $\dbinom{12}{4,4,4}$ arrangements. Therefore, the probability is $$p=\dfrac{5\times\dbinom{20}{5}\times4\times\dbinom{15}{3}\times\dbinom{12}{4,4,4}}{5^{20}}=\dfrac{5\times4\times\dbinom{20}{5,3,4,4,4}}{5^{20}}$$ The ratio is \[\dfrac pq=\dfrac{5\times4\times\dbinom{20}{5,3,4,4,4}}{\dbinom{20}{4,4,4,4,4}}=\dfrac{5\times4\times\dfrac{20!}{5!\times3!\times4!\times4!\times4!}}{\dfrac{20!}{4!\times4!\times4!\times4!\times4!}}=\dfrac{5\times4\times(4!)^5}{5!\times3!\times(4!)^3}=\dfrac{5\times4\times4}{5}=16\]

Inside a right circular cone with base radius $5$ and height $12$ are three congruent spheres with radius $r$. Each sphere is tangent to the other two spheres and also tangent to the base and side of the cone. What is $r$?

$\textbf{(A)}\ \dfrac{3}{2} \qquad\textbf{(B)}\ \dfrac{90-40\sqrt{3}}{11} \qquad\textbf{(C)}\ 2 \qquad\textbf{(D)}\ \dfrac{144-25\sqrt{3}}{44} \qquad\textbf{(E)}\ \dfrac{5}{2}$

$\textbf{B}$

The three-dimensional diagram is drawn below.

We can take half of a cross section of the sphere, as such:

Notice that we choose a cross section where one of the spheres was tangent to the lateral surface of the cone at $D$. It is well-worth noticing that the sphere is $\textbf{NOT}$ tangent to $BC$.

To evaluate $r$, we will find $AE$ and $EC$ in terms of $r$; we also know that $AE+EC = 5$, so with this, we can solve $r$. Firstly, to find $EC$, we can take a bird's eye view of the cone:

Note that $C$ is the centroid of equilateral triangle $EXY$. Also, since all of the medians of an equilateral triangle are also altitudes, we want to find two-thirds of the altitude from $E$ to $XY$; this is because medians cut each other into a $2$ to $1$ ratio. This equilateral triangle has a side length of $2r$. Therefore, it has an altitude of length $\sqrt{3}r$; two thirds of this is $\dfrac{2\sqrt{3}}{3}r$, so $EC = \dfrac{2\sqrt{3}}{3}r.$

To evaluate $AE$ in terms of $r$, first we see that $\tan\angle BAC=\dfrac{BC}{AC}=\dfrac{12}{5}$. Since $AO$ bisects $\angle BAC$, by the formula $\tan2\theta=\dfrac{2\tan\theta}{1-\tan^2\theta}$, we get $\tan\angle OAE=\dfrac{OE}{AE}=\dfrac{r}{AE}=\dfrac{2}{3}$. So $AE=\dfrac32r$.

Finally we have $$AC=AE+CE=\dfrac32r+\dfrac{2\sqrt{3}}{3}r=5\rightarrow r=\dfrac{90-40\sqrt{3}}{11}$$

For each positive integer $n$, let $f_1(n)$ be twice the number of positive integer divisors of $n$, and for $j \ge 2$, let $f_j(n) = f_1(f_{j-1}(n))$. For how many values of $n \le 50$ is $f_{50}(n) = 12?$

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

$\textbf{D}$

If $f_1(n)=12$, then $n$ must have $6$ divisors, which means its prime factorization is in the form $pq^2$ or $p^5$, where $p$ and $q$ are prime numbers. Listing out values less than $50$ which have these forms of prime factorization, we find $12,18,20,28,44,45,50$ for $pq^2$, and just $32$ for $p^5$. Here $12$ is a special number. If $n=12$, then we have\begin{align*}

f_1(n)&=12\\

f_2(n)&=f_1(f_1(n))=f_1(12)=12\\

&\vdots\\

f_{50}(n)&=12

\end{align*}

If $n=18,20,28,32,44,45,50$, we also have \begin{align*}

f_1(n)&=12\\

f_2(n)&=f_1(f_1(n))=f_1(12)=12\\

&\vdots\\

f_{50}(n)&=12

\end{align*}

Furthermore, if $f_1(n)=18,20,28,32,44,45,50$, it also works. Now $n$ has $9,10,14,16,22,22.5$, or $15$ divisors. Looking through the forms of prime factorization like what we did for $f_1(n)=12$ above, we see that $f_1(n)$ could only possibly be equal to 18 and 20, and still have $n$ less than or equal to $50$. This means $n$ must have 9 or 10 divisors, then $n$ will be in the form $p^2q^2$ or $p^4q$. The only two values less than or equal to $50$ would be 36 and 48 respectively.

Together, we find 10 possible values for $n$: $12,18,20,28,32,36,44,45,48,50.$

Each of the $12$ edges of a cube is labeled $0$ or $1$. Two labelings are considered different even if one can be obtained from the other by a sequence of one or more rotations and/or reflections. For how many such labelings is the sum of the labels on the edges of each of the $6$ faces of the cube equal to $2$?

$\textbf{(A) } 8 \qquad\textbf{(B) } 10 \qquad\textbf{(C) } 12 \qquad\textbf{(D) } 16 \qquad\textbf{(E) } 20$

$\textbf{E}$

For simplicity, we will name this cube $ABCDEFGH$ by vertices, as shown below.

Note that for each face of this cube, two edges are labeled $0,$ and two edges are labeled $1.$ For all twelve edges of this cube, we conclude that six edges are labeled $0,$ and six edges are labeled $1.$

We apply casework to face $ABCD.$ Recall that there are $\dbinom42=6$ ways to label its edges:

$\textbf{Case 1: Opposite edges have the same label.}$

There are $2$ ways to label the edges of $ABCD.$ We will consider one of the ways, then multiply the count by $2.$ Without loss of generality, we assume that $\overline{AB},\overline{BC},\overline{CD},\overline{DA}$ are labeled $1,0,1,0,$ respectively:

We apply casework to the label of $\overline{AE},$ as shown below.

We have $2\cdot2=4$ such labelings for this case.

$\textbf{Case 2: Opposite edges have different labels.}$

There are $4$ ways to label the edges of $ABCD.$ We will consider one of the ways, then multiply the count by $4.$ Without loss of generality, we assume that $\overline{AB},\overline{BC},\overline{CD},\overline{DA}$ are labeled $1,1,0,0,$ respectively:

We apply casework to the labels of $\overline{AE}$ and $\overline{BF},$ as shown below.

We have $4\cdot4=16$ such labelings for this case.

Therefore, we have $4+16=20$ such labelings in total.

A quadratic polynomial with real coefficients and leading coefficient $1$ is called $\textit{disrespectful}$ if the equation $p(p(x))=0$ is satisfied by exactly three real numbers. Among all the disrespectful quadratic polynomials, there is a unique such polynomial $\tilde{p}(x)$ for which the sum of the roots is maximized. What is $\tilde{p}(1)$?

$\textbf{(A) } \dfrac{5}{16} \qquad\textbf{(B) } \dfrac{1}{2} \qquad\textbf{(C) } \dfrac{5}{8} \qquad\textbf{(D) } 1 \qquad\textbf{(E) } \dfrac{9}{8}$

$\textbf{A}$

Let $r_1$ and $r_2$ be the roots of $\tilde{p}(x)$. Then $\tilde{p}(x)=(x-r_1)(x-r_2)=x^2-(r_1+r_2)x+r_1r_2$. The solutions to $\tilde{p}(\tilde{p}(x))=0$ is the union of the solutions to \[\tilde{p}(x)-r_1=x^2-(r_1+r_2)x+(r_1r_2-r_1)=0\] and \[\tilde{p}(x)-r_2=x^2-(r_1+r_2)x+(r_1r_2-r_2)=0\] Note that one of these two quadratics has one solution (a double root) and the other has two as there are exactly three solutions. WLOG, assume that the quadratic with one solution is $x^2-(r_1+r_2)x+(r_1r_2-r_1)=0$. Then the discriminant is $0$, so $(r_1+r_2)^2 = 4r_1r_2 - 4r_1$. Thus, $$r_1-r_2=\pm 2\sqrt{-r_1}$$ Since $x^2-(r_1+r_2)x+(r_1r_2-r_2)=0$ has two solutions, the discriminant $(r_1+r_2)^2-4r_1r_2+4r_2$ must be positive. From here, we get that $(r_1-r_2)^2>-4r_2$, so $-4r_1>-4r_2 \implies r_1<r_2$. Hence, $r_1-r_2$ is negative, so $$r_1-r_2=-2\sqrt{-r_1}$$ It follows that the sum of the roots of $\tilde{p}(x)$ is $r_1+r_2=2r_1 + 2\sqrt{-r_1}$, whose maximum value occurs when $r_1 = - \dfrac{1}{4}$. Solving for $r_2$ yields $r_2 = \dfrac{3}{4}$. Therefore, $$\tilde{p}(x)=x^2 - \dfrac{1}{2} x - \dfrac{3}{16}$$ so $\tilde{p}(1)=\dfrac{5}{16}$.