# 2021 AMC10A真题及答案 高清文字版

## Problem1

What is the value of

$$(2^2-2)-(3^2-3)+(4^2-4)$$

$\textbf{(A) } 1 \qquad\textbf{(B) } 2 \qquad\textbf{(C) } 5 \qquad\textbf{(D) } 8 \qquad\textbf{(E) } 12$

## Problem2

Portia's high school has $3$ times as many students as Lara's high school. The two high schools have a total of $2600$ students. How many students does Portia's high school have?

$\textbf{(A)} ~600 \qquad\textbf{(B)} ~650 \qquad\textbf{(C)} ~1950 \qquad\textbf{(D)} ~2000\qquad\textbf{(E)} ~2050$

## Problem3

The sum of two natural numbers is $17{,}402$. One of the two numbers is divisible by $10$. If the units digit of that number is erased, the other number is obtained. What is the difference of these two numbers?

$\textbf{(A)} ~10{,}272\qquad\textbf{(B)} ~11{,}700\qquad\textbf{(C)} ~13{,}362\qquad\textbf{(D)} ~14{,}238\qquad\textbf{(E)} ~15{,}426$

## Problem4

A cart rolls down a hill, travelling $5$ inches the first second and accelerating so that during each successive $1$-second time interval, it travels $7$inches more than during the previous $1$-second interval. The cart takes $30$ seconds to reach the bottom of the hill. How far, in inches, does it travel?

$\textbf{(A)} ~215 \qquad\textbf{(B)} ~360\qquad\textbf{(C)} ~2992\qquad\textbf{(D)} ~3195\qquad\textbf{(E)} ~3242$

## Problem5

The quiz scores of a class with $k > 12$ students have a mean of $8$. The mean of a collection of $12$ of these quiz scores is $14$. What is the mean of the remaining quiz scores of terms of $k$?

$\textbf{(A)} ~\frac{14-8}{k-12} \qquad\textbf{(B)} ~\frac{8k-168}{k-12} \qquad\textbf{(C)} ~\frac{14}{12} - \frac{8}{k} \qquad\textbf{(D)} ~\frac{14(k-12)}{k^2} \qquad\textbf{(E)} ~\frac{14(k-12)}{8k}$

## Problem6

Chantal and Jean start hiking from a trailhead toward a fire tower. Jean is wearing a heavy backpack and walks slower. Chantal starts walking at $4$ miles per hour. Halfway to the tower, the trail becomes really steep, and Chantal slows down to $2$ miles per hour. After reaching the tower, she immediately turns around and descends the steep part of the trail at $3$ miles per hour. She meets Jean at the halfway point. What was Jean's average speed, in miles per hour, until they meet?

$\textbf{(A)} ~\frac{12}{13} \qquad\textbf{(B)} ~1 \qquad\textbf{(C)} ~\frac{13}{12} \qquad\textbf{(D)} ~\frac{24}{13} \qquad\textbf{(E)} ~2$

## Problem7

Tom has a collection of $13$ snakes, $4$ of which are purple and $5$ of which are happy. He observes that all of his happy snakes can add, none of his purple snakes can subtract, and all of his snakes that can't subtract also can't add. Which of these conclusions can be drawn about Tom's snakes?

$\textbf{(A) }$ Purple snakes can add.

$\textbf{(B) }$ Purple snakes are happy.

$\textbf{(C) }$ Snakes that can add are purple.

$\textbf{(D) }$ Happy snakes are not purple.

$\textbf{(E) }$ Happy snakes can't subtract.

## Problem8

A sequence of numbers is defined by $D_0=0,D_1=0,D_2=1$ and $D_n=D_{n-1}+D_{n-3}$ for $n\ge 3$. What are the parities (evenness or oddness) of the triple of numbers $(D_{2021},D_{2022},D_{2023})$, where $E$ denotes even and $O$ denotes odd?

$\textbf{(A) }(O,E,O) \qquad \textbf{(B) }(E,E,O) \qquad \textbf{(C) }(E,O,E) \qquad \textbf{(D) }(O,O,E) \qquad \textbf{(E) }(O,O,O)$

## Problem9

Which of the following is equivalent to$$(2+3)(2^2+3^2)(2^4+3^4)(2^8+3^8)(2^{16}+3^{16})(2^{32}+3^{32})(2^{64}+3^{64})?$$$\textbf{(A)} ~3^{127} + 2^{127} \qquad\textbf{(B)} ~3^{127} + 2^{127} + 2 \cdot 3^{63} + 3 \cdot 2^{63} \qquad\textbf{(C)} ~3^{128}-2^{128} \qquad\textbf{(D)} ~3^{128} + 3^{128} \qquad\textbf{(E)} ~5^{127}$

## Problem10

Which of the following is equivalent to$$(2+3)(2^2+3^2)(2^4+3^4)(2^8+3^8)(2^{16}+3^{16})(2^{32}+3^{32})(2^{64}+3^{64})?$$$\textbf{(A)} ~3^{127} + 2^{127} \qquad\textbf{(B)} ~3^{127} + 2^{127} + 2 \cdot 3^{63} + 3 \cdot 2^{63} \qquad\textbf{(C)} ~3^{128}-2^{128} \qquad\textbf{(D)} ~3^{128} + 3^{128} \qquad\textbf{(E)} ~5^{127}$

## Problem11

For which of the following integers $b$ is the base-$b$ number $2021_b - 221_b$ not divisible by $3$?

$\textbf{(A)} ~3 \qquad\textbf{(B)} ~4\qquad\textbf{(C)} ~6\qquad\textbf{(D)} ~7\qquad\textbf{(E)} ~8$

## Problem12

Two right circular cones with vertices facing down as shown in the figure below contains the same amount of liquid. The radii of the tops of the liquid surfaces are $3$ cm and $6$ cm. Into each cone is dropped a spherical marble of radius $1$ cm, which sinks to the bottom and is completely submerged without spilling any liquid. What is the ratio of the rise of the liquid level in the narrow cone to the rise of the liquid level in the wide cone?

$[asy] size(350); defaultpen(linewidth(0.8)); real h1 = 10, r = 3.1, s=0.75; pair P = (r,h1), Q = (-r,h1), Pp = s * P, Qp = s * Q; path e = ellipse((0,h1),r,0.9), ep = ellipse((0,h1*s),r*s,0.9); draw(ellipse(origin,r*(s-0.1),0.8)); fill(ep,gray(0.8)); fill(origin--Pp--Qp--cycle,gray(0.8)); draw((-r,h1)--(0,0)--(r,h1)^^e); draw(subpath(ep,0,reltime(ep,0.5)),linetype("4 4")); draw(subpath(ep,reltime(ep,0.5),reltime(ep,1))); draw(Qp--(0,Qp.y),Arrows(size=8)); draw(origin--(0,12),linetype("4 4")); draw(origin--(r*(s-0.1),0)); label("3",(-0.9,h1*s),N,fontsize(10)); real h2 = 7.5, r = 6, s=0.6, d = 14; pair P = (d+r-0.05,h2-0.15), Q = (d-r+0.05,h2-0.15), Pp = s * P + (1-s)*(d,0), Qp = s * Q + (1-s)*(d,0); path e = ellipse((d,h2),r,1), ep = ellipse((d,h2*s+0.09),r*s,1); draw(ellipse((d,0),r*(s-0.1),0.8)); fill(ep,gray(0.8)); fill((d,0)--Pp--Qp--cycle,gray(0.8)); draw(P--(d,0)--Q^^e); draw(subpath(ep,0,reltime(ep,0.5)),linetype("4 4")); draw(subpath(ep,reltime(ep,0.5),reltime(ep,1))); draw(Qp--(d,Qp.y),Arrows(size=8)); draw((d,0)--(d,10),linetype("4 4")); draw((d,0)--(d+r*(s-0.1),0)); label("6",(d-r/4,h2*s-0.06),N,fontsize(10)); [/asy]$

$\textbf{(A) }1 \qquad \textbf{(B) }\frac{47}{43} \qquad \textbf{(C) }2 \qquad \textbf{(D) }\frac{40}{13} \qquad \textbf{(E) }4$

## Problem13

What is the volume of tetrahedron $ABCD$ with edge lengths $AB = 2$$AC = 3$$AD = 4$$BC = \sqrt{13}$$BD = 2\sqrt{5}$, and $CD = 5$ ?

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

## Problem14

All the roots of the polynomial $z^6-10z^5+Az^4+Bz^3+Cz^2+Dz+16$ are positive integers, possibly repeated. What is the value of $B$?

$\textbf{(A) }-88 \qquad \textbf{(B) }-80 \qquad \textbf{(C) }-64 \qquad \textbf{(D) }-41\qquad \textbf{(E) }-40$

## Problem15

Values for $A,B,C,$ and $D$ are to be selected from $\{1, 2, 3, 4, 5, 6\}$ without replacement (i.e. no two letters have the same value). How many ways are there to make such choices so that the two curves $y=Ax^2+B$ and $y=Cx^2+D$ intersect? (The order in which the curves are listed does not matter; for example, the choices $A=3, B=2, C=4, D=1$ is considered the same as the choices $A=4, B=1, C=3, D=2.$)

$\textbf{(A) }30 \qquad \textbf{(B) }60 \qquad \textbf{(C) }90 \qquad \textbf{(D) }180 \qquad \textbf{(E) }360$

## Problem16

In the following list of numbers, the integer $n$ appears $n$ times in the list for $1 \leq n \leq 200$.$$1, 2, 2, 3, 3, 3, 4, 4, 4, 4, \ldots, 200, 200, \ldots , 200$$What is the median of the numbers in this list?

$\textbf{(A)} ~100.5 \qquad\textbf{(B)} ~134 \qquad\textbf{(C)} ~142 \qquad\textbf{(D)} ~150.5 \qquad\textbf{(E)} ~167$

## Problem17

Trapezoid $ABCD$ has $\overline{AB}\parallel\overline{CD},BC=CD=43$, and $\overline{AD}\perp\overline{BD}$. Let $O$ be the intersection of the diagonals $\overline{AC}$ and $\overline{BD}$, and let $P$ be the midpoint of $\overline{BD}$. Given that $OP=11$, the length of $AD$ can be written in the form $m\sqrt{n}$, where $m$ and $n$ are positive integers and $n$ is not divisible by the square of any prime. What is $m+n$?

$\textbf{(A) }65 \qquad \textbf{(B) }132 \qquad \textbf{(C) }157 \qquad \textbf{(D) }194\qquad \textbf{(E) }215$

## Problem18

Let $f$ be a function defined on the set of positive rational numbers with the property that $f(a\cdot b) = f(a)+f(b)$ for all positive rational numbers $a$ and $b$. Furthermore, suppose that $f$ also has the property that $f(p)=p$ for every prime number $p$. For which of the following numbers $x$ is $f(x) < 0$?

$\textbf{(A) }\frac{17}{32}\qquad\textbf{(B) }\frac{11}{16}\qquad\textbf{(C) }\frac{7}{9}\qquad\textbf{(D) }\frac{7}{6}\qquad\textbf{(E) }\frac{25}{11}\qquad$

## Problem19

The area of the region bounded by the graph of$$x^2+y^2 = 3|x-y| + 3|x+y|$$is $m+n\pi$, where $m$ and $n$ are integers. What is $m + n$?

$\textbf{(A)} ~18\qquad\textbf{(B)} ~27\qquad\textbf{(C)} ~36\qquad\textbf{(D)} ~45\qquad\textbf{(E)} ~54$

## Problem20

In how many ways can the sequence $1,2,3,4,5$ be rearranged so that no three consecutive terms are increasing and no three consecutive terms are decreasing?

$\textbf{(A)} ~10\qquad\textbf{(B)} ~18\qquad\textbf{(C)} ~24 \qquad\textbf{(D)} ~32 \qquad\textbf{(E)} ~44$

## Problem21

Let $ABCDEF$ be an equiangular hexagon. The lines $AB, CD,$ and $EF$ determine a triangle with area $192\sqrt{3}$, and the lines $BC, DE,$ and $FA$ determine a triangle with area $324\sqrt{3}$. The perimeter of hexagon $ABCDEF$ can be expressed as $m +n\sqrt{p}$, where $m, n,$ and $p$ are positive integers and $p$ is not divisible by the square of any prime. What is $m + n + p$?

$\textbf{(A)} ~47\qquad\textbf{(B)} ~52\qquad\textbf{(C)} ~55\qquad\textbf{(D)} ~58\qquad\textbf{(E)} ~63$

## Problem22

Hiram's algebra notes are $50$ pages long and are printed on $25$ sheets of paper; the first sheet contains pages $1$ and $2$, the second sheet contains pages $3$ and $4$, and so on. One day he leaves his notes on the table before leaving for lunch, and his roommate decides to borrow some pages from the middle of the notes. When Hiram comes back, he discovers that his roommate has taken a consecutive set of sheets from the notes and that the average (mean) of the page numbers on all remaining sheets is exactly $19$. How many sheets were borrowed?

$\textbf{(A)} ~10\qquad\textbf{(B)} ~13\qquad\textbf{(C)} ~15\qquad\textbf{(D)} ~17\qquad\textbf{(E)} ~20$

## Problem23

Frieda the frog begins a sequence of hops on a $3 \times 3$ grid of squares, moving one square on each hop and choosing at random the direction of each hop-up, down, left, or right. She does not hop diagonally. When the direction of a hop would take Frieda off the grid, she "wraps around" and jumps to the opposite edge. For example if Frieda begins in the center square and makes two hops "up", the first hop would place her in the top row middle square, and the second hop would cause Frieda to jump to the opposite edge, landing in the bottom row middle square. Suppose Frieda starts from the center square, makes at most four hops at random, and stops hopping if she lands on a corner square. What is the probability that she reaches a corner square on one of the four hops?

$\textbf{(A)} ~\frac{9}{16}\qquad\textbf{(B)} ~\frac{5}{8}\qquad\textbf{(C)} ~\frac{3}{4}\qquad\textbf{(D)} ~\frac{25}{32}\qquad\textbf{(E)} ~\frac{13}{16}$

## Problem24

The interior of a quadrilateral is bounded by the graphs of $(x+ay)^2 = 4a^2$ and $(ax-y)^2 = a^2$, where $a$ a positive real number. What is the area of this region in terms of $a$, valid for all $a > 0$?

$\textbf{(A)} ~\frac{8a^2}{(a+1)^2}\qquad\textbf{(B)} ~\frac{4a}{a+1}\qquad\textbf{(C)} ~\frac{8a}{a+1}\qquad\textbf{(D)} ~\frac{8a^2}{a^2+1}\qquad\textbf{(E)} ~\frac{8a}{a^2+1}$

## Problem25

How many ways are there to place $3$ indistinguishable red chips, $3$ indistinguishable blue chips, and $3$ indistinguishable green chips in the squares of a $3 \times 3$ grid so that no two chips of the same color are directly adjacent to each other, either vertically or horizontally.

$\textbf{(A)} ~12\qquad\textbf{(B)} ~18\qquad\textbf{(C)} ~24\qquad\textbf{(D)} ~30\qquad\textbf{(E)} ~36$

## Solution 1

Call the different colors A,B,C. There are $3!=6$ ways to rearrange these colors to these three letters, so $6$ must be multiplied after the letters are permuted in the grid. WLOG assume that A is in the center.$$\begin{tabular}{ c c c } ? & ? & ? \\ ? & A & ? \\ ? & ? & ? \end{tabular}$$In this configuration, there are two cases, either all the A's lie on the same diagonal:$$\begin{tabular}{ c c c } ? & ? & A \\ ? & A & ? \\ A & ? & ? \end{tabular}$$or all the other two A's are on adjacent corners:$$\begin{tabular}{ c c c } A & ? & A \\ ? & A & ? \\ ? & ? & ? \end{tabular}$$In the first case there are two ways to order them since there are two diagonals, and in the second case there are four ways to order them since there are four pairs of adjacent corners.

In each case there is only one way to put the three B's and the three C's as shown in the diagrams.$$\begin{tabular}{ c c c } C & B & A \\ B & A & C \\ A & C & B \end{tabular}$$$$\begin{tabular}{ c c c } A & B & A \\ C & A & C \\ B & C & B \end{tabular}$$This means that there are $4+2=6$ ways to arrange A,B, and C in the grid, and there are 6 ways to rearrange the colors. Therefore, there are $6\cdot6=36$ ways in total, which is $\boxed{\text{E}}$.

-happykeeper

## Solution 2 (Casework)

Without the loss of generality, we place a red ball in the top-left square. There are two cases:

Case (1): The two balls adjacent to the top-left red ball have different colors.$$\begin{tabular}{|c|c|c|} \hline R & B & ? \\ \hline G & R & ? \\ \hline ? & ? & ? \\ \hline \end{tabular}$$Each placement has $6$ permutations, as there are $3!=6$ ways to permute RBG.

There are three sub-cases for Case (1):$$\begin{tabular}{ccccccc} \begin{tabular}{|c|c|c|} \hline R & B & R \\ \hline G & R & G \\ \hline B & G & B \\ \hline \end{tabular} & & & \begin{tabular}{|c|c|c|} \hline R & B & G \\ \hline G & R & B \\ \hline R & B & G \\ \hline \end{tabular} & & & \begin{tabular}{|c|c|c|} \hline R & B & G \\ \hline G & R & B \\ \hline B & G & R \\ \hline \end{tabular} \end{tabular}$$So, Case (1) has $3\cdot6=18$ ways.

Case (2): The two balls adjacent to the top-left red ball have the same color.$$\begin{tabular}{|c|c|c|} \hline R & B & ? \\ \hline B & ? & ? \\ \hline ? & ? & ? \\ \hline \end{tabular}$$Each placement has $6$ permutations, as there are $\binom32\binom21=6$ ways to choose three balls consisting of exactly two colors (RBB, RGG, BRR, BGG, GRR, GBB). There are three sub-cases for Case (2):$$\begin{tabular}{ccccccc} \begin{tabular}{|c|c|c|} \hline R & B & R \\ \hline B & G & B \\ \hline G & R & G \\ \hline \end{tabular} & & & \begin{tabular}{|c|c|c|} \hline R & B & G \\ \hline B & G & R \\ \hline R & B & G \\ \hline \end{tabular} & & & \begin{tabular}{|c|c|c|} \hline R & B & G \\ \hline B & G & R \\ \hline G & R & B \\ \hline \end{tabular} \end{tabular}$$So, Case (2) has $3\cdot6=18$ ways.

Together, the answer is $18+18=\boxed{\textbf{(E)} ~36}.$

~MRENTHUSIASM

## Solution 3 (Casework and Derangements)

Case (1): We have a permutation of R, B, and G as all of the rows. There are $3!$ ways to rearrange these three colors. After finishing the first row, we move onto the second. Notice how the second row must be a derangement of the first one. By the derangement formula, $\frac{3!}{e} \approx 2$, so there are two possible permutations of the second row. (Note: You could have also found the number of derangements of PIE). Finally, there are $2$possible permutations for the last row. Thus, there are $3!\cdot2\cdot2=24$ possibilities.

Case (2): All of the rows have two balls that are the same color and one that is different. There are obviously $3$ possible configurations for the first row, $2$ for the second, and $2$ for the third. $3\cdot2\cdot2=12$.

Therefore, our answer is $24+12=\boxed{\textbf{(E)} ~36}.$

2

## Solution 1

The following system of equations can be formed with $p$ representing the number of students in Portia's high school and $l$ representing the number of students in Lara's high school.$$p=3q$$$$p+q=2600$$Substituting $p$ with $3q$ we get $4q=2600$. Solving for $q$, we get $q=650$. Since we need to find $p$ we multiply $650$ by 3 to get $p=1950$, which is $\boxed{\text{C}}$

-happykeeper

## Solution 2 (One Variable)

Suppose Lara's high school has $x$ students. It follows that Portia's high school has $3x$ students. We know that $x+3x=2600,$ or $4x=2600.$Our answer is$$3x=2600\left(\frac 34\right)=650(3)=\boxed{\textbf{(C)} ~1950}.$$

~MRENTHUSIASM

## Solution 3 (Arithmetics)

Clearly, $2600$ students is $4$ times as many students as Lara's high school. Therefore, Lara's high school has $2600\div4=650$ students, and Portia's high school has $650\cdot3=\boxed{\textbf{(C)} ~1950}$ students.

~MRENTHUSIASM

### Solution 4.1 (Quick Inspection)

The number of students in Portia's high school must be a multiple of $3.$ This eliminates $\textbf{(B)},\textbf{(D)},$ and $\textbf{(E)}$. Since $\textbf{(A)}$ is too small (as $600+600/3<2600$ is clearly true), we are left with $\boxed{\textbf{(C)} ~1950}.$

~MRENTHUSIASM

### Solution 4.2 (Plug in the Answer Choices)

For $\textbf{(A)},$ we have $600+\frac{600}{3}=800\neq2600.$ So, $\textbf{(A)}$ is incorrect.

For $\textbf{(B)},$ we have $650+\frac{650}{3}=866\frac{2}{3}\neq2600.$ So, $\textbf{(B)}$ is incorrect.

For $\textbf{(C)},$ we have $1950+\frac{1950}{3}=2600.$ So, $\boxed{\textbf{(C)} ~1950}$ is correct. For completeness, we will check choices $\textbf{(D)}$ and $\textbf{(E)}.$

For $\textbf{(D)},$ we have $2000+\frac{2000}{3}=2666\frac{2}{3}\neq2600.$ So, $\textbf{(D)}$ is incorrect.

For $\textbf{(E)},$ we have $2050+\frac{2050}{3}=2733\frac{1}{3}\neq2600.$ So, $\textbf{(E)}$ is incorrect.

3.

## Solution 1

The units digit of a multiple of $10$ will always be $0$. We add a $0$ whenever we multiply by $10$. So, removing the units digit is equal to dividing by $10$.

Let the smaller number (the one we get after removing the units digit) be $a$. This means the bigger number would be $10a$.

We know the sum is $10a+a = 11a$ so $11a=17402$. So $a=1582$. The difference is $10a-a = 9a$. So, the answer is $9(1582) = 14238 = \boxed{\textbf{(D)}}$.

--abhinavg0627

## Solution 2 (Lazy Speed)

Since the ones place of a multiple of $10$ is $0$, this implies the other integer has to end with a $2$ since both integers sum up to a number that ends with a $2$. Thus, the ones place of the difference has to be $10-2=8$, and the only answer choice that ends with an $8$ is $\boxed{\textbf{(D)}~14238}$

~CoolJupiter 2021

## Solution 3 (Vertical Addition and Logic)

Let the larger number be $\overline{AB,CD0}.$ It follows that the smaller number is $\overline{A,BCD}.$ Adding vertically, we have$$\begin{array}{cccccc} & A & B & C & D & 0 \\ + & & A & B & C & D \\ \hline & 1 & 7 & 4 & 0 & 2 \\ \end{array}$$Working from right to left, we have$$D=2\Rightarrow C=8 \Rightarrow B=5 \Rightarrow A=1.$$The larger number is $15,820$ and the smaller number is $1,582.$ Their difference is $15,820-1,582=\boxed{\textbf{(D)} ~14{,}238}.$

4.

## Solution 1 (Arithmetic Series)

Since$$\text{Distance}=\text{Speed}\times\text{Time},$$we seek the sum$$5(1)+12(1)+19(1)+26(1)+\cdots=5+12+19+26+\cdots,$$in which there are 30 addends. The last addend is $5+7(30-1)=208.$ Therefore, the requested sum is$$5+12+19+26+\cdots+208=\frac{(5+208)(30)}{2}=\boxed{\textbf{(D)} ~3195}.$$Recall that to find the sum of an arithmetic series, we take the average of the first and last terms, then multiply by the number of terms, namely$$\frac{\text{First}+\text{Last}}{2}\cdot\text{Count}.$$~MRENTHUSIASM

## Solution 2 (Answer Choices and Modular Arithmetic)

From the $30$-term sum$$5+12+19+26+\cdots$$in the previous solution, taking modulo $10$ gives$$5+12+19+26+\cdots \equiv 3(0+1+2+3+4+5+6+7+8+9) = 3(45)\equiv5 \pmod{10}.$$The only answer choices that are $5\mod{10}$ are $\textbf{(A)}$ and $\textbf{(D)}.$ By a quick estimate, $\textbf{(A)}$ is too small, leaving us with $\boxed{\textbf{(D)} ~3195}.$

5.

## Solution 1 (Generalized)

The total score in the class is $8k.$ The total score on the $12$ quizzes is $12\cdot14=168.$ Therefore, for the remaining quizzes ($k-12$ of them), the total score is $8k-168.$ Their mean score is $\boxed{\textbf{(B)} ~\frac{8k-168}{k-12}}.$

~MRENTHUSIASM

## Solution 2 (Convenient Values and Observations)

Set $k=13.$ The answer is the same as the last student's quiz score, which is $8\cdot13-14\cdot12<0.$ From the answer choices, only $\boxed{\textbf{(B)} ~\frac{8k-168}{k-12}}$ yields a negative value for $k=13.$

6.

## Solution 1 (Generalized Distance)

Let $2d$ miles be the distance from the start to the fire tower. When Chantal meets Jean, she has traveled for$$\frac d4 + \frac d2 + \frac d3 = d\left(\frac 14 + \frac 12 + \frac 13\right) =d\left(\frac{3}{12} + \frac{6}{12} + \frac{4}{12}\right)=\frac{13}{12}d$$hours. Jean also has traveled for $\frac{13}{12}d$ hours, and he travels for $d$ miles. So, his average speed is$$\frac{d}{\left(\frac{13}{12}d\right)}=\boxed{\textbf{(A)} ~\frac{12}{13}}$$miles per hour.

~MRENTHUSIASM

## Solution 2 (Convenient Distance)

We use the same template as shown in Solution 1, except that we replace $d$ with a concrete number.

Let $24$ miles be the distance from the start to the fire tower. When Chantal meets Jean, she travels for$$\frac{12}{4} + \frac{12}{2}+\frac{12}{3}=3+6+4=13$$hours. Jean also has traveled for $13$ hours, and he travels for $12$ miles. So, his average speed is $\boxed{\textbf{(A)} ~\frac{12}{13}}$ miles per hour.

7.

## Solution 1

We know that purple snakes cannot subtract, thus they cannot add either. Since happy snakes must be able to add, the purple snakes cannot be happy. Therefore, we know that the happy snakes are not purple and the answer is $\boxed{\textbf{(D)}}$.

--abhinavg0627

## Solution 2 (Explains Solution 1 Using Arrows)

We are given that

$\text{(1) Happy}\Rightarrow\text{can add}$

$\text{(2) Purple}\Rightarrow\text{cannot subtract}$

$\text{(3) Cannot subtract}\Rightarrow\text{cannot add}$

Combining $\text{(2)}$ and $\text{(3)}$ into $\text{(*)}$ below, we have

$\text{(1) Happy}\Rightarrow\text{can add}$

$\text{(*) Purple}\Rightarrow\text{cannot subtract}\Rightarrow\text{cannot add}$

Clearly, the answer is $\boxed{\textbf{(D)}}.$

8.

## Solution

It is known that $0.\overline{ab}=\frac{ab}{99}$ and $0.ab=\frac{ab}{100}$. Let $\overline {ab} = x$. We have that $66(1+\frac{x}{100})+0.5=66(1+\frac{x}{99})$. Solving gives that $100x-75=99x$ so $x=\boxed{\text{(E)} 75}$. ~aop2014

9.

## Solution 1

Expanding, we get that the expression is $x^2+2xy+y^2+x^2y^2-2xy+1$ or $x^2+y^2+x^2y^2+1$. By the trivial inequality(all squares are nonnegative) the minimum value for this is $\boxed{\text{(D)} 1}$, which can be achieved at $x=y=0$. ~aop2014

## Solution 2 (Beyond Overkill)

Like solution 1, expand and simplify the original equation to $x^2+y^2+x^2y^2+1$ and let $f(x, y) = x^2+y^2+x^2y^2+1$. To find local extrema, find where $\nabla f(x, y) = \boldsymbol{0}$. First, find the first partial derivative with respect to x and y and find where they are $0$:$$\frac{\partial f}{\partial x} = 2x + 2xy^{2} = 2x(1 + y^{2}) = 0 \implies x = 0$$$$\frac{\partial f}{\partial y} = 2y + 2yx^{2} = 2y(1 + x^{2}) = 0 \implies y = 0$$

Thus, there is a local extreme at $(0, 0)$. Because this is the only extreme, we can assume that this is a minimum because the problem asks for the minimum (though this can also be proven using the partial second derivative test) and the global minimum since it's the only minimum, meaning $f(0, 0)$ is the minimum of $f(x, y)$. Plugging $(0, 0)$ into $f(x, y)$, we find 1 $\implies \boxed{\bold{(D)} \ 1}$

10.

## Solution 1

All you need to do is multiply the entire equation by $(3-2)$. Then all the terms will easily simplify by difference of squares and you will get $3^{128}-2^{128}$ or $\boxed{C}$ as your final answer. Notice you don't need to worry about $3-2$ because that's equal to $1$.

-Lemonie

## Solution 2

If you weren't able to come up with the $(3 - 2)$ insight, then you could just notice that the answer is divisible by $(2 + 3) = 5$, and $(2^2 + 3^2) = 13$. We can then use Fermat's Little Theorem for $p = 5, 13$ on the answer choices to determine which of the answer choices are divisible by both $5$ and $13$. This is $\boxed{C}$.

-MEWTO

## Solution 3

After expanding the first few terms, the result after each term appears to be $2^{2^n-1} + 2^{2^n-2}\cdot{3^1} + 2^{2^n-3}\cdot{3^2} + ... + 2^1\cdot{3^{2^n-2}} + 3^{2^n-1}$where n is the number of terms expanded. We can prove this using mathematical induction. The base step is trivial. When expanding another term, all of the previous terms multiplied by $2^{2^{n-1}}$ would give $2^{2^n-1} + 2^{2^n-2}\cdot{3^1} + 2^{2^n-3}\cdot{3^2} + ... + 2^{2^{n-1}+1}\cdot{3^{2^{n-1}-1}} + 2^{2^{n-1}}\cdot{3^{2^{n-1}}}$, and all the previous terms multiplied by $3^{2^{n-1}}$ would give $3^{2^n-1} + 3^{2^n-2}\cdot{2^1} + 3^{2^n-3}\cdot{2^2} + ... + 3^{2^{n-1}+1}\cdot{2^{2^{n-1}-1}} + 3^{2^{n-1}}\cdot{2^{2^{n-1}}}$. Their sum is equal to $2^{2^n-1} + 2^{2^n-2}\cdot{3^1} + 2^{2^n-3}\cdot{3^2} + ... + 2^1\cdot{3^{2^n-2}} + 3^{2^n-1}$, so the proof is complete. Since $\frac{3^{2^n}-2^{2^n}}{3-2}$ is equal to $2^{2^n-1} + 2^{2^n-2}\cdot{3^1} + 2^{2^n-3}\cdot{3^2} + ... + 2^1\cdot{3^{2^n-2}} + 3^{2^n-1}$, the answer is $\frac{3^{2^7}-2^{2^7}}{3-2}=\boxed{C}$.

-SmileKat32

## Solution 4 (Engineer's Induction)

We can compute some of the first few partial products, and notice that $\prod_{k = 0}^{2^n} (2^{2^n}+3^{2^n}) = 3^{2^{n+1}} - 2^{2^{n+1}}$. As we don't have to prove this, we get the product is $3^{2^7} - 2^{2^7} = 3^{128} - 2^{128}$, and smugly click $\boxed{\textbf{(C)} ~3^{128} - 2^{128}}$. ~rocketsri

10.

## Solution 1

All you need to do is multiply the entire equation by $(3-2)$. Then all the terms will easily simplify by difference of squares and you will get $3^{128}-2^{128}$ or $\boxed{C}$ as your final answer. Notice you don't need to worry about $3-2$ because that's equal to $1$.

-Lemonie

## Solution 2

If you weren't able to come up with the $(3 - 2)$ insight, then you could just notice that the answer is divisible by $(2 + 3) = 5$, and $(2^2 + 3^2) = 13$. We can then use Fermat's Little Theorem for $p = 5, 13$ on the answer choices to determine which of the answer choices are divisible by both $5$ and $13$. This is $\boxed{C}$.

-MEWTO

## Solution 3

After expanding the first few terms, the result after each term appears to be $2^{2^n-1} + 2^{2^n-2}\cdot{3^1} + 2^{2^n-3}\cdot{3^2} + ... + 2^1\cdot{3^{2^n-2}} + 3^{2^n-1}$where n is the number of terms expanded. We can prove this using mathematical induction. The base step is trivial. When expanding another term, all of the previous terms multiplied by $2^{2^{n-1}}$ would give $2^{2^n-1} + 2^{2^n-2}\cdot{3^1} + 2^{2^n-3}\cdot{3^2} + ... + 2^{2^{n-1}+1}\cdot{3^{2^{n-1}-1}} + 2^{2^{n-1}}\cdot{3^{2^{n-1}}}$, and all the previous terms multiplied by $3^{2^{n-1}}$ would give $3^{2^n-1} + 3^{2^n-2}\cdot{2^1} + 3^{2^n-3}\cdot{2^2} + ... + 3^{2^{n-1}+1}\cdot{2^{2^{n-1}-1}} + 3^{2^{n-1}}\cdot{2^{2^{n-1}}}$. Their sum is equal to $2^{2^n-1} + 2^{2^n-2}\cdot{3^1} + 2^{2^n-3}\cdot{3^2} + ... + 2^1\cdot{3^{2^n-2}} + 3^{2^n-1}$, so the proof is complete. Since $\frac{3^{2^n}-2^{2^n}}{3-2}$ is equal to $2^{2^n-1} + 2^{2^n-2}\cdot{3^1} + 2^{2^n-3}\cdot{3^2} + ... + 2^1\cdot{3^{2^n-2}} + 3^{2^n-1}$, the answer is $\frac{3^{2^7}-2^{2^7}}{3-2}=\boxed{C}$.

-SmileKat32

## Solution 4 (Engineer's Induction)

We can compute some of the first few partial products, and notice that $\prod_{k = 0}^{2^n} (2^{2^n}+3^{2^n}) = 3^{2^{n+1}} - 2^{2^{n+1}}$. As we don't have to prove this, we get the product is $3^{2^7} - 2^{2^7} = 3^{128} - 2^{128}$, and smugly click $\boxed{\textbf{(C)} ~3^{128} - 2^{128}}$. ~rocketsri

11.

## Solution

We have$$2021_b - 221_b = 2000_b - 200_b = 2b^3 - 2b^2 = 2b^2(b-1).$$This expression is divisible by $3$ unless $b\equiv2\pmod{3}.$ The only choice congruent to $2$ modulo $3$ is $\boxed{\textbf{(E)} ~8}.$

12.

## Solution 1 (Use Tables to Organize Information)

Initial Scenario

$$\begin{array}{cccc} & \textbf{Base} & \textbf{Height} & \textbf{Volume} \\ [2ex] \textbf{Narrow Cone} & 3 & h_1 & \frac13\pi(3)^2h_1=3\pi h_1 \\ [2ex] \textbf{Wide Cone} & 6 & h_2 & \frac13\pi(6)^2h_2=12\pi h_2 \end{array}$$By similar triangles:

For the narrow cone, the ratio of base radius to height is $\frac{3}{h_1},$ which remains constant.

For the wide cone, the ratio of base radius to height is $\frac{6}{h_2},$ which remains constant.

Equating the initial volumes gives $3\pi h_1=12\pi h_2,$ which simplifies to $\frac{h_1}{h_2}=4.$

Final Scenario (Two solutions follow from here.)

### Solution 1.1 (Fraction Trick)

Let the base radii of the narrow cone and the wide cone be $3x$ and $6y,$ respectively, where $x,y>1.$ We have the following table:$$\begin{array}{cccc} & \textbf{Base} & \textbf{Height} & \textbf{Volume} \\ [2ex] \textbf{Narrow Cone} & 3x & h_1x & \frac13\pi(3x)^2h_1=3\pi h_1 x^3 \\ [2ex] \textbf{Wide Cone} & 6y & h_2y & \frac13\pi(6y)^2h_2=12\pi h_2 y^3 \end{array}$$

Equating the final volumes gives $3\pi h_1 x^3=12\pi h_2 y^3,$ which simplifies to $x^3=y^3,$ or $x=y.$

Lastly, the requested ratio is$$\frac{h_1 x - h_1}{h_2 y - h_2}=\frac{h_1 (x-1)}{h_2 (y-1)}=\frac{h_1}{h_2}=\boxed{\textbf{(E) }4}.$$

PS:

1. This problem uses the following fraction trick:

For unequal positive numbers $a,b,c$ and $d,$ if $\frac ab = \frac cd = k,$ then $\frac{a\pm c}{b\pm d}=k.$

Quick Proof

From $\frac ab = \frac cd = k,$ we know that $a=bk$ and $c=dk$. Therefore,$$\frac{a\pm c}{b\pm d}=\frac{bk\pm dk}{b\pm d}=\frac{(b\pm d)k}{b\pm d}=k.$$

2. The work above shows that, regardless of the shape or the volume of the solid dropped in, as long as the solid sinks to the bottom and is completely submerged without spilling any liquid, the answer will remain unchanged.

~MRENTHUSIASM

### Solution 1.2 (Bash)

Let the base radii of the narrow cone and the wide cone be $r_1$ and $r_2,$ respectively.

Let the rises of the liquid levels of the narrow cone and the wide cone be $\Delta h_1$ and $\Delta h_2,$ respectively. We have the following table:$$\begin{array}{cccc} & \textbf{Base} & \textbf{Height} & \textbf{Volume} \\ [2ex] \textbf{Narrow Cone} & r_1 & h_1+\Delta h_1 & \frac13\pi r_1^2(h_1+\Delta h_1) \\ [2ex] \textbf{Wide Cone} & r_2 & h_2+\Delta h_2 & \frac13\pi r_2^2(h_2+\Delta h_2) \end{array}$$

By similar triangles discussed above, we have$$\begin{array}{cccc} \frac{3}{h_1}=\frac{r_1}{h_1+\Delta h_1} &\Rightarrow &r_1=\frac{3}{h_1}(h_1+\Delta h_1) & \ \ \ \ \ \ \ (1) \\ [2ex] \frac{6}{h_2}=\frac{r_2}{h_2+\Delta h_2} &\Rightarrow &r_2=\frac{6}{h_2}(h_2+\Delta h_2) & \ \ \ \ \ \ \ (2) \end{array}$$

The volume of the marble dropped in is $\frac43\pi(1)^3=\frac43\pi.$

Now, we set up an equation for the volume of the narrow cone and solve for $\Delta h_1:$\begin{align*} \frac13\pi r_1^2(h_1+\Delta h_1) &= 3\pi h_1+\frac43\pi \\ \frac13\pi{\underbrace{\left(\frac{3}{h_1}(h_1+\Delta h_1)\right)}_{\text{by (1)}}}^2(h_1+\Delta h_1) &= 3\pi h_1+\frac43\pi \\ \frac{3}{h_1^2}(h_1+\Delta h_1)^3 &= 3h_1+\frac43 \\ (h_1+\Delta h_1)^3 &= h_1^3 + \frac{4h_1^2}{9} \\ \Delta h_1 &= \sqrt[3]{h_1^3 + \frac{4h_1^2}{9}}-h_1. \end{align*}

Next, we set up an equation for the volume of the wide cone $\Delta h_2:$$$\frac13\pi r_2^2(h_2+\Delta h_2) = 12\pi h_2+\frac43\pi.$$Using the exact same process from above (but with different numbers), we get$$\Delta h_2 = \sqrt[3]{h_2^3+\frac{h_2^2}{9}}-h_2.$$Recall that $\frac{h_1}{h_2}=4.$ Therefore, the requested ratio is\begin{align*} \frac{\Delta h_1}{\Delta h_2}&=\frac{\sqrt[3]{h_1^3 + \frac{4h_1^2}{9}}-h_1}{\sqrt[3]{h_2^3+\frac{h_2^2}{9}}-h_2} \\ &=\frac{\sqrt[3]{(4h_2)^3 + \frac{4(4h_2)^2}{9}}-4h_2}{\sqrt[3]{h_2^3+\frac{h_2^2}{9}}-h_2} \\ &=\frac{\sqrt[3]{4^3\left(h_2^3 + \frac{h_2^2}{9}\right)}-4h_2}{\sqrt[3]{h_2^3+\frac{h_2^2}{9}}-h_2} \\ &=\frac{4\sqrt[3]{h_2^3+\frac{h_2^2}{9}}-4h_2}{\sqrt[3]{h_2^3+\frac{h_2^2}{9}}-h_2} \\ &=\boxed{\textbf{(E) }4}. \end{align*}

~MRENTHUSIASM

## Solution 2 (Quick and dirty)

The heights of the cones are not given, so suppose the heights are very large (i.e. tending towards infinity) in order to approximate the cones as cylinders with base radii 3 and 6 and infinitely large height. Then the base area of the wide cylinder is 4 times that of the narrow cylinder. Since we are dropping a ball of the same volume into each cylinder, the water level in the narrow cone/cylinder should rise $\boxed{\textbf{(E) } 4}$ times as much.

13.

## Solution

Drawing the tetrahedron out and testing side lengths, we realize that the triangles ABD and ABC are right triangles. It is now easy to calculate the volume of the tetrahedron using the formula for the volume of a pyramid: $\frac{3\cdot4\cdot2}{3\cdot2}=4$, so we have an answer of $\boxed{C}$. ~IceWolf10

14.

## Solution 1:

By Vieta's formulae, the sum of the 6 roots is 10 and the product of the 6 roots is 16. By inspection, we see the roots are 1, 1, 2, 2, 2, and 2, so the function is $(z-1)^2(z-2)^4=(z^2-2z+1)(z^4-8z^3+24z^2-32z+16)$. Therefore, $B = -32 - 48 - 8 = \boxed{\textbf{(A)} -88}$. ~JHawk0224

## Solution 2:

Using the same method as Solution 1, we find that the roots are $2, 2, 2, 2, 1,$ and $1$. Note that $B$ is the negation of the 3rd symmetric sum of the roots. Using casework on the number of 1's in each of the $\binom {6}{3} = 20$ products $r_a \cdot r_b \cdot r_c,$ we obtain$$B= - \left(\binom {4}{3} \binom {2}{0} \cdot 2^{3} + \binom {4}{2} \binom{2}{1} \cdot 2^{2} \cdot 1 + \binom {4}{1} \binom {2}{2} \cdot 2 \right) = -\left(32+48+8 \right) = \boxed{\textbf{(A)} -88}.$$

15.

## Solution 1 (Intuition):

Visualizing the two curves, we realize they are both parabolas with the same axis of symmetry. Now assume that the first equation is above the second, since order doesn't matter. Then $C>A$ and $B>D$. Therefore the number of ways to choose the four integers is $\tbinom{6}{2}\tbinom{4}{2}=90$, and the answer is $\boxed{C}$. ~IceWolf10

## Solution 2 (Algebra):

Setting $y = Ax^2+B = Cx^2+D$, we find that $Ax^2-Cx^2 = x^2(A-C) = D-B$, so $x^2 = \frac {D-B}{A-C} \ge 0$ by the trivial inequality. This implies that $D-B$ and $A-C$ must both be positive or negative. If two distinct values are chosen for $(A, C)$ and $(B, D)$ respectively, there are $2$ ways to order them so that both the numerator and denominator are positive/negative (increasing and decreasing). We must divide by $2$ at the end, however, since the $2$ curves aren't considered distinct. Calculating, we get$$\frac {1}{2} \cdot \binom {6}{2} \binom {4}{2} \cdot 2 = \boxed{\textbf{(C) }90}.$$

16.

## Solution 1

There are $1+2+..+199+200=\frac{(200)(201)}{2}=20100$ numbers in total. Let the median be $k$. We want to find the median $k$ such that$$\frac{k(k+1)}{2}=20100/2,$$or$$k(k+1)=20100.$$Note that $\sqrt{20100} \approx 142$. Plugging this value in as $k$ gives$$\frac{1}{2}(142)(143)=10153.$$$10153-142<10050$, so $142$ is the $152$nd and $153$rd numbers, and hence, our desired answer. $\fbox{(C) 142}$.

Note that we can derive $\sqrt{20100} \approx 142$ through the formula$$\sqrt{n} = \sqrt{a+b} \approx \sqrt{a} + \frac{b}{2\sqrt{a} + 1},$$where $a$ is a perfect square less than or equal to $n$. We set $a$ to $19600$, so $\sqrt{a} = 140$, and $b = 500$. We then have $n \approx 140 + \frac{500}{2(140)+1} \approx 142$. ~approximation by ciceronii

## Solution 2

The $x$th number of this sequence is $\left\lceil\frac{-1\pm\sqrt{1+8x}}{2}\right\rceil$ via the quadratic formula. We can see that if we halve $x$ we end up getting $\left\lceil\frac{-1\pm\sqrt{1+4x}}{2}\right\rceil$. This is approximately the number divided by $\sqrt{2}$$\frac{200}{\sqrt{2}} = 141.4$ and since $142$ looks like the only number close to it, it is answer $\boxed{(C) 142}$ ~Lopkiloinm

We can look at answer choice $C$, which is $142$ first. That means that the number of numbers from $1$ to $142$ is roughly the number of numbers from $143$ to $200$.

The number of numbers from $1$ to $142$ is $\frac{142(142+1)}{2}$ which is approximately $10000.$ The number of numbers from $143$ to $200$ is $\frac{200(200+1)}{2}-\frac{142(142+1)}{2}$ which is approximately $10000$ as well. Therefore, we can be relatively sure the answer choice is $\boxed{(C) \text{ } 142}.$

17.

## Solution 1

Angle chasing reveals that $\triangle BPC\sim\triangle BDA$, therefore$$2=\frac{BD}{BP}=\frac{AB}{BC}=\frac{AB}{43}$$$$AB=86$$Additional angle chasing shows that $\triangle ABO \sim\triangle CDO$, therefore$$2=\frac{AB}{CD}=\frac{BP}{PD}=\frac{\frac{BD}{2}+11}{\frac{BD}{2}-11}$$$$BD=66$$Since $\triangle ADB$ is right, the Pythagorean theorem implies that$$AD=\sqrt{86^2-66^2}$$$$AD=4\sqrt{190}$$$4\sqrt{190}\implies 4 + 190 = \boxed{\textbf{D) } 194}$

~mn28407

## Solution 2 (One Pair of Similar Triangles, then Areas)

Since $\triangle BCD$ is isosceles with legs $\overline{CB}$ and $\overline{CD},$ it follows that the median $\overline{CP}$ is also an altitude of $\triangle BCD.$ Let $DO=x$ and $CP=h.$We have $PB=x+11.$

Since $\triangle ADO\sim\triangle CPO$ by AA, we have$$AD=CP\cdot\frac{DO}{PO}=h\cdot\frac{x}{11}.$$

Let the brackets denote areas. Notice that $[ADO]=[BCO]$ (By the same base/height, $[ADC]=[BCD].$ Subtracting $[OCD]$ from both sides gives $[ADO]=[BCO].$). Doubling both sides, we have\begin{align*} 2[ADO]&=2[BCO] \\ \frac{x^2 h}{11}&=(x+22)h \\ x^2&=11x+11\cdot22 \\ (x-22)(x+11)&=0 \\ x&=22. \end{align*}

In $\triangle CPB,$ we have$$h=\sqrt{43^2-33^2}=\sqrt{76\cdot10}=2\sqrt{190}$$and$$AD=h\cdot\frac{x}{11}=4\sqrt{190}.$$Finally, $4+190=\boxed{\textbf{(D) }194}.$

~MRENTHUSIASM

## Solution 3 (short)

Let $CP = y$ and $CP$ is perpendicular bisector of $DB.$ Let $DO = x,$ so $DP = PB = 11+x.$

(1) $\triangle CPO \sim \triangle ADO,$ so we get $\frac{AD}{x} = \frac{y}{11},$ or $AD = \frac{xy}{11}.$

(2) pythag on $\triangle CDP$ gives $(11+x)^2 + y^2 = 43^2.$

(3) $\triangle BPC \sim \triangle BDA$ with ratio $1:2,$ so $AD = 2y.$

Thus, $xy/11 = 2y,$ or $x = 22.$ And $y = \sqrt{43^2 - 33^2} = 2 \sqrt{190},$ so $AD = 4 \sqrt{190}$ and the answer is $\boxed{194}.$

~ ccx09

## Solution 4 - Extending the line

Observe that $\triangle BPC$ is congruent to $\triangle DPC$; both are similar to $\triangle BDA$. Let's extend $\overline{AD}$ and $\overline{BC}$ past points $D$ and $C$ respectively, such that they intersect at a point $E$. Observe that $\angle BDE$ is $90$ degrees, and that $\angle DBE \cong \angle PBC \cong \angle DBA \implies \angle DBE \cong \angle DBA$. Thus, by ASA, we know that $\triangle ABD \cong \triangle EBD$, thus, $AD = ED$, meaning $D$ is the midpoint of $AE$. Let $M$ be the midpoint of $\overline{DE}$. Note that $\triangle CME$ is congruent to $\triangle BPC$, thus $BC = CE$, meaning $C$ is the midpoint of $\overline{BE}.$

Therefore, $\overline{AC}$ and $\overline{BD}$ are both medians of $\triangle ABE$. This means that $O$ is the centroid of $\triangle ABE$; therefore, because the centroid divides the median in a 2:1 ratio, $\frac{BO}{2} = DO = \frac{BD}{3}$. Recall that $P$ is the midpoint of $BD$$DP = \frac{BD}{2}$. The question tells us that $OP = 11$$DP-DO=11$; we can write this in terms of $DB$$\frac{DB}{2}-\frac{DB}{3} = \frac{DB}{6} = 11 \implies DB = 66$.

We are almost finished. Each side length of $\triangle ABD$ is twice as long as the corresponding side length $\triangle CBP$ or $\triangle CPD$, since those triangles are similar; this means that $AB = 2 \cdot 43 = 86$. Now, by Pythagorean theorem on $\triangle ABD$$AB^{2} - BD^{2} = AD^{2} \implies 86^{2}-66^{2} = AD^{2} \implies AD = \sqrt{3040} \implies AD = 4 \sqrt{190}$$4+190 = \boxed{194, \textbf{D}}$

18.

## Solution 1

Looking through the solutions we can see that $f(\frac{25}{11})$ can be expressed as $f(\frac{25}{11} \cdot 11) = f(11) + f(\frac{25}{11})$ so using the prime numbers to piece together what we have we can get $10=11+f(\frac{25}{11})$, so $f(\frac{25}{11})=-1$ or $\boxed{E}$.

-Lemonie

$f(\frac{25}{11} \cdot 11) = f(25) = f(5) + f(5) = 10$

- awesomediabrine

## Solution 2

We know that $f(p) = f(p \cdot 1) = f(p) + f(1)$. By transitive, we have$$f(p) = f(p) + f(1).$$Subtracting $f(p)$ from both sides gives $0 = f(1).$ Also$$f(2)+f\left(\frac{1}{2}\right)=f(1)=0 \implies 2+f\left(\frac{1}{2}\right)=0 \implies f\left(\frac{1}{2}\right) = -2$$$$f(3)+f\left(\frac{1}{3}\right)=f(1)=0 \implies 3+f\left(\frac{1}{3}\right)=0 \implies f\left(\frac{1}{3}\right) = -3$$$$f(11)+f\left(\frac{1}{11}\right)=f(1)=0 \implies 11+f\left(\frac{1}{11}\right)=0 \implies f\left(\frac{1}{11}\right) = -11$$In $\textbf{(A)}$ we have $f\left(\frac{17}{32}\right)=17+5f\left(\frac{1}{2}\right)=17-5(2)=7$.

In $\textbf{(B)}$ we have $f\left(\frac{11}{16}\right)=11+4f\left(\frac{1}{2}\right)=11-4(2)=3$.

In $\textbf{(C)}$ we have $f\left(\frac{7}{9}\right)=7+2f\left(\frac{1}{3}\right)=7-2(3)=1$.

In $\textbf{(D)}$ we have $f\left(\frac{7}{6}\right)=7+f\left(\frac{1}{2}\right)+f\left(\frac{1}{3}\right)=7-2-3=2$.

In $\textbf{(E)}$ we have $f\left(\frac{25}{11}\right)=10+f\left(\frac{1}{11}\right)=10-11=-1$.

Thus, our answer is $\boxed{\textbf{(E)} \frac{25}{11}}$

~JHawk0224 ~awesomediabrine

## Solution 3 (Deeper)

Consider the rational $\frac{a}{b}$, for $a,b$ integers. We have $f(a)=f\left(\frac{a}{b}\cdot b\right)=f\left(\frac{a}{b}\right)+f(b)$. So $f\left(\frac{a}{b}\right)=f(a)-f(b)$. Let $p$ be a prime. Notice that $f(p^k)=kf(p)$. And $f(p)=p$. So if $a=p_1^{a_1}p_2^{a_2}\cdots p_k^{a_k}$$f(a)=a_1p_1+a_2p_2+....+a_kp_k$. We simply need this to be greater than what we have for $f(b)$. Notice that for answer choices $A,B,C,$ and $D$, the numerator $(a)$ has less prime factors than the denominator, and so they are less likely to work. We check $E$ first, and it works, therefore the answer is $\boxed{\textbf{(E)}}$.

~yofro

## Solution 4 (Most Comprehensive, Similar to Solution 3)

We have the following important results:

$(1) \ f\left(\prod_{k=1}^{n}a_k\right)=\sum_{k=1}^{n}f(a_k)$ for all positive integers $k$

$(2) \ f\left(a^n\right)=nf(a)$ for all positive rational numbers $a$

$(3) \ f(1)=0$

$(4) \ f\left({\frac 1a}\right)=-f(a)$ for all positive rational numbers $a$

Proofs

Result $(1)$ can be shown by induction.

Result $(2):$ Since powers are just repeated multiplication, we will use result $(1)$ to prove result $(2):$$$f\left(a^n\right)=f\left(\prod_{k=1}^{n}a\right)=\sum_{k=1}^{n}f(a)=nf(a).$$

Result $(3):$ For all positive rational numbers $a,$ we have$$f(a)=f(a\cdot1)=f(a)+f(1).$$Therefore, we get $f(1)=0.$ So, result $(3)$ is true.

Result $(4):$ For all positive rational numbers $a,$ we have$$f(a)+f\left(\frac1a\right)=f\left(a\cdot\frac1a\right)=f(1)=0.$$It follows that $f\left({\frac 1a}\right)=-f(a),$ and result $(4)$ is true.

For all positive integers $x$ and $y,$ suppose $\prod_{k=1}^{m}p_k^{e_k}$ and $\prod_{k=1}^{n}q_k^{d_k}$ are their prime factorizations, respectively, we have\begin{align*} f\left(\frac xy\right)&=f(x)+f\left(\frac 1y\right) \\ &=f(x)-f(y) \\ &=f\left(\prod_{k=1}^{m}p_k^{e_k}\right)-f\left(\prod_{k=1}^{n}q_k^{d_k}\right) \\ &=\left[\sum_{k=1}^{m}f\left(p_k^{e_k}\right)\right]-\left[\sum_{k=1}^{n}f\left(q_k^{d_k}\right)\right] \\ &=\left[\sum_{k=1}^{m}e_k f\left(p_k\right)\right]-\left[\sum_{k=1}^{n}d_k f\left(q_k\right)\right] \\ &=\left[\sum_{k=1}^{m}e_k p_k \right]-\left[\sum_{k=1}^{n}d_k q_k \right]. \end{align*}

We apply function $f$ on each fraction in the choices:

$$\begin{array}{cccccccc} \textbf{(A) } & f\left(\frac{17}{32}\right) & = & f\left(\frac{17^1}{2^5}\right) & = & [1(17)]-[5(2)] & = & 7 \\ [2ex] \textbf{(B) } & f\left(\frac{11}{16}\right) & = & f\left(\frac{11^1}{2^4}\right) & = & [1(11)]-[4(2)] & = & 3 \\ [2ex] \textbf{(C) } & f\left(\frac{7}{9}\right) & = & f\left(\frac{7^1}{3^2}\right) & = & [1(7)]-[2(3)] & = & 1 \\ [2ex] \textbf{(D) } & f\left(\frac{7}{6}\right) & = & f\left(\frac{7^1}{2^1\cdot3^1}\right) & = & [1(7)]-[1(2)+1(3)] & = & 2 \\ [2ex] \textbf{(E) } & f\left(\frac{25}{11}\right) & = & f\left(\frac{5^2}{11^1}\right) & = & [2(5)]-[1(11)] & = & -1. \end{array}$$Therefore, the answer is $\boxed{\textbf{(E) }\frac{25}{11}}.$

~MRENTHUSIASM

## Solution 5

The problem gives us that f(p)=p. If we let a=p and b=1, we get f(p)=f(p)+f(1), which implies f(1)=0. Notice that the answer choices are all fractions, which means we will have to multiply an integer by a fraction to be able to solve it. Therefore, let's try plugging in fractions and try to solve them. Note that if we plug in a=p and b=1/p, we get f(1)=f(p)+f(1/p). We can solve for f(1/p) as -f(p)! This gives us the information we need to solve the problem. Testing out the answer choices gives us the answer of E.

19.

## Solution 1

In order to attack this problem, we need to consider casework:

Case 1: $|x-y|=x-y, |x+y|=x+y$

Substituting and simplifying, we have $x^2-6x+y^2=0$, i.e. $(x-3)^2+y^2=3^2$, which gives us a circle of radius $3$ centered at $(3,0)$.

Case 2: $|x-y|=y-x, |x+y|=x+y$

Substituting and simplifying again, we have $x^2+y^2-6y=0$, i.e. $x^2+(y-3)^2=3^2$. This gives us a circle of radius $3$ centered at $(0,3)$.

Case 3: $|x-y|=x-y, |x+y|=-x-y$

Doing the same process as before, we have $x^2+y^2+6y=0$, i.e. $x^2+(y+3)^2=3^2$. This gives us a circle of radius $3$ centered at $(0,-3)$.

Case 4: $|x-y|=y-x, |x+y|=-x-y$

One last time: we have $x^2+y^2+6x=0$, i.e. $(x+3)^2+y^2=3^2$. This gives us a circle of radius $3$ centered at $(-3,0)$.

After combining all the cases and drawing them on the Cartesian Plane, this is what the diagram looks like:

$[asy] size(10cm); Label f; f.p=fontsize(7); xaxis(-8,8,Ticks(f, 1.0)); yaxis(-8,8,Ticks(f, 1.0)); draw(arc((-3,0),3,90,270) -- cycle, gray); draw(arc((0,3),3,0,180) -- cycle, gray); draw(arc((3,0),3,-90,90) -- cycle, gray); draw(arc((0,-3),3,-180,0) -- cycle, gray); draw((-3,3)--(3,3)--(3,-3)--(-3,-3)--cycle, grey); [/asy]$Now, the area of the shaded region is just a square with side length $6$ with four semicircles of radius $3$. The area is $6\cdot6+4\cdot \frac{9\pi}{2} = 36+18\pi$. The answer is $36+18$ which is $\boxed{\textbf{(E) }54}$

20.

## Solution 1 (Bashing)

We write out the $120$ cases. These cases are the ones that work: $13254,14253,14352,15243,15342,21435,21534,23154,24153,24351,25143,25341,\linebreak 31425,31524,32415,32451,34152,34251,35142,35241,41325,41523,42315,42513,\linebreak 43512,45132,45231,51324,51423,52314,52413,53412. \linebreak$ We count these out and get $\boxed{\text{D: }32}$permutations that work. ~contactbibliophile

## Solution 2 (Casework)

Reading the terms from left to right, we have two cases:

$\text{Case \#1: }+,-,+,-$

$\text{Case \#2: }-,+,-,+$

($+$ stands for increase and $-$ stands for decrease.)

For $\text{Case \#1},$ note that for the second and fourth terms, one of which must be a $5,$ and the other one must be a $3$ or $4.$ We have four sub-cases:

$(1) \ \underline{\hspace{3mm}}3\underline{\hspace{3mm}}5\underline{\hspace{3mm}}$

$(2) \ \underline{\hspace{3mm}}5\underline{\hspace{3mm}}3\underline{\hspace{3mm}}$

$(3) \ \underline{\hspace{3mm}}4\underline{\hspace{3mm}}5\underline{\hspace{3mm}}$

$(4) \ \underline{\hspace{3mm}}5\underline{\hspace{3mm}}4\underline{\hspace{3mm}}$

For $(1),$ the first two blanks must be $1$ and $2$ in some order, and the last blank must be a $4,$ for a total of $2$ possibilities. Similarly, $(2)$ also has $2$possibilities.

For $(3),$ there are no restrictions for the numbers $1, 2,$ and $3.$ So, we have $3!=6$ possibilities. Similarly, $(4)$ also has $6$ possibilities.

Together, $\text{Case \#1}$ has $2+2+6+6=16$ possibilities. By symmetry, $\text{Case \#2}$ also has $16$ possibilities. Together, the answer is $16+16=\boxed{\textbf{(D)} ~32}.$

21.

## Solution (Misplaced problem?)

Note that the extensions of the given lines will determine an equilateral triangle because the hexagon is equiangular. The area of the first triangle is $192\sqrt{3}$, so the side length is $\sqrt{192\cdot 4}=16\sqrt{3}$. The area of the second triangle is $324\sqrt{3}$, so the side length is $\sqrt{4\cdot 324}=36$. We can set the first value equal to $AB+CD+EF$ and the second equal to $BC+DE+FA$ by substituting some lengths in with different sides of the same equilateral triangle. The perimeter of the hexagon is just the sum of these two, which is $16\sqrt{3}+36$ and $16+3+36=\boxed{55~\textbf{(C)}}$

22.

## Solution

Suppose the roommate took pages $a$ through $b$, or equivalently, page numbers $2a-1$ through $2b$. Because there are $(2b-2a+2)$ numbers taken,$$\frac{(2a-1+2b)(2b-2a+2)}{2}+19(50-(2b-2a+2))=\frac{50*51}{2} \implies (2a+2b-39)(b-a+1)=\frac{50*13}{2}=25*13.$$The first possible solution that comes to mind is if $2a+2b-39=25, b-a+1=13 \implies a+b=32, b-a=12$, which indeed works, giving $b=22$ and $a=10$. The answer is $22-10+1=\boxed{(\textbf{B})13}$

~Lcz

## Solution 2 (Different Variable Choice, Similar Logic)

Suppose the smallest page number removed is $k,$ and $n$ pages are removed. It follows that the largest page number removed is $k+n-1.$

Remarks:

1. $n$ pages are removed means that $\frac{n}{2}$ sheets are removed, from which $n$ must be even.

2. $k$ must be odd, as the smallest page number removed is on the right side (odd-numbered).

3. $1+2+3+\cdots+50=\frac{51(50)}{2}=1275.$

4. The sum of the page numbers removed is $\frac{(2k+n-1)n}{2}.$

Together, we have\begin{align*} \frac{1275-\frac{(2k+n-1)n}{2}}{50-n}&=19 \\ 1275-\frac{(2k+n-1)n}{2}&=19(50-n) \\ 2550-(2k+n-1)n&=38(50-n) \\ 2550-(2k+n-1)n&=1900-38n \\ 650&=(2k+n-39)n. \end{align*}The factors of $650$ are$$1,2,5,10,13,25,26,50,65,130,325,650.$$Since $n$ is even, we only have a few cases to consider:

$$\begin{array}{ c c c } \boldsymbol{n} & \boldsymbol{2k+n-39} & \boldsymbol{k} \\ \hline 2 & 325 & 181 \\ 10 & 65 & 47 \\ 26 & 25 & 19 \\ 50 & 13 & 1 \\ 130 & 5 & \text{negative} \\ 650 & 1 & \text{negative} \\ \end{array}$$

Since $1\leq k \leq 50,$ only $k=47,19,1$ are possible:

If $k=47,$ then the notebook will run out if we take $10$ pages starting from page $47.$

If $k=1,$ then the average page number of the remaining pages will be undefined, as there is no page remaining (after taking $50$ pages starting from page $1$).

So, the only possibility is $k=19,$ from which $n=26$ pages are taken out, which is $\frac n2=\boxed{\textbf{(B)} ~13}$ sheets.

23.

## Solution 1 (complementary counting)

We will use complementary counting. First, the frog can go left with probability $\frac14$. We observe symmetry, so our final answer will be multiplied by 4 for the 4 directions, and since $4\cdot \frac14=1$, we will ignore the leading probability.

From the left, she either goes left to another edge ($\frac14$) or back to the center ($\frac14$). Time for some casework.

$\textbf{Case 1:}$ She goes back to the center.

Now, she can go in any 4 directions, and then has 2 options from that edge. This gives $\frac12$. --End case 1

$\textbf{Case 2:}$ She goes to another edge (rightmost).

Subcase 1: She goes back to the left edge. She now has 2 places to go, giving $\frac12$

Subcase 2: She goes to the center. Now any move works.

$\frac14 \cdot \frac12 + \frac14 \cdot 1=\frac18 + \frac 14=\frac38$ for this case. --End case 2

She goes back to the center in Case 1 with probability $\frac14$, and to the right edge with probability $\frac14$

So, our answer is $\frac14 \cdot \frac12 + \frac14 \cdot \frac38=\frac14 (\frac12+\frac38)=\frac14 \cdot \frac78 = \frac7{32}$

But, don't forget complementary counting. So, we get $1-\frac7{32}=\frac{25}{32} \implies \boxed{D}$. ~ firebolt360

Video Solution for those who prefer: https://youtu.be/ude2rzO1cmk ~ firebolt360

## Solution 2 (direct counting and probability states)

We can draw a state diagram with three states: center, edge, and corner. Denote center by M, edge by E, and corner by C. There are a few ways Frieda can reach a corner in four or less moves: EC, EEC, EEEC, EMEC. Then, calculating the probabilities of each of these cases happening, we have $1\cdot\tfrac{1}{2}+1\cdot\tfrac{1}{4}\cdot\tfrac{1}{2}+1\cdot\tfrac{1}{4}\cdot\tfrac{1}{4}\cdot\tfrac{1}{2}+1\cdot\tfrac{1}{4}\cdot1\cdot\tfrac{1}{2}=\tfrac{25}{32}$, so the answer is $\boxed{D}$. ~IceWolf10

## Solution 3 (Similar to Solution 2, but Finds the Numerator and Denominator Separately)

Denominator

There are $4^4=256$ ways to make $4$ hops without restrictions.

Numerator (Casework)

Suppose Frieda makes $4$ hops without stopping. We perform casework on which hop reaches a corner for the first time.

$(1)$ Hop $\#2$ (Hops $\#3$ and $\#4$ have no restrictions)

The $4$ independent hops have $4, 2, 4, 4$ options, respectively. So, this case has $4\cdot2\cdot4\cdot4=128$ ways.

$(2)$ Hop $\#3$ (Hop $\#4$ has no restriction)

No matter which direction the first hop takes, the second hop must "wrap around".

The $4$ independent hops have $4, 1, 2, 4$ options, respectively. So, this case has $4\cdot1\cdot2\cdot4=32$ ways.

$(3)$ Hop $\#4$

Two sub-cases:

$(3.1)$ The second hop "wraps around". It follows that the third hop also "wraps around".

The $4$ independent hops have $4, 1, 1, 2$ options, respectively. So, this sub-case has $4\cdot1\cdot1\cdot2=8$ ways.

$(3.2)$ The second hop backs to the center.

The $4$ independent hops have $4, 1, 4, 2$ options, respectively. So, this sub-case has $4\cdot1\cdot4\cdot2=32$ ways.

Together, Case $(3)$ has $8+32=40$ ways.

The numerator is $128+32+40=200.$

Probability

$\frac{200}{256}=\boxed{\textbf{(D)} ~\frac{25}{32}}.$

This problem is quite similar to 1995 AIME Problem 3: https://artofproblemsolving.com/wiki/index.php/1995_AIME_Problems/Problem_3

~MRENTHUSIASM

## Solution 4

Let $C_n$ be the probability that Frieda is on the central square after n moves, $E_n$ be the probability that Frieda is on one of the four squares on the middle of the edges after n moves, and $V_n$ (V for vertex) be the probability that Frieda is on a corner after n moves. The only way to reach the center is by moving in $1$ specific direction out of $4$ total directions from the middle of an edge, so $C_{n+1}=\frac{E_n}{4}$. The ways to reach the middle of an edge are by moving in any direction from the center or by moving in $1$ specific direction from the middle of an edge, so $E_{n+1}=C_n+\frac{E_n}{4}$. The ways to reach a corner are by simply staying there after reaching there in a previous move or by moving in $2$ specific directions from the middle of an edge, so $V_{n+1}=V_n+\frac{E_n}{2}$. Since Frieda always start from the center, $C_0=1$$E_0=0$, and $V_0=0$. We use the previous formulas to work out $V_4$ and find it to be $\boxed{\textbf{(D)} ~\frac{25}{32}}$.

## Solution 5

Imagine an infinite grid of $2$ by $2$ squares such that there is a $2$ by $2$ square centered at $(3x, 3y)$ for all ordered pairs of integers $(x, y).$

$[asy] dot((0,0)); draw((-1,-1)--(-1,1)--(1,1)--(1,-1)--cycle); draw((-4,-1)--(-4,1)--(-2,1)--(-2,-1)--cycle); draw((2,-1)--(2,1)--(4,1)--(4,-1)--cycle); draw((-1,2)--(-1,4)--(1,4)--(1,2)--cycle); draw((-1,-4)--(-1,-2)--(1,-2)--(1,-4)--cycle); draw((-4,2)--(-2,2)--(-2,4)--(-4,4)--cycle); draw((4,-2)--(2,-2)--(2,-4)--(4,-4)--cycle); draw((4,2)--(2,2)--(2,4)--(4,4)--cycle); draw((-4,-2)--(-2,-2)--(-2,-4)--(-4,-4)--cycle); draw((-3,-3)--(3,-3)--(3,3)--(-3,3)--cycle); draw((-4,0)--(4,0)); draw((0,4)--(0,-4)); [/asy]$

It is easy to see that the problem is equivalent to Frieda moving left, right, up, or down on this infinite grid starting at $(0, 0)$. (minus the teleportations) Since counting the complement set is easier, we'll count the number of $4$-step paths such that Frieda never reaches a corner point.

In other words, since the reachable corner points are $(\pm 1, \pm 1), (\pm 1, \pm 2), (\pm 2, \pm 1),$ and $(\pm 2, \pm 2),$ Frieda can only travel along the collection of points included in $S$, where $S$ is all points on $x=0$ and $y=0$ such that $|y|<4$ and $|x|<4$, respectively, plus all points on the big square with side length $6$ centered at $(0, 0).$ We then can proceed with casework:

Case $1$: Frieda never reaches $(0, \pm 3)$ nor $(\pm 3, 0).$

When Frieda only moves horizontally or vertically for her four moves, she can do so in $2^4 - 4 = 12$ ways for each case . Thus, $12 \cdot 2$ total paths for the subcase of staying in one direction. (For instance, all length $4$ combinations of $F$ and $B$ except $FFFF$$BBBB$$FFFB$, and $BBBF$ for the horizontal direction.)

There is another subcase where she changes directions during her path. There are four symmetric cases for this subcase depending on which of the four quadrants Frieda hugs. For the first quadrant, the possible paths are $FBUD$$FBUU$$UDFB$, and $UDFF.$ Thus, a total of $4 \cdot 4 = 16$ ways for this subcase.

Total for Case $1$$24 + 16 = 40$

Case $2$: Frieda reaches $(0, \pm 3)$ or $(\pm 3, 0)$.

Once Frieda reaches one of the points listed above (by using three moves), she has four choices for her last move. Thus, a total of $4 \cdot 4 = 16$paths for this case.

Our total number of paths never reaching coroners is thus $16+40=56,$ making for an answer of$$\frac{4^4-56}{4^4} = \boxed{\textbf{(D)} ~\frac{25}{32}}.$$

-fidgetboss_4000

## Solution 6 (Casework)

We take cases on the number of hops needed to reach a corner. For simplicity, denote $E$ as a move that takes Frieda to an edge, $W$ as wrap-around move and $C$ as a corner move. Also, denote $O$ as a move that takes us to the center.

2 Hops

Then, Frieda will have to $(E, C)$ as her set of moves. There are $4$ ways to move to an edge, and $2$ corners to move to, for a total of $4 \cdot 2 = 8$cases here. Then, there are $4$ choices for each move, for a probability of $\frac{8}{4 \cdot 4} = \frac{1}{2}$.

3 Hops

In this case, Frieda must wrap-around. There's only one possible combination, just $(E, W, C)$. There are $4$ ways to move to an edge, $1$ way to wrap-around (you must continue in the same direction) and $2$ corners, for a total of $4 \cdot 1 \cdot 2 = 8$ cases here. Then, there are $4$ choices for each move, for a probability of $\frac{8}{4 \cdot 4 \cdot 4} = \frac{1}{8}$.

4 Hops

Lastly, there are two cases we must consider here. The first case is $(E, O, E, C)$, and the second is $(E, W, W, C)$. For the first case, there are $4$ ways to move to an edge, $1$ way to return to the center, $4$ ways to move to an edge once again, and $2$ ways to move to a corner. Hence, there is a total of $4 \cdot 1 \cdot 4 \cdot 2 = 32$ cases here. Then, for the second case, there are $4$ ways to move to a corner, $1$ way to wrap-around, $1$ way to wrap-around again, and $2$ ways to move to a corner. This implies there are $4 \cdot 1 \cdot 1 \cdot 2 = 8$ cases here. Then, there is a total of $8+32 = 40$ cases, out of a total of $4^4 = 256$ cases, for a probability of $\frac{40}{256} = \frac{5}{32}$.

Then, the total probability that Frieda ends up on a corner is $\frac{1}{2} + \frac{1}{8} + \frac{5}{32} = \frac{25}{32}$, corresponding to choice $\boxed{\textbf{(D)} ~\frac{25}{32}}$. ~rocketsri

## Solution 7

I denote 3x3 grid by

- HOME square (x1)

- CORN squares (x4)

- SIDE squares (x4)

Transitions:

- HOME always move to SIDE

- CORN is DONE

- SIDE move to CORN with $p=1/4,$ move to SIDE with $p=1/4,$ and move to CORN with $p=1/2.$

After one move, will be on $\text{SIDE}$ square

After two moves, will be $1/2 + 1/4 (\text{HOME}) + 1/4 (\text{SIDE})$

After three moves, will be $1/2 + 1/4 (\text{SIDE}) + 1/4 (1/2 + 1/4(\text{SIDE}) + 1/4(\text{HOME}))$

After four moves, probability on CORN will be $1/2 + 1/4 (1/2) + 1/4( 1/2 + 1/4(1/2)) = 1/2 + 1/8 + 5/32 = 25 / 32.$

24.

## Solution 1

The conditions $(x+ay)^2 = 4a^2$ and $(ax-y)^2 = a^2$ give $|x+ay| = |2a|$ and $|ax-y| = |a|$ or $x+ay = \pm 2a$ and $ax-y = \pm a$. The slopes here are perpendicular, so the quadrilateral is a rectangle. Plug in $a=1$ and graph it. We quickly see that the area is $2\sqrt{2} \cdot \sqrt{2} = 4$, so the answer can't be $A$ or $B$ by testing the values they give (test it!). Now plug in $a=2$. We see using a ruler that the sides of the rectangle are about $\frac74$ and $\frac72$. So the area is about $\frac{49}8 = 6.125$. Testing $C$ we get $\frac{16}3$ which is clearly less than $6$, so it is out. Testing $D$ we get $\frac{32}5$ which is near our answer, so we leave it. Testing $E$ we get $\frac{16}5$, way less than $6$, so it is out. So, the only plausible answer is $\boxed{D}$ ~firebolt360

## Solution 2 (Casework)

For the equation $(x+ay)^2 = 4a^2,$ the cases are

$(1) \ x+ay=2a.$ This is a line with $x$-intercept $2a,y$-intercept $2,$ and slope $-\frac 1a.$

$(2) \ x+ay=-2a.$ This is a line with $x$-intercept $-2a,y$-intercept $-2,$ and slope $-\frac 1a.$

For the equation $(ax-y)^2 = a^2,$ the cases are

$(1') \ ax-y=a.$ This is a line with $x$-intercept $1,y$-intercept $-a,$ and slope $a.$

$(2') \ ax-y=-a.$ This is a line with $x$-intercept $-1,y$-intercept $a,$ and slope $a.$

Plugging $a=2$ into the choices gives

$\textbf{(A)} ~\frac{32}{9}\qquad\textbf{(B)} ~\frac{8}{3}\qquad\textbf{(C)} ~\frac{16}{3}\qquad\textbf{(D)} ~\frac{32}{5}\qquad\textbf{(E)} ~\frac{16}{5}$

Plugging $a=2$ into the four above equations and solving systems of equations for intersecting lines [$(1)$ and $(1'), (1)$ and $(2'), (2)$ and $(1'), (2)$ and $(2')$], we get the respective solutions$$(x,y)=\left(\frac 85, \frac 65\right), (0,2), \left(-\frac 85, -\frac 65\right), (0,-2).$$

### Solution 2.1 (Rectangle)

Since the slopes of the intersecting lines (from the four above equations) are negative reciprocals, the quadrilateral is a rectangle. Finally, by the Distance Formula, the length and width of the rectangle are $\frac{8\sqrt5}{5}$ and $\frac{4\sqrt5}{5}.$ The area we seek is$$\left(\frac{8\sqrt5}{5}\right)\left(\frac{4\sqrt5}{5}\right)=\frac{32}{5}.$$

The answer is $\boxed{\textbf{(D)} ~\frac{8a^2}{a^2+1}}.$

~MRENTHUSIASM

### Solution 2.2 (Shoelace Formula)

Even if we do not recognize that the solutions form the vertices of a rectangle, we can apply the Shoelace Formula on consecutive vertices\begin{align*} (x_1,y_1) &= \left(\frac 85, \frac 65\right), \\ (x_2,y_2) &= (0,2), \\ (x_3,y_3) &= \left(-\frac 85, -\frac 65\right), \\ (x_4,y_4) &= (0,-2). \end{align*}

The area formula is\begin{align*} A &= \frac{1}{2} \left|(x_1y_2 + x_2y_3 + x_3y_4 + x_4y_1) - (y_1x_2 + y_2x_3 + y_3x_4 + y_4x_1)\right| \\ &= \frac{1}{2} \left|\left[\frac85\cdot2+0\cdot\left(-\frac65\right)+\left(-\frac{8}{5}\right)\cdot(-2)+0\cdot\frac65\right] - \left[\frac65\cdot0+2\cdot\left(-\frac85\right)+\left(-\frac65\right)\cdot0+(-2)\cdot\frac85\right]\right| \\ &= \frac{1}{2} \left|\left[\frac{16}{5}+\frac{16}{5}\right]-\left[-\frac{16}{5}-\frac{16}{5}\right]\right| \\ &= \frac{1}{2} \left|\frac{64}{5}\right| \\ &= \frac{32}{5}. \end{align*}Therefore, the answer is $\boxed{\textbf{(D)} ~\frac{8a^2}{a^2+1}}.$

Suggested Reading for the Shoelace Formula: https://artofproblemsolving.com/wiki/index.php/Shoelace_Theorem

~MRENTHUSIASM

## Solution 3 (Geometry)

Similar to Solution 2, we will use the equations of the four cases:

(1) $x+ay=2a.$ This is a line with $x$-intercept $2a$$y$-intercept $2$, and slope $-\frac 1a.$

(2) $x+ay=-2a.$ This is a line with $x$-intercept $-2a$$y$-intercept $-2$, and slope $-\frac 1a.$

(3)* $ax-y=a.$ This is a line with $x$-intercept $1$$y$-intercept $-a$, and slope $a.$

(4)* $ax-y=-a.$ This is a line with $x$-intercept $-1$$y$-intercept $a$, and slope $a.$

The area of the rectangle created by the four equations can be written as $2a\cdot \cos A\cdot4\sin A$

$8a\cos A \cdot \sin A$

$8a\cdot~\frac{1}{\sqrt{a^2+1}}\cdot~\frac{a}{\sqrt{a^2+1}}$

$\boxed{\textbf{(D)} ~\frac{8a^2}{a^2+1}}.$

(Note: $\tan A=$ slope $a$)

-fnothing4994

## Solution 4 (bruh moment solution)

Trying $a = 1$ narrows down the choices to options $\textbf{(C)}$$\textbf{(D)}$ and $\textbf{(E)}$. Trying $a = 2$ and $a = 3$ eliminates $\textbf{(C)}$ and $\textbf{(E)}$, to obtain $\boxed{\textbf{(D)} ~\frac{8a^2}{a^2+1}}.$ as our answer. -¢

25.

## Solution 1

Call the different colors A,B,C. There are $3!=6$ ways to rearrange these colors to these three letters, so $6$ must be multiplied after the letters are permuted in the grid. WLOG assume that A is in the center.$$\begin{tabular}{ c c c } ? & ? & ? \\ ? & A & ? \\ ? & ? & ? \end{tabular}$$In this configuration, there are two cases, either all the A's lie on the same diagonal:$$\begin{tabular}{ c c c } ? & ? & A \\ ? & A & ? \\ A & ? & ? \end{tabular}$$or all the other two A's are on adjacent corners:$$\begin{tabular}{ c c c } A & ? & A \\ ? & A & ? \\ ? & ? & ? \end{tabular}$$In the first case there are two ways to order them since there are two diagonals, and in the second case there are four ways to order them since there are four pairs of adjacent corners.

In each case there is only one way to put the three B's and the three C's as shown in the diagrams.$$\begin{tabular}{ c c c } C & B & A \\ B & A & C \\ A & C & B \end{tabular}$$$$\begin{tabular}{ c c c } A & B & A \\ C & A & C \\ B & C & B \end{tabular}$$This means that there are $4+2=6$ ways to arrange A,B, and C in the grid, and there are 6 ways to rearrange the colors. Therefore, there are $6\cdot6=36$ ways in total, which is $\boxed{\text{E}}$.

-happykeeper

## Solution 2 (Casework)

Without the loss of generality, we place a red ball in the top-left square. There are two cases:

Case (1): The two balls adjacent to the top-left red ball have different colors.$$\begin{tabular}{|c|c|c|} \hline R & B & ? \\ \hline G & R & ? \\ \hline ? & ? & ? \\ \hline \end{tabular}$$Each placement has $6$ permutations, as there are $3!=6$ ways to permute RBG.

There are three sub-cases for Case (1):$$\begin{tabular}{ccccccc} \begin{tabular}{|c|c|c|} \hline R & B & R \\ \hline G & R & G \\ \hline B & G & B \\ \hline \end{tabular} & & & \begin{tabular}{|c|c|c|} \hline R & B & G \\ \hline G & R & B \\ \hline R & B & G \\ \hline \end{tabular} & & & \begin{tabular}{|c|c|c|} \hline R & B & G \\ \hline G & R & B \\ \hline B & G & R \\ \hline \end{tabular} \end{tabular}$$So, Case (1) has $3\cdot6=18$ ways.

Case (2): The two balls adjacent to the top-left red ball have the same color.$$\begin{tabular}{|c|c|c|} \hline R & B & ? \\ \hline B & ? & ? \\ \hline ? & ? & ? \\ \hline \end{tabular}$$Each placement has $6$ permutations, as there are $\binom32\binom21=6$ ways to choose three balls consisting of exactly two colors (RBB, RGG, BRR, BGG, GRR, GBB). There are three sub-cases for Case (2):$$\begin{tabular}{ccccccc} \begin{tabular}{|c|c|c|} \hline R & B & R \\ \hline B & G & B \\ \hline G & R & G \\ \hline \end{tabular} & & & \begin{tabular}{|c|c|c|} \hline R & B & G \\ \hline B & G & R \\ \hline R & B & G \\ \hline \end{tabular} & & & \begin{tabular}{|c|c|c|} \hline R & B & G \\ \hline B & G & R \\ \hline G & R & B \\ \hline \end{tabular} \end{tabular}$$So, Case (2) has $3\cdot6=18$ ways.

Together, the answer is $18+18=\boxed{\textbf{(E)} ~36}.$

~MRENTHUSIASM

## Solution 3 (Casework and Derangements)

Case (1): We have a permutation of R, B, and G as all of the rows. There are $3!$ ways to rearrange these three colors. After finishing the first row, we move onto the second. Notice how the second row must be a derangement of the first one. By the derangement formula, $\frac{3!}{e} \approx 2$, so there are two possible permutations of the second row. (Note: You could have also found the number of derangements of PIE). Finally, there are $2$possible permutations for the last row. Thus, there are $3!\cdot2\cdot2=24$ possibilities.

Case (2): All of the rows have two balls that are the same color and one that is different. There are obviously $3$ possible configurations for the first row, $2$ for the second, and $2$ for the third. $3\cdot2\cdot2=12$.

Therefore, our answer is $24+12=\boxed{\textbf{(E)} ~36}.$