2021 AMC 12A Fall Solutions | LIVE by Po-Shen Loh

1.

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

$7$

$21$

$49$

$64$

$91$

Concepts:factoring perfect square

Difficulty rating: 890

Small Hint:

$2112 - 2021 = 91$

Big Hint:

$91 = 7 \times 13$ and $169 = 13^2$

Solution:

Since $2112 - 2021 = 91 = 7 \cdot 13$ and $169 = 13^2,$ the fraction is $\dfrac{(7 \cdot 13)^2}{13^2} = 7^2 = 49.$

Thus, the correct answer is C.

2.

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?

$16$

$17$

$18$

$19$

$20$

Concepts:rectangle area

Difficulty rating: 1020

Small Hint:

Shortening a side by $1$ to get area $18$ means $3 \times 6 = 18,$ so the $4$ -inch side was the one reduced

Big Hint:

Instead reduce the $6$ -inch side by $1,$ giving $4 \times 5$

Solution:

The original card is $4 \times 6.$ Shortening a side by $1$ inch gives area $18,$ which requires $3 \times 6 = 18,$ so the reduced side was the $4$ -inch side.

Shortening the other side by $1$ inch instead gives $4 \times 5 = 20$ square inches.

Thus, the correct answer is E.

3.

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 $\tfrac{1}{2}$ -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?

$2 \tfrac{3}{4}$

$3 \tfrac{3}{4}$

$4 \tfrac{1}{2}$

$5 \tfrac{1}{2}$

$6 \tfrac{3}{4}$

Concepts:distance rate and time unit conversion

Difficulty rating: 1130

Small Hint:

Route A takes $\dfrac{6}{30}$ hour; convert to minutes

Big Hint:

Route B splits into $4.5$ miles at $40$ mph and $0.5$ mile at $20$ mph

Solution:

Route A takes $\dfrac{6}{30} = \dfrac{1}{5}$ hour $= 12$ minutes.

Route B has $4.5$ miles at $40$ mph and $0.5$ mile at $20$ mph, taking $\dfrac{4.5}{40} + \dfrac{0.5}{20}$ $= 0.1125 + 0.025$ $= 0.1375$ hour $= 8.25$ minutes.

The difference is $12 - 8.25 = 3.75 = 3\tfrac{3}{4}$ minutes.

Thus, the correct answer is B.

4.

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 $A?$

$1$

$3$

$5$

$7$

$9$

Concepts:prime divisibility

Difficulty rating: 1200

Small Hint:

The number is $202100 + A;$ even $A$ makes it even, and $A = 5$ makes it a multiple of $5$

Big Hint:

For $A = 1$ and $A = 7$ the digit sum is a multiple of $3,$ and $202103 = 11 \cdot 18373$

Solution:

The number is $202100 + A.$ Any even $A$ makes it even, and $A = 5$ makes it divisible by $5,$ so $A$ must be odd and not $5.$

For $A = 1$ the digit sum is $6$ (divisible by $3$ ); for $A = 7$ the digit sum is $12$ (divisible by $3$ ); and $202103 = 11 \cdot 18373.$ Only $202109$ survives all tests, and it is prime.

Thus, the correct answer is E.

5.

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?

$6$

$8$

$10$

$11$

$15$

Concepts:fencepost counting ratio and proportion

Difficulty rating: 1270

Small Hint:

From the first pole to the $41$ st pole there are $40$ gaps, so each gap is $\dfrac{5280}{40}$ feet

Big Hint:

Elmer’s stride is one gap divided by $44;$ Oscar’s leap is one gap divided by $12$

Solution:

There are $40$ gaps between the first and $41$ st poles, so each gap is $\dfrac{5280}{40} = 132$ feet.

Elmer’s stride is $\dfrac{132}{44} = 3$ feet and Oscar’s leap is $\dfrac{132}{12} = 11$ feet, a difference of $11 - 3 = 8$ feet.

Thus, the correct answer is B.

6.

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 $\angle AFE?$

$160$

$164$

$166$

$170$

$174$

Concepts:angle chasing isosceles triangle square (geometry)

Difficulty rating: 1350

Small Hint:

Around point $D,$ the triangle’s angle is $360^\circ$ minus $\angle ADC+\angle CDE$

Big Hint:

Triangle $DFE$ is isosceles with $DF = DE,$ so its two base angles are equal

Solution:

Because $ABCD$ is a square, $\angle ADC = 90^\circ.$ Since $E$ and $A$ lie on opposite sides of line $CD,$ ray $DE$ is swung past $DC,$ so the angle of triangle $DFE$ at $D$ (with $F$ on $\overline{AD}$ ) is $\angle FDE = 360^\circ$ $- (\angle ADC + \angle CDE)$ $= 360^\circ - (90^\circ + 110^\circ)$ $= 160^\circ.$

Since $DF = DE,$ triangle $DFE$ is isosceles with base angles $\angle DFE = \tfrac{180^\circ - 160^\circ}{2} = 10^\circ.$

As $A,$ $F,$ $D$ are collinear, $\angle AFE = 180^\circ - \angle DFE = 170^\circ.$

Thus, the correct answer is D.

7.

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 $t - s?$

$-18.5$

$-13.5$

$0$

$13.5$

$18.5$

Concepts:mean expected value

Difficulty rating: 1410

Small Hint:

$t$ is a plain average over the $5$ classes: $\dfrac{50+20+20+5+5}{5}$

Big Hint:

For $s,$ each class of size $n$ is counted $n$ times: $s = \dfrac{\sum n^2}{100}$

Solution:

The teacher average is $t = \dfrac{50 + 20 + 20 + 5 + 5}{5}$ $= \dfrac{100}{5} = 20.$

The student average weights each class size by how many students are in it: $\begin{aligned} s &= \small \frac{50^2 + 20^2 + 20^2 + 5^2 + 5^2}{100} \\ &= \small \frac{2500 + 400 + 400 + 25 + 25}{100} \\ &= 33.5. \end{aligned}$

So $t - s = 20 - 33.5 = -13.5.$

Thus, the correct answer is B.

8.

Let $M$ be the least common multiple of all the integers $10$ through $30,$ inclusive. Let $N$ be the least common multiple of $M,$ $32,$ $33,$ $34,$ $35,$ $36,$ $37,$ $38,$ $39,$ and $40.$ What is the value of $\dfrac{N}{M}?$

$1$

$2$

$37$

$74$

$2886$

Concepts:least common multiple prime factorization

Difficulty rating: 1440

Small Hint:

$M$ already contains $2^4, 3^3, 5^2, 7,$ and all primes up to $29;$ check what $32$ – $40$ add

Big Hint:

$32 = 2^5$ raises the power of $2,$ and $37$ is a new prime; nothing else is new

Solution:

$M = \operatorname{lcm}(10, \ldots, 30)$ contains $2^4$ (from $16$ ), $3^3$ (from $27$ ), $5^2$ (from $25$ ), $7,$ and every prime up to $29.$

Among $32, \ldots, 40,$ the only new contributions are $32 = 2^5,$ which raises the power of $2$ from $2^4$ to $2^5,$ and the new prime $37.$ Everything else factors into primes and powers already in $M.$

Therefore $\dfrac{N}{M} = 2 \cdot 37 = 74.$

Thus, the correct answer is D.

9.

A right rectangular prism whose surface area and volume are numerically equal has edge lengths $\log_2 x,$ $\log_3 x,$ and $\log_4 x.$ What is $x?$

$2\sqrt{6}$

$6\sqrt{6}$

$24$

$48$

$576$

Concepts:surface area volume logarithm

Difficulty rating: 1500

Small Hint:

With edges $a, b, c,$ the equation $abc = 2(ab + bc + ca)$ divides to $1 = 2\left(\tfrac{1}{a} + \tfrac{1}{b} + \tfrac{1}{c}\right)$

Big Hint:

$\dfrac{1}{\log_2 x}$ $+ \dfrac{1}{\log_3 x}$ $+ \dfrac{1}{\log_4 x}$ $= \log_x 2 + \log_x 3 + \log_x 4$ $= \log_x 24$

Solution:

Let $a = \log_2 x,$ $b = \log_3 x,$ $c = \log_4 x.$ Surface area equals volume gives $2(ab + bc + ca) = abc.$ Dividing by $abc,$ $1 = 2\left(\frac{1}{c} + \frac{1}{a} + \frac{1}{b}\right).$

Since $\dfrac{1}{a} = \log_x 2,$ etc., the sum is $\log_x 2 + \log_x 3 + \log_x 4 = \log_x 24.$ Thus $1 = 2\log_x 24,$ so $\log_x 24 = \tfrac{1}{2},$ meaning $x^{\frac{1}{2}} = 24$ and $x = 576.$

Thus, the correct answer is E.

10.

The base-nine representation of the number $N$ is $27{,}006{,}000{,}052_9.$ What is the remainder when $N$ is divided by $5?$

$0$

$1$

$2$

$3$

$4$

Concepts:modular arithmetic number base

Difficulty rating: 1560

Small Hint:

$9 \equiv -1 \pmod 5,$ so $9^k \equiv (-1)^k \pmod 5$

Big Hint:

Take the alternating sum of the base-nine digits (from the right)

Solution:

Since $9 \equiv -1 \pmod 5,$ each power $9^k \equiv (-1)^k,$ so $N$ is congruent to the alternating sum of its base-nine digits.

The nonzero digits, with their positions from the right, are $2$ (position $0$ ), $5$ (position $1$ ), $6$ (position $6$ ), $7$ (position $9$ ), and $2$ (position $10$ ). The alternating sum is $2 - 5 + 6 - 7 + 2 = -2$ $\equiv 3 \pmod 5.$

Thus, the correct answer is D.

11.

Consider two concentric circles of radius $17$ and $19.$ The larger circle has a chord, half of which lies inside the smaller circle. What is the length of the chord in the larger circle?

$12\sqrt{2}$

$10\sqrt{3}$

$\sqrt{17 \cdot 19}$

$18$

$8\sqrt{6}$

Concepts:chord Pythagorean Theorem

Difficulty rating: 1590

Small Hint:

Let the chord be at distance $d$ from the center. Its half-length is $\sqrt{19^2 - d^2},$ and the part inside the small circle has half-length $\sqrt{17^2 - d^2}$

Big Hint:

“Half lies inside” means the inside portion equals half the whole chord: $2\sqrt{17^2 - d^2} = \sqrt{19^2 - d^2}$

Solution:

Let the chord lie at distance $d$ from the common center. Its total length is $2\sqrt{361 - d^2},$ and the portion inside the smaller circle has length $2\sqrt{289 - d^2}.$

Since half the chord lies inside, $2\sqrt{289 - d^2} = \tfrac{1}{2}\cdot 2\sqrt{361 - d^2}.$ Squaring gives $4(289 - d^2) = 361 - d^2,$ so $3d^2 = 795$ and $d^2 = 265.$

The chord length is $2\sqrt{361 - 265} = 2\sqrt{96} = 8\sqrt{6}.$

Thus, the correct answer is E.

12.

What is the number of terms with rational coefficients among the $1001$ terms in the expansion of $\left(x\sqrt[3]{2} + y\sqrt{3}\right)^{1000}?$

$0$

$166$

$167$

$500$

$501$

Concepts:binomial theorem divisibility

Difficulty rating: 1630

Small Hint:

The term for index $k$ carries $2^{\frac{1000-k}{3}}\,3^{\frac{k}{2}};$ rationality needs $1000 - k$ to be a multiple of $3$ and $k$ to be a multiple of $2$

Big Hint:

These force $k \equiv 1 \pmod 3$ and $k$ even, i.e. $k \equiv 4 \pmod 6$

Solution:

The general term is $\binom{1000}{k}(x\sqrt[3]{2})^{1000-k}(y\sqrt{3})^{k},$ whose coefficient contains $2^{\frac{1000-k}{3}}$ and $3^{\frac{k}{2}}.$ This is rational exactly when $(1000 - k)$ is a multiple of $3$ and $k$ is even.

Since $1000 \equiv 1 \pmod 3,$ we need $k \equiv 1 \pmod 3$ and $k$ even, which combine to $k \equiv 4 \pmod 6.$ The valid values $k = 4, 10, \ldots, 1000$ number $\dfrac{1000 - 4}{6} + 1 = 167.$

Thus, the correct answer is C.

13.

The angle bisector of the acute angle formed at the origin by the graphs of the lines $y = x$ and $y = 3x$ has equation $y = kx.$ What is $k?$

$\dfrac{1 + \sqrt{5}}{2}$

$\dfrac{1 + \sqrt{7}}{2}$

$\dfrac{2 + \sqrt{3}}{2}$

$2$

$\dfrac{2 + \sqrt{5}}{2}$

Concepts:angle bisector vector rationalizing denominator

Difficulty rating: 1660

Small Hint:

A bisector direction is the sum of unit vectors along the two lines: $\dfrac{(1,1)}{\sqrt2} + \dfrac{(1,3)}{\sqrt{10}}$

Big Hint:

$k$ is the ratio (second component)/(first component); rationalize the result

Solution:

The bisector points along the sum of the unit vectors of the two lines: $\dfrac{(1,1)}{\sqrt2} + \dfrac{(1,3)}{\sqrt{10}}.$ Its slope is $k = \frac{\tfrac{1}{\sqrt2} + \tfrac{3}{\sqrt{10}}}{\tfrac{1}{\sqrt2} + \tfrac{1}{\sqrt{10}}} = \frac{\sqrt5 + 3}{\sqrt5 + 1}.$

Multiplying numerator and denominator by $\sqrt5 - 1$ gives $\dfrac{(\sqrt5 + 3)(\sqrt5 - 1)}{4}$ $= \dfrac{2 + 2\sqrt5}{4}$ $= \dfrac{1 + \sqrt5}{2}.$

Thus, the correct answer is A.

14.

In the figure, equilateral hexagon $ABCDEF$ has three nonadjacent acute interior angles that each measure $30^\circ.$ The enclosed area of the hexagon is $6\sqrt{3}.$ What is the perimeter of the hexagon?

$4$

$4\sqrt{3}$

$12$

$18$

$12\sqrt{3}$

Concepts:area decomposition trigonometry

Difficulty rating: 1730

Small Hint:

Let the side be $s.$ The three $30^\circ$ tips are isosceles triangles with legs $s;$ they sit on the sides of the inner triangle formed by the three reflex vertices

Big Hint:

Each tip has area $\tfrac12 s^2 \sin 30^\circ,$ and the inner equilateral triangle has side $2s\sin 15^\circ$

Solution:

Let the common side length be $s.$ The three acute vertices are the tips of isosceles triangles with two sides $s$ and apex $30^\circ;$ each has area $\tfrac12 s^2 \sin 30^\circ = \tfrac{s^2}{4}.$

The three reflex vertices form an inner equilateral triangle with side $2s\sin 15^\circ,$ whose area is $\sqrt3\,s^2\sin^2 15^\circ.$ Using $\sin^2 15^\circ = \tfrac{2 - \sqrt3}{4},$ the total area is $\frac{3s^2}{4} + \sqrt3\,s^2\cdot\frac{2 - \sqrt3}{4} = \frac{s^2\sqrt3}{2}.$

Setting $\tfrac{s^2\sqrt3}{2} = 6\sqrt3$ gives $s^2 = 12,$ so $s = 2\sqrt3$ and the perimeter is $6s = 12\sqrt3.$

Thus, the correct answer is E.

15.

Recall that the conjugate of the complex number $w = a + bi,$ where $a$ and $b$ are real numbers and $i = \sqrt{-1},$ is the complex number $\overline{w} = a - bi.$ For any complex number $z,$ let $f(z) = 4i\overline{z}.$ The polynomial $P(z) = z^4 + 4z^3 + 3z^2 + 2z + 1$ has four complex roots: $z_1,$ $z_2,$ $z_3,$ and $z_4.$ Let $\begin{aligned} &Q(z) = z^4 + Az^3 + Bz^2 \\ &\quad {}+ Cz + D \end{aligned}$ be the polynomial whose roots are $f(z_1),$ $f(z_2),$ $f(z_3),$ and $f(z_4),$ where the coefficients $A,$ $B,$ $C,$ and $D$ are complex numbers. What is $B + D?$

$-304$

$-208$

$12i$

$208$

$304$

Concepts:Vieta’s Formulas complex number

Difficulty rating: 1800

Small Hint:

The elementary symmetric functions of $\overline{z_j}$ are the conjugates of those of $z_j;$ all of $P$ ’s coefficients are real, so those symmetric sums are real

Big Hint:

$B = (4i)^2 \sum_{i\lt j}\overline{z_i}\,\overline{z_j}$ and $D = (4i)^4 \prod \overline{z_j}$

Solution:

By Vieta on $P,$ $\sum_{i\lt j} z_iz_j = 3$ and $\prod z_j = 1,$ both real, so their conjugates are also $3$ and $1.$

The roots of $Q$ are $4i\overline{z_j}.$ Then $B$ is the sum of products of pairs: $B = (4i)^2 \sum_{i\lt j}\overline{z_i}\,\overline{z_j}$ $= -16 \cdot 3 = -48.$ And $D = (4i)^4 \prod \overline{z_j} = 256 \cdot 1 = 256.$

So $B + D = -48 + 256 = 208.$

Thus, the correct answer is D.

16.

An organization has $30$ employees, $20$ of whom have a brand A computer while the other $10$ have a brand B computer. For security, the computers can only be connected to each other and only by cables. The cables can only connect a brand A computer to a brand B computer. Employees can communicate with each other if their computers are directly connected by a cable or by relaying messages through a series of connected computers. Initially, no computer is connected to any other. A technician arbitrarily selects one computer of each brand and installs a cable between them, provided there is not already a cable between that pair. The technician stops once every employee can communicate with each other. What is the maximum possible number of cables used?

$190$

$191$

$192$

$195$

$196$

Concepts:graph theory extremal argument

Difficulty rating: 1840

Small Hint:

To delay connectivity, keep the graph disconnected as long as possible, then one final cable joins everything

Big Hint:

The most edges while still disconnected is a complete bipartite graph on $19$ brand A and $10$ brand B, leaving one brand A computer isolated

Solution:

The technician keeps adding cables until the graph becomes connected. To maximize the count, keep the network disconnected for as long as possible: leave a single brand A computer isolated and fully connect the remaining $19$ brand A computers to all $10$ brand B computers.

That uses $19 \cdot 10 = 190$ cables while still disconnected. The next cable connects the last brand A computer, joining everyone, for a total of $190 + 1 = 191.$

Thus, the correct answer is B.

17.

For how many ordered pairs $(b, c)$ of positive integers does neither $x^2 + bx + c = 0$ nor $x^2 + cx + b = 0$ have two distinct real solutions?

$4$

$6$

$8$

$12$

$16$

Concepts:quadratic inequality

Difficulty rating: 1910

Small Hint:

“No two distinct real solutions” means each discriminant is $\le 0:$ $b^2 \le 4c$ and $c^2 \le 4b$

Big Hint:

Multiplying gives $b^2c^2 \le 16bc,$ so $bc \le 16;$ test small positive integers

Solution:

Neither quadratic has two distinct real roots exactly when both discriminants are nonpositive: $b^2 \le 4c$ and $c^2 \le 4b.$

Combining $c\ge \frac{b^2}{4}$ with $c\le2\sqrt b$ gives $b^{\frac{3}{2}}\le8,$ so $b\le4.$ Checking: $b = 1$ gives $c \in \{1,2\};$ $b = 2$ gives $c \in \{1,2\};$ $b = 3$ gives $c = 3;$ and $b = 4$ gives $c = 4.$

That is $(1,1),$ $(1,2),$ $(2,1),$ $(2,2),$ $(3,3),$ $(4,4)$ — $6$ ordered pairs.

Thus, the correct answer is B.

18.

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 $\dfrac{p}{q}?$

$1$

$4$

$8$

$12$

$16$

Concepts:basic probability multiset permutations

Difficulty rating: 1990

Small Hint:

Both probabilities share the same $5^{20}$ denominator, so $\dfrac{p}{q}$ is a ratio of counts

Big Hint:

$q$ uses $\dfrac{20!}{(4!)^5};$ $p$ uses $5\cdot 4$ choices of the special bins times $\dfrac{20!}{3!\,5!\,(4!)^3}$

Solution:

Both probabilities divide by $5^{20},$ so $\dfrac{p}{q}$ is a ratio of arrangement counts.

For $q,$ all bins have $4:$ $\dfrac{20!}{(4!)^5}.$ For $p,$ choose which bin has $3$ and which has $5$ in $5\cdot 4 = 20$ ways, times $\dfrac{20!}{3!\,5!\,(4!)^3}.$ Therefore $\begin{aligned} \frac{p}{q} &= 20 \cdot \frac{(4!)^5}{3!\,5!\,(4!)^3} \\ &= 20 \cdot \frac{(4!)^2}{3!\,5!} \\ &= 20 \cdot \frac{576}{720} \\ &= 16. \end{aligned}$

Thus, the correct answer is E.

19.

Let $x$ be the least real number greater than $1$ such that $\sin(x) = \sin(x^2),$ where the arguments are in degrees. What is $x$ rounded up to the closest integer?

$10$

$13$

$14$

$19$

$20$

Concepts:trigonometry quadratic

Difficulty rating: 2040

Small Hint:

$\sin x = \sin x^2$ requires $x^2 = x + 360k$ or $x^2 = 180 - x + 360k$ for an integer $k$

Big Hint:

The smallest $x \gt 1$ comes from $x^2 = 180 - x$ (with $k = 0$ )

Solution:

Equal sines require $x^2 = x + 360k$ or $x^2 = 180 - x + 360k$ for some integer $k.$

The family $x^2 = x + 360k$ first exceeds $1$ at $k = 1,$ giving $x \approx 19.5.$ The family $x^2 = 180 - x + 360k$ with $k = 0$ gives $x^2 + x - 180 = 0,$ so $x = \dfrac{-1 + \sqrt{721}}{2} \approx 12.93,$ which is smaller.

Rounded up, $x = 13.$

Thus, the correct answer is B.

20.

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 $f_{50}(n) = 12?$

$7$

$8$

$9$

$10$

$11$

Concepts:factor counting recursion

Difficulty rating: 2110

Small Hint:

$f_1(12) = 2\cdot 6 = 12,$ so $12$ is a fixed point; $f_1(8) = 2\cdot 4 = 8$ is another

Big Hint:

After enough steps every $n$ lands on $8$ or $12;$ count those whose orbit reaches $12$

Solution:

Both $8$ and $12$ are fixed. For $n\le50,$ the first value $f_1(n)=2d(n)$ is at most $20.$ Checking the even values through $20$ shows that an orbit reaches $12$ exactly when its first value is $12,18,$ or $20.$ Thus we need $d(n)=6,9,$ or $10.$

The numbers at most $50$ with $6$ divisors are $12,18,20,28,32,44,45,50;$ the only one with $9$ divisors is $36,$ and the only one with $10$ divisors is $48.$ These $10$ values all reach the fixed point $12.$

Thus, the correct answer is D.

21.

Let $ABCD$ be an isosceles trapezoid with $\overline{BC} \parallel \overline{AD}$ and $AB = CD.$ Points $X$ and $Y$ lie on diagonal $\overline{AC}$ with $X$ between $A$ and $Y,$ as shown in the figure. Suppose $\angle AXD = \angle BYC = 90^\circ,$ $AX = 3,$ $XY = 1,$ and $YC = 2.$ What is the area of $ABCD?$

$15$

$5\sqrt{11}$

$3\sqrt{35}$

$18$

$7\sqrt{7}$

Concepts:coordinate geometry trapezoid shoelace formula

Difficulty rating: 2170

Small Hint:

Place $A, X, Y, C$ on a horizontal line with $AX = 3,$ $XY = 1,$ $YC = 2.$ Then $D$ is vertically below $X$ and $B$ vertically above $Y$

Big Hint:

Use $\overline{BC}\parallel\overline{AD}$ and $AB = CD$ to relate the heights of $B$ and $D,$ then apply the shoelace formula

Solution:

Put $A = (0,0),$ $X = (3,0),$ $Y = (4,0),$ $C = (6,0).$ The right angles give $D = (3, t)$ and $B = (4, s)$ on opposite sides of $AC.$

Parallelism $\overline{AD}\parallel\overline{BC}$ forces $t = -\tfrac{3}{2}s,$ and $AB = CD$ gives $16 + s^2 = 9 + t^2,$ so $t^2 - s^2 = 7.$ Substituting yields $s^2 = \tfrac{28}{5}.$

The shoelace formula gives area $= 3\,|t - s|$ $= 3\cdot\tfrac{5}{2}s$ $= \tfrac{15}{2}s$ $= \tfrac{15}{2}\sqrt{\tfrac{28}{5}}$ $= 3\sqrt{35}.$

Thus, the correct answer is C.

22.

Azar and Carl play a game of tic-tac-toe. Azar places an $X$ in one of the boxes in a $3$ -by- $3$ array of boxes, then Carl places an $O$ in one of the remaining boxes. After that, Azar places an $X$ in one of the remaining boxes, and so on until all $9$ boxes are filled or one of the players has $3$ of their symbols in a row — horizontal, vertical, or diagonal — whichever comes first, in which case that player wins the game. Suppose the players make their moves at random, rather than trying to follow a rational strategy, and that Carl wins the game when he places his third $O.$ How many ways can the board look after the game is over?

$36$

$112$

$120$

$148$

$160$

Concepts:casework combinations complementary counting

Difficulty rating: 2270

Small Hint:

The final board has three $O$ s forming a line and three $X$ s that must not themselves form a line

Big Hint:

Split into cases by whether $O$ ’s line is a row/column or a diagonal, since that changes how many $X$ placements accidentally make a row

Solution:

Carl wins on his third $O,$ so the board has three $O$ s forming one of the $8$ lines and three $X$ s in the other six cells. The $X$ s must not form a line (else Azar would have won first).

If the $O$ line is a row or column ( $6$ choices), the remaining six cells contain two full lines, so valid $X$ placements number $\binom{6}{3} - 2 = 18.$ If the $O$ line is a diagonal ( $2$ choices), the remaining six cells contain no full line, giving $\binom{6}{3} = 20.$

The total is $6\cdot 18 + 2\cdot 20 = 108 + 40 = 148.$

Thus, the correct answer is D.

23.

A quadratic polynomial with real coefficients and leading coefficient $1$ is called 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 $\tilde{p}(1)?$

$\dfrac{5}{16}$

$\dfrac{1}{2}$

$\dfrac{5}{8}$

$1$

$\dfrac{9}{8}$

Concepts:polynomial quadratic optimization

Difficulty rating: 2380

Small Hint:

If $p$ has roots $r, s,$ then $p(p(x)) = 0$ means $p(x) = r$ or $p(x) = s;$ exactly three roots requires one of these to have a double root

Big Hint:

With $(r - s)^2 + 4s = 0$ and letting $u = r - s,$ maximize $r + s = -\tfrac{u^2}{2} + u$

Solution:

Let $p$ have roots $r$ and $s.$ Then $p(p(x)) = 0$ splits into $p(x) = r$ and $p(x) = s,$ with discriminants $(r - s)^2 + 4r$ and $(r - s)^2 + 4s.$ Exactly three real roots means one discriminant is $0$ and the other positive.

Take $(r - s)^2 + 4s = 0$ and set $u = r - s.$ Then $s = -\tfrac{u^2}{4}$ and the other discriminant is $4u,$ so $u>0.$ Also $r+s=-\tfrac{u^2}{2}+u,$ maximized at $u=1,$ giving $r=\tfrac34$ and $s=-\tfrac14.$

So $\tilde{p}(x) = \left(x - \tfrac34\right)\left(x + \tfrac14\right),$ and $\tilde{p}(1) = \tfrac14\cdot\tfrac54 = \tfrac{5}{16}.$

Thus, the correct answer is A.

24.

Convex quadrilateral $ABCD$ has $AB = 18,$ $\angle A = 60^\circ,$ and $\overline{AB} \parallel \overline{CD}.$ In some order, the lengths of the four sides form an arithmetic progression, and side $AB$ is a side of maximum length. The length of another side is $a.$ What is the sum of all possible values of $a?$

$24$

$42$

$60$

$66$

$84$

Concepts:trapezoid arithmetic sequence casework

Difficulty rating: 2520

Small Hint:

With $AB = 18$ the maximum, the four sides are $18, 18 - d, 18 - 2d, 18 - 3d;$ place $A = (0,0),$ $B = (18,0),$ and $D$ at angle $60^\circ$

Big Hint:

For each assignment of the three smaller lengths to $BC, CD, DA,$ impose that $\overline{CD}$ is horizontal and solve for $d;$ also include common difference $d = 0$

Solution:

Since $AB = 18$ is the largest, the four sides are $18, 18 - d, 18 - 2d, 18 - 3d.$ Placing $A = (0,0),$ $B = (18,0),$ and $D = \left(\tfrac{m}{2}, \tfrac{m\sqrt3}{2}\right)$ with $m=DA,$ let $n=CD$ and $\ell=BC.$ Then $C=(\tfrac m2+n,\tfrac{m\sqrt3}{2}),$ so $\begin{aligned} \ell^2&=m^2+(18-n)^2 \\ &\quad {}-m(18-n). \end{aligned}$

Write $u_j=18-jd$ for $j=1,2,3.$ Substituting the permutations $123,132,213,231,312,321$ of $(u_1,u_2,u_3)$ for $(m,n,\ell)$ gives the nonzero candidates $d=18,2,9,5,6,6,$ respectively. Positivity of $u_3$ requires $d<6,$ leaving only $d=2$ and $d=5.$ The case $d=0$ also gives a valid rhombus.

Thus the side sets are $\{18,16,14,12\},$ $\{18,13,8,3\},$ and $\{18,18,18,18\}.$

The possible values of a non- $AB$ side length are $\{3, 8, 12, 13, 14, 16, 18\},$ whose sum is $84.$

Thus, the correct answer is E.

25.

Let $m \ge 5$ be an odd integer, and let $D(m)$ denote the number of quadruples $(a_1, a_2, a_3, a_4)$ of distinct integers with $1 \le a_i \le m$ for all $i$ such that $m$ divides $a_1 + a_2 + a_3 + a_4.$ There is a polynomial $q(x) = c_3x^3 + c_2x^2 + c_1x + c_0$ such that $D(m) = q(m)$ for all odd integers $m \ge 5.$ What is $c_1?$

$-6$

$-1$

$4$

$6$

$11$

Concepts:modular arithmetic roots of unity polynomial

Difficulty rating: 2650

Small Hint:

First count all ordered residue quadruples summing to $0\pmod m,$ then use inclusion-exclusion on equal coordinates

Big Hint:

Group collision patterns as one pair, two pairs, a triple, or all four equal; odd $m$ makes multiplication by $2$ and $4$ invertible

Solution:

Regard $1,2,\ldots,m$ as all residues modulo $m.$ Without the distinctness condition, the first three entries are arbitrary and the fourth is determined, giving $m^3$ ordered quadruples.

Inclusion-exclusion over equal-coordinate partitions gives the remaining terms. There are $6$ choices of one equal pair, each leaving $m^2$ solutions, for $-6m^2.$ The $3$ two-pair partitions contribute $+3m.$ The $4$ triple-and-single partitions have inclusion-exclusion coefficient $2$ and contribute $+8m.$ Finally, the all-equal partition has coefficient $-6$ and one solution. (Here odd $m$ makes $2$ and $4$ invertible modulo $m.$ ) Therefore $\begin{aligned} D(m)&=m^3-6m^2 \\ &\quad {}+11m-6 \\ &=(m-1)(m-2) \\ &\quad {}\cdot(m-3). \end{aligned}$ Hence $c_1=11.$

Thus, the correct answer is E.