2019 AMC 12A — Problems & Online Timed Exam

Timed

1:15:00

1.

The area of a pizza with radius $4$ inches is $N$ percent larger than the area of a pizza with radius $3$ inches. What is the integer closest to $N?$

$25$

$33$

$44$

$66$

$78$

Answer: E

Concepts:circle area percentage

Difficulty rating: 770

Small Hint:

The areas are in the ratio $4^2 : 3^2$

Big Hint:

The percent increase is $\left(\dfrac{16}{9} - 1\right) \times 100$

Solution:

The areas are proportional to the squares of the radii, so the ratio of the larger area to the smaller is $\dfrac{16}{9}.$

The percent increase is $\left(\dfrac{16}{9} - 1\right) \times 100 = \dfrac{700}{9} \approx 77.8.$ The closest integer is $78.$

Thus, the correct answer is E.

2.

Suppose $a$ is $150\%$ of $b.$ What percent of $a$ is $3b?$

$50$

$66\dfrac{2}{3}$

$150$

$200$

$450$

Answer: D

Concepts:percentage ratio and proportion

Difficulty rating: 770

Small Hint:

Write $a = 1.5b$

Big Hint:

Compute $\dfrac{3b}{a}$ and convert to a percentage

Solution:

Since $a = 1.5b,$ we have $\dfrac{3b}{a} = \dfrac{3b}{1.5b} = 2.$

As a percentage, $3b$ is $200\%$ of $a.$

Thus, the correct answer is D.

3.

A box contains $28$ red balls, $20$ green balls, $19$ yellow balls, $13$ blue balls, $11$ white balls, and $9$ black balls. What is the minimum number of balls that must be drawn from the box without replacement to guarantee that at least $15$ balls of a single color will be drawn?

$75$

$76$

$79$

$84$

$91$

Answer: B

Concepts:pigeonhole principle

Difficulty rating: 1020

Small Hint:

Consider the worst case: draw as many as possible without ever reaching $15$ of one color

Big Hint:

Take $14$ of each color that has at least $14,$ and all balls of the smaller colors

Solution:

In the worst case, we draw $14$ each of red, green, and yellow, plus all of the blue $(13),$ white $(11),$ and black $(9),$ without reaching $15$ of any color.

That is $14 + 14 + 14 + 13 + 11 + 9 = 75$ balls.

The next ball must complete a set of $15,$ so $76$ balls are needed.

Thus, the correct answer is B.

4.

What is the greatest number of consecutive integers whose sum is $45?$

$9$

$25$

$45$

$90$

$120$

Answer: D

Concepts:arithmetic sequence symmetry

Difficulty rating: 1170

Small Hint:

The integers are allowed to be negative

Big Hint:

Adding $-44, -43, \ldots, 44$ contributes $0,$ so extend the run just past $45$

Solution:

Negative integers are allowed. The integers from $-44$ to $44$ sum to $0,$ so the integers from $-44$ to $45$ sum to $45.$

This run has $45 - (-44) + 1 = 90$ integers. Conversely, if $L$ consecutive integers have sum $45,$ then twice their sum is $L$ times an integer, so $90$ is a multiple of $L.$ Hence $L\le90,$ proving that this run is longest.

Thus, the correct answer is D.

5.

Two lines with slopes $\dfrac{1}{2}$ and $2$ intersect at $(2, 2).$ What is the area of the triangle enclosed by these two lines and the line $x + y = 10?$

$4$

$4\sqrt{2}$

$6$

$8$

$6\sqrt{2}$

Answer: C

Concepts:coordinate geometry triangle area shoelace formula

Difficulty rating: 1280

Small Hint:

Find where each line meets $x + y = 10$

Big Hint:

The three vertices are $(2, 2),$ $(6, 4),$ and $(4, 6);$ use the shoelace formula

Solution:

The two lines are $y = \tfrac{1}{2}x + 1$ and $y = 2x - 2.$ Intersecting each with $x + y = 10$ gives the points $(6, 4)$ and $(4, 6).$

The triangle has vertices $(2, 2),$ $(6, 4),$ and $(4, 6).$ By the shoelace formula,

$\begin{aligned} &\small \tfrac{1}{2}\left| 2(4 - 6) + 6(6 - 2) + 4(2 - 4) \right| \\ &= \tfrac{1}{2}\left| -4 + 24 - 8 \right| \\ &= 6. \end{aligned}$

Thus, the correct answer is C.

6.

The figure below shows line $\ell$ with a regular, infinite, recurring pattern of squares and line segments.

How many of the following four kinds of rigid motion transformations of the plane in which this figure is drawn, other than the identity transformation, will transform this figure into itself?

• some rotation around a point of line $\ell$

• some translation in the direction parallel to line $\ell$

• the reflection across line $\ell$

• some reflection across a line perpendicular to line $\ell$

$0$

$1$

$2$

$3$

$4$

Answer: C

Concepts:transformation symmetry

Difficulty rating: 1310

Small Hint:

A translation by one period clearly works

Big Hint:

Test a $180^\circ$ rotation about a point on $\ell,$ and check that neither reflection maps the diagonal segments correctly

Solution:

A translation by one full period maps the figure to itself, so translation works.

A $180^\circ$ rotation about a suitable point on $\ell$ sends each square above the line to the square below it, with the diagonal segments matching, so this rotation works.

Reflection across $\ell$ sends the top-right diagonals to top-right diagonals below the line, but the actual below-line diagonals point to the bottom-left, so it fails. A reflection across a perpendicular line fails for the same reason. Only $2$ of the four transformations work.

Thus, the correct answer is C.

7.

Melanie computes the mean $\mu,$ the median $M,$ and the modes of the $365$ values that are the dates in the months of $2019.$ Thus her data consist of $12$ $1$ s, $12$ $2$ s, $\ldots,$ $12$ $28$ s, $11$ $29$ s, $11$ $30$ s, and $7$ $31$ s. Let $d$ be the median of the modes. Which of the following statements is true?

$\mu \lt d \lt M$

$M \lt d \lt \mu$

$d = M = \mu$

$d \lt M \lt \mu$

$d \lt \mu \lt M$

Answer: E

Concepts:mean median (data)mode

Difficulty rating: 1330

Small Hint:

The modes are $1$ through $28,$ each appearing $12$ times, so $d$ is their median

Big Hint:

The median $M$ is the $183$ rd value; compute $\mu$ directly from the totals

Solution:

The values $1$ through $28$ each appear $12$ times and are the modes, so $d = \dfrac{14 + 15}{2} = 14.5.$

The $183$ rd of the $365$ ordered values is the median. Values $1$ through $15$ fill the first $180$ positions, so position $183$ is $16;$ thus $M = 16.$

The total of all values is $12(1 + \cdots + 28)$ $+ 11(29 + 30)$ $+ 7 \cdot 31 = 5738,$ so $\mu = \dfrac{5738}{365} \approx 15.7.$

Therefore $d \lt \mu \lt M.$

Thus, the correct answer is E.

8.

For a set of four distinct lines in a plane, there are exactly $N$ distinct points that lie on two or more of the lines. What is the sum of all possible values of $N?$

$14$

$16$

$18$

$19$

$21$

Answer: D

Concepts:counting intersections casework

Difficulty rating: 1380

Small Hint:

The maximum is $\binom{4}{2} = 6$ when the lines are in general position

Big Hint:

Enumerate configurations using parallel classes and concurrences; check which counts from $0$ to $6$ are achievable

Solution:

The values $0$ and $1$ occur when all four lines are parallel or all four are concurrent. Three parallel lines crossed by a fourth give $3$ points. Three concurrent lines together with a fourth line not through their common point give $4.$ One parallel pair with no three concurrent gives $5,$ and four lines in general position give $6.$

The value $2$ is impossible. Once two lines meet at $P,$ a third line not through $P$ creates at least one new intersection; a fourth distinct line must then create another new point. If every remaining line passes through $P,$ there is only one intersection point instead. Thus the achievable values are $0,1,3,4,5,6,$ whose sum is $19.$

Thus, the correct answer is D.

9.

A sequence of numbers is defined recursively by $a_1 = 1,$ $a_2 = \dfrac{3}{7},$ and

$a_n = \dfrac{a_{n-2} \cdot a_{n-1}}{2a_{n-2} - a_{n-1}}$

for all $n \ge 3.$ Then $a_{2019}$ can be written as $\dfrac{p}{q},$ where $p$ and $q$ are relatively prime positive integers. What is $p + q?$

$2020$

$4039$

$6057$

$6061$

$8078$

Answer: E

Concepts:recursion substitution arithmetic sequence

Difficulty rating: 1500

Small Hint:

Take reciprocals of both sides

Big Hint:

Setting $b_n = \dfrac{1}{a_n}$ gives $b_n = 2b_{n-1} - b_{n-2},$ an arithmetic sequence

Solution:

Taking reciprocals, $\begin{aligned} \dfrac{1}{a_n} &= \dfrac{2a_{n-2} - a_{n-1}}{a_{n-2}a_{n-1}} \\ &= \dfrac{2}{a_{n-1}} - \dfrac{1}{a_{n-2}}. \end{aligned}$

Let $b_n = \dfrac{1}{a_n}.$ Then $b_n = 2b_{n-1} - b_{n-2},$ so $b_n$ is arithmetic with $b_1 = 1,$ $b_2 = \dfrac{7}{3},$ and common difference $\dfrac{4}{3}.$

Thus $b_{2019} = 1 + 2018 \cdot \dfrac{4}{3} = \dfrac{8075}{3},$ so $a_{2019} = \dfrac{3}{8075}.$ Since these are relatively prime, $p + q = 8078.$

Thus, the correct answer is E.

10.

The figure below shows $13$ circles of radius $1$ within a larger circle. All the intersections occur at points of tangency. What is the area of the region, shaded in the figure, inside the larger circle but outside all the circles of radius $1?$

$4\pi\sqrt{3}$

$7\pi$

$\pi(3\sqrt{3} + 2)$

$10\pi(\sqrt{3} - 1)$

$\pi(\sqrt{3} + 6)$

Answer: A

Concepts:circle area tangent circles area decomposition

Difficulty rating: 1500

Small Hint:

A central circle is ringed by six circles, then six more sit in the outer notches

Big Hint:

The outermost centers are at distance $2\sqrt{3}$ from the center, so the big radius is $2\sqrt{3} + 1$

Solution:

Place a unit circle at the center, six around it with centers at distance $2$ (a hexagon), and six more with centers at distance $2\sqrt{3}$ in the outer gaps. That is $1 + 6 + 6 = 13$ circles.

The outermost circles are tangent to the big circle, whose radius is therefore $2\sqrt{3} + 1.$ Its area is $\pi(2\sqrt{3} + 1)^2 = \pi(13 + 4\sqrt{3}).$

Subtracting the $13$ unit circles leaves $\pi(13 + 4\sqrt{3}) - 13\pi = 4\pi\sqrt{3}.$

Thus, the correct answer is A.

11.

For some positive integer $k,$ the repeating base- $k$ representation of the (base-ten) fraction $\dfrac{7}{51}$ is $0.\overline{23}_k = 0.232323\ldots_k.$ What is $k?$

$13$

$14$

$15$

$16$

$17$

Answer: D

Concepts:number base repeating decimal quadratic

Difficulty rating: 1440

Small Hint:

A two-digit repeating block in base $k$ equals $\dfrac{2k + 3}{k^2 - 1}$

Big Hint:

Set $\dfrac{2k + 3}{k^2 - 1} = \dfrac{7}{51}$ and clear denominators

Solution:

The repeating block gives $0.\overline{23}_k = \dfrac{2k + 3}{k^2 - 1} = \dfrac{7}{51}.$

Cross-multiplying, $51(2k + 3) = 7(k^2 - 1),$ so $7k^2 - 102k - 160 = 0.$

The quadratic formula gives $\begin{aligned} k &= \dfrac{102 + \sqrt{14884}}{14} \\ &= \dfrac{102 + 122}{14} = 16. \end{aligned}$

Thus, the correct answer is D.

12.

Positive real numbers $x \ne 1$ and $y \ne 1$ satisfy $\log_2 x = \log_y 16$ and $xy = 64.$ What is $\left(\log_2 \dfrac{x}{y}\right)^2?$

$\dfrac{25}{2}$

$20$

$\dfrac{45}{2}$

$25$

$32$

Answer: B

Concepts:logarithm system of equations

Difficulty rating: 1560

Small Hint:

Let $a = \log_2 x$ and $b = \log_2 y;$ then $\log_y 16 = \dfrac{4}{b}$

Big Hint:

You get $ab = 4$ and $a + b = 6;$ note $\left(\log_2\tfrac{x}{y}\right)^2 = (a - b)^2$

Solution:

Let $a = \log_2 x$ and $b = \log_2 y.$ Then $\log_y 16 = \dfrac{4}{b},$ so $a = \dfrac{4}{b},$ giving $ab = 4.$

Since $xy = 64,$ we have $a + b = 6.$

Therefore $\begin{aligned} \left(\log_2 \tfrac{x}{y}\right)^2 &= (a - b)^2 \\ &= (a + b)^2 - 4ab \\ &= 36 - 16 = 20. \end{aligned}$

Thus, the correct answer is B.

13.

How many ways are there to paint each of the integers $2,$ $3,$ $\ldots,$ $9$ either red, green, or blue so that each number has a different color from each of its proper divisors?

$144$

$216$

$256$

$384$

$432$

Answer: E

Concepts:graph theory divisibility casework

Difficulty rating: 1630

Small Hint:

Numbers $5$ and $7$ are unconstrained; handle the chain $2, 4, 8$ and the constraints on $6, 9$

Big Hint:

The color of $6$ must avoid both $2$ and $3,$ so count depends on whether $2$ and $3$ share a color

Solution:

The primes $5$ and $7$ have no proper divisors here, giving $3$ choices each.

Along the chain $2 \to 4 \to 8,$ there are $3 \cdot 2 \cdot 1 = 6$ colorings. Number $9$ must differ from $3,$ giving $2$ choices once $3$ is set.

Number $6$ must differ from both $2$ and $3.$ Summing over the colors of $2$ and $3$ (equal in $3$ pairs, unequal in $6$ pairs), the combined factor for $4, 8, 9, 6$ totals $2 \cdot 2 \cdot (3 \cdot 2 + 6 \cdot 1) = 48.$

Multiplying by the $9$ ways for $5$ and $7$ gives $48 \cdot 9 = 432.$

Thus, the correct answer is E.

14.

For a certain complex number $c,$ the polynomial

$\begin{aligned} P(x) &= (x^2 - 2x + 2) \\ &\quad {}\cdot (x^2 - cx + 4) \\ &\quad {}\cdot (x^2 - 4x + 8) \end{aligned}$

has exactly $4$ distinct roots. What is $|c|?$

$2$

$\sqrt{6}$

$2\sqrt{2}$

$3$

$\sqrt{10}$

Answer: E

Concepts:complex number polynomial Vieta’s Formulas

Difficulty rating: 1690

Small Hint:

The first and third factors give roots $1 \pm i$ and $2 \pm 2i$

Big Hint:

For only $4$ distinct roots, $x^2 - cx + 4$ must reuse two of these; its roots multiply to $4$

Solution:

The factors $x^2 - 2x + 2$ and $x^2 - 4x + 8$ have roots $1 \pm i$ and $2 \pm 2i,$ which are $4$ distinct values.

For $P$ to have exactly $4$ distinct roots, the roots of $x^2 - cx + 4$ must lie among these. Their product must equal $4,$ and the only such pair is one root from each factor, for example $(1 + i)(2 - 2i) = 4.$

Then $c = (1 + i) + (2 - 2i) = 3 - i,$ so $|c| = \sqrt{3^2 + 1^2} = \sqrt{10}.$

Thus, the correct answer is E.

15.

Positive real numbers $a$ and $b$ have the property that

$\begin{aligned} &\sqrt{\log a} + \sqrt{\log b} \\ &\quad {}+ \log \sqrt{a} + \log \sqrt{b} = 100 \end{aligned}$

and all four terms on the left are positive integers, where $\log$ denotes the base $10$ logarithm. What is $ab?$

$10^{52}$

$10^{100}$

$10^{144}$

$10^{164}$

$10^{200}$

Answer: D

Concepts:logarithm Diophantine Equation casework

Difficulty rating: 1730

Small Hint:

Let $\sqrt{\log a} = p$ and $\sqrt{\log b} = q;$ then $\log\sqrt{a} = \tfrac{p^2}{2}$ must be an integer

Big Hint:

With $p = 2m,$ $q = 2n,$ the equation becomes $m(m+1) + n(n+1) = 50$

Solution:

Let $\sqrt{\log a} = p$ and $\sqrt{\log b} = q,$ so $\log a = p^2$ and $\log\sqrt{a} = \dfrac{p^2}{2}.$ For this to be an integer, $p$ is even; likewise $q.$

Writing $p = 2m,$ $q = 2n,$ the equation $p + q + \dfrac{p^2}{2} + \dfrac{q^2}{2} = 100$ becomes $m(m+1) + n(n+1) = 50.$

The only solution is $\{m, n\} = \{4, 5\},$ giving $\log(ab) = p^2 + q^2$ $= 4(16 + 25) = 164.$

Therefore $ab = 10^{164}.$

Thus, the correct answer is D.

16.

The numbers $1,$ $2,$ $\ldots,$ $9$ are randomly placed into the $9$ squares of a $3 \times 3$ grid. Each square gets one number, and each of the numbers is used once. What is the probability that the sum of the numbers in each row and each column is odd?

$\dfrac{1}{21}$

$\dfrac{1}{14}$

$\dfrac{5}{63}$

$\dfrac{2}{21}$

$\dfrac{1}{7}$

Answer: B

Concepts:parity basic probability permutations

Difficulty rating: 1800

Small Hint:

There are $5$ odd and $4$ even numbers; each row and column needs an odd count of odd numbers

Big Hint:

The odd entries must fill one entire row and one entire column, forming a plus shape of $5$ cells

Solution:

There are $5$ odd and $4$ even numbers. Each row and column must contain an odd number of odd entries.

The only way to place $5$ odd entries with every row and column odd is to fill one complete row and one complete column (a plus shape of $3 + 3 - 1 = 5$ cells). There are $3 \cdot 3 = 9$ such patterns.

Each pattern admits $5!$ placements of the odd numbers and $4!$ of the even numbers, so the probability is $\dfrac{9 \cdot 5! \cdot 4!}{9!} = \dfrac{1}{14}.$

Thus, the correct answer is B.

17.

Let $s_k$ denote the sum of the $k$ th powers of the roots of the polynomial $x^3 - 5x^2 + 8x - 13.$ In particular, $s_0 = 3,$ $s_1 = 5,$ and $s_2 = 9.$ Let $a,$ $b,$ and $c$ be real numbers such that $s_{k+1} = a\,s_k + b\,s_{k-1} + c\,s_{k-2}$ for $k = 2,$ $3,$ $\ldots.$ What is $a + b + c?$

$-6$

$0$

$6$

$10$

$26$

Answer: D

Concepts:Newton’s Sums polynomial recursion

Difficulty rating: 1860

Small Hint:

Each root $r$ satisfies $r^3 = 5r^2 - 8r + 13$

Big Hint:

Multiply by $r^{k-2}$ and sum over the roots to read off $a, b, c$

Solution:

Every root $r$ satisfies $r^3 = 5r^2 - 8r + 13,$ so $r^{k+1} = 5r^k - 8r^{k-1} + 13r^{k-2}.$

Summing over the three roots gives $s_{k+1} = 5s_k - 8s_{k-1} + 13s_{k-2},$ so $a = 5,$ $b = -8,$ $c = 13.$

Therefore $a + b + c = 5 - 8 + 13 = 10.$

Thus, the correct answer is D.

18.

A sphere with center $O$ has radius $6.$ A triangle with sides of length $15,$ $15,$ and $24$ is situated in space so that each of its sides is tangent to the sphere. What is the distance between $O$ and the plane determined by the triangle?

$2\sqrt{3}$

$4$

$3\sqrt{2}$

$2\sqrt{5}$

$5$

Answer: D

Concepts:sphere incircle, incenter, and inradius Pythagorean Theorem

Difficulty rating: 1910

Small Hint:

The sphere meets the plane in a circle tangent to all three sides, so that circle is the incircle

Big Hint:

If $d$ is the distance, the cross-circle has radius $\sqrt{36 - d^2},$ which equals the inradius

Solution:

The sphere intersects the triangle’s plane in a circle of radius $\sqrt{36 - d^2},$ where $d$ is the distance from $O$ to the plane. Since each side is tangent to the sphere, this circle is the triangle’s incircle.

The triangle has area $\tfrac{1}{2} \cdot 24 \cdot 9 = 108$ and semiperimeter $27,$ so its inradius is $\dfrac{108}{27} = 4.$

Thus $\sqrt{36 - d^2} = 4,$ giving $d^2 = 20$ and $d = 2\sqrt{5}.$

Thus, the correct answer is D.

19.

In $\triangle ABC$ with integer side lengths,

$\begin{aligned} \cos A &= \dfrac{11}{16}, \\ \cos B &= \dfrac{7}{8}, \\ \cos C &= -\dfrac{1}{4}. \end{aligned}$

What is the least possible perimeter for $\triangle ABC?$

$9$

$12$

$23$

$27$

$44$

Answer: A

Concepts:law of sines ratio and proportion triangle inequality

Difficulty rating: 2000

Small Hint:

Find $\sin A, \sin B, \sin C$ from the given cosines

Big Hint:

By the Law of Sines the sides are proportional to $\sin A : \sin B : \sin C = 3 : 2 : 4$

Solution:

Each sine is $\sqrt{1 - \cos^2}:$ $\sin A = \dfrac{3\sqrt{15}}{16},$ $\sin B = \dfrac{2\sqrt{15}}{16},$ $\sin C = \dfrac{4\sqrt{15}}{16}.$

By the Law of Sines the sides are in ratio $3 : 2 : 4.$ The smallest integer sides are $3, 2, 4,$ which satisfy the triangle inequality.

The least perimeter is $3 + 2 + 4 = 9.$

Thus, the correct answer is A.

20.

Real numbers between $0$ and $1,$ inclusive, are chosen in the following manner. A fair coin is flipped. If it lands heads, then it is flipped again and the chosen number is $0$ if the second flip is heads and $1$ if the second flip is tails. On the other hand, if the first coin flip is tails, then the number is chosen uniformly at random from the closed interval $[0, 1].$ Two random numbers $x$ and $y$ are chosen independently in this manner. What is the probability that $|x - y| \gt \dfrac{1}{2}?$

$\dfrac{1}{3}$

$\dfrac{7}{16}$

$\dfrac{1}{2}$

$\dfrac{9}{16}$

$\dfrac{2}{3}$

Answer: B

Concepts:geometric probability casework

Difficulty rating: 2070

Small Hint:

Each number is $0$ with probability $\tfrac14,$ $1$ with probability $\tfrac14,$ and uniform on $[0,1]$ with probability $\tfrac12$

Big Hint:

Split into the nine type-combinations; for two uniforms, $P\!\left(|x - y| \gt \tfrac12\right) = \tfrac14$

Solution:

Each variable equals $0$ with probability $\tfrac14,$ equals $1$ with probability $\tfrac14,$ and is uniform on $[0, 1]$ with probability $\tfrac12.$

Considering the nine combinations of types: the pairs $(0, 1)$ and $(1, 0)$ each contribute $\tfrac{1}{16}.$ Each of the four point-versus-uniform cases contributes $\tfrac{1}{16}.$ The uniform-versus-uniform case contributes $\tfrac14 \cdot \tfrac14 = \tfrac{1}{16}.$

The total is $\dfrac{2 + 4 + 1}{16} = \dfrac{7}{16}.$

Thus, the correct answer is B.

21.

Let $z = \dfrac{1 + i}{\sqrt{2}}.$ What is

$\begin{aligned} &\left(z^{1^2} + z^{2^2} + z^{3^2} + \cdots + z^{12^2}\right) \\ &\quad {}\cdot \scriptsize \left(\dfrac{1}{z^{1^2}} + \dfrac{1}{z^{2^2}} + \dfrac{1}{z^{3^2}} + \cdots + \dfrac{1}{z^{12^2}}\right)? \end{aligned}$

$18$

$72 - 36\sqrt{2}$

$36$

$72$

$72 + 36\sqrt{2}$

Answer: C

Concepts:roots of unity complex number modular arithmetic

Difficulty rating: 2160

Small Hint:

$z = e^{\frac{i\pi}{4}},$ so $z^{k^2}$ depends only on $k^2 \bmod 8$

Big Hint:

Over $k = 1, \ldots, 12,$ the exponent $k^2 \bmod 8$ is $1$ six times, $4$ three times, and $0$ three times

Solution:

Since $z = e^{\frac{i\pi}{4}},$ we have $z^{k^2} = e^{\frac{i\pi k^2}{4}},$ depending only on $k^2 \bmod 8.$

For $k = 1$ to $12,$ the residue $k^2 \bmod 8$ is $1$ (giving $z$ ) six times, $4$ (giving $-1$ ) three times, and $0$ (giving $1$ ) three times. So the first sum is $6z - 3 + 3 = 6z.$

The second sum is likewise $\dfrac{6}{z} - 3 + 3 = \dfrac{6}{z}.$ Their product is $6z \cdot \dfrac{6}{z} = 36.$

Thus, the correct answer is C.

22.

Circles $\omega$ and $\gamma,$ both centered at $O,$ have radii $20$ and $17,$ respectively. Equilateral triangle $ABC,$ whose interior lies in the interior of $\omega$ but in the exterior of $\gamma,$ has vertex $A$ on $\omega,$ and the line containing side $BC$ is tangent to $\gamma.$ Segments $AO$ and $BC$ intersect at $P,$ and $\dfrac{BP}{CP} = 3.$ Then $AB$ can be written in the form $\dfrac{m}{\sqrt{n}} - \dfrac{p}{\sqrt{q}}$ for positive integers $m,$ $n,$ $p,$ $q$ with $\gcd(m, n) = \gcd(p, q) = 1.$ What is $m + n + p + q?$

$42$

$86$

$92$

$114$

$130$

Answer: E

Concepts:coordinate geometry equilateral triangle tangent line

Difficulty rating: 2310

Small Hint:

Let $s = AB.$ Since $\tfrac{BP}{CP} = 3,$ we have $BP = \tfrac{3s}{4}$ and $CP = \tfrac{s}{4}$

Big Hint:

Place $P$ at the origin with $BC$ horizontal; $O$ lies on line $AP$ at distance $17$ from $BC$ and $20$ from $A$

Solution:

Let $s = AB.$ Since $\dfrac{BP}{CP} = 3,$ we have $BP = \dfrac{3s}{4}$ and $CP = \dfrac{s}{4}.$ Put $P$ at the origin with $BC$ on the $x$ -axis, $B = \left(-\tfrac{3s}{4}, 0\right),$ $C = \left(\tfrac{s}{4}, 0\right),$ and apex $A = \left(-\tfrac{s}{4}, \tfrac{s\sqrt{3}}{2}\right).$

Points $P, O, A$ are collinear, so $O = t \cdot A$ for some scalar $t.$ Two conditions pin it down: $O$ is at distance $17$ from line $BC,$ giving $|t| \cdot \dfrac{s\sqrt{3}}{2} = 17,$ and $A$ is on $\omega,$ giving $|t - 1| \cdot \dfrac{s\sqrt{13}}{4} = 20$ since $|A| = \dfrac{s\sqrt{13}}{4}.$

Solving, $|t| s = \dfrac{34}{\sqrt{3}}$ and $|t - 1| s = \dfrac{80}{\sqrt{13}}.$ The valid configuration has $O$ and $A$ on opposite sides of $P,$ so $|t-1|s=|t|s+s.$ Therefore $AB = s = \dfrac{80}{\sqrt{13}} - \dfrac{34}{\sqrt{3}}.$

Then $m + n + p + q = 80 + 13$ $+ 34 + 3 = 130.$

Thus, the correct answer is E.

23.

Define binary operations $\diamondsuit$ and $\heartsuit$ by

$a \diamondsuit b = a^{\log_7(b)}$ and $a \heartsuit b = a^{\frac{1}{\log_7(b)}}$

for all real numbers $a$ and $b$ for which these expressions are defined. The sequence $(a_n)$ is defined recursively by $a_3 = 3 \heartsuit 2$ and $a_n = (n \heartsuit (n - 1)) \diamondsuit a_{n-1}$ for all integers $n \ge 4.$ To the nearest integer, what is $\log_7(a_{2019})?$

$8$

$9$

$10$

$11$

$12$

Answer: D

Concepts:custom operation logarithm telescoping

Difficulty rating: 2240

Small Hint:

Apply $\log_7:$ then $\log_7(a \diamondsuit b) = \log_7 a \cdot \log_7 b$

Big Hint:

Writing $L(x) = \log_7 x,$ the recurrence becomes $L(a_n) = L(a_{n-1}) \cdot \dfrac{L(n)}{L(n-1)}$

Solution:

Let $L(x) = \log_7 x.$ Then $L(a \diamondsuit b) = L(a)L(b)$ and $L(a \heartsuit b) = \dfrac{L(a)}{L(b)}.$

So $L(a_3) = \dfrac{L(3)}{L(2)},$ and $L(a_n) = \dfrac{L(n)}{L(n-1)} \cdot L(a_{n-1}).$ The product telescopes: $\begin{aligned} L(a_N) &= \dfrac{L(3)}{L(2)} \cdot \dfrac{L(N)}{L(3)} \\ &= \dfrac{L(N)}{L(2)}. \end{aligned}$

Hence $L(a_{2019}) = \dfrac{\log_7 2019}{\log_7 2}$ $= \log_2 2019 \approx 10.98,$ which rounds to $11.$

Thus, the correct answer is D.

24.

For how many integers $n$ between $1$ and $50,$ inclusive, is $\dfrac{(n^2 - 1)!}{(n!)^n}$ an integer? (Recall that $0! = 1.$ )

$31$

$32$

$33$

$34$

$35$

Answer: D

Concepts:Legendre’s Formula prime digits

Difficulty rating: 2420

Small Hint:

Compare the exponent of a prime $p$ in numerator and denominator using $v_p(m!) = \dfrac{m - s_p(m)}{p - 1}$

Big Hint:

Failure requires $n$ to be a prime power $p^a$ with $p^a - 1 \lt 2a(p - 1)$

Solution:

Fix a prime $p\le n.$ By Legendre’s formula, the difference between the exponent of $p$ in the numerator and its exponent in the denominator is $\begin{aligned} D_p &=\sum_{k\ge1}\left\lfloor\dfrac{n^2-1}{p^k}\right\rfloor\\ &\quad-n\sum_{k\ge1}\left\lfloor\dfrac{n}{p^k}\right\rfloor. \end{aligned}$ If $r_k$ is the remainder of $n$ modulo $p^k,$ the $k$ th summand is $\left\lfloor\dfrac{nr_k-1}{p^k}\right\rfloor.$

Let $a=v_p(n).$ The first $a$ summands are $-1.$ If $n$ is not a power of $p,$ write $n=p^a m$ with $m\ge2.$ When $a\ge1,$ the next summand is at least $\dfrac{n}{p}-1\ge a,$ and all later summands are nonnegative; when $a=0,$ every summand is already nonnegative. Thus $D_p\ge0$ unless $n$ is a power of $p.$

For $n=p^a,$ Legendre’s formula reduces the requirement to $p^a-1\ge2a(p-1).$ Among prime powers at most $50,$ this fails exactly when $a=1$ (so $n$ is prime) and when $n=2^2=4.$ There are $15$ primes at most $50,$ plus $n=4,$ giving $16$ failures. Hence $50 - 16 = 34$ values of $n$ work.

Thus, the correct answer is D.

25.

Let $\triangle A_0 B_0 C_0$ be a triangle whose angle measures are exactly $59.999^\circ,$ $60^\circ,$ and $60.001^\circ.$ For each positive integer $n$ define $A_n$ to be the foot of the altitude from $A_{n-1}$ to line $B_{n-1}C_{n-1}.$ Likewise, define $B_n$ to be the foot of the altitude from $B_{n-1}$ to line $A_{n-1}C_{n-1},$ and $C_n$ to be the foot of the altitude from $C_{n-1}$ to line $A_{n-1}B_{n-1}.$ What is the least positive integer $n$ for which $\triangle A_n B_n C_n$ is obtuse?

$10$

$11$

$13$

$14$

$15$

Answer: E

Concepts:altitude recursion angle chasing

Difficulty rating: 2520

Small Hint:

For an acute triangle, the orthic triangle has angles $180^\circ - 2\alpha$

Big Hint:

Track the deviation from $60^\circ:$ each step multiplies it by $-2,$ and the triangle turns obtuse when a deviation exceeds $30^\circ$

Solution:

For an acute triangle, the orthic triangle (feet of the altitudes) has angles $180^\circ - 2\alpha$ for each original angle $\alpha.$

Writing an angle as $60^\circ + x,$ the new angle is $60^\circ - 2x,$ so each deviation from $60^\circ$ is multiplied by $-2.$ The initial deviations are $\pm 0.001^\circ.$

After $n$ steps a deviation has magnitude $0.001 \cdot 2^n$ degrees. The triangle first becomes obtuse when this exceeds $30^\circ,$ i.e. $2^n \gt 30000.$ Since $2^{14} = 16384$ and $2^{15} = 32768,$ the least such $n$ is $15.$

Thus, the correct answer is E.