2018 AMC 12A — Problems & Online Timed Exam

Timed

1:15:00

1.

A large urn contains $100$ balls, of which $36\%$ are red and the rest are blue. How many of the blue balls must be removed so that the percentage of red balls in the urn will be $72\%?$ (No red balls are to be removed.)

$28$

$32$

$36$

$50$

$64$

Answer: D

Concepts:percentage

Difficulty rating: 890

Small Hint:

The number of red balls never changes; only blue balls leave

Big Hint:

If $36$ red balls are to be $72\%$ of the urn, the urn must hold $36 \div 0.72$ balls

Solution:

There are $36$ red balls, and this count stays fixed. For the red balls to be $72\%$ of the urn, the urn must contain $36 \div 0.72 = 50$ balls. Since $100 - 50 = 50$ , exactly $50$ blue balls are removed.

Thus, the correct answer is D.

2.

While exploring a cave, Carl comes across a collection of $5$ -pound rocks worth $ $14$ each, $4$ -pound rocks worth $ $11$ each, and $1$ -pound rocks worth $ $2$ each. There are at least $20$ of each size. He can carry at most $18$ pounds. What is the maximum value, in dollars, of the rocks he can carry out of the cave?

$48$

$49$

$50$

$51$

$52$

Answer: C

Concepts:optimization casework

Difficulty rating: 1020

Small Hint:

Compare the value per pound: $\$2.80,$ $\$2.75,$ and $\$2$ for the three rock sizes

Big Hint:

Filling the full $18$ pounds matters more than always grabbing the best per-pound rock; try two $5$ -pound and two $4$ -pound rocks

Solution:

The rocks are worth $ $2.80,$ $ $2.75,$ and $ $2$ per pound, respectively. For each possible number of $5$ -pound rocks, use as many $4$ -pound rocks as fit and fill any leftover capacity with $1$ -pound rocks. Taking $0,1,2,$ or $3$ of the $5$ -pound rocks gives maximum values $ $48,$ $ $49,$ $ $50,$ and $ $48,$ respectively.

Thus two $5$ -pound and two $4$ -pound rocks use all $18$ pounds, and the maximum value is $ $50.$ Thus, the correct answer is C.

3.

How many ways can a student schedule $3$ mathematics courses—algebra, geometry, and number theory—in a $6$ -period day if no two mathematics courses can be taken in consecutive periods? (What courses the student takes during the other $3$ periods is of no concern here.)

$3$

$6$

$12$

$18$

$24$

Answer: E

Concepts:arrangements with restrictions permutations

Difficulty rating: 1130

Small Hint:

First count the sets of $3$ periods out of $6$ with no two chosen periods adjacent

Big Hint:

There are $4$ such sets, and the three distinct courses can be assigned in $3!$ ways

Solution:

The choices of three non-consecutive periods are $\{1,3,5\},$ $\{1,3,6\},$ $\{1,4,6\},$ and $\{2,4,6\},$ a total of $4.$ The three distinct courses can be placed into any such set in $3! = 6$ orders, giving $4 \cdot 6 = 24$ schedules.

Thus, the correct answer is E.

4.

Alice, Bob, and Charlie were on a hike and were wondering how far away the nearest town was. When Alice said, “We are at least $6$ miles away,” Bob replied, “We are at most $5$ miles away.” Charlie then remarked, “Actually the nearest town is at most $4$ miles away.” It turned out that none of the three statements was true. Let $d$ be the distance in miles to the nearest town. Which of the following intervals is the set of all possible values of $d?$

$(0, 4)$

$(4, 5)$

$(4, 6)$

$(5, 6)$

$(5, \infty)$

Answer: D

Concepts:logical deduction inequality

Difficulty rating: 1200

Small Hint:

Each statement is false, so replace each with its negation

Big Hint:

“At least $6$ ” false means $d \lt 6;$ “at most $5$ ” false means $d \gt 5;$ “at most $4$ ” false means $d \gt 4$

Solution:

Negating the three false statements gives $d \lt 6,$ $d \gt 5,$ and $d \gt 4.$ The intersection of these conditions is $5 \lt d \lt 6,$ that is, the interval $(5, 6).$

Thus, the correct answer is D.

5.

What is the sum of all possible values of $k$ for which the polynomials $x^2 - 3x + 2$ and $x^2 - 5x + k$ have a root in common?

$3$

$4$

$5$

$6$

$10$

Answer: E

Concepts:quadratic factoring

Difficulty rating: 1270

Small Hint:

Factor $x^2 - 3x + 2$ to find its two roots

Big Hint:

A common root must be $1$ or $2;$ substitute each into $x^2 - 5x + k$ and solve for $k$

Solution:

Since $x^2 - 3x + 2 = (x-1)(x-2),$ its roots are $1$ and $2.$ If $1$ is a shared root then $1 - 5 + k = 0,$ so $k = 4.$ If $2$ is a shared root then $4 - 10 + k = 0,$ so $k = 6.$ The sum of possible values is $4 + 6 = 10.$

Thus, the correct answer is E.

6.

For positive integers $m$ and $n$ such that $m + 10 \lt n + 1,$ both the mean and the median of the set $\{m, m + 4, m + 10,$ $n + 1,$ $n + 2,$ $2n\}$ are equal to $n.$ What is $m + n?$

$20$

$21$

$22$

$23$

$24$

Answer: B

Concepts:mean median (data)system of equations

Difficulty rating: 1350

Small Hint:

The condition $m + 10 \lt n + 1$ forces the six values to be listed in increasing order

Big Hint:

Set the median $\frac{(m+10)+(n+1)}{2} = n$ and set the mean equal to $n,$ then solve the two equations

Solution:

Because $m + 10 \lt n + 1,$ the six numbers are already increasing, so the median is the average of the middle two: $\frac{(m+10)+(n+1)}{2} = n,$ giving $m = n - 11.$ The mean condition is $\tiny \frac{(n-11)+(n-7)+(n-1)+(n+1)+(n+2)+2n}{6} = n,$ so $7n - 16 = 6n$ and $n = 16.$ Then $m = 5,$ and $m + n = 21.$

Thus, the correct answer is B.

7.

For how many (not necessarily positive) integer values of $n$ is the value of $4000 \cdot \left(\tfrac{2}{5}\right)^n$ an integer?

$3$

$4$

$6$

$8$

$9$

Answer: E

Concepts:prime factorization exponent

Difficulty rating: 1380

Small Hint:

Write $4000 = 2^5 \cdot 5^3,$ so the expression is $2^{5+n} \cdot 5^{3-n}$

Big Hint:

This is an integer exactly when both exponents $5 + n$ and $3 - n$ are nonnegative

Solution:

Since $4000 = 2^5 \cdot 5^3,$ the expression equals $2^{5+n} \cdot 5^{3-n}.$ This is an integer exactly when both $5 + n \ge 0$ and $3 - n \ge 0,$ that is, $-5 \le n \le 3.$ There are $3 - (-5) + 1 = 9$ such integers.

Thus, the correct answer is E.

8.

All of the triangles in the diagram below are similar to isosceles triangle $ABC,$ in which $AB = AC.$ Each of the $7$ smallest triangles has area $1,$ and $\triangle ABC$ has area $40.$ What is the area of trapezoid $DBCE?$

$16$

$18$

$20$

$22$

$24$

Answer: E

Difficulty rating: 1440

Small Hint:

The base $DE$ of triangle $ADE$ spans $4$ small-triangle bases

Big Hint:

Area scales as the square of length, so $[ADE] = 4^2 \cdot 1;$ subtract this from $[ABC] = 40$

Solution:

The base $DE$ of $\triangle ADE$ is $4$ times the base of a smallest triangle, so by the square scaling of similar areas, $[ADE] = 4^2 \cdot 1 = 16.$ The trapezoid $DBCE$ is what remains of $\triangle ABC,$ so its area is $40 - 16 = 24.$

Thus, the correct answer is E.

9.

Which of the following describes the largest subset of values of $y$ within the closed interval $[0, \pi]$ for which $\sin(x + y) \le \sin(x) + \sin(y)$ for every $x$ between $0$ and $\pi,$ inclusive?

$y = 0$

$0 \le y \le \tfrac{\pi}{4}$

$0 \le y \le \tfrac{\pi}{2}$

$0 \le y \le \tfrac{3\pi}{4}$

$0 \le y \le \pi$

Answer: E

Concepts:trigonometric identity inequality

Difficulty rating: 1500

Small Hint:

Expand $\sin(x+y)$ $= \sin x \cos y + \cos x \sin y$

Big Hint:

For $x, y \in [0, \pi]$ both $\sin x, \sin y \ge 0$ and $\cos x, \cos y \le 1;$ compare term by term

Solution:

For $0 \le x \le \pi$ and $0 \le y \le \pi$ we have $\sin x \ge 0,$ $\sin y \ge 0,$ $\cos x \le 1,$ and $\cos y \le 1.$ Hence $\begin{aligned} \sin(x+y) &= \sin x \cos y \\ &\quad {}+ \cos x \sin y \\ &\le \sin x + \sin y. \end{aligned}$ The inequality therefore holds for every $y$ with $0 \le y \le \pi.$

Thus, the correct answer is E.

10.

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

$x + 3y = 3$ $\big|\,|x| - |y|\,\big| = 1$

$1$

$2$

$3$

$4$

$8$

Answer: C

Concepts:absolute value system of equations casework

Difficulty rating: 1560

Small Hint:

The equation $\big|\,|x| - |y|\,\big| = 1$ means $|x| - |y| = \pm 1,$ which unfolds into $x = \pm y \pm 1$

Big Hint:

Solve $x + 3y = 3$ together with each of $x = y + 1,$ $x = y - 1,$ $x = -y + 1,$ $x = -y - 1,$ then discard repeats

Solution:

The second equation gives $|x| - |y| = \pm 1,$ equivalently $x = \pm y \pm 1.$ Substituting into $x + 3y = 3:$

If $x = y + 1,$ then $(x, y) = \left(\tfrac32, \tfrac12\right).$ If $x = y - 1,$ then $(x, y) = (0, 1).$ If $x = -y + 1,$ then again $(x, y) = (0, 1).$ If $x = -y - 1,$ then $(x, y) = (-3, 2).$

The distinct solutions are $(-3, 2),$ $(0, 1),$ and $\left(\tfrac32, \tfrac12\right),$ all of which check, so there are $3.$

Thus, the correct answer is C.

11.

A paper triangle with sides of lengths $3,$ $4,$ and $5$ inches, as shown, is folded so that point $A$ falls on point $B.$ What is the length in inches of the crease?

$1 + \tfrac12 \sqrt{2}$

$\sqrt{3}$

$\tfrac{7}{4}$

$\tfrac{15}{8}$

$2$

Answer: D

Concepts:paper folding perpendicular bisector similarity

Difficulty rating: 1570

Small Hint:

Folding $A$ onto $B$ creases along the perpendicular bisector of $AB;$ since $AC \gt BC$ it meets $AC$

Big Hint:

With $D$ the midpoint of $AB$ and $E$ on $AC,$ triangle $ADE \sim$ triangle $ACB,$ so $DE = AD \cdot \tfrac{CB}{AC}$

Solution:

The crease lies along the perpendicular bisector of $AB,$ meeting $AC$ at $E$ because $AC \gt BC.$ Let $D$ be the midpoint of $AB,$ so $AD = \tfrac52$ and $\triangle ADE$ is right-angled at $D.$ Since $\triangle ADE \sim \triangle ACB,$ we have $\tfrac{DE}{AD} = \tfrac{CB}{AC} = \tfrac34,$ so $DE = \frac52 \cdot \frac34 = \frac{15}{8}.$

Thus, the correct answer is D.

12.

Let $S$ be a set of $6$ integers taken from $\{1, 2, \ldots, 12\}$ with the property that if $a$ and $b$ are elements of $S$ with $a \lt b,$ then $b$ is not a multiple of $a.$ What is the least possible value of an element of $S?$

$2$

$3$

$4$

$5$

$7$

Answer: C

Concepts:divisibility pigeonhole principle extremal argument

Difficulty rating: 1630

Small Hint:

Group $\{1, \ldots, 12\}$ into chains where each number divides the next: $\{1,2,4,8\},$ $\{3,6,12\},$ $\{5,10\},$ $\{7\},$ $\{9\},$ $\{11\}$

Big Hint:

At most one element comes from each chain; with $6$ chains, $S$ must use exactly one from each, forcing $7, 9, 11 \in S$

Solution:

Partition $\{1, \ldots, 12\}$ into the six divisibility chains $\{1,2,4,8\},$ $\{3,6,12\},$ $\{5,10\},$ $\{7\},$ $\{9\},$ $\{11\}.$ Since no element of $S$ may divide another, at most one comes from each chain; needing $6$ elements forces exactly one from each, so $7, 9, 11 \in S.$

Because $9 \in S,$ $3 \notin S,$ so the second chain contributes $6$ or $12,$ and then neither $1$ nor $2$ can be chosen from the first chain (they divide $6$ and $12$ ). Taking $4$ from the first chain works: $S = \{4, 5, 6, 7, 9, 11\}$ has the property. Hence the least possible element is $4.$

Thus, the correct answer is C.

13.

How many nonnegative integers can be written in the form

$\begin{aligned} &a_7 \cdot 3^7 + a_6 \cdot 3^6 + a_5 \cdot 3^5 \\ &\quad {}+ a_4 \cdot 3^4 + a_3 \cdot 3^3 + a_2 \cdot 3^2 \\ &\quad {}+ a_1 \cdot 3^1 + a_0 \cdot 3^0, \end{aligned}$

where $a_i \in \{-1, 0, 1\}$ for $0 \le i \le 7?$

$512$

$729$

$1094$

$3281$

$59{,}048$

Answer: D

Concepts:number base bijection symmetry

Difficulty rating: 1660

Small Hint:

Shifting each $a_i$ from $\{-1, 0, 1\}$ to $\{0, 1, 2\}$ makes an ordinary base- $3$ numeral

Big Hint:

The $3^8$ representable integers are symmetric about $0;$ count $0$ together with half of the remaining values

Solution:

Adding $1$ to every $a_i$ gives a bijection between these expressions and the base- $3$ numerals for $0$ through $3^8 - 1,$ so exactly $3^8 = 6561$ distinct integers occur. They are symmetric about $0$ (negating all $a_i$ negates the value), so besides $0$ itself, half are positive: $1 + \tfrac12(6561 - 1) = 3281$ nonnegative integers, namely $0$ through $3280.$

Thus, the correct answer is D.

14.

The solution to the equation $\log_{3x} 4 = \log_{2x} 8,$ where $x$ is a positive real number other than $\tfrac13$ or $\tfrac12,$ can be written as $\tfrac{p}{q},$ where $p$ and $q$ are relatively prime positive integers. What is $p + q?$

$5$

$13$

$17$

$31$

$35$

Answer: D

Concepts:logarithm

Difficulty rating: 1730

Small Hint:

Rewrite both sides with a common base: $\log_{3x} 4 = \tfrac{\log 4}{\log 3x},$ and similarly for $\log_{2x} 8$

Big Hint:

Cross-multiplying reduces the equation to $(2x)^2 = (3x)^3;$ solve for $x$ as a reduced fraction

Solution:

Writing both logarithms in base $2:$ $\tfrac{2}{\log_2 3x} = \tfrac{3}{\log_2 2x},$ so $2 \log_2 2x = 3 \log_2 3x,$ i.e. $(2x)^2 = (3x)^3.$ Then $4x^2 = 27x^3,$ giving $x = \tfrac{4}{27}.$ Since $\gcd(4, 27) = 1,$ we get $p + q = 4 + 27 = 31.$

Thus, the correct answer is D.

15.

A scanning code consists of a $7 \times 7$ grid of squares, with some of its squares colored black and the rest colored white. There must be at least one square of each color in this grid of $49$ squares. A scanning code is called symmetric if its look does not change when the entire square is rotated by a multiple of $90^\circ$ counterclockwise around its center, nor when it is reflected across a line joining opposite corners or a line joining midpoints of opposite sides. What is the total number of possible symmetric scanning codes?

$510$

$1022$

$8190$

$8192$

$65{,}534$

Answer: B

Concepts:symmetry basic counting

Difficulty rating: 1800

Small Hint:

The required symmetries force every square in one rotation-reflection orbit to share a color

Big Hint:

There are $10$ independent orbits, each freely black or white; exclude the all-black and all-white codes

Solution:

Under the symmetry group of the square, the $49$ cells break into orbits, and every cell in an orbit must have the same color. Give a cell coordinates $(i,j)$ relative to the center, where $-3\le i,j\le3.$ Rotations and reflections can change signs and interchange the coordinates, so each orbit has one representative with $0\le i\le j\le3.$ There are $1+2+3+4=10$ such pairs. Each orbit is black or white, giving $2^{10}$ colorings, but the all-black and all-white grids are excluded. So there are $2^{10} - 2 = 1022$ symmetric scanning codes.

Thus, the correct answer is B.

16.

Which of the following describes the set of values of $a$ for which the curves $x^2 + y^2 = a^2$ and $y = x^2 - a$ in the real $xy$ -plane intersect at exactly $3$ points?

$a = \tfrac14$

$\tfrac14 \lt a \lt \tfrac12$

$a \gt \tfrac14$

$a = \tfrac12$

$a \gt \tfrac12$

Answer: E

Concepts:circle parabola quadratic

Difficulty rating: 1840

Small Hint:

From $y = x^2 - a$ substitute $x^2 = y + a$ into the circle equation to get a quadratic in $y$

Big Hint:

One intersection is the vertex tangency at $x = 0;$ two more require $x^2 = 2a - 1$ to have real nonzero solutions

Solution:

Substituting $x^2 = y + a$ into $x^2 + y^2 = a^2$ gives $y^2 + y + (a - a^2) = 0,$ which factors as $(y + 1 - a)(y + a) = 0,$ so $y = a - 1$ or $y = -a.$ These correspond to $x^2 = 2a - 1$ and $x^2 = 0.$

The equation $x^2 = 0$ always gives the single point $(0, -a),$ the vertex of the parabola. The equation $x^2 = 2a - 1$ gives two more points exactly when $2a - 1 \gt 0,$ i.e. $a \gt \tfrac12.$ So there are $3$ intersection points precisely when $a \gt \tfrac12.$

Thus, the correct answer is E.

17.

Farmer Pythagoras has a field in the shape of a right triangle. The right triangle’s legs have lengths of $3$ and $4$ units. In the corner where those sides meet at a right angle, he leaves a small unplanted square $S$ so that from the air it looks like the right angle symbol. The rest of the field is planted. The shortest distance from $S$ to the hypotenuse is $2$ units. What fraction of the field is planted?

$\tfrac{25}{27}$

$\tfrac{26}{27}$

$\tfrac{73}{75}$

$\tfrac{145}{147}$

$\tfrac{74}{75}$

Answer: D

Concepts:coordinate geometry distance formula area

Difficulty rating: 1910

Small Hint:

Put the right angle at the origin so the hypotenuse is the line $3x + 4y = 12$

Big Hint:

The far corner $(s, s)$ of the square is distance $2$ from that line; solve $\tfrac{|3s + 4s - 12|}{5} = 2$ and keep the valid root

Solution:

Place the right angle at the origin with legs on the axes, so the vertices are $(4, 0),$ $(0, 3),$ $(0, 0),$ and the square $S$ is $[0, s] \times [0, s].$ The hypotenuse is $3x + 4y - 12 = 0,$ and the distance from its nearest corner $(s, s)$ is $\frac{|3s + 4s - 12|}{\sqrt{3^2 + 4^2}} = \frac{|7s - 12|}{5} = 2.$ This gives $s = \tfrac{22}{7}$ or $s = \tfrac27;$ only $s = \tfrac27$ keeps the square inside the triangle.

The field has area $\tfrac12 \cdot 3 \cdot 4 = 6$ and the unplanted square has area $\left(\tfrac27\right)^2 = \tfrac{4}{49}.$ The planted fraction is $1 - \frac{\frac{4}{49}}{6} = 1 - \frac{2}{147} = \frac{145}{147}.$

Thus, the correct answer is D.

18.

Triangle $ABC$ with $AB = 50$ and $AC = 10$ has area $120.$ Let $D$ be the midpoint of $\overline{AB},$ and let $E$ be the midpoint of $\overline{AC}.$ The angle bisector of $\angle BAC$ intersects $\overline{DE}$ and $\overline{BC}$ at $F$ and $G,$ respectively. What is the area of quadrilateral $FDBG?$

$60$

$65$

$70$

$75$

$80$

Answer: D

Concepts:angle bisector theorem area ratio midpoint

Difficulty rating: 1990

Small Hint:

Since $D, E$ are midpoints, $[ADE] = \tfrac14 [ABC],$ so trapezoid $EDBC$ is the remaining $\tfrac34$

Big Hint:

By the Angle Bisector Theorem the bisector cuts $DE$ and $BC$ in ratio $AB : AC = 5 : 1,$ making $FDBG$ equal to $\tfrac56$ of $EDBC$

Solution:

Since $D$ and $E$ are midpoints, $\triangle ADE$ has $\tfrac14$ the area of $\triangle ABC,$ namely $30,$ so trapezoid $EDBC$ has area $120 - 30 = 90.$

By the Angle Bisector Theorem, $G$ divides $BC$ with $BG = \tfrac{AB}{AB + AC} \cdot BC = \tfrac56 BC,$ and likewise $F$ divides $DE$ so that $DF = \tfrac56 DE.$ Because $FDBG$ and $EDBC$ share the same height, the area of $FDBG$ is $\tfrac56$ of the area of $EDBC:$ $\tfrac56 \cdot 90 = 75.$

Thus, the correct answer is D.

19.

Let $A$ be the set of positive integers that have no prime factors other than $2,$ $3,$ or $5.$ The infinite sum $\begin{aligned} &\frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} \\ &\quad {}+ \frac{1}{5} + \frac{1}{6} + \frac{1}{8} + \frac{1}{9} \\ &\quad {}+ \frac{1}{10} + \frac{1}{12} + \frac{1}{15} + \frac{1}{16} \\ &\quad {}+ \frac{1}{18} + \frac{1}{20} + \cdots \end{aligned}$ of the reciprocals of all the elements of $A$ can be expressed as $\tfrac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. What is $m + n?$

$16$

$17$

$19$

$23$

$36$

Answer: C

Concepts:geometric sequence summation prime factorization

Difficulty rating: 1930

Small Hint:

Every element is $2^i 3^j 5^k,$ so the sum factors as a product of three separate geometric series

Big Hint:

Evaluate $\dfrac{1}{1 - \frac12} \cdot \dfrac{1}{1 - \frac13} \cdot \dfrac{1}{1 - \frac15}$ and reduce the result

Solution:

Each element of $A$ is uniquely $2^i 3^j 5^k$ with $i, j, k \ge 0,$ so summing all reciprocals factors as $\begin{aligned} &\left(\sum_{i \ge 0} \tfrac{1}{2^i}\right) \\ &\quad {}\cdot \left(\sum_{j \ge 0} \tfrac{1}{3^j}\right) \\ &\quad {}\cdot \left(\sum_{k \ge 0} \tfrac{1}{5^k}\right) \\ &= \frac{1}{1 - \frac12} \cdot \frac{1}{1 - \frac13} \\ &\quad {}\cdot \frac{1}{1 - \frac15}. \end{aligned}$ This equals $2 \cdot \tfrac32 \cdot \tfrac54 = \tfrac{15}{4}.$ With $\gcd(15, 4) = 1,$ $m + n = 15 + 4 = 19.$

Thus, the correct answer is C.

20.

Triangle $ABC$ is an isosceles right triangle with $AB = AC = 3.$ Let $M$ be the midpoint of hypotenuse $\overline{BC}.$ Points $I$ and $E$ lie on sides $\overline{AC}$ and $\overline{AB},$ respectively, so that $AI \gt AE$ and $AIME$ is a cyclic quadrilateral. Given that triangle $EMI$ has area $2,$ the length $CI$ can be written as $\tfrac{a - \sqrt{b}}{c},$ where $a,$ $b,$ and $c$ are positive integers and $b$ is not divisible by the square of any prime. What is the value of $a + b + c?$

$9$

$10$

$11$

$12$

$13$

Answer: D

Concepts:cyclic quadrilateral law of cosines triangle area

Difficulty rating: 2110

Small Hint:

Because $AIME$ is cyclic and $\angle A = 90^\circ,$ the opposite angle $\angle IME = 90^\circ$ as well

Big Hint:

With $x = CI,$ $y = BE,$ use the Law of Cosines for $IM, ME$ (base angles $45^\circ$ ), the area $\tfrac12 IM \cdot ME = 2,$ and the relation $x + y = 3$

Solution:

Since $\triangle ABC$ is an isosceles right triangle, $CM = BM = \tfrac32 \sqrt2$ and the base angles at $B, C$ are $45^\circ.$ As $AIME$ is cyclic with right angle at $A,$ angle $\angle IME = 90^\circ.$ Let $x = CI$ and $y = BE.$ By the Law of Cosines in $\triangle MCI,$ $\begin{aligned} IM^2 &= x^2 + \tfrac92 \\ &\quad {}- 2 \cdot x \cdot \tfrac32\sqrt2 \cdot \cos 45^\circ \\ &= x^2 - 3x + \tfrac92, \end{aligned}$ and similarly $ME^2 = y^2 - 3y + \tfrac92.$

The Pythagorean Theorem in right triangles $EMI$ and $IAE$ gives $IM^2 + ME^2$ $= (3-x)^2$ $+ (3-y)^2,$ which simplifies to $x + y = 3.$ The area condition $\tfrac12 IM \cdot ME = 2$ means $IM^2 \cdot ME^2 = 16.$ Substituting $y = 3 - x$ makes $ME^2 = x^2 - 3x + \tfrac92 = IM^2,$ so $\left(x^2 - 3x + \tfrac92\right)^2 = 16,$ hence $x^2 - 3x + \tfrac92 = 4,$ i.e. $x^2 - 3x + \tfrac12 = 0.$

Since $AI \gt AE$ forces $y \gt x,$ we take the smaller root $x = \tfrac{3 - \sqrt{7}}{2}.$ Then $a + b + c = 3 + 7 + 2 = 12.$

Thus, the correct answer is D.

21.

Which of the following polynomials has the greatest real root?

$x^{19} + 2018x^{11} + 1$

$x^{17} + 2018x^{11} + 1$

$x^{19} + 2018x^{13} + 1$

$x^{17} + 2018x^{13} + 1$

$2019x + 2018$

Answer: B

Concepts:polynomial inequality bounding to limit cases

Difficulty rating: 2210

Small Hint:

By Descartes’ Rule of Signs each polynomial has exactly one real root, and it lies in $(-1, 0)$

Big Hint:

On $(-1, 0)$ a polynomial with larger values has a smaller root; note $x^{19} \gt x^{17}$ and $x^{13} \gt x^{11}$ there

Solution:

Each polynomial in choices A–D has no positive root and exactly one negative root, which lies in $(-1, 0)$ (it is positive at $0$ and negative at $-1$ ) and is increasing there. On the interval $(-1, 0),$ $x^{19} \gt x^{17}$ and $x^{13} \gt x^{11}.$ Thus each of A, C, and D has a larger value than B at every point of this interval, so each crosses $0$ to the left of B. Therefore B has the greatest root among A–D.

The linear choice E has root $e=-\tfrac{2018}{2019}.$ Since $\tfrac{2018}{2019}\gt\tfrac9{10},$ we have $e^{17}+2018e^{11}+1$ $\lt-2018\left(\tfrac9{10}\right)^{11}+1$ $\lt0.$ Because B is increasing and is negative at $e,$ its root lies to the right of E’s. Hence B has the greatest real root.

Thus, the correct answer is B.

22.

The solutions to the equations $z^2 = 4 + 4\sqrt{15}\,i$ and $z^2 = 2 + 2\sqrt{3}\,i,$ where $i = \sqrt{-1},$ form the vertices of a parallelogram in the complex plane. The area of this parallelogram can be written in the form $p\sqrt{q} - r\sqrt{s},$ where $p,$ $q,$ $r,$ and $s$ are positive integers and neither $q$ nor $s$ is divisible by the square of any prime number. What is $p + q + r + s?$

$20$

$21$

$22$

$23$

$24$

Answer: A

Concepts:complex number shoelace formula

Difficulty rating: 2270

Small Hint:

Solve $z^2 = 4 + 4\sqrt{15}\,i$ by writing $z = a + bi$ and matching real and imaginary parts; do the same for the other equation

Big Hint:

The four vertices are $\pm z_1$ and $\pm z_2;$ apply the shoelace formula to their coordinates

Solution:

Writing $z = a + bi$ with $(a + bi)^2 = 4 + 4\sqrt{15}\,i$ gives $a^2 - b^2 = 4$ and $2ab = 4\sqrt{15}.$ Then $a^4 - 4a^2 - 60 = 0,$ so $(a^2 - 10)(a^2 + 6) = 0,$ yielding $a = \pm\sqrt{10},$ $b = \pm\sqrt{6}.$ The vertices from the first equation are $\pm(\sqrt{10} + \sqrt{6}\,i).$ The same method on $z^2 = 2 + 2\sqrt{3}\,i$ gives $\pm(\sqrt{3} + i).$

Applying the shoelace formula to $(\sqrt{10}, \sqrt{6}),$ $(\sqrt{3}, 1),$ $(-\sqrt{10}, -\sqrt{6}),$ $(-\sqrt{3}, -1)$ gives area $6\sqrt{2} - 2\sqrt{10}.$ Thus $p + q + r + s$ $= 6 + 2 + 2 + 10$ $= 20.$

Thus, the correct answer is A.

23.

In $\triangle PAT,$ $\angle P = 36^\circ,$ $\angle A = 56^\circ,$ and $PA = 10.$ Points $U$ and $G$ lie on sides $\overline{TP}$ and $\overline{TA},$ respectively, so that $PU = AG = 1.$ Let $M$ and $N$ be the midpoints of segments $\overline{PA}$ and $\overline{UG},$ respectively. What is the degree measure of the acute angle formed by lines $MN$ and $PA?$

$76$

$77$

$78$

$79$

$80$

Answer: E

Concepts:angle chasing midpoint parallelogram

Difficulty rating: 2370

Small Hint:

Extend $PN$ to $Q$ with $PN = NQ;$ then $UPGQ$ is a parallelogram, so $GQ \parallel PT$ and $GQ = PU = AG$

Big Hint:

Triangle $QGA$ is isosceles, and $MN$ is a midline of triangle $QPA,$ so $\angle NMA = \angle QAP$

Solution:

Extend $PN$ through $N$ to $Q$ with $PN = NQ.$ Since $N$ is the midpoint of $UG$ and of $PQ,$ the quadrilateral $UPGQ$ is a parallelogram, so $GQ \parallel PT$ and $GQ = PU = 1 = AG.$ Then $\angle QGA$ $= 180^\circ - \angle T$ $= \angle P + \angle A$ $= 36^\circ + 56^\circ = 92^\circ,$ and the isosceles triangle $QGA$ gives $\angle QAG = \tfrac12(180^\circ - 92^\circ) = 44^\circ.$

Because $M, N$ are midpoints, $MN$ is a midline of $\triangle QPA,$ so $MN \parallel AQ$ and $\begin{aligned} \angle NMP &= \angle QAP \\ &= \angle QAG + \angle GAP \\ &= 44^\circ + 56^\circ = 100^\circ. \end{aligned}$ The acute angle between line $MN$ and $PA$ is therefore $180^\circ - 100^\circ = 80^\circ.$

Thus, the correct answer is E.

24.

Alice, Bob, and Carol play a game in which each of them chooses a real number between $0$ and $1.$ The winner of the game is the one whose number is between the numbers chosen by the other two players. Alice announces that she will choose her number uniformly at random from all the numbers between $0$ and $1,$ and Bob announces that he will choose his number uniformly at random from all the numbers between $\tfrac12$ and $\tfrac23.$ Armed with this information, what number should Carol choose to maximize her chance of winning?

$\tfrac12$

$\tfrac{13}{24}$

$\tfrac{7}{12}$

$\tfrac58$

$\tfrac23$

Answer: B

Concepts:geometric probability casework optimization

Difficulty rating: 2520

Small Hint:

Split into cases by where Carol’s number $c$ sits relative to $\tfrac12$ and $\tfrac23;$ for $\tfrac12 \lt c \lt \tfrac23$ she can win in two ways

Big Hint:

In that middle range her win probability is $-12c^2 + 13c - 3;$ maximize this downward quadratic at its vertex $c = \tfrac{-b}{2a}$

Solution:

If $c \le \tfrac12,$ Carol beats Bob automatically, so she wins only if Alice is below $c,$ probability $c \le \tfrac12.$ If $c \ge \tfrac23,$ she wins with probability $1 - c \le \tfrac13.$ Neither case exceeds $\tfrac12.$

For $\tfrac12 \lt c \lt \tfrac23,$ the chance Bob’s number exceeds $c$ is $\frac{\frac{2}{3} - c}{\frac{2}{3} - \frac{1}{2}} = 4 - 6c,$ so the probability Carol is above Alice and below Bob is $c(4 - 6c);$ the reverse ordering has probability $(1 - c)(6c - 3).$ Adding, $\begin{aligned} &c(4 - 6c) + (1 - c)(6c - 3) \\ &= -12c^2 + 13c - 3. \end{aligned}$ This downward parabola is maximized at $c = \tfrac{13}{24},$ which lies in $\left(\tfrac12, \tfrac23\right),$ and its value exceeds $\tfrac12.$

Thus, the correct answer is B.

25.

For a positive integer $n$ and nonzero digits $a,$ $b,$ and $c,$ let $A_n$ be the $n$ -digit integer each of whose digits is equal to $a;$ let $B_n$ be the $n$ -digit integer each of whose digits is equal to $b;$ and let $C_n$ be the $2n$ -digit (not $n$ -digit) integer each of whose digits is equal to $c.$ What is the greatest possible value of $a + b + c$ for which there are at least two values of $n$ such that $C_n - B_n = A_n^2?$

$12$

$14$

$16$

$18$

$20$

Answer: D

Concepts:digits algebraic manipulation

Difficulty rating: 2650

Small Hint:

Write $A_n = a \cdot \tfrac{10^n - 1}{9},$ and likewise $B_n$ and $C_n = c \cdot \tfrac{10^{2n} - 1}{9},$ then substitute into $C_n - B_n = A_n^2$

Big Hint:

Requiring it for two values of $n$ forces the coefficient of $10^n$ to vanish: $9c = a^2$ and $9b - 9c = a^2$

Solution:

Using $A_n = a \cdot \tfrac{10^n - 1}{9},$ $B_n = b \cdot \tfrac{10^n - 1}{9},$ and $C_n = c \cdot \tfrac{10^{2n} - 1}{9},$ the equation $C_n - B_n = A_n^2$ becomes, after dividing by $10^n - 1$ and clearing fractions, $(9c - a^2) \cdot 10^n = 9b - 9c - a^2.$ For this to hold at two different $n,$ the coefficient of $10^n$ must be zero, so $9c = a^2$ and hence $9b - 9c - a^2 = 0.$

Then $c = \tfrac{a^2}{9}$ and $b = 2c.$ So $a \in \{3, 6, 9\}$ with $c \in \{1, 4, 9\}$ and $b \in \{2, 8, 18\};$ the case $b = 18$ is not a digit. The valid triples are $(a, b, c) = (3, 2, 1)$ and $(6, 8, 4),$ and indeed $4444 - 88 = 4356 = 66^2.$ The greater digit sum is $6 + 8 + 4 = 18.$

Thus, the correct answer is D.