2020 AMC 12B — Problems & Online Timed Exam

Timed

1:15:00

1.

What is the value in simplest form of the following expression?

$\begin{aligned} &\sqrt{1} + \sqrt{1+3} + \sqrt{1+3+5} \\ &\quad {}+ \sqrt{1+3+5+7} \end{aligned}$

$5$

$4 + \sqrt{7} + \sqrt{10}$

$10$

$15$

$4 + 3\sqrt{3} + 2\sqrt{5} + \sqrt{7}$

Answer: C

Concepts:sum of first n odd numbers perfect square radical

Difficulty rating: 890

Small Hint:

The sum of the first $k$ odd numbers is $k^2$

Big Hint:

Each term simplifies to $\sqrt{1},$ $\sqrt{4},$ $\sqrt{9},$ and $\sqrt{16}$

Solution:

The sum of the first $k$ odd numbers equals $k^2,$ so each radicand is a perfect square: $\begin{aligned} &\sqrt{1} + \sqrt{4} + \sqrt{9} + \sqrt{16} \\ &= 1 + 2 + 3 + 4 = 10. \end{aligned}$

Thus, the correct answer is C.

2.

What is the value of the following expression?

$\frac{100^2 - 7^2}{70^2 - 11^2} \cdot \frac{(70 - 11)(70 + 11)}{(100 - 7)(100 + 7)}$

$1$

$\dfrac{9951}{9950}$

$\dfrac{4780}{4779}$

$\dfrac{108}{107}$

$\dfrac{81}{80}$

Answer: A

Concepts:difference of squares

Difficulty rating: 1020

Small Hint:

$a^2 - b^2 = (a - b)(a + b)$

Big Hint:

Factor $100^2 - 7^2$ and $70^2 - 11^2$ , then look for cancellation

Solution:

Using the difference of squares, $100^2 - 7^2 = (100 - 7)(100 + 7)$ and $70^2 - 11^2 = (70 - 11)(70 + 11).$ The expression becomes $\begin{aligned} &\frac{(100 - 7)(100 + 7)}{(70 - 11)(70 + 11)} \\ &\quad {}\cdot \frac{(70 - 11)(70 + 11)}{(100 - 7)(100 + 7)} = 1. \end{aligned}$

Thus, the correct answer is A.

3.

The ratio of $w$ to $x$ is $4 : 3,$ the ratio of $y$ to $z$ is $3 : 2,$ and the ratio of $z$ to $x$ is $1 : 6.$ What is the ratio of $w$ to $y?$

$4 : 3$

$3 : 2$

$8 : 3$

$4 : 1$

$16 : 3$

Answer: E

Concepts:ratio and proportion

Difficulty rating: 1130

Small Hint:

Pick a convenient value such as $x = 6$ and work out $w, z,$ then $y$

Big Hint:

From $z : x = 1 : 6,$ take $z = 1, x = 6;$ then $w = \tfrac43 x$ and $y = \tfrac32 z$

Solution:

Let $x = 6.$ From $z : x = 1 : 6,$ we get $z = 1.$ From $w : x = 4 : 3,$ we get $w = \tfrac43 \cdot 6 = 8.$ From $y : z = 3 : 2,$ we get $y = \tfrac32 \cdot 1 = \tfrac32.$

Therefore $w : y = 8 : \tfrac32 = 16 : 3.$

Thus, the correct answer is E.

4.

The acute angles of a right triangle are $a^\circ$ and $b^\circ,$ where $a \gt b$ and both $a$ and $b$ are prime numbers. What is the least possible value of $b?$

$2$

$3$

$5$

$7$

$11$

Answer: D

Concepts:prime angle sum

Difficulty rating: 1220

Small Hint:

The acute angles satisfy $a + b = 90$

Big Hint:

Try the smallest primes for $b$ and check whether $90 - b$ is also prime

Solution:

Since the angles are complementary, $a + b = 90.$ To minimize $b,$ try small primes and require $90 - b$ to be prime as well.

For $b = 2, 3, 5,$ the value $90 - b = 88, 87, 85$ is not prime. For $b = 7,$ we get $90 - 7 = 83,$ which is prime. So the least possible value is $b = 7.$

Thus, the correct answer is D.

5.

Teams $A$ and $B$ are playing in a basketball league where each game results in a win for one team and a loss for the other team. Team $A$ has won $\tfrac23$ of its games and team $B$ has won $\tfrac58$ of its games. Also, team $B$ has won $7$ more games and lost $7$ more games than team $A.$ How many games has team $A$ played?

$21$

$27$

$42$

$48$

$63$

Answer: C

Concepts:system of equations fraction

Difficulty rating: 1290

Small Hint:

Let $A$ play $a$ games and $B$ play $b$ games, then write both the win and loss conditions

Big Hint:

Wins: $\tfrac58 b = \tfrac23 a + 7;$ losses: $\tfrac38 b = \tfrac13 a + 7$

Solution:

Let $a$ be the number of games team $A$ played and $b$ the number team $B$ played. Team $A$ wins $\tfrac23 a$ and loses $\tfrac13 a;$ team $B$ wins $\tfrac58 b$ and loses $\tfrac38 b.$ The conditions give $\tfrac58 b = \tfrac23 a + 7$ and $\tfrac38 b = \tfrac13 a + 7.$

Subtracting the equations gives $\tfrac14 b = \tfrac13 a,$ so $b = \tfrac43 a.$ Substituting into the loss equation: $\tfrac38 \cdot \tfrac43 a = \tfrac13 a + 7,$ i.e. $\tfrac12 a = \tfrac13 a + 7,$ so $\tfrac16 a = 7$ and $a = 42.$

Thus, the correct answer is C.

6.

For all integers $n \ge 9,$ the value of

$\frac{(n + 2)! - (n + 1)!}{n!}$

is always which of the following?

a multiple of $4$

a multiple of $10$

a prime number

a perfect square

a perfect cube

Answer: D

Concepts:factorial factoring perfect square

Difficulty rating: 1270

Small Hint:

Factor $(n + 1)!$ out of the numerator

Big Hint:

$(n + 2)! - (n + 1)!$ $= (n + 1)!\,\big[(n + 2) - 1\big]$ $= (n + 1)!\,(n + 1)$

Solution:

Factor $(n + 1)!$ from the numerator: $\begin{gathered} (n + 2)! - (n + 1)! \\ {}= (n + 1)!\,\big[(n + 2) - 1\big] \\ {}= (n + 1)!\,(n + 1). \end{gathered}$

Dividing by $n!$ leaves $\dfrac{(n + 1)!\,(n + 1)}{n!}$ $= (n + 1)(n + 1)$ $= (n + 1)^2,$ which is always a perfect square.

Thus, the correct answer is D.

7.

Two nonhorizontal, non-vertical lines in the $xy$ -coordinate plane intersect to form a $45^\circ$ angle. One line has slope equal to $6$ times the slope of the other line. What is the greatest possible value of the product of the slopes of the two lines?

$\dfrac16$

$\dfrac23$

$\dfrac32$

$3$

$6$

Answer: C

Concepts:slope quadratic optimization

Difficulty rating: 1410

Small Hint:

The tangent of the angle between lines of slopes $m_1, m_2$ is $\left|\dfrac{m_1 - m_2}{1 + m_1 m_2}\right|$

Big Hint:

With slopes $m$ and $6m,$ set $\left|\dfrac{5m}{1 + 6m^2}\right| = 1$ and maximize $6m^2$

Solution:

Let the slopes be $m$ and $6m.$ The angle between the lines satisfies $\left|\frac{6m - m}{1 + 6m^2}\right| = \tan 45^\circ = 1,$ so $5m = \pm(1 + 6m^2),$ giving $6m^2 - 5m + 1 = 0$ or $6m^2 + 5m + 1 = 0.$

The first yields $m = \tfrac12$ or $m = \tfrac13;$ the second yields the negatives of these. The product of the slopes is $6m^2,$ which is largest when $m = \tfrac12,$ giving $6 \cdot \tfrac14 = \tfrac32.$

Thus, the correct answer is C.

8.

How many ordered pairs of integers $(x, y)$ satisfy the equation

$x^{2020} + y^2 = 2y?$

$1$

$2$

$3$

$4$

infinitely many

Answer: D

Concepts:completing the square Diophantine Equation bounding to limit cases

Difficulty rating: 1410

Small Hint:

Complete the square in $y:$ $x^{2020} + (y - 1)^2 = 1$

Big Hint:

Since $x^{2020} \ge 0$ and $(y - 1)^2 \ge 0,$ only $x \in \{-1, 0, 1\}$ can work

Solution:

Completing the square gives $x^{2020} + (y - 1)^2 = 1.$ Both terms are nonnegative, so $x^{2020} \le 1,$ forcing $x \in \{-1, 0, 1\}.$

If $x = 0,$ then $(y - 1)^2 = 1,$ giving $y = 0$ or $y = 2.$ If $x = \pm 1,$ then $x^{2020} = 1,$ so $(y - 1)^2 = 0$ and $y = 1.$ The solutions are $(0, 0), (0, 2), (1, 1),$ and $(-1, 1)$ — four in all.

Thus, the correct answer is D.

9.

A three-quarter sector of a circle of radius $4$ inches together with its interior can be rolled up to form the lateral surface of a right circular cone by taping together along the two radii shown. What is the volume of the cone in cubic inches?

$3\pi \sqrt{5}$

$4\pi \sqrt{3}$

$3\pi \sqrt{7}$

$6\pi \sqrt{3}$

$6\pi \sqrt{7}$

Answer: C

Concepts:cone volume Pythagorean Theorem

Difficulty rating: 1470

Small Hint:

The arc length of the sector becomes the circumference of the cone’s base

Big Hint:

The sector radius $4$ becomes the slant height; find the base radius, then the cone’s height

Solution:

The sector’s arc length is $\tfrac34 \cdot 2\pi \cdot 4 = 6\pi,$ which becomes the base circumference: $2\pi r = 6\pi,$ so $r = 3.$

The slant height is the sector radius $4,$ so the height is $h = \sqrt{4^2 - 3^2} = \sqrt{7}.$ The volume is $\frac13 \pi r^2 h = \frac13 \pi \cdot 9 \cdot \sqrt{7} = 3\pi\sqrt{7}.$

Thus, the correct answer is C.

10.

In unit square $ABCD,$ the inscribed circle $\omega$ intersects $\overline{CD}$ at $M,$ and $\overline{AM}$ intersects $\omega$ at a point $P$ different from $M.$ What is $AP?$

$\dfrac{\sqrt{5}}{12}$

$\dfrac{\sqrt{5}}{10}$

$\dfrac{\sqrt{5}}{9}$

$\dfrac{\sqrt{5}}{8}$

$\dfrac{2\sqrt{5}}{15}$

Answer: B

Concepts:coordinate geometry circle power of a point

Difficulty rating: 1560

Small Hint:

Place $A = (0, 0),$ so the circle has center $\left(\tfrac12, \tfrac12\right)$ and $M = \left(\tfrac12, 1\right)$

Big Hint:

Substitute the line $AM$ into the circle’s equation; the two roots give $M$ and $P$

Solution:

Let $A = (0, 0),$ $B = (1, 0),$ $C = (1, 1),$ $D = (0, 1).$ The inscribed circle has center $\left(\tfrac12, \tfrac12\right)$ and radius $\tfrac12,$ touching $\overline{CD}$ at $M = \left(\tfrac12, 1\right).$

Line $AM$ is $y = 2x.$ Substituting into $\left(x - \tfrac12\right)^2 + \left(y - \tfrac12\right)^2 = \tfrac14$ gives $20x^2 - 12x + 1 = 0,$ with roots $x = \tfrac12$ (point $M$ ) and $x = \tfrac{1}{10}$ (point $P$ ).

So $P = \left(\tfrac{1}{10}, \tfrac15\right)$ and $AP = \sqrt{\left(\tfrac{1}{10}\right)^2 + \left(\tfrac15\right)^2} = \frac{\sqrt5}{10}.$

Thus, the correct answer is B.

11.

As shown in the figure below, six semicircles lie in the interior of a regular hexagon with side length $2$ so that the diameters of the semicircles coincide with the sides of the hexagon. What is the area of the shaded region—inside the hexagon but outside all of the semicircles?

$6\sqrt{3} - 3\pi$

$\dfrac{9\sqrt{3}}{2} - 2\pi$

$\dfrac{3\sqrt{3}}{2} - \dfrac{\pi}{3}$

$3\sqrt{3} - \pi$

$\dfrac{9\sqrt{3}}{2} - \pi$

Answer: D

Concepts:regular polygon area decomposition inclusion-exclusion

Difficulty rating: 1590

Small Hint:

The hexagon area is $\dfrac{3\sqrt3}{2}s^2$ with $s = 2,$ and each semicircle has radius $1$

Big Hint:

Adjacent semicircles overlap; subtract the union, accounting for the six overlapping lenses

Solution:

The hexagon has area $\tfrac{3\sqrt3}{2}\cdot 2^2 = 6\sqrt3.$ Each semicircle has radius $1$ and area $\tfrac{\pi}{2},$ totaling $3\pi.$

Adjacent semicircle centers (side midpoints) are a distance $\sqrt3$ apart, so each adjacent pair overlaps in a lens of area $2\cos^{-1}\!\left(\tfrac{\sqrt3}{2}\right) - \tfrac{\sqrt3}{2} = \tfrac{\pi}{3} - \tfrac{\sqrt3}{2}.$ There are six such lenses.

The union of the semicircles is $3\pi - 6\left(\frac{\pi}{3} - \frac{\sqrt3}{2}\right) = \pi + 3\sqrt3.$ Subtracting from the hexagon gives the shaded area $6\sqrt3 - (\pi + 3\sqrt3) = 3\sqrt3 - \pi.$

Thus, the correct answer is D.

12.

Let $\overline{AB}$ be a diameter in a circle of radius $5\sqrt2.$ Let $\overline{CD}$ be a chord in the circle that intersects $\overline{AB}$ at a point $E$ such that $BE = 2\sqrt5$ and $\angle AEC = 45^\circ.$ What is $CE^2 + DE^2?$

$96$

$98$

$44\sqrt{5}$

$70\sqrt{2}$

$100$

Answer: E

Concepts:coordinate geometry chord Vieta’s Formulas

Difficulty rating: 1630

Small Hint:

Put the center at the origin with $\overline{AB}$ on the $x$ -axis and parametrize the chord from $E$ at $45^\circ$

Big Hint:

If $t_1, t_2$ are the signed distances from $E$ to $C, D,$ then $CE^2 + DE^2 = t_1^2 + t_2^2$

Solution:

Place the center at the origin with $\overline{AB}$ on the $x$ -axis; the radius is $R = 5\sqrt2,$ so $R^2 = 50.$ Then $E = (x_E, 0)$ with $x_E = R - 2\sqrt5$ (its exact value is not needed).

Parametrize the chord as $E + t\left(\tfrac{1}{\sqrt2}, \tfrac{1}{\sqrt2}\right).$ Substituting into $x^2 + y^2 = 50$ gives $t^2 + \sqrt2\,x_E\, t + (x_E^2 - 50) = 0,$ whose roots are the signed distances $t_1, t_2$ to $C$ and $D.$

By Vieta, $t_1 + t_2 = -\sqrt2\,x_E$ and $t_1 t_2 = x_E^2 - 50,$ so $\begin{gathered} CE^2 + DE^2 = t_1^2 + t_2^2 \\ {}= (t_1 + t_2)^2 - 2t_1 t_2 \\ {}= 2x_E^2 - 2(x_E^2 - 50) \\ {}= 100. \end{gathered}$

Thus, the correct answer is E.

13.

Which of the following is the value of $\sqrt{\log_2 6 + \log_3 6}?$

$1$

$\sqrt{\log_5 6}$

$2$

$\sqrt{\log_2 3} + \sqrt{\log_3 2}$

$\sqrt{\log_2 6} + \sqrt{\log_3 6}$

Answer: D

Concepts:logarithm algebraic manipulation

Difficulty rating: 1590

Small Hint:

Write $\log_2 6 = 1 + \log_2 3$ and $\log_3 6 = 1 + \log_3 2$

Big Hint:

Compare with $\left(\sqrt{\log_2 3} + \sqrt{\log_3 2}\right)^2,$ using $\log_2 3 \cdot \log_3 2 = 1$

Solution:

Let $a = \log_2 3,$ so $\log_3 2 = \tfrac1a.$ Then $\begin{gathered} \log_2 6 + \log_3 6 \\ {}= (1 + \log_2 3) \\ \quad {}+ (1 + \log_3 2) \\ {}= 2 + a + \frac1a. \end{gathered}$

Meanwhile $\left(\sqrt{\log_2 3} + \sqrt{\log_3 2}\right)^2$ $= a + \frac1a + 2\sqrt{a \cdot \tfrac1a}$ $= a + \frac1a + 2,$ which equals the expression above.

Taking square roots, $\sqrt{\log_2 6 + \log_3 6}$ $= \sqrt{\log_2 3} + \sqrt{\log_3 2}.$

Thus, the correct answer is D.

14.

Bela and Jenn play the following game on the closed interval $[0, n]$ of the real number line, where $n$ is a fixed integer greater than $4.$ They take turns playing, with Bela going first. At his first turn, Bela chooses any real number in the interval $[0, n].$ Thereafter, the player whose turn it is chooses a real number that is more than one unit away from all numbers previously chosen by either player. A player unable to choose such a number loses. Using optimal strategy, which player will win the game?

Bela will always win.

Jenn will always win.

Bela will win if and only if $n$ is odd.

Jenn will win if and only if $n$ is odd.

Jenn will win if and only if $n \gt 8.$

Answer: A

Concepts:combinatorial game symmetry

Difficulty rating: 1500

Small Hint:

Consider what happens if Bela plays exactly at the midpoint of the interval first

Big Hint:

After claiming the center, Bela can mirror each of Jenn’s moves through the midpoint

Solution:

Bela first plays the midpoint $\tfrac{n}{2}.$ This choice makes the configuration symmetric about the center of the interval.

Thereafter, whenever Jenn picks a number $x,$ Bela responds with its mirror image $n - x.$ Since Bela has already chosen $\tfrac n2,$ Jenn’s legal move satisfies $\left|x-\tfrac n2\right|>1.$ Therefore $|x-(n-x)|=2\left|x-\tfrac n2\right|>2,$ so Bela’s response is far enough from Jenn’s new point. Symmetry shows that it is also far enough from every earlier point. Thus Bela always has a move whenever Jenn does, so Jenn is the first to be stuck. Bela always wins.

Thus, the correct answer is A.

15.

There are $10$ people standing equally spaced around a circle. Each person knows exactly $3$ of the other $9$ people: the $2$ people standing next to her or him, as well as the person directly across the circle. How many ways are there for the $10$ people to split up into $5$ pairs so that the members of each pair know each other?

$11$

$12$

$13$

$14$

$15$

Answer: C

Concepts:graph theory casework

Difficulty rating: 1730

Small Hint:

Count perfect matchings using edges that are either adjacent (neighbor) pairs or diameters (across)

Big Hint:

Organize by how many diameter pairs are used; only $0, 1, 3,$ or $5$ of them can occur

Solution:

Label the people $0$ through $9.$ Allowed pairings use neighbor edges $(i, i + 1)$ or diameter edges $(i, i + 5).$ Count perfect matchings by the number of diameter edges used.

Using no diameters, the ten people split into adjacent pairs in $2$ ways (all “even” edges or all “odd” edges). Using exactly one diameter, choose it in $5$ ways; the remaining two arcs of four people each pair up uniquely, giving $5.$ Using all five diameters gives $1$ matching.

If at least one diameter is used, following the forced adjacent pairings around the circle shows that the number of diameters must be odd. With exactly three diameters, the four remaining people must be two opposite adjacent pairs. The position of one such opposite pair determines the matching, giving $5$ possibilities. Hence the total is $2+5+5+1=13.$

Thus, the correct answer is C.

16.

An urn contains one red ball and one blue ball. A box of extra red and blue balls lies nearby. George performs the following operation four times: he draws a ball from the urn at random and then takes a ball of the same color from the box and returns those two matching balls to the urn. After the four iterations the urn contains six balls. What is the probability that the urn contains three balls of each color?

$\dfrac16$

$\dfrac15$

$\dfrac14$

$\dfrac13$

$\dfrac12$

Answer: B

Concepts:basic probability combinations

Difficulty rating: 1660

Small Hint:

Reaching three of each color means exactly two of the four added balls are red

Big Hint:

Any specific order of two red and two blue additions has the same probability, so the number of red balls is uniform

Solution:

To end with three of each color, exactly two of the four added balls must be red. Consider any sequence of draws. When the urn holds $k$ balls, drawing a particular color with count $c$ has probability $\tfrac{c}{k}.$

Any ordering with two red and two blue additions gives the same product $\tfrac{1\cdot 2\cdot 1\cdot 2}{2\cdot 3\cdot 4\cdot 5},$ and there are $\binom42 = 6$ such orderings, for probability $\tfrac{6\cdot 4}{120} = \tfrac15.$ (Equivalently, the number of red balls after four steps is uniform on $\{1, 2, 3, 4, 5\},$ so $3$ red occurs with probability $\tfrac15.$ )

Thus, the correct answer is B.

17.

How many polynomials of the form $x^5 + ax^4 + bx^3 + cx^2 + dx$ $+ 2020,$ where $a,$ $b,$ $c,$ and $d$ are real numbers, have the property that whenever $r$ is a root, so is $\dfrac{-1 + i\sqrt3}{2}\cdot r?$ (Note that $i = \sqrt{-1}.$ )

$0$

$1$

$2$

$3$

$4$

Answer: C

Concepts:roots of unity complex number polynomial

Difficulty rating: 1960

Small Hint:

$\omega = \dfrac{-1 + i\sqrt3}{2}$ is a primitive cube root of unity, so nonzero roots come in triples $\{r, \omega r, \omega^2 r\}$

Big Hint:

With five roots the distinct roots form one triple with multiplicities summing to $5;$ real coefficients force conjugate symmetry

Solution:

Here $\omega = \tfrac{-1 + i\sqrt3}{2}$ is a primitive cube root of unity. Since $0$ is not a root, the set of distinct roots is closed under multiplication by $\omega,$ so it consists of triples $\{r, \omega r, \omega^2 r\}$ equally spaced in argument. Five roots cannot fill two such triples, so there is exactly one triple, with multiplicities $m_1, m_2, m_3 \ge 1$ summing to $5.$

Real coefficients require the root multiset to be closed under conjugation. This is possible only when the triple’s arguments are symmetric about the real axis, which happens for the two configurations $\{0^\circ, 120^\circ, 240^\circ\}$ and $\{60^\circ, 180^\circ, 300^\circ\}.$

The product of the roots must equal $-2020.$ In the first configuration the real root is positive, forcing a positive product, which is impossible. In the second, the real root is negative and the product is $-\rho^5;$ setting $\rho^5 = 2020$ works, and the two conjugate-symmetric multiplicity patterns $(1, 3, 1)$ and $(2, 1, 2)$ each give a valid polynomial. Hence there are $2.$

Thus, the correct answer is C.

18.

In square $ABCD,$ points $E$ and $H$ lie on $\overline{AB}$ and $\overline{DA},$ respectively, so that $AE = AH.$ Points $F$ and $G$ lie on $\overline{BC}$ and $\overline{CD},$ respectively, and points $I$ and $J$ lie on $\overline{EH}$ so that $\overline{FI} \perp \overline{EH}$ and $\overline{GJ} \perp \overline{EH}.$ See the figure below. Triangle $AEH,$ quadrilateral $BFIE,$ quadrilateral $DHJG,$ and pentagon $FCGJI$ each has area $1.$ What is $FI^2?$

$\dfrac73$

$8 - 4\sqrt{2}$

$1 + \sqrt{2}$

$\dfrac74 \sqrt{2}$

$2\sqrt{2}$

Answer: B

Concepts:coordinate geometry area decomposition special right triangle

Difficulty rating: 1910

Small Hint:

The four regions tile the square, so its area is $4$ and its side is $2$

Big Hint:

$\triangle AEH$ is an isosceles right triangle of area $1,$ so $AE = \sqrt2;$ set coordinates and use the area of $BFIE$

Solution:

The four regions have total area $4,$ so the square has side $2.$ Put $A = (0, 0),$ $B = (2, 0),$ $C = (2, 2),$ $D = (0, 2).$ Since $\triangle AEH$ is an isosceles right triangle with area $1,$ we get $AE = AH = \sqrt2,$ so $E = (\sqrt2, 0)$ and $H = (0, \sqrt2).$ Line $EH$ is $x + y = \sqrt2.$

Let $F = (2, t).$ Its perpendicular distance to line $EH$ is $FI = \tfrac{2 + t - \sqrt2}{\sqrt2}.$ Write $s=\frac{FI}{\sqrt2}=\tfrac{2+t-\sqrt2}{2},$ so the foot of the perpendicular is $I=(2-s,t-s).$ The shoelace formula on $B=(2,0),$ $F,$ $I,$ and $E=(\sqrt2,0)$ gives $[BFIE]=s^2-(3-2\sqrt2).$ Since this area is $1,$ we get $s^2=4-2\sqrt2.$

Then $FI^2 = 2s^2 = 8 - 4\sqrt2.$

Thus, the correct answer is B.

19.

Square $ABCD$ in the coordinate plane has vertices at the points $A(1, 1),$ $B(-1, 1),$ $C(-1, -1),$ and $D(1, -1).$ Consider the following four transformations:

$L,$ a rotation of $90^\circ$ counterclockwise around the origin;

$R,$ a rotation of $90^\circ$ clockwise around the origin;

$H,$ a reflection across the $x$ -axis; and

$V,$ a reflection across the $y$ -axis.

Each of these transformations maps the square onto itself, but the positions of the labeled vertices will change. For example, applying $R$ and then $V$ would send the vertex $A$ at $(1, 1)$ to $(-1, -1)$ and would send the vertex $B$ at $(-1, 1)$ to itself. How many sequences of $20$ transformations chosen from $\{L, R, H, V\}$ will send all of the labeled vertices back to their original positions? (For example, $R,$ $R,$ $V,$ $H$ is one sequence of $4$ transformations that will send the vertices back to their original positions.)

$2^{37}$

$3 \cdot 2^{36}$

$2^{38}$

$3 \cdot 2^{37}$

$2^{39}$

Answer: C

Concepts:transformation casework

Difficulty rating: 2000

Small Hint:

The first $19$ transformations can be anything; the last one is then forced to be the unique inverse

Big Hint:

The forced last move must be one of the four allowed transformations; find the probability that it is

Solution:

Label the vertices $0,1,2,3$ cyclically. Each allowed transformation has the form $j\mapsto \varepsilon j+\delta\pmod 4,$ where $\varepsilon\in\{1,-1\}$ and $\delta\in\{1,-1\}.$ These four choices give exactly the two quarter-turns and the two stated reflections.

Under composition, the parity of $\delta$ changes at every move. Thus a composition of $19$ allowed transformations again has odd $\delta,$ and so is one of the four allowed transformations. Its inverse is also allowed. Consequently every sequence of the first $19$ moves has exactly one choice for the final move, giving $4^{19}=2^{38}$ successful sequences.

Thus, the correct answer is C.

20.

Two different cubes of the same size are to be painted, with the color of each face being chosen independently and at random to be either black or white. What is the probability that after they are painted, the cubes can be rotated to be identical in appearance?

$\dfrac{9}{64}$

$\dfrac{289}{2048}$

$\dfrac{73}{512}$

$\dfrac{147}{1024}$

$\dfrac{589}{4096}$

Answer: D

Concepts:Burnside’s Lemma casework

Difficulty rating: 2040

Small Hint:

The probability equals $\dfrac{1}{64^2}\sum_{\mathcal O} |\mathcal O|^2$ , where the sum is over rotation-equivalence classes $\mathcal O$ of colorings

Big Hint:

Group colorings by number of black faces and find each class’s orbit size (e.g. two black faces: opposite gives size $3,$ adjacent gives size $12$ )

Solution:

For a fixed first cube, the number of second cubes matching it (up to rotation) equals the size of its rotation orbit. So the desired probability is $\tfrac{1}{64^2}\sum_{\mathcal O} |\mathcal O|^2.$

Grouping by black-face count, the orbit sizes are: $0$ or $6$ black $\to 1;$ $1$ or $5$ black $\to 6;$ $2$ or $4$ black $\to 3$ (opposite) and $12$ (adjacent); $3$ black $\to 8$ (corner) and $12$ (band). Then $\begin{gathered} \sum_{\mathcal O} |\mathcal O|^2 \\ {}= 1 + 36 + (9 + 144) \\ \quad {}+ (64 + 144) + (9 + 144) \\ \quad {}+ 36 + 1 = 588. \end{gathered}$

The probability is $\tfrac{588}{4096} = \tfrac{147}{1024}.$

Thus, the correct answer is D.

21.

How many positive integers $n$ satisfy

$\frac{n + 1000}{70} = \lfloor \sqrt{n} \rfloor?$

(Recall that $\lfloor x \rfloor$ is the greatest integer not exceeding $x.$ )

$2$

$4$

$6$

$30$

$32$

Answer: C

Concepts:floor and ceiling functions substitution inequality

Difficulty rating: 1800

Small Hint:

Let $k = \lfloor \sqrt{n} \rfloor,$ so $n = 70k - 1000$ must satisfy $k^2 \le n \lt (k + 1)^2$

Big Hint:

This gives $k^2 - 70k + 1000 \le 0$ and $k^2 - 68k + 1001 \gt 0;$ count integer $k$

Solution:

The right side is an integer, so let $k = \lfloor \sqrt{n} \rfloor.$ Then $n = 70k - 1000,$ and $k = \lfloor \sqrt{n} \rfloor$ requires $k^2 \le n \lt (k + 1)^2.$

The lower bound $k^2 \le 70k - 1000$ gives $k^2 - 70k + 1000 \le 0,$ i.e. $20 \le k \le 50.$ The upper bound $70k - 1000 \lt (k + 1)^2$ gives $k^2 - 68k + 1001 \gt 0,$ i.e. $k \le 21$ or $k \ge 47.$

Intersecting, $k \in \{20, 21, 47, 48, 49, 50\},$ giving $6$ values of $n.$

Thus, the correct answer is C.

22.

What is the maximum value of

$\frac{(2^t - 3t)\,t}{4^t}$

for real values of $t?$

$\dfrac{1}{16}$

$\dfrac{1}{15}$

$\dfrac{1}{12}$

$\dfrac{1}{10}$

$\dfrac19$

Answer: C

Concepts:substitution quadratic optimization

Difficulty rating: 1860

Small Hint:

Let $u = \dfrac{t}{2^t};$ then $\dfrac{t^2}{4^t} = u^2$

Big Hint:

The expression becomes $u - 3u^2,$ a downward parabola in $u$

Solution:

Split the fraction: $\dfrac{(2^t - 3t)t}{4^t} = \dfrac{t}{2^t} - \dfrac{3t^2}{4^t}.$ Let $u = \dfrac{t}{2^t},$ so $\dfrac{t^2}{4^t} = u^2$ and the expression is $u - 3u^2.$

This parabola has maximum at $u = \tfrac16,$ with value $\tfrac16 - 3\cdot\tfrac1{36} = \tfrac1{12}.$ Since $u = \tfrac{t}{2^t}$ is continuous and attains the value $\tfrac16,$ the maximum is achieved.

Thus, the correct answer is C.

23.

How many integers $n \ge 2$ are there such that whenever $z_1,$ $z_2,$ $\ldots,$ $z_n$ are complex numbers such that $|z_1| = |z_2| = \cdots = |z_n| = 1$ and $z_1 + z_2 + \cdots + z_n = 0,$ then the numbers $z_1,$ $z_2,$ $\ldots,$ $z_n$ are equally spaced on the unit circle in the complex plane?

$1$

$2$

$3$

$4$

$5$

Answer: B

Concepts:roots of unity complex number counterexample

Difficulty rating: 2100

Small Hint:

Check small cases: $n = 2$ forces antipodal points and $n = 3$ forces an equilateral triangle

Big Hint:

For $n \ge 4,$ build a counterexample by combining a smaller balanced configuration with an antipodal pair

Solution:

For $n = 2,$ $z_1 + z_2 = 0$ forces $z_2 = -z_1,$ which is equally spaced. For $n = 3,$ three unit vectors summing to zero must form an equilateral triangle, so they are equally spaced.

For every even $n\ge4,$ choose $\frac{n}{2}$ antipodal pairs at generic angles that do not form a regular $n$ -gon. For every odd $n\ge5,$ choose the vertices of an equilateral triangle together with $\frac{n-3}{2}$ generic antipodal pairs. Each construction has sum $0$ but is not equally spaced.

Hence only $n = 2$ and $n = 3$ work, giving $2$ values.

Thus, the correct answer is B.

24.

Let $D(n)$ denote the number of ways of writing the positive integer $n$ as a product $n = f_1 \cdot f_2 \cdots f_k,$ where $k \ge 1,$ the $f_i$ are integers strictly greater than $1,$ and the order in which the factors are listed matters (that is, two representations that differ only in the order of the factors are counted as distinct). For example, the number $6$ can be written as $6,$ $2 \cdot 3,$ and $3 \cdot 2,$ so $D(6) = 3.$ What is $D(96)?$

$112$

$128$

$144$

$172$

$184$

Answer: A

Concepts:recursive counting factor

Difficulty rating: 2220

Small Hint:

Use the recursion $D(n) = \sum_{d \mid n,\, d \gt 1} D(\frac{n}{d}),$ with $D(1) = 1$

Big Hint:

Since $96 = 2^5 \cdot 3,$ build up $D$ on the divisors $2,$ $3,$ $4,$ $6,$ $8,$ $12,$ $16,$ $24,$ $32,$ $48,$ $96$

Solution:

The first factor $f_1$ can be any divisor $d \gt 1,$ after which the rest is an ordered factorization of $\frac{n}{d}.$ So $D(n) = \sum_{d \mid n,\, d \gt 1} D(\frac{n}{d}),$ with $D(1) = 1.$

Computing over the divisors of $96 = 2^5\cdot 3:$ $D(2) = D(3) = 1,$ $D(4) = 2,$ $D(6) = 3,$ $D(8) = 4,$ $D(12) = 8,$ $D(16) = 8,$ $D(24) = 20,$ $D(32) = 16,$ $D(48) = 48.$

Finally $\begin{gathered} D(96) = D(48) + D(32) \\ {}+ D(24) + D(16) + D(12) \\ {}+ D(8) + D(6) + D(4) \\ {}+ D(3) + D(2) + D(1) \\ = 48 + 16 + 20 + 8 \\ {}+ 8 + 4 + 3 + 2 \\ {}+ 1 + 1 + 1 = 112. \end{gathered}$

Thus, the correct answer is A.

25.

For each real number $a$ with $0 \le a \le 1,$ let numbers $x$ and $y$ be chosen independently at random from the intervals $[0, a]$ and $[0, 1],$ respectively, and let $P(a)$ be the probability that $\sin^2(\pi x) + \sin^2(\pi y) \gt 1.$ What is the maximum value of $P(a)?$

$\dfrac{7}{12}$

$2 - \sqrt{2}$

$\dfrac{1 + \sqrt{2}}{4}$

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

$\dfrac58$

Answer: B

Concepts:geometric probability optimization calculus

Difficulty rating: 2540

Small Hint:

The condition is $|\sin \pi x| \gt |\cos \pi y|;$ for fixed $x,$ the probability over $y$ is a simple triangular function

Big Hint:

Express $P(a) = \tfrac1a \int_0^a g(x)\,dx$ where $g$ rises then falls, then maximize over $a$

Solution:

Since $\sin^2(\pi y) = 1 - \cos^2(\pi y),$ the condition is $|\sin \pi x| \gt |\cos \pi y|.$ For fixed $x,$ the probability over $y \in [0, 1]$ is $g(x) = 2x$ for $0 \le x \le \tfrac12$ and $g(x) = 2 - 2x$ for $\tfrac12 \le x \le 1.$

Then $P(a) = \tfrac1a \int_0^a g(x)\,dx.$ For $a \le \tfrac12,$ $P(a) = a,$ increasing to $\tfrac12.$ For $a \ge \tfrac12,$ $\begin{gathered} P(a) = \frac{2a - a^2 - \tfrac12}{a} \\ {}= 2 - a - \frac{1}{2a}. \end{gathered}$

Setting the derivative to zero gives $2a^2 = 1,$ so $a = \tfrac{1}{\sqrt2},$ and $P\!\left(\tfrac{1}{\sqrt2}\right) = 2 - \sqrt2.$

Thus, the correct answer is B.