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

1.

What is the value of $1234 + 2341 + 3412 + 4123?$

$10{,}000$

$10{,}010$

$10{,}110$

$11{,}000$

$11{,}110$

Concepts:place value symmetry

Difficulty rating: 720

Small Hint:

Each of the digits $1, 2, 3, 4$ appears once in every place value

Big Hint:

Every column adds to $1 + 2 + 3 + 4$

Solution:

Each of the digits $1, 2, 3, 4$ appears exactly once in the thousands, hundreds, tens, and units columns. So each column sums to $1 + 2 + 3 + 4 = 10.$

The total is therefore $10 \cdot 1111 = 11110.$

Thus, the correct answer is E.

2.

What is the area of the shaded figure shown below?

$4$

$6$

$8$

$10$

$12$

Concepts:triangle area area decomposition

Difficulty rating: 810

Small Hint:

The shaded region is a large triangle with a triangular notch removed from its base

Big Hint:

Both triangles have base $4$ along the $x$ -axis

Solution:

The outer triangle has vertices $(1, 0),$ $(3, 5),$ and $(5, 0),$ giving base $4$ and height $5,$ so its area is $\tfrac12 \cdot 4 \cdot 5 = 10.$

Removed from it is the triangle with vertices $(1, 0),$ $(3, 2),$ and $(5, 0),$ which has base $4$ and height $2,$ so area $\tfrac12 \cdot 4 \cdot 2 = 4.$

The shaded area is $10 - 4 = 6.$

Thus, the correct answer is B.

3.

At noon on a certain day, Minneapolis is $N$ degrees warmer than St. Louis. At $4{:}00$ the temperature in Minneapolis has fallen by $5$ degrees while the temperature in St. Louis has risen by $3$ degrees, at which time the temperatures in the two cities differ by $2$ degrees. What is the product of all possible values of $N?$

$10$

$30$

$60$

$100$

$120$

Concepts:absolute value casework

Difficulty rating: 1090

Small Hint:

At noon the difference is $N$ ; track how each change affects the gap

Big Hint:

The new difference is $|N - 5 - 3| = 2$

Solution:

At noon the gap is $N.$ Minneapolis then loses $5$ degrees and St. Louis gains $3,$ so the gap changes by $8.$ The new absolute difference is $|N - 8| = 2.$

This gives $N = 10$ or $N = 6,$ whose product is $60.$

Thus, the correct answer is C.

4.

Let $n = 8^{2022}.$ Which of the following is equal to $\dfrac{n}{4}?$

$4^{1010}$

$2^{2022}$

$8^{2018}$

$4^{3031}$

$4^{3032}$

Concepts:exponent

Difficulty rating: 1150

Small Hint:

Rewrite everything as a power of $2$

Big Hint:

$\dfrac{n}{4} = \dfrac{2^{6066}}{2^{2}}$

Solution:

Write $n = 8^{2022} = 2^{6066}.$ Then $\dfrac{n}{4} = \dfrac{2^{6066}}{2^{2}} = 2^{6064}.$

Since $2^{6064} = 4^{3032},$ this matches the last option.

Thus, the correct answer is E.

5.

Call a fraction $\dfrac{a}{b},$ not necessarily in the simplest form, special if $a$ and $b$ are positive integers whose sum is $15.$ How many distinct integers can be written as the sum of two, not necessarily different, special fractions?

$9$

$10$

$11$

$12$

$13$

Concepts:fraction casework

Difficulty rating: 1400

Small Hint:

List the special fractions and note which are integers, half-integers, or quarter-integers

Big Hint:

Two specials sum to an integer only when their fractional parts cancel

Solution:

Listing $\frac{a}{15-a}$ for $1\le a\le14,$ the integers are $2,4,14;$ the fractions with fractional part $\frac{1}{2}$ are $\tfrac12,\tfrac32,\tfrac{13}{2};$ and $\tfrac14,\tfrac{11}{4}$ have complementary fractional parts $\frac{1}{4},\frac{3}{4}.$

The remaining fractional parts are $\frac{1}{14},\frac{2}{13},\frac{4}{11},\frac{2}{3},\frac{7}{8},\frac{1}{7},$ and none of their complements occurs. Thus only the integer, half-integer, and quarter cases can give integer sums.

Integer pairs give $4, 6, 8, 16, 18, 28.$ Half-integer pairs $\left(\tfrac12, \tfrac32, \tfrac{13}{2}\right)$ give $1, 2, 3, 7, 8, 13.$ The quarter pair $\tfrac14 + \tfrac{11}{4}$ gives $3.$

The distinct integers are $1, 2, 3, 4, 6, 7, 8, 13, 16, 18, 28,$ a total of $11.$

Thus, the correct answer is C.

6.

The greatest prime number that is a divisor of $16{,}384$ is $2$ because $16{,}384 = 2^{14}.$ What is the sum of the digits of the greatest prime number that is a divisor of $16{,}383?$

$3$

$7$

$10$

$16$

$22$

Concepts:difference of squares prime factorization

Difficulty rating: 1280

Small Hint:

$16{,}383 = 2^{14} - 1$

Big Hint:

Factor as $(2^{7} - 1)(2^{7} + 1)$

Solution:

We have $16{,}383 = 2^{14} - 1$ $= (2^{7} - 1)(2^{7} + 1)$ $= 127 \cdot 129.$ Then $129 = 3 \cdot 43,$ so $16{,}383 = 3 \cdot 43 \cdot 127.$

The greatest prime factor is $127,$ whose digit sum is $1 + 2 + 7 = 10.$

Thus, the correct answer is C.

7.

Which of the following conditions is sufficient to guarantee that integers $x,$ $y,$ and $z$ satisfy the equation $\begin{aligned} &x(x - y) + y(y - z) \\ &\quad {}+ z(z - x) = 1? \end{aligned}$

$x \gt y$ and $y = z$

$x = y - 1$ and $y = z - 1$

$x = z + 1$ and $y = x + 1$

$x = z$ and $y - 1 = x$

$x + y + z = 1$

Concepts:algebraic manipulation casework

Difficulty rating: 1400

Small Hint:

Multiply by $2$ : the left side equals $(x - y)^2 + (y - z)^2 + (z - x)^2$

Big Hint:

For integers, that sum equals $2$ only when two differences are $\pm 1$ and one is $0$

Solution:

The expression satisfies $\small \begin{aligned} &2\bigl[x(x-y) + y(y-z) + z(z-x)\bigr] \\ &= (x-y)^2 + (y-z)^2 + (z-x)^2. \end{aligned}$ So the equation holds exactly when this sum of squares equals $2.$

Since the three differences sum to $0,$ this requires two of them to be $\pm 1$ and one to be $0.$

Option D gives $z - x = 0,$ $x - y = -1,$ and $y - z = 1,$ so the squares are $1 + 1 + 0 = 2.$ This works for all such integers.

Thus, the correct answer is D.

8.

The product of the lengths of the two congruent sides of an obtuse isosceles triangle is equal to the product of the base and twice the triangle’s height to the base. What is the measure, in degrees, of the vertex angle of this triangle?

$105$

$120$

$135$

$150$

$165$

Concepts:isosceles triangle triangle area trigonometry

Difficulty rating: 1530

Small Hint:

Write the area two ways using the legs and the vertex angle $\theta$

Big Hint:

The area is both $\tfrac12 b h$ and $\tfrac12 s^2 \sin\theta$

Solution:

Let the congruent sides have length $s,$ the base be $b,$ and the height to the base be $h.$ The given condition is $s^2 = 2bh.$

The area equals $\tfrac12 bh$ and also $\tfrac12 s^2 \sin\theta,$ where $\theta$ is the vertex angle. So $bh = s^2 \sin\theta.$

Substituting $s^2 = 2bh$ gives $bh = 2bh\sin\theta,$ so $\sin\theta = \tfrac12.$ Since the triangle is obtuse, $\theta = 150^\circ.$

Thus, the correct answer is D.

9.

Triangle $ABC$ is equilateral with side length $6.$ Suppose that $O$ is the center of the inscribed circle of this triangle. What is the area of the circle passing through $A,$ $O,$ and $C?$

$9\pi$

$12\pi$

$18\pi$

$24\pi$

$27\pi$

Concepts:equilateral triangle circumcircle, circumcenter, and circumradius law of sines

Difficulty rating: 1610

Small Hint:

In an equilateral triangle the incenter coincides with the circumcenter

Big Hint:

In triangle $AOC,$ angle $AOC = 120^\circ$ and $AC = 6$ ; apply the law of sines

Solution:

For an equilateral triangle, $O$ is also the circumcenter, so $OA = OC = \dfrac{6}{\sqrt3} = 2\sqrt3.$ The central angle $\angle AOC = 120^\circ.$

In triangle $AOC,$ side $AC = 6$ is opposite the $120^\circ$ angle, so the circumradius $R'$ of this triangle satisfies $2R' = \dfrac{6}{\sin 120^\circ} = 4\sqrt3,$ giving $R' = 2\sqrt3.$

The area of the circle is $\pi (2\sqrt3)^2 = 12\pi.$

Thus, the correct answer is B.

10.

What is the sum of all possible values of $t$ between $0$ and $360$ such that the triangle in the coordinate plane whose vertices are $(\cos 40^\circ, \sin 40^\circ),\$ $(\cos 60^\circ, \sin 60^\circ),$ and $\ \ (\cos t^\circ, \sin t^\circ)$ is isosceles?

$100$

$150$

$330$

$360$

$380$

Concepts:circle chord isosceles triangle

Difficulty rating: 1820

Small Hint:

All three points lie on the unit circle, so equal sides mean equal chords, i.e. equal arc separations

Big Hint:

Consider the three cases where the third point matches distance to $A,$ to $B,$ or lies on their perpendicular bisector

Solution:

The three points lie on the unit circle at angles $40^\circ,$ $60^\circ,$ and $t^\circ.$ A chord’s length depends only on the angular separation of its endpoints.

If the third point is equidistant from the other two, it lies on the perpendicular bisector: $t = 50$ or $t = 230.$

If its distance to $40^\circ$ equals the fixed chord (separation $20^\circ$ ), then $t = 20$ (since $t = 60$ is degenerate). If its distance to $60^\circ$ matches, then $t = 80$ (since $t = 40$ is degenerate).

The valid values are $50, 230, 20, 80,$ summing to $380.$

Thus, the correct answer is E.

11.

Una rolls $6$ standard $6$ -sided dice simultaneously and calculates the product of the $6$ numbers obtained. What is the probability that the product is divisible by $4?$

$\dfrac{3}{4}$

$\dfrac{57}{64}$

$\dfrac{59}{64}$

$\dfrac{187}{192}$

$\dfrac{63}{64}$

Concepts:dice (probability)complementary probability

Difficulty rating: 1650

Small Hint:

Use complementary counting: the product has fewer than two factors of $2$

Big Hint:

Track whether each die contributes zero, one, or two factors of $2$

Solution:

The product fails to be divisible by $4$ when it has at most one factor of $2.$ Each die is odd with probability $\tfrac12,$ contributes exactly one factor of $2$ (a $2$ or $6$ ) with probability $\tfrac13,$ and two factors (a $4$ ) with probability $\tfrac16.$

All six odd: $\left(\tfrac12\right)^6 = \tfrac{1}{64}.$ Exactly one die a $2$ or $6$ and the rest odd: $6 \cdot \tfrac13 \cdot \left(\tfrac12\right)^5 = \tfrac{1}{16} = \tfrac{4}{64}.$

The complement is $\tfrac{1}{64} + \tfrac{4}{64} = \tfrac{5}{64},$ so the answer is $1 - \tfrac{5}{64} = \tfrac{59}{64}.$

Thus, the correct answer is C.

12.

For $n$ a positive integer, let $f(n)$ be the quotient obtained when the sum of all positive divisors of $n$ is divided by $n.$ For example, $\begin{aligned} f(14) &= (1 + 2 + 7 + 14) \div 14 \\ &= \dfrac{12}{7}. \end{aligned}$ What is $f(768) - f(384)?$

$\dfrac{1}{768}$

$\dfrac{1}{192}$

$1$

$\dfrac{4}{3}$

$\dfrac{8}{3}$

Concepts:sum of factors prime factorization

Difficulty rating: 1760

Small Hint:

Factor $768 = 2^8 \cdot 3$ and $384 = 2^7 \cdot 3,$ then use $\sigma$ of a prime power

Big Hint:

Put both over a common denominator of $192$

Solution:

Since $768 = 2^8 \cdot 3,$ its divisor sum is $(2^9 - 1)(1 + 3)$ $= 511 \cdot 4 = 2044,$ so $f(768) = \dfrac{2044}{768} = \dfrac{511}{192}.$

Since $384 = 2^7 \cdot 3,$ its divisor sum is $(2^8 - 1)(1 + 3)$ $= 255 \cdot 4 = 1020,$ so $f(384) = \dfrac{1020}{384} = \dfrac{510}{192}.$

The difference is $\dfrac{511 - 510}{192} = \dfrac{1}{192}.$

Thus, the correct answer is B.

13.

Let $c = \dfrac{2\pi}{11}.$ What is the value of $\small \dfrac{\sin 3c \cdot \sin 6c \cdot \sin 9c \cdot \sin 12c \cdot \sin 15c}{\sin c \cdot \sin 2c \cdot \sin 3c \cdot \sin 4c \cdot \sin 5c}?$

$-1$

$-\dfrac{\sqrt{11}}{5}$

$\dfrac{\sqrt{11}}{5}$

$\dfrac{10}{11}$

$1$

Concepts:trigonometric identity symmetry

Difficulty rating: 1900

Small Hint:

Each angle is a multiple of $\dfrac{2\pi}{11}$ ; reduce the numerator angles modulo $2\pi$

Big Hint:

Use $\sin(2\pi - x) = -\sin x$ to relate leftover numerator factors to denominator factors

Solution:

Write each angle as $kc = \dfrac{2\pi k}{11}.$ Reducing modulo $2\pi,$ $\sin 12c = \sin c$ and $\sin 15c = \sin 4c.$

So the numerator is $\sin 3c \cdot \sin 6c \cdot \sin 9c$ $\cdot \sin c \cdot \sin 4c.$ Cancelling the common factors $\sin c,$ $\sin 3c,$ $\sin 4c$ leaves $\dfrac{\sin 6c \cdot \sin 9c}{\sin 2c \cdot \sin 5c}.$

Now $\sin 9c = \sin\left(2\pi - 2c\right) = -\sin 2c$ and $\sin 6c = \sin\left(2\pi - 5c\right) = -\sin 5c,$ so the ratio equals $\dfrac{(-\sin 5c)(-\sin 2c)}{\sin 2c \cdot \sin 5c} = 1.$

Thus, the correct answer is E.

14.

Suppose that $P(z),$ $Q(z),$ and $R(z)$ are polynomials with real coefficients, having degrees $2,$ $3,$ and $6,$ respectively, and constant terms $1,$ $2,$ and $3,$ respectively. Let $N$ be the number of distinct complex numbers $z$ that satisfy the equation $P(z) \cdot Q(z) = R(z).$ What is the minimum possible value of $N?$

$0$

$1$

$2$

$3$

$5$

Concepts:polynomial extremal argument

Difficulty rating: 1850

Small Hint:

Consider $D(z) = P(z)Q(z) - R(z)$ ; what is its degree and constant term?

Big Hint:

A degree- $6$ polynomial can have a single root of multiplicity $6$

Solution:

Let $D(z) = P(z)Q(z) - R(z).$ Since $P Q$ has degree $5$ and $R$ has degree $6,$ the degree of $D$ is $6.$ Its constant term is $1 \cdot 2 - 3 = -1 \neq 0.$

Because $R$ is otherwise unconstrained, $D$ can be made equal to any real degree- $6$ polynomial with constant term $-1,$ for instance $-(z - 1)^6.$

Such a polynomial has a single distinct root, so the minimum is $N = 1.$

Thus, the correct answer is B.

15.

Three identical square sheets of paper each with side length $6$ are stacked on top of each other. The middle sheet is rotated clockwise $30^\circ$ about its center and the top sheet is rotated clockwise $60^\circ$ about its center, resulting in the $24$ -sided polygon shown in the figure below. The area of this polygon can be expressed in the form $a - b\sqrt{c},$ where $a,$ $b,$ and $c$ are positive integers, and $c$ is not divisible by the square of any prime. What is $a + b + c?$

$75$

$93$

$96$

$129$

$147$

Concepts:area decomposition trigonometry symmetry

Difficulty rating: 2100

Small Hint:

By symmetry the polygon’s vertices alternate between outer square-corners and inner edge-crossings, every $15^\circ$

Big Hint:

The outer vertices are at distance $3\sqrt2$ from the center and the inner ones at distance $2\sqrt3$

Solution:

Because the three squares are rotated by $0^\circ,$ $30^\circ,$ and $60^\circ,$ the figure has $12$ -fold symmetry. Its $24$ vertices alternate every $15^\circ$ : outer vertices are the square corners at distance $3\sqrt2$ from the center, and inner vertices are edge crossings at distance $2\sqrt3.$

Connecting the center to all $24$ vertices splits the polygon into $24$ triangles, each with sides $3\sqrt2$ and $2\sqrt3$ and included angle $15^\circ.$ The total area is $\begin{aligned} &24 \cdot \tfrac12 (3\sqrt2)(2\sqrt3)\sin 15^\circ \\ &= 72\sqrt6 \cdot \dfrac{\sqrt6 - \sqrt2}{4}. \end{aligned}$

This simplifies to $18\sqrt6(\sqrt6 - \sqrt2) = 108 - 36\sqrt3,$ so $a + b + c = 108 + 36 + 3 = 147.$

Thus, the correct answer is E.

16.

Suppose $a,$ $b,$ $c$ are positive integers such that $a + b + c = 23$ and $\begin{aligned} &\gcd(a, b) + \gcd(b, c) \\ &\quad {}+ \gcd(c, a) = 9. \end{aligned}$ What is the sum of all possible distinct values of $a^2 + b^2 + c^2?$

$259$

$438$

$516$

$625$

$687$

Concepts:greatest common divisor parity casework

Difficulty rating: 2100

Small Hint:

By symmetry assume $a\le b\le c;$ then $a\le7$ and $c=23-a-b$

Big Hint:

Use $\gcd(b,c)=\gcd(b,23-a)$ and check the short range $a\le b\le\frac{23-a}{2}$

Solution:

The condition is symmetric, so assume $a\le b\le c.$ Then $a\le7,$ and for each fixed $a$ we have $a\le b\le\lfloor\frac{23-a}{2}\rfloor$ and $c=23-a-b.$ The gcd sum can be evaluated as $\begin{aligned} &\gcd(a,b)+\gcd(b,23-a) \\ &\quad {}+\gcd(a,23-b). \end{aligned}$

Checking these short ranges, the possible gcd sums for $a=1,2,\ldots,7$ are, respectively, $\{3,4,13\},$ $\{4,6,10\},$ $\{3,5,6,8,9,12\},$ $\{4,6\},$ $\{3,7,8,11\},$ $\{4,6,8\},$ and $\{9,10\}.$ The value $9$ occurs only at $(a,b,c)=(3,5,15)$ and $(7,7,9).$

The triple $(7, 7, 9)$ has $\gcd$ sum $7 + 1 + 1 = 9$ and $a^2 + b^2 + c^2 = 49 + 49 + 81$ $= 179.$ The triple $(3, 5, 15)$ has $\gcd$ sum $1 + 5 + 3 = 9$ and $a^2 + b^2 + c^2 = 9 + 25$ $+ 225 = 259.$

The sum of the distinct values is $179 + 259 = 438.$

Thus, the correct answer is B.

17.

A bug starts at a vertex of a grid made of equilateral triangles of side length $1.$ At each step the bug moves in one of the $6$ possible directions along the grid lines randomly and independently with equal probability. What is the probability that after $5$ moves the bug never will have been more than $1$ unit away from the starting position?

$\dfrac{13}{108}$

$\dfrac{7}{54}$

$\dfrac{29}{216}$

$\dfrac{4}{27}$

$\dfrac{1}{16}$

Concepts:random walk recursive probability

Difficulty rating: 2230

Small Hint:

The bug must always be at the start or one of its $6$ neighbors

Big Hint:

From a neighbor, only $3$ of the $6$ moves stay within distance $1$ : back to start or to the two adjacent neighbors

Solution:

Staying within distance $1$ means the bug is always at the origin or one of its $6$ neighbors. From the origin, all $6$ moves are allowed. From a neighbor, only $3$ moves keep it in range: back to the origin, or to either of the two adjacent neighbors.

Let $a_k$ and $b_k$ count valid $k$ -step paths ending at the origin and at a neighbor. Then $a_{k+1} = b_k$ and $b_{k+1} = 6a_k + 2b_k,$ starting from $a_0 = 1,$ $b_0 = 0.$

Iterating gives $b_1 = 6,$ then $(a_2, b_2) = (6, 12),$ $(a_3, b_3) = (12, 60),$ $(a_4, b_4) = (60, 192),$ $(a_5, b_5) = (192, 744).$ The total number of valid paths is $192 + 744 = 936.$

The probability is $\dfrac{936}{6^5} = \dfrac{936}{7776} = \dfrac{13}{108}.$

Thus, the correct answer is A.

18.

Set $u_0 = \dfrac{1}{4},$ and for $k \ge 0$ let $u_{k+1}$ be determined by the recurrence $u_{k+1} = 2u_k - 2u_k^2.$ This sequence tends to a limit; call it $L.$ What is the least value of $k$ such that $|u_k - L| \le \dfrac{1}{2^{1000}}?$

$10$

$87$

$123$

$329$

$401$

Concepts:recursion substitution

Difficulty rating: 2230

Small Hint:

The limit is $L = \tfrac12$ ; substitute $v_k = 1 - 2u_k$

Big Hint:

The substitution gives $v_{k+1} = v_k^2,$ so $v_k$ squares repeatedly

Solution:

The limit satisfies $L = 2L - 2L^2,$ giving $L = \tfrac12.$ Let $v_k = 1 - 2u_k.$ Then $\begin{aligned} v_{k+1} &= 1 - 2u_{k+1} \\ &= 1 - 4u_k + 4u_k^2 \\ &= (1 - 2u_k)^2 = v_k^2. \end{aligned}$

Since $v_0 = 1 - 2 \cdot \tfrac14 = \tfrac12,$ we get $v_k = \left(\tfrac12\right)^{2^k},$ so $|u_k - L| = \dfrac{|v_k|}{2} = 2^{-2^k - 1}.$

We need $2^k + 1 \ge 1000,$ i.e. $2^k \ge 999.$ The least such $k$ is $10,$ since $2^{10} = 1024.$

Thus, the correct answer is A.

19.

Regular polygons with $5,$ $6,$ $7,$ and $8$ sides are inscribed in the same circle. No two of the polygons share a vertex, and no three of their sides intersect at a common point. At how many points inside the circle do two of their sides intersect?

$52$

$56$

$60$

$64$

$68$

Concepts:regular polygon counting intersections

Difficulty rating: 2320

Small Hint:

Consider each pair of polygons separately

Big Hint:

Two inscribed polygons with $m$ and $n$ sides in general position cross at $2\min(m, n)$ points

Solution:

For two convex polygons inscribed in the same circle with no shared vertices, each side of the smaller polygon crosses the larger polygon’s boundary exactly twice, so they meet at $2\min(m, n)$ points.

Summing over all pairs: $(5,6), (5,7), (5,8)$ give $10$ each; $(6,7), (6,8)$ give $12$ each; $(7,8)$ gives $14.$

The total is $3 \cdot 10 + 2 \cdot 12 + 14 = 68.$

Thus, the correct answer is E.

20.

A cube is constructed from $4$ white unit cubes and $4$ blue unit cubes. How many different ways are there to construct the $2 \times 2 \times 2$ cube using these smaller cubes? (Two constructions are considered the same if one can be rotated to match the other.)

$7$

$8$

$9$

$10$

$11$

Concepts:Burnside’s Lemma cube geometry

Difficulty rating: 2380

Small Hint:

Use Burnside’s lemma over the $24$ rotations of the cube acting on the $8$ corner positions

Big Hint:

For each rotation type, count the colorings with exactly $4$ blue cubes that it fixes

Solution:

By Burnside’s lemma, the count is the average number of $4$ -blue colorings fixed by each of the $24$ rotations acting on the $8$ cubies.

The identity fixes $\binom{8}{4} = 70.$ The $6$ face quarter-turns fix $2$ each $(12).$ The $3$ face half-turns fix $6$ each $(18).$ The $8$ vertex rotations fix $4$ each $(32).$ The $6$ edge half-turns fix $6$ each $(36).$

The total is $70 + 12 + 18 + 32 + 36 = 168,$ and $\dfrac{168}{24} = 7.$

Thus, the correct answer is A.

21.

For real numbers $x,$ let $\begin{aligned} P(x) &= 1 + \cos(x) + i\sin(x) \\ &\quad {}- \cos(2x) - i\sin(2x) \\ &\quad {}+ \cos(3x) + i\sin(3x) \end{aligned}$ where $i = \sqrt{-1}.$ For how many values of $x$ with $0 \le x \lt 2\pi$ does $P(x) = 0?$

$0$

$1$

$2$

$3$

$4$

Concepts:complex number trigonometric identity

Difficulty rating: 2420

Small Hint:

Write $P(x) = 1 + e^{ix} - e^{2ix} + e^{3ix}$ and require both the real and imaginary parts to vanish

Big Hint:

The imaginary part factors as $\sin(2x)(2\cos x - 1)$

Solution:

Group by Euler’s formula: $P(x) = 1 + e^{ix} - e^{2ix} + e^{3ix}.$ The imaginary part is $\begin{gathered} \sin x - \sin 2x + \sin 3x \\ = (\sin x + \sin 3x) - \sin 2x \\ = \sin 2x(2\cos x - 1). \end{gathered}$

This vanishes when $\sin 2x = 0$ (so $x = 0, \tfrac{\pi}{2}, \pi, \tfrac{3\pi}{2}$ ) or $\cos x = \tfrac12$ (so $x = \tfrac{\pi}{3}, \tfrac{5\pi}{3}$ ).

Checking the real part $1 + \cos x - \cos 2x + \cos 3x$ at each of these values gives $\pm 2$ or $1,$ never $0.$ So no $x$ makes $P(x) = 0.$

Thus, the correct answer is A.

22.

Right triangle $ABC$ has side lengths $BC = 6,$ $AC = 8,$ and $AB = 10.$ A circle centered at $O$ is tangent to line $BC$ at $B$ and passes through $A.$ A circle centered at $P$ is tangent to line $AC$ at $A$ and passes through $B.$ What is $OP?$

$\dfrac{23}{8}$

$\dfrac{29}{10}$

$\dfrac{35}{12}$

$\dfrac{73}{25}$

$3$

Concepts:coordinate geometry tangent line distance formula

Difficulty rating: 2490

Small Hint:

Place $C$ at the origin with legs along the axes

Big Hint:

A tangent point’s radius is perpendicular to the tangent line, so $O$ lies directly above $B$ and $P$ directly across from $A$

Solution:

Place $C = (0, 0),$ $B = (6, 0),$ and $A = (0, 8),$ so the right angle is at $C.$

Circle $O$ is tangent to line $BC$ (the $x$ -axis) at $B,$ so $O = (6, k).$ Setting $OA = OB$ gives $36 + (k - 8)^2 = k^2,$ so $k = \tfrac{25}{4}$ and $O = \left(6, \tfrac{25}{4}\right).$

Circle $P$ is tangent to line $AC$ (the $y$ -axis) at $A,$ so $P = (h, 8).$ Setting $PB = PA$ gives $(h - 6)^2 + 64 = h^2,$ so $h = \tfrac{25}{3}$ and $P = \left(\tfrac{25}{3}, 8\right).$

Then $OP = \sqrt{\left(\tfrac73\right)^2 + \left(\tfrac74\right)^2}$ $= 7\sqrt{\tfrac{25}{144}}$ $= \dfrac{35}{12}.$

Thus, the correct answer is C.

23.

What is the average number of pairs of consecutive integers in a randomly selected subset of $5$ distinct integers chosen from the set $\{1, 2, 3, \ldots, 30\}?$ (For example the set $\{1, 17, 18, 19, 30\}$ has $2$ pairs of consecutive integers.)

$\dfrac{2}{3}$

$\dfrac{29}{36}$

$\dfrac{5}{6}$

$\dfrac{29}{30}$

$1$

Concepts:expected value sampling without replacement

Difficulty rating: 2190

Small Hint:

Use linearity of expectation over the $29$ possible consecutive pairs $(i, i+1)$

Big Hint:

The probability that both $i$ and $i+1$ are chosen is $\dfrac{5}{30} \cdot \dfrac{4}{29}$

Solution:

For each of the $29$ adjacent pairs $(i, i+1),$ let an indicator be $1$ if both are in the subset. The probability of this is $\dfrac{5}{30} \cdot \dfrac{4}{29} = \dfrac{2}{87}.$

By linearity of expectation, the expected number of consecutive pairs is $29 \cdot \dfrac{2}{87} = \dfrac{2}{3}.$

Thus, the correct answer is A.

24.

Triangle $ABC$ has side lengths $AB = 11,$ $BC = 24,$ and $CA = 20.$ The bisector of $\angle BAC$ intersects $\overline{BC}$ in point $D,$ and intersects the circumcircle of $\triangle ABC$ in point $E \neq A.$ The circumcircle of $\triangle BED$ intersects the line $AB$ in points $B$ and $F \neq B.$ What is $CF?$

$28$

$20\sqrt{2}$

$30$

$32$

$20\sqrt{3}$

Concepts:power of a point angle bisector similarity

Difficulty rating: 2650

Small Hint:

Apply power of the point $A$ to the circle through $B,$ $E,$ $D$ : $AB \cdot AF = AD \cdot AE$

Big Hint:

Because $\triangle ABE \sim \triangle ADC,$ we get $AD \cdot AE = AB \cdot AC,$ so $AF = AC$

Solution:

Points $A,$ $D,$ $E$ are collinear on the bisector, and $A,$ $B,$ $F$ are collinear on line $AB.$ The power of $A$ with respect to the circle through $B,$ $E,$ $D$ gives $AB \cdot AF = AD \cdot AE.$

Since $\angle BAE = \angle DAC$ and $\angle AEB = \angle ACB$ (subtending $AB$ ), triangles $ABE$ and $ADC$ are similar, so $AD \cdot AE = AB \cdot AC.$ Therefore $AF = AC = 20.$

Place $A = (0, 0),$ $B = (11, 0).$ From $CA = 20,$ $CB = 24,$ point $C = \left(-\tfrac52, \tfrac{\sqrt{1575}}{2}\right).$ Point $F$ lies on ray $AB$ with $AF = 20,$ so $F = (20, 0).$

Then $CF^2 = \left(20 + \tfrac52\right)^2 + \tfrac{1575}{4}$ $= \tfrac{2025}{4} + \tfrac{1575}{4}$ $= 900,$ so $CF = 30.$

Thus, the correct answer is C.

25.

For $n$ a positive integer, let $R(n)$ be the sum of the remainders when $n$ is divided by $2,$ $3,$ $4,$ $5,$ $6,$ $7,$ $8,$ $9,$ and $10.$ For example, $\begin{aligned} &R(15) = 1 + 0 + 3 + 0 + 3 \\ &\quad {}+ 1 + 7 + 6 + 5 = 26. \end{aligned}$ How many two-digit positive integers $n$ satisfy $R(n) = R(n + 1)?$

$0$

$1$

$2$

$3$

$4$

Concepts:modular arithmetic divisibility

Difficulty rating: 2800

Small Hint:

When $n$ increases by $1,$ each remainder rises by $1$ unless $n+1$ is divisible by $m,$ where it drops to $0$

Big Hint:

$R(n+1) - R(n)$ $= 9 - \sum_{m \mid n+1} m,$ so you need the divisors of $n+1$ among $2, \ldots, 10$ to sum to $9$

Solution:

Going from $n$ to $n + 1,$ each remainder $n \bmod m$ increases by $1$ unless $n + 1$ is divisible by $m,$ in which case it drops from $m - 1$ to $0.$ So $\begin{aligned} &R(n+1) - R(n) \\ &= 9 - \sum_{\substack{2 \le m \le 10 \\ m \mid n+1}} m. \end{aligned}$

Distinct integers from $2$ through $10$ can sum to $9$ only as $\{9\},\{2,7\},\{3,6\},\{4,5\},$ or $\{2,3,4\}.$ Divisibility closure eliminates every case except $\{2,7\}:$ for example, a multiple of $9$ also has divisor $3,$ and a multiple of $6$ also has divisors $2$ and $3.$ Thus $n+1$ must be a multiple of $14$ with no other divisor in $\{2,\ldots,10\}.$

Testing the multiples $14,28,42,56,70,84,98$ in the required range leaves only $14$ and $98.$ Hence $n=13$ or $n=97,$ giving $2$ values.

Thus, the correct answer is C.