2024 AMC 12B — Problems & Online Timed Exam

Timed

1:15:00

1.

In a long line of people arranged left to right, the $1013$ th person from the left is also the $1010$ th person from the right. How many people are in the line?

$2021$

$2022$

$2023$

$2024$

$2025$

Answer: B

Concepts:counting integers in a range basic counting

Difficulty rating: 890

Small Hint:

The people to the left of the target person and the people to the right, plus the person themself, make up the whole line

Big Hint:

There are $1012$ people to the left and $1009$ to the right; add these to the target person

Solution:

The target person has $1013 - 1 = 1012$ people to the left and $1010 - 1 = 1009$ people to the right. Including the person themself, the line has $1012 + 1009 + 1 = 2022$ people.

Thus, the correct answer is B.

2.

What is $10! - 7! \cdot 6!?$

$-120$

$0$

$120$

$600$

$720$

Answer: B

Concepts:factorial

Difficulty rating: 1020

Small Hint:

Write $7! \cdot 6!$ and see whether it can be rebuilt into $10!$

Big Hint:

Since $6! = 720 = 8 \cdot 9 \cdot 10,$ we get $7! \cdot 6! = 7! \cdot 8 \cdot 9 \cdot 10 = 10!$

Solution:

Note that $6! = 720 = 8 \cdot 9 \cdot 10.$ Therefore $7! \cdot 6! = 7! \cdot (8 \cdot 9 \cdot 10) = 10!.$ So $10! - 7! \cdot 6! = 10! - 10! = 0.$

Thus, the correct answer is B.

3.

For how many integer values of $x$ is $|2x| \le 7\pi?$

$16$

$17$

$19$

$20$

$21$

Answer: E

Concepts:absolute value counting integers in a range estimation

Difficulty rating: 1130

Small Hint:

Rewrite as $|x| \le \dfrac{7\pi}{2}$ and estimate $\dfrac{7\pi}{2}$ as a decimal

Big Hint:

Since $\dfrac{7\pi}{2} \approx 10.99,$ count the integers from $-10$ to $10$

Solution:

The inequality $|2x| \le 7\pi$ is equivalent to $|x| \le \dfrac{7\pi}{2} \approx 10.99.$ The integers satisfying this run from $-10$ to $10,$ which is $10 + 10 + 1 = 21$ values.

Thus, the correct answer is E.

4.

Balls numbered $1,$ $2,$ $3,$ $\ldots$ are deposited in $5$ bins, labeled $A,$ $B,$ $C,$ $D,$ and $E,$ using the following procedure. Ball $1$ is deposited in bin $A,$ and balls $2$ and $3$ are deposited in $B.$ The next three balls are deposited in bin $C,$ the next $4$ in bin $D,$ and so on, cycling back to bin $A$ after balls are deposited in bin $E.$ (For example, $22,$ $23,$ $\ldots,$ $28$ are deposited in bin $B$ at step $7$ of this process.) In which bin is ball $2024$ deposited?

$A$

$B$

$C$

$D$

$E$

Answer: D

Concepts:triangular number modular arithmetic

Difficulty rating: 1270

Small Hint:

At step $k,$ exactly $k$ balls are deposited, so after $k$ steps a total of $\dfrac{k(k+1)}{2}$ balls have been placed

Big Hint:

Find the step containing ball $2024,$ then note that step $k$ uses the bin at position $(k-1)\bmod 5$ in the cycle $A, B, C, D, E$

Solution:

Step $k$ deposits $k$ balls, so after step $k$ a total of $\dfrac{k(k+1)}{2}$ balls have been placed. Since $\dfrac{63 \cdot 64}{2} = 2016$ and $\dfrac{64 \cdot 65}{2} = 2080,$ ball $2024$ falls in step $64.$

The steps cycle through the bins $A, B, C, D, E,$ so step $k$ uses position $(k-1) \bmod 5.$ Here $(64 - 1) \bmod 5 = 63 \bmod 5 = 3,$ which is the fourth bin, $D.$

Thus, the correct answer is D.

5.

In the following expression, Melanie changed some of the plus signs to minus signs:

$1 + 3 + 5 + 7 + \cdots + 97 + 99$

When the new expression was evaluated, it was negative. What is the least number of plus signs that Melanie could have changed to minus signs?

$14$

$15$

$16$

$17$

$18$

Answer: B

Concepts:arithmetic sequence optimization

Difficulty rating: 1340

Small Hint:

The original sum is $1 + 3 + \cdots + 99 = 50^2 = 2500;$ flipping a term of value $v$ lowers the total by $2v$

Big Hint:

To go negative the flipped terms must total more than $1250;$ flip the largest odd numbers first, where the largest $k$ of them sum to $k(100-k)$

Solution:

The original expression sums the first $50$ odd numbers, giving $50^2 = 2500.$ Changing a term of value $v$ from $+$ to $-$ decreases the total by $2v,$ so to make the result negative the flipped terms must total more than $\dfrac{2500}{2} = 1250.$

To use as few terms as possible, flip the largest odd numbers $99, 97, 95, \ldots$ The largest $k$ of them sum to $k(100 - k).$ With $k = 14$ this is $14 \cdot 86 = 1204 \le 1250,$ but with $k = 15$ it is $15 \cdot 85 = 1275 \gt 1250.$ So $15$ sign changes suffice and $14$ do not.

Thus, the correct answer is B.

6.

The national debt of the United States is on track to reach $5 \cdot 10^{13}$ dollars by $2033.$ How many digits does this number of dollars have when written as a numeral in base $5?$ (The approximation of $\log_{10} 5$ as $0.7$ is sufficient for this problem.)

$18$

$20$

$22$

$24$

$26$

Answer: B

Concepts:logarithm number base

Difficulty rating: 1370

Small Hint:

The number of base- $5$ digits of $N$ is $\lfloor \log_5 N \rfloor + 1$

Big Hint:

Use $\log_5 N = \dfrac{\log_{10} N}{\log_{10} 5}$ with $\log_{10} 5 = 0.7$ and $\log_{10}(5\cdot 10^{13}) = 13 + \log_{10} 5$

Solution:

The number of digits of $N$ in base $5$ is $\lfloor \log_5 N \rfloor + 1.$ With $N = 5 \cdot 10^{13},$ $\log_{10} N = 13 + \log_{10} 5 = 13.7.$ Converting bases, $\log_5 N$ $= \dfrac{13.7}{\log_{10} 5}$ $= \dfrac{13.7}{0.7}$ $= 19.57\ldots$

Thus the number of digits is $\lfloor 19.57 \rfloor + 1 = 19 + 1 = 20.$

Thus, the correct answer is B.

7.

In the figure below $WXYZ$ is a rectangle with $WX = 4$ and $WZ = 8.$ Point $M$ lies on $\overline{XY},$ point $A$ lies on $\overline{YZ},$ and $\angle WMA$ is a right angle. The areas of $\triangle WXM$ and $\triangle WAZ$ are equal. What is the area of $\triangle WMA?$

$13$

$14$

$15$

$16$

$17$

Answer: C

Concepts:coordinate geometry triangle area

Difficulty rating: 1420

Small Hint:

Place $X = (0,0),$ $W = (0,4),$ $Y = (8,0),$ $Z = (8,4),$ with $M = (m, 0)$ and $A = (8, a)$

Big Hint:

The right angle gives $\overrightarrow{MW} \cdot \overrightarrow{MA} = 0,$ and the equal-area condition gives a second equation; solve for $m$ and $a$

Solution:

Set $X = (0,0),$ $W = (0,4),$ $Y = (8,0),$ $Z = (8,4),$ with $M = (m, 0)$ on $\overline{XY}$ and $A = (8, a)$ on $\overline{YZ}.$

Since $\angle WMA = 90^\circ,$ $\overrightarrow{MW} \cdot \overrightarrow{MA}$ $= (-m)(8-m)$ $+ 4a = 0,$ so $4a = m(8-m).$ The areas give $[\triangle WXM] = \tfrac12 \cdot 4 \cdot m = 2m$ and $[\triangle WAZ]$ $= \tfrac12 \cdot 8 \cdot (4 - a)$ $= 4(4 - a).$ Setting these equal yields $m = 8 - 2a.$

Substituting $a = \tfrac{8-m}{2}$ into $4a = m(8-m)$ gives $2(8-m) = m(8-m),$ so $m = 2$ and $a = 3.$ The other algebraic root, $m=8,$ gives $M=A=Y$ and no defined angle $\angle WMA,$ so it is invalid. Then with $W = (0,4),$ $M = (2,0),$ $A = (8,3),$ $\begin{aligned} [\triangle WMA] &= \tfrac12\,\bigl|2(3 - 4) + 8(4 - 0)\bigr| \\ &= \tfrac12 (30) = 15. \end{aligned}$

Thus, the correct answer is C.

8.

What value of $x$ satisfies

$\frac{\log_2 x \cdot \log_3 x}{\log_2 x + \log_3 x} = 2?$

$25$

$32$

$36$

$42$

$48$

Answer: C

Concepts:logarithm

Difficulty rating: 1460

Small Hint:

Divide numerator and denominator by $\log_2 x \cdot \log_3 x$ to turn the left side into $\dfrac{1}{\frac{1}{\log_2 x} + \frac{1}{\log_3 x}}$

Big Hint:

Since $\dfrac{1}{\log_2 x} = \log_x 2$ and $\dfrac{1}{\log_3 x} = \log_x 3,$ the equation becomes $\log_x 6 = \dfrac12$

Solution:

Dividing top and bottom by $\log_2 x \cdot \log_3 x,$ the left side becomes $\begin{gathered} \frac{1}{\dfrac{1}{\log_2 x} + \dfrac{1}{\log_3 x}} \\ = \frac{1}{\log_x 2 + \log_x 3} \\ = \frac{1}{\log_x 6}. \end{gathered}$ So $\dfrac{1}{\log_x 6} = 2,$ meaning $\log_x 6 = \dfrac12,$ i.e. $x^{\frac{1}{2}} = 6.$

Therefore $x = 36.$

Thus, the correct answer is C.

9.

A dartboard is the region $B$ in the coordinate plane consisting of points $(x, y)$ such that $|x| + |y| \le 8.$ A target $T$ is the region where $(x^2 + y^2 - 25)^2 \le 49.$ A dart is thrown and lands at a random point in $B.$ The probability that the dart lands in $T$ can be expressed as $\dfrac{m}{n} \cdot \pi,$ where $m$ and $n$ are relatively prime positive integers. What is $m + n?$

$39$

$71$

$73$

$75$

$135$

Answer: B

Concepts:geometric probability annulus

Difficulty rating: 1540

Small Hint:

The region $B$ is a square with diagonals of length $16,$ and $(x^2+y^2-25)^2 \le 49$ means $18 \le x^2 + y^2 \le 32$

Big Hint:

The annulus $T$ has area $\pi(32 - 18);$ check that its outer radius $\sqrt{32}$ exactly reaches the sides of $B,$ so $T$ lies entirely inside $B$

Solution:

The dartboard $|x| + |y| \le 8$ is a square with diagonals $16,$ so its area is $\tfrac12 \cdot 16 \cdot 16 = 128.$ The target condition $(x^2 + y^2 - 25)^2 \le 49$ means $-7 \le x^2 + y^2 - 25 \le 7,$ i.e. $18 \le x^2 + y^2 \le 32,$ an annulus of area $\pi(32 - 18) = 14\pi.$

The distance from the origin to a side of the square (for instance $x + y = 8$ ) is $\dfrac{8}{\sqrt2} = \sqrt{32},$ exactly the annulus’s outer radius. So the annulus is tangent to the square and lies entirely within $B.$ The probability is $\dfrac{14\pi}{128} = \dfrac{7}{64}\pi,$ giving $m + n = 7 + 64 = 71.$

Thus, the correct answer is B.

10.

A list of $9$ real numbers consists of $1,$ $2.2,$ $3.2,$ $5.2,$ $6.2,$ and $7,$ as well as $x,$ $y,$ $z$ with $x \le y \le z.$ The range of the list is $7,$ and the mean and median are both positive integers. How many ordered triples $(x, y, z)$ are possible?

$1$

$2$

$3$

$4$

infinitely many

Answer: C

Concepts:mean median (data)range casework

Difficulty rating: 1600

Small Hint:

The six fixed numbers sum to $24.8,$ so the mean is an integer only when $x + y + z$ ends in $0.2$

Big Hint:

The range condition gives three cases: $(\min,\max)=(0,7),\ (1,8),$ or $(t,t+7)$ with $0\lt t\lt1$

Solution:

The six fixed numbers sum to $24.8$ and span $[1,7].$ There are three possible arrangements of the overall extremes.

If the extremes are $0$ and $7,$ then $x=0$ and $z\le7.$ The only possible integer means are $3$ and $4,$ requiring $y+z=2.2$ or $11.2.$ The first makes the median $2.2.$ In the second, an integer median forces $y=5,$ hence $z=6.2.$ This gives $(0,5,6.2).$

If the extremes are $1$ and $8,$ then $z=8$ and the integer mean forces $x+y=3.2$ or $12.2.$ The first makes the median $3.2;$ the second has an integer median only for $x=6,\ y=6.2.$ This gives $(6,6.2,8).$

Finally, if both extremes are new, write $x=t,\ z=t+7$ with $0\lt t\lt1.$ The mean must be $4,$ so $y=4.2-2t.$ The median is the fourth number among $1,2.2,3.2,5.2,6.2,7,y;$ it is an integer only when $y=4,$ giving $t=0.1.$ Thus the third triple is $(0.1,4,7.1),$ and there are exactly $3$ in all.

Thus, the correct answer is C.

11.

Let $x_n = \sin^2(n^\circ).$ What is the mean of $x_1,$ $x_2,$ $x_3,$ $\ldots,$ $x_{90}?$

$\dfrac{11}{45}$

$\dfrac{22}{45}$

$\dfrac{89}{180}$

$\dfrac{1}{2}$

$\dfrac{91}{180}$

Answer: E

Concepts:trigonometric identity mean symmetry

Difficulty rating: 1610

Small Hint:

Use $\sin^2\theta = \dfrac{1 - \cos 2\theta}{2}$ to rewrite the sum

Big Hint:

The cosine terms $\cos 2^\circ, \cos 4^\circ, \ldots, \cos 180^\circ$ cancel in pairs via $\cos(180^\circ - \theta) = -\cos\theta,$ leaving only $\cos 180^\circ = -1$

Solution:

Using $\sin^2\theta = \dfrac{1 - \cos 2\theta}{2},$ $\begin{aligned} \sum_{n=1}^{90} \sin^2(n^\circ) &= \frac{90}{2} \\ &\quad {}- \frac12 \sum_{n=1}^{90}\cos(2n^\circ). \end{aligned}$ In the cosine sum, the terms for $n$ and $90 - n$ satisfy $\cos(2n^\circ) + \cos(180^\circ - 2n^\circ) = 0,$ and $\cos 90^\circ = 0,$ so everything cancels except $\cos 180^\circ = -1.$

Hence the sum is $45 - \tfrac12(-1) = 45.5,$ and the mean is $\dfrac{45.5}{90} = \dfrac{91}{180}.$

Thus, the correct answer is E.

12.

Suppose $z$ is a complex number with positive imaginary part, with real part greater than $1,$ and with $|z| = 2.$ In the complex plane, the four points $0,$ $z,$ $z^2,$ and $z^3$ are the vertices of a quadrilateral with area $15.$ What is the imaginary part of $z?$

$\dfrac{3}{4}$

$1$

$\dfrac{4}{3}$

$\dfrac{3}{2}$

$\dfrac{5}{3}$

Answer: D

Concepts:complex number shoelace formula

Difficulty rating: 1670

Small Hint:

The signed area of the quadrilateral $0 \to z \to z^2 \to z^3$ is $\tfrac12\,\bigl|\operatorname{Im}(\bar z z^2 + \overline{z^2}\, z^3)\bigr|$

Big Hint:

Since $\bar z z^2 = |z|^2 z$ and $\overline{z^2} z^3 = |z|^4 z,$ the area is $\tfrac12(|z|^2 + |z|^4)\operatorname{Im}(z);$ plug in $|z| = 2$

Solution:

For vertices $0, z, z^2, z^3$ the shoelace formula gives area $\begin{aligned} &\tfrac12\,\bigl|\operatorname{Im}(\bar z z^2 + \overline{z^2}\,z^3)\bigr| \\ &= \tfrac12\,\bigl|\operatorname{Im}\bigl((|z|^2 + |z|^4)z\bigr)\bigr| \\ &= \tfrac12(|z|^2 + |z|^4)\operatorname{Im}(z). \end{aligned}$

With $|z| = 2,$ this is $\tfrac12(4 + 16)\operatorname{Im}(z) = 10\operatorname{Im}(z).$ Setting $10\operatorname{Im}(z) = 15$ gives $\operatorname{Im}(z) = \dfrac32.$ (Then $\operatorname{Re}(z) = \sqrt{4 - \tfrac94} = \tfrac{\sqrt7}{2} \gt 1,$ as required.)

Thus, the correct answer is D.

13.

There are real numbers $x,$ $y,$ $h,$ and $k$ that satisfy the system of equations

$x^2 + y^2 - 6x - 8y = h$

$x^2 + y^2 - 10x + 4y = k.$

What is the minimum possible value of $h + k?$

$-54$

$-46$

$-34$

$-16$

$16$

Answer: C

Concepts:completing the square optimization

Difficulty rating: 1640

Small Hint:

Add the two equations to get $h + k$ as a single expression in $x$ and $y$

Big Hint:

Complete the square: $h + k = 2(x-4)^2$ $+ 2(y-1)^2 - 34,$ which is smallest when both squares vanish

Solution:

Adding the equations, $\begin{aligned} h + k &= 2x^2 + 2y^2 - 16x - 4y \\ &= 2(x - 4)^2 \\ &\quad {}+ 2(y - 1)^2 - 34. \end{aligned}$ Both squared terms are nonnegative, so the minimum occurs at $x = 4,$ $y = 1,$ giving $h + k = -34.$

Thus, the correct answer is C.

14.

How many different remainders can result when the $100$ th power of an integer is divided by $125?$

$1$

$2$

$5$

$25$

$125$

Answer: B

Concepts:Euler’s Totient Function modular exponentiation

Difficulty rating: 1760

Small Hint:

Split into integers coprime to $5$ and integers divisible by $5$

Big Hint:

Since $\varphi(125) = 100,$ Euler’s theorem gives $n^{100} \equiv 1 \pmod{125}$ when $\gcd(n,5)=1;$ a multiple of $5$ has $n^{100}$ divisible by $125$

Solution:

If $n$ is coprime to $5,$ then since $\varphi(125) = 100,$ Euler’s theorem gives $n^{100} \equiv 1 \pmod{125}.$ If $n$ is a multiple of $5,$ then $n^{100}$ is divisible by $5^{100},$ hence by $125,$ leaving remainder $0.$

So the only possible remainders are $0$ and $1,$ which is $2$ distinct values.

Thus, the correct answer is B.

15.

A triangle in the coordinate plane has vertices $A(\log_2 1, \log_2 2),$ $B(\log_2 3, \log_2 4),$ and $C(\log_2 7, \log_2 8).$ What is the area of $\triangle ABC?$

$\log_2 \dfrac{\sqrt3}{7}$

$\log_2 \dfrac{3}{\sqrt7}$

$\log_2 \dfrac{7}{\sqrt3}$

$\log_2 \dfrac{11}{\sqrt7}$

$\log_2 \dfrac{11}{\sqrt3}$

Answer: B

Concepts:shoelace formula logarithm triangle area

Difficulty rating: 1800

Small Hint:

The vertices are $(0, 1),$ $(\log_2 3, 2),$ and $(\log_2 7, 3)$

Big Hint:

Apply the shoelace formula and simplify with $2\log_2 3 - \log_2 7 = \log_2 \dfrac{9}{7},$ then halve the logarithm

Solution:

The vertices are $A = (0, 1),$ $B = (\log_2 3, 2),$ $C = (\log_2 7, 3).$ By the shoelace formula, $\begin{gathered} [\triangle ABC]\\ {}=\tfrac12\,\bigl|2\log_2 3-\log_2 7\bigr|. \end{gathered}$

This equals $\tfrac12\log_2 \dfrac{9}{7} = \log_2 \sqrt{\tfrac{9}{7}} = \log_2 \dfrac{3}{\sqrt7}.$

Thus, the correct answer is B.

16.

A group of $16$ people will be partitioned into $4$ indistinguishable $4$ -person committees. Each committee will have one chairperson and one secretary. The number of different ways to make these assignments can be written as $3^r M,$ where $r$ and $M$ are positive integers and $M$ is not divisible by $3.$ What is $r?$

$5$

$6$

$7$

$8$

$9$

Answer: A

Concepts:multiplication principle Legendre’s Formula

Difficulty rating: 1860

Small Hint:

The number of assignments is $\dfrac{16!}{(4!)^4\, 4!} \cdot 12^4,$ where each committee contributes $4 \cdot 3 = 12$ chair-and-secretary choices

Big Hint:

Track only powers of $3:$ $16!$ gives $3^6,$ the denominator $(4!)^4\,4!$ gives $3^5,$ and $12^4$ gives $3^4$

Solution:

The number of ways to split $16$ people into $4$ indistinguishable groups of $4$ is $\dfrac{16!}{(4!)^4\, 4!}.$ Each committee then chooses a chairperson and a secretary in $4 \cdot 3 = 12$ ways, contributing $12^4.$ So the total is $\dfrac{16!}{(4!)^4\,4!}\cdot 12^4.$

Counting factors of $3:$ $16!$ contributes $\lfloor \frac{16}{3}\rfloor + \lfloor \frac{16}{9}\rfloor = 6.$ The denominator $(4!)^4\,4!$ contributes $4 + 1 = 5.$ And $12^4 = (2^2\cdot 3)^4$ contributes $4.$ Thus $r = 6 - 5 + 4 = 5.$

Thus, the correct answer is A.

17.

Integers $a$ and $b$ are randomly chosen without replacement from the set of integers with absolute value not exceeding $10.$ What is the probability that the polynomial $x^3 + ax^2 + bx + 6$ has $3$ distinct integer roots?

$\dfrac{1}{240}$

$\dfrac{1}{221}$

$\dfrac{1}{105}$

$\dfrac{1}{84}$

$\dfrac{1}{63}$

Answer: C

Concepts:Vieta’s Formulas basic probability casework

Difficulty rating: 1910

Small Hint:

If the roots are distinct integers $p, q, r,$ then $pqr = -6,$ $a = -(p+q+r),$ and $b = pq + qr + rp$

Big Hint:

List the ways to write $-6$ as a product of three distinct integers, discard any giving $|a|\gt 10$ or $|b| \gt 10,$ and divide by $21 \cdot 20$ ordered choices

Solution:

The set has $21$ integers, so there are $21 \cdot 20 = 420$ ordered choices of $(a, b).$ If the polynomial has distinct integer roots $p, q, r,$ then $pqr = -6,$ $a = -(p + q + r),$ and $b = pq + qr + rp.$

The triples of distinct integers with product $-6$ are $\{1, 2, -3\},$ $\{1, -2, 3\},$ $\{-1, 2, 3\},$ $\{-1, -2, -3\},$ and $\{1, -1, 6\}.$ These give $(a, b) = (0, -7),$ $(-2, -5),$ $(-4, 1),$ $(6, 11),$ and $(-6, -1).$ The fourth has $b = 11 \gt 10,$ so it is invalid; the other four are valid and distinct.

The probability is $\dfrac{4}{420} = \dfrac{1}{105}.$

Thus, the correct answer is C.

18.

The Fibonacci numbers are defined by $F_1 = 1,$ $F_2 = 1,$ and $F_n = F_{n-1} + F_{n-2}$ for $n \ge 3.$ What is

$\frac{F_2}{F_1} + \frac{F_4}{F_2} + \frac{F_6}{F_3} + \cdots + \frac{F_{20}}{F_{10}}?$

$318$

$319$

$320$

$321$

$322$

Answer: B

Concepts:Fibonacci summation

Difficulty rating: 1930

Small Hint:

Each term simplifies: $\dfrac{F_{2k}}{F_k}$ is an integer; compute the first few, such as $\dfrac{F_2}{F_1} = 1,$ $\dfrac{F_4}{F_2} = 3,$ $\dfrac{F_6}{F_3} = 4$

Big Hint:

These values are the Lucas numbers $L_k = 1, 3, 4, 7, 11, \ldots;$ sum $L_1$ through $L_{10}$

Solution:

Since $F_{2k} = F_k L_k$ where $L_k$ is the $k$ th Lucas number, each term $\dfrac{F_{2k}}{F_k} = L_k.$ The sum is $\begin{gathered} L_1 + L_2 + \cdots + L_{10} \\ = 1 + 3 + 4 + 7 + 11 + 18 \\ {}+ 29 + 47 + 76 + 123 \\ = 319. \end{gathered}$ (Equivalently, $L_1 + \cdots + L_{10}$ $= L_{12} - 3$ $= 322 - 3 = 319.$ )

Thus, the correct answer is B.

19.

Equilateral $\triangle ABC$ with side length $14$ is rotated about its center by angle $\theta,$ where $0 \lt \theta \lt 60^\circ,$ to form $\triangle DEF.$ See the figure. The area of hexagon $ADBECF$ is $91\sqrt3.$ What is $\tan\theta?$

$\dfrac{3}{4}$

$\dfrac{5\sqrt3}{11}$

$\dfrac{4}{5}$

$\dfrac{11}{13}$

$\dfrac{7\sqrt3}{13}$

Answer: B

Concepts:trigonometric identity area transformation

Difficulty rating: 2040

Small Hint:

All six vertices lie on the circle of radius $R = \dfrac{14}{\sqrt3};$ the hexagon’s central angles alternate $\theta$ and $120^\circ - \theta$

Big Hint:

The area is $\tfrac12 R^2\bigl(3\sin\theta + 3\sin(120^\circ - \theta)\bigr)$ $= 98\bigl(\sin\theta + \sin(120^\circ - \theta)\bigr);$ simplify with sum-to-product to solve for $\theta$

Solution:

The six vertices lie on the circumcircle of radius $R = \dfrac{14}{\sqrt3},$ so $R^2 = \dfrac{196}{3}.$ Going around, the central angles alternate between $\theta$ (three times) and $120^\circ - \theta$ (three times). The cyclic-hexagon area is $\begin{aligned} &\tfrac12 R^2\bigl(3\sin\theta + 3\sin(120^\circ - \theta)\bigr) \\ &= 98\bigl(\sin\theta + \sin(120^\circ - \theta)\bigr). \end{aligned}$

By sum-to-product, $\sin\theta + \sin(120^\circ - \theta)$ $= 2\sin 60^\circ\cos(\theta - 60^\circ)$ $= \sqrt3\cos(60^\circ - \theta).$ Setting the area to $91\sqrt3$ gives $\sqrt3\cos(60^\circ - \theta)$ $= \dfrac{91\sqrt3}{98} = \dfrac{13\sqrt3}{14},$ so $\cos(60^\circ - \theta) = \dfrac{13}{14}$ and $\sin(60^\circ - \theta) = \dfrac{3\sqrt3}{14}.$

Then $\tan(60^\circ - \theta) = \dfrac{3\sqrt3}{13},$ and $\begin{aligned} &\tan\theta = \tan\bigl(60^\circ - (60^\circ - \theta)\bigr) \\ &= \frac{\sqrt3 - \frac{3\sqrt3}{13}}{1 + \sqrt3\cdot\frac{3\sqrt3}{13}} \\ &= \frac{\frac{10\sqrt3}{13}}{\frac{22}{13}} \\ &= \frac{5\sqrt3}{11}. \end{aligned}$

Thus, the correct answer is B.

20.

Suppose $A,$ $B,$ and $C$ are points in the plane with $AB = 40$ and $AC = 42,$ and let $x$ be the length of the line segment from $A$ to the midpoint of $\overline{BC}.$ Define a function $f$ by letting $f(x)$ be the area of $\triangle ABC.$ Then the domain of $f$ is an open interval $(p, q),$ and the maximum value $r$ of $f(x)$ occurs at $x = s.$ What is $p + q + r + s?$

$909$

$910$

$911$

$912$

$913$

Answer: C

Concepts:median (geometry)triangle inequality optimization

Difficulty rating: 2110

Small Hint:

The median satisfies $x^2 = \dfrac{2\cdot 40^2 + 2\cdot 42^2 - a^2}{4},$ where $a = BC;$ the triangle inequality $2 \lt a \lt 82$ determines the domain

Big Hint:

The area is largest when $\angle A = 90^\circ;$ find that maximum area and the corresponding $x$

Solution:

Let $a = BC.$ The median length gives $x^2$ $= \dfrac{2\cdot 1600 + 2\cdot 1764 - a^2}{4}$ $= \dfrac{6728 - a^2}{4}.$ The triangle inequality requires $2 \lt a \lt 82,$ i.e. $4 \lt a^2 \lt 6724,$ which translates to $1 \lt x \lt 41.$ So $(p, q) = (1, 41).$

With $AB = 40$ and $AC = 42$ fixed, the area $\tfrac12\cdot 40\cdot 42\sin A$ is largest when $\angle A = 90^\circ,$ giving $r = 840.$ Then $a^2 = 40^2 + 42^2 = 3364,$ so $x^2 = \dfrac{6728 - 3364}{4} = 841,$ i.e. $s = 29.$

Thus $p + q + r + s$ $= 1 + 41 + 840 + 29$ $= 911.$

Thus, the correct answer is C.

21.

The measures of the smallest angles of three different right triangles sum to $90^\circ.$ All three triangles have side lengths that are primitive Pythagorean triples. Two of them are $3$ - $4$ - $5$ and $5$ - $12$ - $13.$ What is the perimeter of the third triangle?

$40$

$126$

$154$

$176$

$208$

Answer: C

Concepts:trigonometric identity Pythagorean Triple

Difficulty rating: 2130

Small Hint:

The smallest angles have tangents $\tfrac34$ and $\tfrac{5}{12};$ the third must satisfy $\alpha + \beta + \gamma = 90^\circ,$ so $\tan\gamma = \cot(\alpha + \beta)$

Big Hint:

Compute $\tan(\alpha + \beta) = \dfrac{56}{33},$ so $\tan\gamma = \dfrac{33}{56};$ find the primitive triple with legs $33$ and $56$

Solution:

The smallest angles $\alpha, \beta$ of the $3$ - $4$ - $5$ and $5$ - $12$ - $13$ triangles have $\tan\alpha = \tfrac34$ and $\tan\beta = \tfrac{5}{12}.$ By the tangent addition formula, $\begin{aligned} &\tan(\alpha + \beta) = \frac{\tfrac34 + \tfrac{5}{12}}{1 - \tfrac34\cdot\tfrac{5}{12}} \\ &= \frac{\tfrac{14}{12}}{\tfrac{33}{48}} \\ &= \frac{56}{33}. \end{aligned}$

The third smallest angle $\gamma$ satisfies $\gamma = 90^\circ - (\alpha + \beta),$ so $\tan\gamma = \dfrac{33}{56}.$ The right triangle with legs $33$ and $56$ has hypotenuse $\sqrt{33^2 + 56^2} = \sqrt{4225} = 65,$ a primitive triple. Its perimeter is $33 + 56 + 65 = 154.$

Thus, the correct answer is C.

22.

Let $\triangle ABC$ be a triangle with integer side lengths and the property that $\angle B = 2\angle A.$ What is the least possible perimeter of such a triangle?

$13$

$14$

$15$

$16$

$17$

Answer: C

Concepts:law of sines Diophantine Equation triangle inequality

Difficulty rating: 2230

Small Hint:

With $a = BC,$ $b = CA,$ $c = AB,$ the condition $\angle B = 2\angle A$ is equivalent to $b^2 = a(a + c)$

Big Hint:

Since the perimeter is $b+\dfrac{b^2}{a}\gt2b,$ a perimeter below $15$ would require $b\le6$ ; check the divisors $a\lt b$ of $b^2$

Solution:

When $\angle B = 2\angle A,$ the side lengths satisfy $b^2 = a(a + c),$ where $a = BC,$ $b = CA,$ $c = AB.$ So $c = \dfrac{b^2 - a^2}{a}$ must be a positive integer, and the sides must form a valid triangle.

Also $b\gt a,$ because $\angle B=2\angle A\gt\angle A.$ The perimeter is $a+b+c=b+\dfrac{b^2}{a}\gt2b.$ Therefore a perimeter below $15$ would require $b\le6.$ For $b=2,3,4,5,6,$ the divisors $a\lt b$ of $b^2$ give the possible pairs $\begin{gathered} (b,a)=(2,1),(3,1),\\ (4,1),(4,2),(5,1),\\ (6,1),(6,2),(6,3),(6,4). \end{gathered}$ Substitution gives a degenerate or invalid triangle in every case except $(a,b)=(4,6),$ which gives $c=5.$ Thus $(4,5,6)$ is the first valid triangle, and its perimeter is $15.$

Thus, the correct answer is C.

23.

A right pyramid has regular octagon $ABCDEFGH$ with side length $1$ as its base and apex $V.$ Segments $\overline{AV}$ and $\overline{DV}$ are perpendicular. What is the square of the height of the pyramid?

$1$

$\dfrac{1 + \sqrt2}{2}$

$\sqrt2$

$\dfrac{3}{2}$

$\dfrac{2 + \sqrt2}{3}$

Answer: B

Concepts:pyramid regular polygon 3D geometry

Difficulty rating: 2300

Small Hint:

Each lateral edge has length $L$ with $L^2 = h^2 + R^2,$ where $R$ is the octagon’s circumradius; the right angle at $V$ gives $AD^2 = 2L^2$

Big Hint:

$A$ and $D$ are three vertices apart, so the central angle is $135^\circ$ and $AD^2 = R^2(2 + \sqrt2);$ also $R^2 = \dfrac{2 + \sqrt2}{2}.$ Solve for $h^2$

Solution:

Let $R$ be the circumradius of the octagon and $L$ the length of each lateral edge, so $L^2 = h^2 + R^2.$ Since $\angle AVD = 90^\circ,$ $AD^2 = 2L^2.$

Vertices $A$ and $D$ are three steps apart, a central angle of $135^\circ,$ so $AD^2$ $= 2R^2(1 - \cos 135^\circ)$ $= R^2(2 + \sqrt2).$ Setting $R^2(2 + \sqrt2) = 2(h^2 + R^2)$ gives $2h^2 = R^2\sqrt2.$

For a regular octagon of side $1,$ $R^2 = \dfrac{1}{2\sin^2(22.5^\circ)} = \dfrac{2 + \sqrt2}{2}.$ Therefore $h^2$ $= \dfrac{R^2\sqrt2}{2}$ $= \dfrac{(2 + \sqrt2)\sqrt2}{4}$ $= \dfrac{2\sqrt2 + 2}{4}$ $= \dfrac{1 + \sqrt2}{2}.$

Thus, the correct answer is B.

24.

What is the number of ordered triples $(a, b, c)$ of positive integers, with $a \le b \le c \le 9,$ such that there exists a (non-degenerate) triangle $\triangle ABC$ with an integer inradius for which $a,$ $b,$ and $c$ are the lengths of the altitudes from $A$ to $\overline{BC},$ $B$ to $\overline{AC},$ and $C$ to $\overline{AB},$ respectively? (Recall that the inradius of a triangle is the radius of the largest possible circle that can be inscribed in the triangle.)

$2$

$3$

$4$

$5$

$6$

Answer: B

Concepts:altitude incircle, incenter, and inradius triangle inequality

Difficulty rating: 2410

Small Hint:

Since each side equals $\dfrac{2[\triangle]}{\text{altitude}},$ the inradius satisfies $\dfrac1r = \dfrac1a + \dfrac1b + \dfrac1c$

Big Hint:

The sides are proportional to $\tfrac1a, \tfrac1b, \tfrac1c,$ so non-degeneracy needs $\tfrac1a \lt \tfrac1b + \tfrac1c;$ seek triples with $\tfrac1a + \tfrac1b + \tfrac1c$ a unit fraction

Solution:

Writing each side as $\dfrac{2[\triangle]}{h},$ the semiperimeter is $[\triangle]\bigl(\tfrac1a + \tfrac1b + \tfrac1c\bigr),$ so the inradius $r = \dfrac{[\triangle]}{s}$ satisfies $\dfrac1r = \dfrac1a + \dfrac1b + \dfrac1c.$ We need this to be $\dfrac1r$ for a positive integer $r,$ with the sides (proportional to $\tfrac1a, \tfrac1b, \tfrac1c$ ) forming a non-degenerate triangle, requiring $\tfrac1a \lt \tfrac1b + \tfrac1c.$

Because $a\le b\le c\le9,$ the reciprocal sum is at least $\frac{3}{c}\ge\frac{1}{3},$ so the integer $r$ is one of $1,2,3.$ Also $\frac{1}{a}\lt\frac{1}{r}\le\frac{3}{a},$ so $r\lt a\le3r.$ For each of these few values of $r,a,$ substitute $b=a,a+1,\ldots,9$ into $c=\frac{abr}{ab-ar-br}.$ Keeping only integral $c$ with $b\le c\le9$ gives the complete list $\begin{array}{c|l} r& (a,b,c)\\ \hline 1&(2,3,6),(2,4,4),(3,3,3)\\ 2&(4,8,8),(6,6,6)\\ 3&(9,9,9). \end{array}$ The triples $(2,3,6),(2,4,4),$ and $(4,8,8)$ have $\tfrac1a=\tfrac1b+\tfrac1c$ and therefore give degenerate triangles. The remaining triples are $(3,3,3),(6,6,6),$ and $(9,9,9),$ so the answer is $3.$

Thus, the correct answer is B.

25.

Pablo will decorate each of $6$ identical white balls with either a striped or a dotted pattern, using either red or blue paint. He will decide on the color and pattern for each ball by flipping a fair coin for each of the $12$ decisions he must make. After the paint dries, he will place the $6$ balls in an urn. Frida will randomly select one ball from the urn and note its color and pattern. The events “the ball Frida selects is red” and “the ball Frida selects is striped” may or may not be independent, depending on the outcome of Pablo’s coin flips. The probability that these two events are independent can be written as $\dfrac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. What is $m?$ (Recall that two events $A$ and $B$ are independent if $P(A \text{ and } B) = P(A)\cdot P(B).$ )

$243$

$245$

$247$

$249$

$251$

Answer: A

Concepts:independent events casework

Difficulty rating: 2510

Small Hint:

Let the six balls include $k$ that are red-and-striped, with $R$ red and $S$ striped total. Independence for Frida’s uniform pick means $\dfrac{k}{6} = \dfrac{R}{6}\cdot\dfrac{S}{6}$

Big Hint:

So the condition is $6k = RS.$ Count assignments of $6$ balls to the $4$ equally likely types satisfying this, over $4^6$ total, using multinomial coefficients

Solution:

Each ball is independently one of four equally likely types: red-striped, red-dotted, blue-striped, blue-dotted. Suppose among the $6$ balls there are $k$ red-striped, with $R$ red and $S$ striped in total. For Frida’s uniform pick, $P(\text{red}) = \tfrac{R}{6},$ $P(\text{striped}) = \tfrac{S}{6},$ and $P(\text{red and striped}) = \tfrac{k}{6}.$ Independence means $\dfrac{k}{6} = \dfrac{R}{6}\cdot\dfrac{S}{6},$ i.e. $6k = RS.$

First count assignments with $R\in\{0,6\}$ or $S\in\{0,6\}.$ Each of these four conditions leaves $2^6=64$ choices for the other attribute, and the four assignments at their pairwise intersections have been counted twice. Their union therefore contributes $4\cdot64-4=252.$

For $1\le R,S\le5,$ the condition that $RS$ be divisible by $6$ leaves only $(R,S)=(2,3),(3,2)$ or $(3,4),(4,3).$ In each case $k=\frac{RS}{6},$ and the four type counts are a permutation of $(1,1,2,2).$ Thus each pair contributes $\dfrac{6!}{1!1!2!2!}=180$ assignments. The favorable count is therefore $252+4\cdot180=972.$ Out of $4^6=4096$ equally likely assignments, the probability is $\dfrac{972}{4096}=\dfrac{243}{1024},$ so $m=243.$

Thus, the correct answer is A.