2021 AMC 12B Spring — Problems & Online Timed Exam

Timed

1:15:00

1.

How many integer values of $x$ satisfy $|x| \lt 3\pi?$

$9$

$10$

$18$

$19$

$20$

Answer: D

Concepts:absolute value estimation

Difficulty rating: 870

Small Hint:

$3\pi$ is a little more than $9$

Big Hint:

Count the integers from $-9$ to $9$ inclusive

Solution:

Since $3\pi \approx 9.42,$ the inequality $|x| \lt 3\pi$ means $-9.42 \lt x \lt 9.42.$

The integers in this range run from $-9$ to $9,$ giving $19$ values.

Thus, the correct answer is D.

2.

At a math contest, $57$ students are wearing blue shirts, and another $75$ students are wearing yellow shirts. The $132$ students are assigned into $66$ pairs. In exactly $23$ of these pairs, both students are wearing blue shirts. In how many pairs are both students wearing yellow shirts?

$23$

$32$

$37$

$41$

$64$

Answer: B

Concepts:basic counting parity

Difficulty rating: 1040

Small Hint:

The $23$ all-blue pairs use $46$ blue students

Big Hint:

Count the blue students left in mixed pairs, then remove that many yellow students

Solution:

The $23$ all-blue pairs account for $46$ blue students, leaving $57 - 46 = 11$ blue students.

Each of those $11$ blue students must be paired with a yellow student, so there are $11$ mixed pairs, using $11$ yellow students.

The remaining $75 - 11 = 64$ yellow students form $64 \div 2 = 32$ all-yellow pairs.

Thus, the correct answer is B.

3.

Suppose $2+\cfrac{1}{1+\cfrac{1}{2+\cfrac{2}{3+x}}}=\dfrac{144}{53}.$

What is the value of $x?$

$\dfrac{3}{4}$

$\dfrac{7}{8}$

$\dfrac{14}{15}$

$\dfrac{37}{38}$

$\dfrac{52}{53}$

Answer: A

Concepts:continued fraction work backwards

Difficulty rating: 1170

Small Hint:

Peel off one layer at a time, starting by subtracting $2$ from $\dfrac{144}{53}$

Big Hint:

After $\dfrac{144}{53}-2=\dfrac{38}{53},$ take reciprocals and continue inward

Solution:

Working from the outside in, $\dfrac{144}{53}-2=\dfrac{38}{53},$ so the inner fraction equals $\dfrac{38}{53}.$

Its reciprocal gives $1+\cfrac{1}{2+\frac{2}{3+x}}=\dfrac{53}{38},$ so $\cfrac{1}{2+\frac{2}{3+x}}=\dfrac{15}{38}.$

Then $2+\dfrac{2}{3+x}=\dfrac{38}{15},$ so $\dfrac{2}{3+x}=\dfrac{8}{15},$ giving $3+x=\dfrac{15}{4}.$

Therefore $x=\dfrac{15}{4}-3=\dfrac{3}{4}.$

Thus, the correct answer is A.

4.

Ms. Blackwell gives an exam to two classes. The mean of the scores of the students in the morning class is $84,$ and the afternoon class’s mean score is $70.$ The ratio of the number of students in the morning class to the number of students in the afternoon class is $\dfrac{3}{4}.$ What is the mean of the scores of all the students?

$74$

$75$

$76$

$77$

$78$

Answer: C

Concepts:mean ratio and proportion

Difficulty rating: 1100

Small Hint:

Take $3$ students in the morning and $4$ in the afternoon

Big Hint:

The overall mean is $\dfrac{3\cdot 84+4\cdot 70}{7}$

Solution:

Suppose there are $3$ students in the morning class and $4$ in the afternoon class.

The total of all scores is $3\cdot 84+4\cdot 70$ $=252+280$ $=532.$

The overall mean is $\dfrac{532}{7}=76.$

Thus, the correct answer is C.

5.

The point $P(a,b)$ in the $xy$ -plane is first rotated counterclockwise by $90^\circ$ around the point $(1,5)$ and then reflected about the line $y=-x.$ The image of $P$ after these two transformations is at $(-6,3).$ What is $b-a?$

$1$

$3$

$5$

$7$

$9$

Answer: D

Concepts:transformation work backwards

Difficulty rating: 1330

Small Hint:

Undo the transformations: reflecting $(-6,3)$ about $y=-x$ reverses the last step

Big Hint:

Reflecting $(x,y)$ about $y=-x$ gives $(-y,-x);$ a $90^\circ$ rotation about $(1,5)$ sends $(a,b)$ to $(6-b,\,4+a)$

Solution:

A $90^\circ$ counterclockwise rotation about $(1,5)$ sends $(a,b)$ to $(1-(b-5),\,5+(a-1))$ $=(6-b,\,4+a).$

Reflecting that about $y=-x$ (which maps $(x,y)$ to $(-y,-x)$ ) gives $(-(4+a),\,-(6-b))$ $=(-4-a,\,b-6).$

Setting this equal to $(-6,3)$ gives $-4-a=-6$ and $b-6=3,$ so $a=2$ and $b=9.$

Therefore $b-a=9-2=7.$

Thus, the correct answer is D.

6.

An inverted cone with base radius $12$ cm and height $18$ cm is full of water. The water is poured into a tall cylinder whose horizontal base has a radius of $24$ cm. What is the height in centimeters of the water in the cylinder?

$1.5$

$3$

$4$

$4.5$

$6$

Answer: A

Concepts:cone cylinder volume

Difficulty rating: 1220

Small Hint:

The volume of a cone is $\dfrac13\pi r^2 h$ and of a cylinder is $\pi r^2 h$

Big Hint:

Set the cone’s volume equal to $\pi\cdot 24^2\cdot h$ and solve for $h$

Solution:

The cone holds $\dfrac13\pi(12)^2(18)=864\pi$ cubic centimeters of water.

Poured into the cylinder, this fills to height $h$ where $\pi(24)^2 h=864\pi.$

Then $576h=864,$ so $h=1.5.$

Thus, the correct answer is A.

7.

Let $N=34\cdot 34\cdot 63\cdot 270.$ What is the ratio of the sum of the odd divisors of $N$ to the sum of the even divisors of $N?$

$1:16$

$1:15$

$1:14$

$1:8$

$1:3$

Answer: C

Concepts:prime factorization sum of factors

Difficulty rating: 1370

Small Hint:

Find the exact power of $2$ dividing $N$

Big Hint:

If $M$ is the odd part of $N$ and $2^k$ the even part, every even divisor is $2^j$ times an odd divisor

Solution:

Factoring, $34=2\cdot 17,$ $63=3^2\cdot 7,$ and $270=2\cdot 3^3\cdot 5,$ so $N=2^3\cdot 3^5\cdot 5\cdot 7\cdot 17^2.$

Let $M$ be the odd part $3^5\cdot 5\cdot 7\cdot 17^2.$ The sum of all divisors is $(1+2+4+8)\,\sigma(M)$ $=15\,\sigma(M).$

The odd divisors sum to $\sigma(M),$ so the even divisors sum to $15\,\sigma(M)-\sigma(M)=14\,\sigma(M).$

The ratio is $\sigma(M):14\,\sigma(M)=1:14.$

Thus, the correct answer is C.

8.

Three equally spaced parallel lines intersect a circle, creating three chords of lengths $38,$ $38,$ and $34.$ What is the distance between two adjacent parallel lines?

$5\tfrac{1}{2}$

$6$

$6\tfrac{1}{2}$

$7$

$7\tfrac{1}{2}$

Answer: B

Concepts:chord Pythagorean Theorem

Difficulty rating: 1500

Small Hint:

Equal chords are equidistant from the center, so the two $38$ chords sit at heights $\pm\tfrac{d}{2}$

Big Hint:

With spacing $d,$ use $r^2-\left(\tfrac{d}{2}\right)^2=19^2$ and $r^2-\left(\tfrac{3d}{2}\right)^2=17^2$

Solution:

Place the center at height $0.$ Two equal chords lie at equal distances from the center, so the three equally spaced lines are at heights $-\tfrac{d}{2},\tfrac{d}{2},\tfrac{3d}{2},$ with the two $38$ -chords at $\pm\tfrac{d}{2}$ and the $34$ -chord at $\tfrac{3d}{2}.$

Half-chord relations give $r^2-\left(\tfrac{d}{2}\right)^2=19^2$ and $r^2-\left(\tfrac{3d}{2}\right)^2=17^2.$

Subtracting, $2d^2=19^2-17^2=72,$ so $d^2=36$ and $d=6.$

Thus, the correct answer is B.

9.

What is the value of the following expression? $\dfrac{\log_2 80}{\log_{40}2}-\dfrac{\log_2 160}{\log_{20}2}$

$0$

$1$

$\dfrac{5}{4}$

$2$

$\log_2 5$

Answer: D

Concepts:logarithm algebraic manipulation

Difficulty rating: 1520

Small Hint:

$\dfrac{1}{\log_{40}2}=\log_2 40,$ and similarly for the other reciprocal

Big Hint:

Write everything in terms of $t=\log_2 5,$ since $\log_2 80=4+t,$ etc.

Solution:

Using $\dfrac{1}{\log_{40}2}=\log_2 40$ and $\dfrac{1}{\log_{20}2}=\log_2 20,$ the expression becomes $(\log_2 80)(\log_2 40)$ $-(\log_2 160)(\log_2 20).$

Let $t=\log_2 5.$ Then $\log_2 80=4+t,$ $\log_2 40=3+t,$ $\log_2 160=5+t,$ $\log_2 20=2+t.$

The value is $(4+t)(3+t)$ $-(5+t)(2+t)$ $=(12+7t+t^2)$ $-(10+7t+t^2)$ $=2.$

Thus, the correct answer is D.

10.

Two distinct numbers are selected from the set $\{1,2,3,4,\ldots,36,37\}$ so that the sum of the remaining $35$ numbers is the product of these two numbers. What is the difference of these two numbers?

$5$

$7$

$8$

$9$

$10$

Answer: E

Concepts:Simon’s Favorite Factoring Trick arithmetic sequence

Difficulty rating: 1530

Small Hint:

The full set sums to $703;$ if the two chosen numbers are $a$ and $b,$ then $703-a-b=ab$

Big Hint:

Rearrange to $(a+1)(b+1)=704$ and factor

Solution:

The sum $1+2+\cdots+37=703.$ If the chosen numbers are $a$ and $b,$ then $703-a-b=ab.$

So $ab+a+b=703,$ and adding $1$ gives $(a+1)(b+1)=704=2^6\cdot 11.$

We need factors $a+1,b+1$ between $2$ and $38.$ The pair $22\cdot 32=704$ works, giving $a=21,$ $b=31.$

Their difference is $31-21=10.$

Thus, the correct answer is E.

11.

Triangle $ABC$ has $AB=13,$ $BC=14,$ and $AC=15.$ Let $P$ be the point on $\overline{AC}$ such that $PC=10.$ There are exactly two points $D$ and $E$ on line $BP$ such that quadrilaterals $ABCD$ and $ABCE$ are trapezoids. What is the distance $DE?$

$\dfrac{42}{5}$

$6\sqrt2$

$\dfrac{84}{5}$

$12\sqrt2$

$18$

Answer: D

Concepts:coordinate geometry parallel lines trapezoid

Difficulty rating: 1690

Small Hint:

Place $A=(0,0),$ $C=(15,0);$ then $B=\left(\tfrac{33}{5},\tfrac{56}{5}\right)$ and $P=(5,0)$

Big Hint:

One trapezoid needs $CD\parallel AB,$ the other needs $AE\parallel BC;$ find where each parallel line meets line $BP$

Solution:

Place $A=(0,0)$ and $C=(15,0).$ Then $B=\left(\tfrac{33}{5},\tfrac{56}{5}\right),$ and since $PC=10,$ $P=(5,0).$ Line $BP$ has slope $7,$ so it is $y=7(x-5).$

For $ABCD$ to be a trapezoid with $D$ on line $BP,$ take $CD\parallel AB.$ The line through $C$ parallel to $AB$ meets line $BP$ at $(\tfrac95,-\tfrac{112}{5}).$

For $ABCE$ with $E$ on line $BP,$ take $AE\parallel BC.$ The line through $A$ parallel to $BC$ meets line $BP$ at $(\tfrac{21}{5},-\tfrac{28}{5}).$

Their coordinate differences are $\tfrac{12}{5}$ and $\tfrac{84}{5},$ so $DE=\tfrac15\sqrt{12^2+84^2}=12\sqrt2.$

Thus, the correct answer is D.

12.

Suppose that $S$ is a finite set of positive integers. If the greatest integer in $S$ is removed from $S,$ then the average value (arithmetic mean) of the integers remaining is $32.$ If the least integer in $S$ is also removed, then the average value of the integers remaining is $35.$ If the greatest integer is then returned to the set, the average value of the integers rises to $40.$ The greatest integer in the original set $S$ is $72$ greater than the least integer in $S.$ What is the average value of all the integers in the set $S?$

$36.2$

$36.4$

$36.6$

$36.8$

$37$

Answer: D

Concepts:mean system of equations

Difficulty rating: 1630

Small Hint:

Let $n=|S|,$ total $T,$ greatest $M,$ least $L;$ write each average as an equation

Big Hint:

Subtracting $\dfrac{T-M}{n-1}=32$ from $\dfrac{T-L}{n-1}=40$ gives $M-L=8(n-1)$

Solution:

Let $n=|S|,$ let $T$ be the total, $M$ the greatest, and $L$ the least. Then $\dfrac{T-M}{n-1}=32,$ $\dfrac{T-M-L}{n-2}=35,$ and $\dfrac{T-L}{n-1}=40.$

Subtracting the first from the third: $\dfrac{M-L}{n-1}=8.$ Since $M-L=72,$ we get $n-1=9,$ so $n=10.$

Then $T-M=288$ and $T-L=360.$ The middle equation gives $T-M-L=35\cdot 8=280,$ so $L=288-280=8$ and $M=80.$

Thus $T=288+80=368,$ and the average is $\dfrac{368}{10}=36.8.$

Thus, the correct answer is D.

13.

How many values of $\theta$ in the interval $0\lt\theta\le 2\pi$ satisfy the following equation? $1-3\sin\theta+5\cos 3\theta=0$

$2$

$4$

$5$

$6$

$8$

Answer: D

Concepts:trigonometry systematic listing

Difficulty rating: 1850

Small Hint:

The term $5\cos 3\theta$ swings between $-5$ and $5,$ dominating the slowly varying $1-3\sin\theta$

Big Hint:

Check $f(\theta)=1-3\sin\theta+5\cos 3\theta$ at multiples of $60^\circ,$ then use $f'$ to rule out extra roots

Solution:

Let $f(\theta)=1-3\sin\theta+5\cos 3\theta.$ At $\theta=0,\tfrac\pi3,\tfrac{2\pi}3,\ldots,2\pi,$ its signs alternate $+,-,+,-,+,-,+.$ Therefore there is at least one root in each of the six intervening intervals.

At any root, $5\cos3\theta=3\sin\theta-1.$ Consequently $\begin{aligned} &25\sin^2 3\theta-\cos^2\theta \\ &\quad =23+6\sin\theta-8\sin^2\theta>0. \end{aligned}$ Hence $15|\sin3\theta|>3|\cos\theta|.$ In the interval $(\tfrac{k\pi}{3},\tfrac{(k+1)\pi}{3}),$ the sign of $f'(\theta)=-3\cos\theta-15\sin3\theta$ is therefore $(-1)^{k+1}$ at every root.

Thus every root in a given interval crosses the axis in the same direction. Two such roots would require an intervening crossing in the opposite direction, so each interval has exactly one root. There are $6$ solutions in all.

Thus, the correct answer is D.

14.

Let $ABCD$ be a rectangle and let $\overline{DM}$ be a segment perpendicular to the plane of $ABCD.$ Suppose that $\overline{DM}$ has integer length, and the lengths of $\overline{MA},$ $\overline{MC},$ and $\overline{MB}$ are consecutive odd positive integers (in this order). What is the volume of pyramid $MABCD?$

$24\sqrt5$

$60$

$28\sqrt5$

$66$

$8\sqrt{70}$

Answer: A

Concepts:3D geometry Pythagorean Theorem pyramid

Difficulty rating: 1790

Small Hint:

With $D$ at the origin, $MA^2=AD^2+DM^2,$ $MC^2=CD^2+DM^2,$ $MB^2=AD^2+CD^2+DM^2$

Big Hint:

So $MB^2=MA^2+MC^2-DM^2;$ with $MA,MC,MB=k,k+2,k+4,$ solve for $k$ and $DM$

Solution:

Place $D$ at the origin with $A,$ $C$ along the rectangle’s edges and $M$ directly above $D.$ Then $MA^2=AD^2+DM^2,$ $MC^2=CD^2+DM^2,$ and $MB^2=AD^2+CD^2+DM^2.$

Thus $MB^2=MA^2+MC^2-DM^2.$ Writing $MA,MC,MB=k,k+2,k+4,$ we get $DM^2=k^2+(k+2)^2$ $-(k+4)^2$ $=k^2-4k-12.$

If $DM=t$ is a positive integer, then $(k-2)^2-t^2=16,$ so $(k-2-t)(k-2+t)=16.$ The only positive same-parity factor pair giving $t>0$ is $(2,8),$ which yields $k=7$ and $t=3.$ Thus $AD^2=49-9=40$ and $CD^2=81-9=72.$

The base area is $AD\cdot CD=\sqrt{40}\cdot\sqrt{72}$ $=\sqrt{2880}=24\sqrt5,$ and the volume is $\tfrac13\cdot 24\sqrt5\cdot 3=24\sqrt5.$

Thus, the correct answer is A.

15.

The figure is constructed from $11$ line segments, each of which has length $2.$ The area of pentagon $ABCDE$ can be written as $\sqrt m+\sqrt n,$ where $m$ and $n$ are positive integers. What is $m+n?$

$20$

$21$

$22$

$23$

$24$

Answer: D

Concepts:circle geometry coordinate geometry shoelace formula

Difficulty rating: 1890

Small Hint:

Each interior point is distance $2$ from three consecutive outer vertices, so it is their circumcenter

Big Hint:

Since chords $AB$ and $BC$ have length equal to the radius $2,$ $AC=2\sqrt3;$ then use symmetric coordinates

Solution:

The right interior point is distance $2$ from $A,$ $B,$ and $C,$ so these three points lie on a circle of radius $2.$ Chords $AB$ and $BC$ also have length $2,$ so each subtends $60^\circ$ at the center. Thus $AC=2\sqrt3.$ Similarly, the other half is its mirror image.

Put $C=(-1,0),$ $D=(1,0),$ and $A=(0,\sqrt{11});$ then $AC=AD=2\sqrt3.$ The other intersection of the radius- $2$ circles centered at $A$ and $C$ is $B=\left(-\tfrac12-\tfrac{\sqrt{11}}{2\sqrt3}, \tfrac{\sqrt{11}}2+\tfrac{1}{2\sqrt3}\right),$ and $E$ is its reflection across the $y$ -axis.

Applying the shoelace formula to $A,B,C,D,E$ gives $[ABCDE]=\sqrt{11}+2\sqrt3,$ which equals $\sqrt{11}+\sqrt{12}.$

So $m+n=11+12=23.$

Thus, the correct answer is D.

16.

Let $g(x)$ be a polynomial with leading coefficient $1,$ whose three roots are the reciprocals of the three roots of $f(x)=x^3+ax^2+bx+c,$ where $1\lt a\lt b\lt c.$ What is $g(1)$ in terms of $a,$ $b,$ and $c?$

$\dfrac{1+a+b+c}{c}$

$1+a+b+c$

$\dfrac{1+a+b+c}{c^2}$

$\dfrac{a+b+c}{c^2}$

$\dfrac{1+a+b+c}{a+b+c}$

Answer: A

Concepts:Vieta’s Formulas polynomial

Difficulty rating: 1720

Small Hint:

If $f$ has roots $r,s,t,$ then $g(1)=\left(1-\tfrac1r\right)\left(1-\tfrac1s\right)\left(1-\tfrac1t\right)$

Big Hint:

This equals $\dfrac{(r-1)(s-1)(t-1)}{rst};$ relate the numerator to $f(1)$ and the denominator to $c$

Solution:

Let $f$ have roots $r,s,t.$ Since $g$ is monic with roots $\tfrac1r,\tfrac1s,\tfrac1t,$ $\begin{aligned} g(1) &= \left(1-\tfrac1r\right)\left(1-\tfrac1s\right) \\ &\quad {}\cdot \left(1-\tfrac1t\right) \\ &= \dfrac{(r-1)(s-1)(t-1)}{rst}. \end{aligned}$

Now $f(1)=(1-r)(1-s)(1-t)$ $=1+a+b+c,$ so $(r-1)(s-1)(t-1)$ $=-(1+a+b+c).$ Also $rst=-c.$

Therefore $g(1)$ $=\dfrac{-(1+a+b+c)}{-c}$ $=\dfrac{1+a+b+c}{c}.$

Thus, the correct answer is A.

17.

Let $ABCD$ be an isosceles trapezoid having parallel bases $\overline{AB}$ and $\overline{CD}$ with $AB\gt CD.$ Line segments from a point inside $ABCD$ to the vertices divide the trapezoid into four triangles whose areas are $2,$ $3,$ $4,$ and $5$ starting with the triangle with base $\overline{CD}$ and moving clockwise as shown in the diagram below. What is the ratio $\dfrac{AB}{CD}?$

$3$

$2+\sqrt2$

$1+\sqrt6$

$2\sqrt3$

$3\sqrt2$

Answer: B

Concepts:trapezoid area ratio quadratic

Difficulty rating: 2010

Small Hint:

The triangles on the bases have areas $4$ (base $AB$ ) and $2$ (base $CD$ ); write them as $\tfrac12 a h_a$ and $\tfrac12 b h_b$

Big Hint:

With $a h_a=8,$ $b h_b=4,$ and $(a+b)(h_a+h_b)=28,$ the cross terms $a h_b$ and $b h_a$ satisfy a quadratic

Solution:

Let $AB=a,$ $CD=b,$ and let the interior point be at heights $h_a$ from $AB$ and $h_b$ from $CD.$ The base triangles give $\tfrac12 a h_a=4$ and $\tfrac12 b h_b=2,$ so $a h_a=8$ and $b h_b=4.$

The total area is $2+3+4+5=14$ $=\tfrac12(a+b)(h_a+h_b),$ so $(a+b)(h_a+h_b)=28.$ Expanding, $a h_a+b h_b+a h_b+b h_a=28,$ giving $a h_b+b h_a=16.$

Let $u=a h_b$ and $v=b h_a.$ Then $u+v=16$ and $uv=(a h_a)(b h_b)=32,$ so $u,v=8\pm 4\sqrt2.$

Finally $\dfrac{AB}{CD}$ $=\dfrac{a}{b}$ $=\dfrac{a h_b}{b h_b}$ $=\dfrac{u}{4}$ $=\dfrac{8+4\sqrt2}{4}$ $=2+\sqrt2.$

Thus, the correct answer is B.

18.

Let $z$ be a complex number satisfying $12|z|^2$ $=2|z+2|^2+|z^2+1|^2+31.$ What is the value of $z+\dfrac{6}{z}?$

$-2$

$-1$

$\dfrac{1}{2}$

$1$

$4$

Answer: A

Concepts:complex number completing the square

Difficulty rating: 1940

Small Hint:

Let $p=|z|^2$ and $s=z+\bar z;$ expand every modulus using $|w|^2=w\bar w$

Big Hint:

The equation collapses to $(p-6)^2+(s+2)^2=0$

Solution:

Let $p=|z|^2=z\bar z$ and $s=z+\bar z.$ Then $|z+2|^2=p+2s+4,$ and $|z^2+1|^2$ $=p^2+(z^2+\bar z^2)+1$ $=p^2+(s^2-2p)+1.$

Substituting, $12p=2(p+2s+4)$ $+p^2+s^2-2p+1+31,$ which simplifies to $p^2-12p+s^2+4s+40=0.$

Completing the square gives $(p-6)^2+(s+2)^2=0,$ so $p=6$ and $s=-2.$

Then $z+\dfrac{6}{z}=z+\dfrac{6\bar z}{|z|^2}=z+\bar z=-2.$

Thus, the correct answer is A.

19.

Two fair dice, each with at least $6$ faces are rolled. On each face of each die is printed a distinct integer from $1$ to the number of faces on that die, inclusive. The probability of rolling a sum of $7$ is $\dfrac34$ of the probability of rolling a sum of $10,$ and the probability of rolling a sum of $12$ is $\dfrac{1}{12}.$ What is the least possible number of faces on the two dice combined?

$16$

$17$

$18$

$19$

$20$

Answer: B

Concepts:dice (probability)optimization

Difficulty rating: 2120

Small Hint:

With at least $6$ faces each, a sum of $7$ always has $6$ ways, so a sum of $10$ has $8$ ways

Big Hint:

Let the dice have $a\le b$ faces; count outcomes for sums $10$ and $12$ and use $\text{ways}(12)=\tfrac{ab}{12}$

Solution:

Let the dice have $a\le b$ faces. Since both have at least $6$ faces, a sum of $7$ occurs in exactly $6$ ways, so a sum of $10$ occurs in $6\div\tfrac34=8$ ways.

The number of ways to roll $10$ is $\min(a,9)$ $-\max(1,10-b)+1=8.$ A sum of $12$ has probability $\tfrac{1}{12},$ so it occurs in $\tfrac{ab}{12}$ ways.

Having $8$ outcomes for a sum of $10$ requires $a\ge8$ and $b\ge9,$ so necessarily $a+b\ge17.$ For $(a,b)=(8,9),$ a sum of $10$ has $8$ outcomes, while a sum of $12$ has the $6$ outcomes with the first die showing $3,4,\ldots,8.$ Since $\frac{6}{8\cdot9}=\frac{1}{12},$ both conditions hold and the lower bound $17$ is attained.

Thus, the correct answer is B.

20.

Let $Q(z)$ and $R(z)$ be the unique polynomials such that $\begin{aligned} &z^{2021}+1 \\ &\quad = (z^2+z+1)Q(z)+R(z) \end{aligned}$ and the degree of $R$ is less than $2.$ What is $R(z)?$

$-z$

$-1$

$2021$

$z+1$

$2z+1$

Answer: A

Concepts:polynomial roots of unity modular arithmetic

Difficulty rating: 1990

Small Hint:

Modulo $z^2+z+1$ we have $z^3\equiv 1,$ since $z^3-1=(z-1)(z^2+z+1)$

Big Hint:

Reduce the exponent $2021$ modulo $3,$ then reduce $z^2$ using $z^2\equiv -z-1$

Solution:

Since $z^3\equiv 1\pmod{z^2+z+1}$ and $2021=3\cdot 673+2,$ we have $z^{2021}\equiv z^2.$

So $z^{2021}+1\equiv z^2+1.$ Reducing further with $z^2\equiv -z-1,$ this is $-z-1+1=-z.$

Therefore $R(z)=-z.$

Thus, the correct answer is A.

21.

Let $S$ be the sum of all positive real numbers $x$ for which $x^{2^{\sqrt2}}=\sqrt2^{\,2^x}.$

Which of the following statements is true?

$S\lt\sqrt2$

$S=\sqrt2$

$\sqrt2\lt S\lt 2$

$2\le S\lt 6$

$S\ge 6$

Answer: D

Concepts:logarithm bounding to limit cases

Difficulty rating: 2260

Small Hint:

Take $\log_2$ of both sides to get $2^{\sqrt2}\log_2 x=2^{x-1}$

Big Hint:

$x=\sqrt2$ is one solution; analyze the sign of $2^{x-1}-2^{\sqrt2}\log_2 x$ to locate the other

Solution:

Taking $\log_2,$ the equation becomes $2^{\sqrt2}\log_2 x=2^{x-1}.$ Substituting $x=\sqrt2$ gives $2^{\sqrt2}\cdot\tfrac12=2^{\sqrt2-1},$ which holds, so $x=\sqrt2$ is a solution.

Let $f(x)=2^{x-1}-2^{\sqrt2}\log_2 x.$ Then $f(1)\gt 0,$ $f(\sqrt2)=0,$ $f(2)\lt 0,$ and $f(4)\gt 0,$ so there is a second root $x_0$ between $2$ and $4.$

To prove there are no others, divide the equation by $2^x$ and consider $h(x)=\frac{\ln x}{2^x}$ for $x\gt1.$ Its derivative has the sign of $\frac{1}{x}-(\ln2)(\ln x),$ a strictly decreasing expression. Thus $h$ increases once and then decreases, so a horizontal line meets its graph at most twice. The two roots already found are all the solutions. Since $2\lt x_0\lt4,$ we have $2\lt S=\sqrt2+x_0\lt6.$

Thus, the correct answer is D.

22.

Arjun and Beth play a game in which they take turns removing one brick or two adjacent bricks from one “wall” among a set of several walls of bricks, with gaps possibly creating new walls. The walls are one brick tall. For example, a set of walls of sizes $4$ and $2$ can be changed into any of the following by one move: $(3,2),$ $\ (2,1,2),$ $\ (4),$ $\ (4,1),$ $\ (2,2),$ or $(1,1,2).$

Arjun plays first, and the player who removes the last brick wins. For which starting configuration is there a strategy that guarantees a win for Beth?

$(6,1,1)$

$(6,2,1)$

$(6,2,2)$

$(6,3,1)$

$(6,3,2)$

Answer: B

Concepts:combinatorial game invariant

Difficulty rating: 2390

Small Hint:

Compute the Grundy value of a single wall of length $n,$ where a move removes $1$ or $2$ adjacent bricks and may split the wall

Big Hint:

Beth (the second player) wins exactly when the XOR of the walls’ Grundy values is $0$

Solution:

Treat each wall as a Nim-like heap with a Grundy value. A move removes $1$ or $2$ adjacent bricks, possibly splitting a wall into lengths $a,b.$ Thus $g(n)$ is the mex of $g(a)\oplus g(b)$ over $a+b=n-1$ or $a+b=n-2.$

Starting with $g(0)=0,$ this recurrence gives $g(1),g(2),\ldots,g(6)$ equal to $1,2,3,1,4,3,$ respectively.

The second player Beth wins exactly when the XOR of the walls’ Grundy values is $0.$ Checking each option, only $(6,2,1)$ gives $g(6)\oplus g(2)\oplus g(1)$ $=3\oplus 2\oplus 1=0.$

Thus, the correct answer is B.

23.

Three balls are randomly and independently tossed into bins numbered with the positive integers so that for each ball, the probability that it is tossed into bin $i$ is $2^{-i}$ for $i=1,$ $2,$ $3,$ $\ldots.$ More than one ball is allowed in each bin. The probability that the balls end up evenly spaced in distinct bins is $\dfrac{p}{q},$ where $p$ and $q$ are relatively prime positive integers. (For example, the balls are evenly spaced if they are tossed into bins $3,$ $17,$ and $10.$ ) What is $p+q?$

$55$

$56$

$57$

$58$

$59$

Answer: A

Concepts:basic probability geometric sequence arithmetic sequence

Difficulty rating: 2390

Small Hint:

Evenly spaced means the bins are $n, n+d, n+2d$ for some $n,d\ge 1;$ their labels sum to $3(n+d)$

Big Hint:

Each such ordered assignment has probability $2^{-3(n+d)},$ and there are $3!$ orderings of the three balls

Solution:

Evenly spaced distinct bins form an arithmetic progression $n,n+d,n+2d$ with $n,d\ge 1.$ The three labels sum to $3(n+d),$ so a fixed assignment of balls to these bins has probability $2^{-3(n+d)}.$

The three balls can be ordered in $3!=6$ ways, so the total probability is $6\sum_{n\ge 1}\sum_{d\ge 1}2^{-3(n+d)}$ $=6\left(\sum_{n\ge 1}\tfrac{1}{8^n}\right)^2$ $=6\cdot\tfrac17\cdot\tfrac17=\tfrac{6}{49}.$

Since $\gcd(6,49)=1,$ we get $p+q=6+49=55.$

Thus, the correct answer is A.

24.

Let $ABCD$ be a parallelogram with area $15.$ Points $P$ and $Q$ are the projections of $A$ and $C,$ respectively, onto the line $BD;$ and points $R$ and $S$ are the projections of $B$ and $D,$ respectively, onto the line $AC.$ See the figure, which also shows the relative locations of these points.

Suppose $PQ=6$ and $RS=8,$ and let $d$ denote the length of $\overline{BD},$ the longer diagonal of $ABCD.$ Then $d^2$ can be written in the form $m+n\sqrt p,$ where $m,$ $n,$ and $p$ are positive integers and $p$ is not divisible by the square of any prime. What is $m+n+p?$

$81$

$89$

$97$

$105$

$113$

Answer: A

Concepts:parallelogram trigonometry area

Difficulty rating: 2480

Small Hint:

Let $\theta$ be the angle between the diagonals; then $PQ=AC\cos\theta$ and $RS=BD\cos\theta$

Big Hint:

The area is $\tfrac12\cdot AC\cdot BD\sin\theta=15,$ which combines with $AC\cos\theta=6,$ $BD\cos\theta=8$

Solution:

Let the diagonals meet at $O$ at angle $\theta.$ The feet of the perpendiculars from $A$ and $C$ to $BD$ are symmetric about $O,$ so $PQ=AC\cos\theta=6;$ likewise $RS=BD\cos\theta=8.$

The parallelogram’s area is $\tfrac12\cdot AC\cdot BD\sin\theta=15,$ so $AC\cdot BD\sin\theta=30.$ Then $\dfrac{48\sin\theta}{\cos^2\theta}=30,$ giving $\dfrac{\sin\theta}{\cos^2\theta}=\dfrac58.$

Writing $s=\sin\theta,$ $8s=5(1-s^2)$ gives $s=\dfrac{-4+\sqrt{41}}{5},$ so $\cos^2\theta=1-s^2=\dfrac{8(\sqrt{41}-4)}{25}.$

Then $d^2=BD^2$ $=\dfrac{64}{\cos^2\theta}$ $=8(\sqrt{41}+4)$ $=32+8\sqrt{41},$ so $m+n+p=32+8+41=81.$

Thus, the correct answer is A.

25.

Let $S$ be the set of lattice points in the coordinate plane, both of whose coordinates are integers between $1$ and $30,$ inclusive. Exactly $300$ points in $S$ lie on or below a line with equation $y=mx.$ The possible values of $m$ lie in an interval of length $\dfrac{a}{b},$ where $a$ and $b$ are relatively prime positive integers. What is $a+b?$

$31$

$47$

$62$

$72$

$85$

Answer: E

Concepts:lattice point floor and ceiling functions

Difficulty rating: 2600

Small Hint:

For a given $m,$ column $x$ contributes $\min(30,\lfloor mx\rfloor)$ points at or below the line

Big Hint:

As $m$ increases, the count jumps at slopes $\tfrac{y}{x};$ find the two consecutive such slopes bracketing a count of $300$

Solution:

For slope $m,$ column $x$ (with $1\le x\le 30$ ) contributes $\min(30,\lfloor mx\rfloor)$ points on or below $y=mx,$ and we need the total to equal $300.$

At $m=\tfrac23,$ the cap at $30$ is inactive and $\sum_{x=1}^{30}\lfloor \frac{2x}{3}\rfloor=300.$ The count remains fixed until the next larger slope $\frac{y}{x}$ with $x\le30.$

If $\frac{y}{x}>\frac{2}{3},$ then $3y-2x$ is a positive integer. The closest possibility has $3y-2x=1;$ maximizing $x\le30$ in this congruence gives $(y,x)=(19,28).$ Any numerator at least $2$ gives a larger gap. Hence the interval is $[\tfrac23,\tfrac{19}{28}),$ whose length is $\tfrac{19}{28}-\tfrac23=\tfrac1{84}.$

Since $\gcd(1,84)=1,$ $a+b=1+84=85.$

Thus, the correct answer is E.