2023 AMC 12B Solutions | LIVE by Po-Shen Loh

1.

Mrs. Jones is pouring orange juice into four identical glasses for her four sons. She fills the first three glasses completely but runs out of juice when the fourth glass is only $\tfrac13$ full. What fraction of a glass must Mrs. Jones pour from each of the first three glasses into the fourth glass so that all four glasses will have the same amount of juice?

$\dfrac{1}{12}$

$\dfrac{1}{4}$

$\dfrac{1}{6}$

$\dfrac{1}{8}$

$\dfrac{2}{9}$

Concepts:fraction mean

Difficulty rating: 890

Small Hint:

The three full glasses plus the partly full one hold a fixed total amount of juice

Big Hint:

The total is $3+\tfrac13$ glasses; divide by $4$ to get each glass’s final level, then subtract from $1$

Solution:

The total juice is $3+\tfrac13=\tfrac{10}{3}$ glasses. Split evenly, each glass should hold $\tfrac{10}{3}\div 4=\tfrac{5}{6}$ of a glass. Each of the first three glasses must therefore give up $1-\tfrac{5}{6}=\tfrac{1}{6}.$

Thus, the correct answer is C.

2.

Carlos went to a sports store to buy running shoes. Running shoes were on sale, with prices reduced by $20\%$ on every pair of shoes. Carlos also knew that he had to pay a $7.5\%$ sales tax on the discounted price. He had $43$ dollars. What is the original (before discount) price of the most expensive shoes he could afford to buy?

$46$

$50$

$48$

$47$

$49$

Concepts:percentage work backwards

Difficulty rating: 1020

Small Hint:

A $20\%$ discount multiplies the price by $0.8,$ then the $7.5\%$ tax multiplies by $1.075$

Big Hint:

Solve $0.8\times 1.075\times p\le 43$ for the largest original price $p$

Solution:

The final cost of a pair with original price $p$ is $0.8\times 1.075\times p=0.86p.$ Setting $0.86p\le 43$ gives $p\le 50,$ so the most expensive affordable pair originally cost $50$ dollars.

Thus, the correct answer is B.

3.

A $3$ - $4$ - $5$ right triangle is inscribed in circle $A,$ and a $5$ - $12$ - $13$ right triangle is inscribed in circle $B.$ What is the ratio of the area of circle $A$ to the area of circle $B?$

$\dfrac{9}{25}$

$\dfrac{1}{9}$

$\dfrac{1}{5}$

$\dfrac{25}{169}$

$\dfrac{4}{25}$

Concepts:right triangle circle area area ratio

Difficulty rating: 1020

Small Hint:

A right triangle inscribed in a circle has its hypotenuse as a diameter

Big Hint:

The circles have diameters $5$ and $13;$ the area ratio is the square of the ratio of radii

Solution:

The hypotenuse of an inscribed right triangle is a diameter, so circle $A$ has diameter $5$ and circle $B$ has diameter $13.$ The ratio of areas is $\left(\tfrac{5}{13}\right)^2= \tfrac{25}{169}.$

Thus, the correct answer is D.

4.

Jackson’s paintbrush makes a narrow strip with a width of $6.5$ millimeters. Jackson has enough paint to make a strip $25$ meters long. How many square centimeters of paper could Jackson cover with paint?

$162{,}500$

$162.5$

$1{,}625$

$1{,}625{,}000$

$16{,}250$

Concepts:unit conversion area

Difficulty rating: 1200

Small Hint:

Convert every length to centimeters before multiplying

Big Hint:

The width is $6.5\text{ mm}=0.65\text{ cm}$ and the length is $25\text{ m}=2500\text{ cm}$

Solution:

Converting units, the strip is $0.65$ cm wide and $2500$ cm long, so its area is $0.65\times 2500=1625$ square centimeters.

Thus, the correct answer is C.

5.

You are playing a game. A $2\times 1$ rectangle covers two adjacent squares (oriented either horizontally or vertically) of a $3\times 3$ grid of squares, but you are not told which two squares are covered. Your goal is to find at least one square that is covered by the rectangle. A “turn” consists of you guessing a square, after which you are told whether that square is covered by the hidden rectangle. What is the minimum number of turns you need to ensure that at least one of your guessed squares is covered by the rectangle?

$3$

$5$

$4$

$8$

$6$

Concepts:combinatorial game extremal argument

Difficulty rating: 1350

Small Hint:

Think about which squares you could leave un-guessed without ever touching the domino

Big Hint:

The un-guessed squares must contain no two adjacent squares; the largest such set in a $3\times 3$ grid has $5$ squares

Solution:

A set of guessed squares is guaranteed to hit the domino if and only if the un-guessed squares contain no two adjacent squares, since otherwise the domino could hide on that adjacent pair. The largest set of pairwise non-adjacent squares in the $3\times 3$ grid is the $5$ -square checkerboard (four corners plus the center). So at most $5$ squares can be left unguessed, and you must guess $9-5=4.$

Thus, the correct answer is C.

6.

When the roots of the polynomial

$\begin{gathered} P(x)=(x-1)^1(x-2)^2 \\ {}\cdot (x-3)^3\cdots \\ {}\cdot (x-10)^{10} \end{gathered}$

are removed from the number line, what remains is the union of $11$ disjoint open intervals. On how many of these intervals is $P(x)$ positive?

$3$

$7$

$6$

$4$

$5$

Concepts:polynomial parity

Difficulty rating: 1380

Small Hint:

The sign of $P(x)$ flips at a root only when that root’s exponent is odd

Big Hint:

For $x\gt 10$ the value is positive; move left, flipping sign only at the odd roots $1,3,5,7,9$

Solution:

The exponent of the factor $(x-k)$ is $k,$ so the sign of $P$ changes at $x=k$ only when $k$ is odd, i.e. at $1,3,5,7,9.$ For $x\gt 10$ every factor is positive, so $P\gt 0.$ Sweeping left and flipping at each odd root, the positive intervals are $(10,\infty),$ $(9,10),$ $(6,7),$ $(5,6),$ $(2,3),$ and $(1,2)$ — six intervals in all.

Thus, the correct answer is C.

7.

For how many integers $n$ does the expression

$\sqrt{\frac{\log(n^2)-(\log n)^2}{\log n-3}}$

represent a real number, where $\log$ denotes the base $10$ logarithm?

$900$

$3$

$902$

$2$

$901$

Concepts:logarithm inequality

Difficulty rating: 1530

Small Hint:

The expression is real exactly when the fraction under the square root is $\ge 0$

Big Hint:

Let $L=\log n;$ the fraction is $\dfrac{L(2-L)}{L-3}\ge 0,$ and $n$ must be a positive integer

Solution:

Write $L=\log n.$ Then $\log(n^2)-(\log n)^2$ $=2L-L^2$ $=L(2-L),$ and the fraction is $\dfrac{L(2-L)}{L-3}.$ A sign chart shows this is $\ge 0$ exactly when $L\le 0$ or $2\le L\lt 3.$ Since $n$ is a positive integer, $L\le 0$ forces $n=1,$ while $2\le L\lt 3$ gives $100\le n\le 999,$ which is $900$ values. In total $1+900=901.$

Thus, the correct answer is E.

8.

How many nonempty subsets $B$ of $\{0,1,2,3,\ldots,12\}$ have the property that the number of elements in $B$ is equal to the least element of $B?$ For example, $B=\{4,6,8,11\}$ satisfies the condition.

$256$

$136$

$108$

$144$

$156$

Concepts:combinations casework

Difficulty rating: 1570

Small Hint:

If the least element is $m,$ then $B$ has exactly $m$ elements, all at least $m$

Big Hint:

Choose the other $m-1$ elements from the $12-m$ integers greater than $m:$ this gives $\binom{12-m}{m-1}$

Solution:

If the least element is $m\ (m\ge 1),$ then $|B|=m$ and the remaining $m-1$ elements come from $\{m+1,\ldots,12\},$ a set of size $12-m.$ The count is $\begin{aligned} \sum_{m\ge 1}\binom{12-m}{m-1} &=\binom{11}{0} \\ &\quad {}+\binom{10}{1} \\ &\quad {}+\binom{9}{2} \\ &\quad {}+\binom{8}{3} \\ &\quad {}+\binom{7}{4} \\ &\quad {}+\binom{6}{5}, \end{aligned}$ which equals $1+10+36+56+35+6$ $=144.$

Thus, the correct answer is D.

9.

What is the area of the region in the coordinate plane defined by

$||x|-1|+||y|-1|\le 1?$

$2$

$8$

$4$

$15$

$12$

Concepts:absolute value area symmetry

Difficulty rating: 1500

Small Hint:

The region is symmetric across both axes, so work with $|x|$ and $|y|$

Big Hint:

In the first quadrant $|x-1|+|y-1|\le 1$ is a diamond of area $2;$ there are $4$ copies

Solution:

Replacing $x,y$ by $|x|,|y|,$ the condition $|x-1|+|y-1|\le 1$ describes a diamond centered at $(1,1)$ with diagonals of length $2,$ hence area $2.$ It lies entirely in the first quadrant (touching the axes only at single points), so reflecting across the two axes produces $4$ disjoint copies. The total area is $4\times 2=8.$

Thus, the correct answer is B.

10.

In the $xy$ -plane, a circle of radius $4$ with center on the positive $x$ -axis is tangent to the $y$ -axis at the origin, and a circle with radius $10$ with center on the positive $y$ -axis is tangent to the $x$ -axis at the origin. What is the slope of the line passing through the two points at which these circles intersect?

$\dfrac{2}{7}$

$\dfrac{3}{7}$

$\dfrac{2}{\sqrt{29}}$

$\dfrac{1}{\sqrt{29}}$

$\dfrac{2}{5}$

Concepts:circle radical axis slope

Difficulty rating: 1440

Small Hint:

Write both circle equations, then subtract to get the line through both intersection points

Big Hint:

The circles simplify to $x^2+y^2=8x$ and $x^2+y^2=20y;$ set the right sides equal

Solution:

The circles are $(x-4)^2+y^2=16$ and $x^2+(y-10)^2=100,$ i.e. $x^2+y^2=8x$ and $x^2+y^2=20y.$ Subtracting gives $8x=20y,$ so the intersection points lie on $y=\tfrac{2}{5}x,$ which has slope $\tfrac{2}{5}.$

Thus, the correct answer is E.

11.

What is the maximum area of an isosceles trapezoid that has legs of length $1$ and one base twice as long as the other?

$\dfrac{5}{4}$

$\dfrac{8}{7}$

$\dfrac{5\sqrt{2}}{4}$

$\dfrac{3}{2}$

$\dfrac{3\sqrt{3}}{4}$

Concepts:trapezoid optimization

Difficulty rating: 1570

Small Hint:

Drop the two heights; each leg spans a horizontal distance of $\frac{\text{long base}-\text{short base}}{2}$

Big Hint:

With shorter base $a,$ the area is $\tfrac{3a}{2}\sqrt{1-\tfrac{a^2}{4}};$ maximize its square

Solution:

Let the bases be $a$ and $2a.$ Each leg has horizontal offset $\tfrac{a}{2},$ so the height is $\sqrt{1-\tfrac{a^2}{4}}$ and the area is $A=\tfrac{3a}{2}\sqrt{1-\tfrac{a^2}{4}}.$ Then $A^2=\tfrac94\left(a^2-\tfrac{a^4}{4}\right),$ maximized when $a^2=2.$ There the height is $\tfrac{1}{\sqrt2}$ and $A=\tfrac{3\sqrt2}{2}\cdot\tfrac{1}{\sqrt2}=\tfrac{3}{2}.$

Thus, the correct answer is D.

12.

For complex numbers $u=a+bi$ and $v=c+di,$ define the binary operation $\otimes$ by

$u\otimes v=ac+bdi.$

Suppose $z$ is a complex number such that $z\otimes z=z^2+40.$ What is $|z|?$

$2$

$5$

$\sqrt{5}$

$\sqrt{10}$

$5\sqrt{2}$

Concepts:custom operation complex number system of equations

Difficulty rating: 1630

Small Hint:

Write $z=a+bi;$ then $z\otimes z=a^2+b^2 i$ while $z^2=a^2-b^2+2abi$

Big Hint:

Match real and imaginary parts of $z\otimes z=z^2+40,$ then compute $a^2+b^2$

Solution:

With $z=a+bi,$ we have $z\otimes z=a^2+b^2 i$ and $z^2+40=(a^2-b^2+40)+2abi.$ The real parts give $a^2=a^2-b^2+40,$ so $b^2=40.$ The imaginary parts give $b^2=2ab,$ so $b=2a$ and $a^2=\tfrac{b^2}{4}=10.$ Then $|z|^2=a^2+b^2=50,$ so $|z|=5\sqrt{2}.$

Thus, the correct answer is E.

13.

A rectangular box $P$ has distinct edge lengths $a,$ $b,$ and $c.$ The sum of the lengths of all $12$ edges of $P$ is $13,$ the sum of the areas of all $6$ faces of $P$ is $\tfrac{11}{2},$ and the volume of $P$ is $\tfrac{1}{2}.$ What is the length of the longest interior diagonal connecting two vertices of $P?$

$2$

$\dfrac{3}{8}$

$\dfrac{9}{8}$

$\dfrac{9}{4}$

$\dfrac{3}{2}$

Concepts:rectangular prism algebraic manipulation

Difficulty rating: 1500

Small Hint:

The diagonal is $\sqrt{a^2+b^2+c^2};$ you are given $a+b+c$ and $ab+bc+ca$

Big Hint:

Use $a^2+b^2+c^2$ $=(a+b+c)^2-2(ab+bc+ca)$

Solution:

From the edges, $4(a+b+c)=13,$ so $a+b+c=\tfrac{13}{4}.$ From the faces, $2(ab+bc+ca)=\tfrac{11}{2},$ so $ab+bc+ca=\tfrac{11}{4}.$ Then $\begin{aligned} a^2+b^2+c^2 &=\left(\tfrac{13}{4}\right)^2-2\cdot\tfrac{11}{4} \\ &=\tfrac{169}{16}-\tfrac{88}{16} \\ &=\tfrac{81}{16}, \end{aligned}$ so the diagonal is $\sqrt{\tfrac{81}{16}}=\tfrac{9}{4}.$

Thus, the correct answer is D.

14.

For how many ordered pairs $(a,b)$ of integers does the polynomial $x^3+ax^2+bx+6$ have $3$ distinct integer roots?

$5$

$6$

$8$

$7$

$4$

Concepts:Vieta’s Formulas factor systematic listing

Difficulty rating: 1630

Small Hint:

If the roots are integers $p,q,r,$ then $pqr=-6$

Big Hint:

List the sets of three distinct integers with product $-6;$ each set gives one pair $(a,b)$

Solution:

By Vieta, the three distinct integer roots multiply to $-6.$ The sets of three distinct integers with product $-6$ are $\{1,2,-3\},$ $\{1,-2,3\},$ $\{-1,2,3\},$ $\{-1,-2,-3\},$ and $\{1,-1,6\}.$ Each set determines $a=-(p+q+r)$ and $b=pq+qr+rp,$ and all five give different pairs, so there are $5$ ordered pairs $(a,b).$

Thus, the correct answer is A.

15.

Suppose $a,$ $b,$ and $c$ are positive integers such that

$\frac{a}{14}+\frac{b}{15}=\frac{c}{210}.$

Which of the following statements are necessarily true?

I. If $\gcd(a,14)=1$ or $\gcd(b,15)=1$ or both, then $\gcd(c,210)=1.$

II. If $\gcd(c,210)=1,$ then $\gcd(a,14)=1$ or $\gcd(b,15)=1$ or both.

III. $\gcd(c,210)=1$ if and only if $\gcd(a,14)=\gcd(b,15)=1.$

$\mathrm{I},$ $\mathrm{II},$ and $\mathrm{III}$

$\mathrm{I}$ only

$\mathrm{I}$ and $\mathrm{II}$ only

$\mathrm{III}$ only

$\mathrm{II}$ and $\mathrm{III}$ only

Concepts:greatest common divisor modular arithmetic logical deduction

Difficulty rating: 1800

Small Hint:

Clearing denominators gives $c=15a+14b,$ so $c\equiv a\pmod{14}$ and $c\equiv -b\pmod{15}$

Big Hint:

Hence $2,7$ divide $c$ iff they divide $a,$ and $3,5$ divide $c$ iff they divide $b$

Solution:

Multiplying by $210$ gives $c=15a+14b.$ Since $15\equiv 1\pmod{14},$ we get $c\equiv a\pmod{14},$ so $\gcd(c,14)=1$ iff $\gcd(a,14)=1.$ Since $14\equiv -1\pmod{15},$ we get $c\equiv -b\pmod{15},$ so $\gcd(c,15)=1$ iff $\gcd(b,15)=1.$ As $210=14\cdot 15$ with $\gcd(14,15)=1,$ statement III follows: $\gcd(c,210)=1$ iff both hold. Statement II is the forward implication of III, hence true. Statement I is false: if $\gcd(a,14)=1$ but $\gcd(b,15)\ne 1,$ then $\gcd(c,15)\ne 1,$ so $\gcd(c,210)\ne 1.$ Only II and III are true.

Thus, the correct answer is E.

16.

In Coinland, there are three types of coins, each worth $6,$ $10,$ and $15.$ What is the sum of the digits of the maximum amount of money that is impossible to have?

$8$

$10$

$7$

$11$

$9$

Concepts:Chicken McNugget Theorem

Difficulty rating: 1660

Small Hint:

List which amounts can be formed from $6,10,15;$ look for the largest that cannot

Big Hint:

Once six consecutive amounts are all attainable, adding more $6$ ’s reaches everything beyond

Solution:

The amounts $30,31,32,33,34,35$ are all attainable (for instance $30=6\cdot 5,$ $31=6+10+15,$ $32=6\cdot 2+10\cdot 2,$ $33=6\cdot 3+15,$ $34=6\cdot 4+10,$ $35=10+10+15$ ). Adding $6$ ’s then reaches every larger amount. Checking below, $29$ is impossible: without a $15$ -coin a sum of $6$ ’s and $10$ ’s is even, while with one $15$ -coin the remaining $14$ cannot be written as $6a+10b;$ two $15$ -coins already exceed $29.$ So the largest impossible amount is $29,$ whose digit sum is $2+9=11.$

Thus, the correct answer is D.

17.

Triangle $ABC$ has side lengths in arithmetic progression, and the smallest side has length $6.$ If the triangle has an angle of $120^\circ,$ what is the area of $ABC?$

$12\sqrt{3}$

$8\sqrt{6}$

$14\sqrt{2}$

$20\sqrt{2}$

$15\sqrt{3}$

Concepts:arithmetic sequence law of cosines triangle area

Difficulty rating: 1630

Small Hint:

Let the sides be $6,\ 6+d,\ 6+2d;$ the $120^\circ$ angle is opposite the longest side

Big Hint:

Apply the law of cosines with $\cos 120^\circ=-\tfrac12$ to find $d,$ then area $=\tfrac12ab\sin 120^\circ$

Solution:

Let the sides be $6,\ 6+d,\ 6+2d.$ The $120^\circ$ angle faces the longest side, so $(6+2d)^2=6^2+(6+d)^2$ $-2\cdot 6\cdot(6+d)\cos 120^\circ.$ Using $\cos 120^\circ=-\tfrac12$ gives $3d^2+6d-72=0,$ so $d=4$ and the sides are $6,10,14.$ The area is $\tfrac12\cdot 6\cdot 10\cdot\sin 120^\circ$ $=30\cdot\tfrac{\sqrt3}{2}$ $=15\sqrt3.$

Thus, the correct answer is E.

18.

Last academic year Yolanda and Zelda took different courses that did not necessarily administer the same number of quizzes during each of the two semesters. Yolanda’s average on all the quizzes she took during the first semester was $3$ points higher than Zelda’s average on all the quizzes she took during the first semester. Yolanda’s average on all the quizzes she took during the second semester was $18$ points higher than her average for the first semester and was again $3$ points higher than Zelda’s average on all the quizzes Zelda took during her second semester. Which one of the following statements cannot possibly be true?

Yolanda’s quiz average for the academic year was $22$ points higher than Zelda’s.

Zelda’s quiz average for the academic year was higher than Yolanda’s.

Yolanda’s quiz average for the academic year was $3$ points higher than Zelda’s.

Zelda’s quiz average for the academic year equaled Yolanda’s.

If Zelda had scored $3$ points higher on each quiz she took, then she would have had the same average for the academic year as Yolanda.

Concepts:weighted mean inequality logical deduction

Difficulty rating: 1800

Small Hint:

Let Zelda’s first-semester average be $0,$ and write the other three semester averages using the given gaps

Big Hint:

If $p,q$ are the fractions of Yolanda’s and Zelda’s quizzes taken in semester two, their yearly-average difference is $3+18(p-q)$

Solution:

Subtracting Zelda’s first-semester average from all four semester averages does not affect the comparison. We may therefore write the semester averages as $0,18$ for Zelda and $3,21$ for Yolanda. Let $p$ and $q$ be the fractions of Yolanda’s and Zelda’s quizzes, respectively, that occurred in the second semester. Their yearly-average difference is $(3+18p)-18q=3+18(p-q).$ Because $0\lt p,q\lt 1,$ this difference is less than $21,$ so it cannot be $22.$

The other options really can occur. Taking $(p,q)=(\tfrac14,\tfrac34)$ makes Zelda’s average higher; taking $p=q$ makes Yolanda’s average $3$ points higher (and also verifies the last option after adding $3$ to every Zelda score); and taking $(p,q)=(\tfrac13,\tfrac12)$ makes the yearly averages equal. Each displayed fraction can be realized by positive integer quiz counts.

Thus, the correct answer is A.

19.

Each of $2023$ balls is placed in one of $3$ bins. Which of the following is closest to the probability that each of the bins will contain an odd number of balls?

$\dfrac{2}{3}$

$\dfrac{3}{10}$

$\dfrac{1}{2}$

$\dfrac{1}{3}$

$\dfrac{1}{4}$

Concepts:roots of unity parity basic probability

Difficulty rating: 1990

Small Hint:

Each ball lands independently; use a parity filter, writing bin $i$ odd as $\tfrac{1-(-1)^{k_i}}{2}$

Big Hint:

Summing over all assignments, the all-odd fraction is $\dfrac{3^{\,n}-3}{4\cdot 3^{\,n}};$ see what it approaches for large $n$

Solution:

Counting assignments where all three bins are odd with the parity filter gives $\scriptsize\frac{1}{8}\sum_{S\subseteq\{1,2,3\}}(-1)^{|S|}(3-2|S|)^n=\frac{3^n-3}{4}$ for odd $n.$ Dividing by the $3^n$ total assignments, the probability is $\dfrac{3^n-3}{4\cdot 3^n},$ which for $n=2023$ is extremely close to $\tfrac14.$

Thus, the correct answer is E.

20.

Cyrus the frog jumps $2$ units in a direction, then $2$ more in another direction. What is the probability that he lands less than $1$ unit away from his starting position?

$\dfrac{1}{6}$

$\dfrac{1}{5}$

$\dfrac{\sqrt{3}}{8}$

$\dfrac{\arctan\tfrac12}{\pi}$

$\dfrac{2\arcsin\tfrac14}{\pi}$

Concepts:geometric probability law of cosines

Difficulty rating: 2110

Small Hint:

Fix the first jump along the $x$ -axis; the second jump makes a uniformly random angle $\theta$

Big Hint:

The landing distance satisfies $R^2=8+8\cos\theta;$ require $R\lt 1,$ i.e. $\cos\theta\lt-\tfrac78$

Solution:

Take the first jump as $(2,0)$ and the second as $(2\cos\theta,2\sin\theta)$ with $\theta$ uniform on $[0,2\pi).$ The landing distance satisfies $R^2=(2+2\cos\theta)^2$ $+(2\sin\theta)^2$ $=8+8\cos\theta.$ We need $R\lt 1,$ i.e. $\cos\theta\lt-\tfrac78.$ The measure of such angles is $2\arccos\tfrac78,$ so the probability is $\dfrac{2\arccos\tfrac78}{2\pi}= \dfrac{\arccos\tfrac78}{\pi}.$ Using $\arccos(1-2x^2)=2\arcsin x$ with $x=\tfrac14$ gives $\arccos\tfrac78=2\arcsin\tfrac14,$ so the probability is $\dfrac{2\arcsin\tfrac14}{\pi}.$

Thus, the correct answer is E.

21.

A lampshade is made in the form of the lateral surface of the frustum of a right circular cone. The height of the frustum is $3\sqrt{3}$ inches, its top diameter is $6$ inches, and its bottom diameter is $12$ inches. A bug is at the bottom of the lampshade and there is a glob of honey on the top edge of the lampshade at the spot farthest from the bug. The bug wants to crawl to the honey, but it must stay on the surface of the lampshade. What is the length in inches of its shortest path to the honey?

$6+3\pi$

$6+6\pi$

$6\sqrt{3}$

$6\sqrt{5}$

$6\sqrt{3}+\pi$

Concepts:cone net (3D geometry)tangent line

Difficulty rating: 2020

Small Hint:

Extend the frustum to a full cone (apex slant distances $6$ and $12$ ); it unrolls to a sector of angle $\pi$

Big Hint:

The straight line to the honey cuts across the missing top cone, so the path runs tangent to the inner circle of radius $6,$ then along its arc

Solution:

Extend the frustum to a full cone. Since the radii are $3$ and $6$ with slant band $6,$ the apex is slant distance $6$ from the top rim and $12$ from the bottom rim. The bottom circumference $12\pi$ unrolls to a sector of radius $12$ and angle $\dfrac{12\pi}{12}=\pi.$ Place the bug at $(12,0)$ in this pattern; the honey, halfway around the base, is at radius $6$ and angle $\tfrac{\pi}{2}.$ The straight chord between them passes within radius $6$ (off the surface), so the geodesic goes tangent to the circle of radius $6:$ the tangent has length $\sqrt{12^2-6^2}=6\sqrt3$ and touches at angle $\tfrac{\pi}{3},$ after which the path follows the arc of angle $\tfrac{\pi}{6}$ on radius $6,$ of length $6\cdot\tfrac{\pi}{6}=\pi.$ The shortest path is $6\sqrt3+\pi.$

Thus, the correct answer is E.

22.

A real-valued function $f$ has the property that for all real numbers $a$ and $b,$

$\begin{gathered} f(a+b)+f(a-b) \\ =2f(a)f(b). \end{gathered}$

Which one of the following cannot be the value of $f(1)?$

$0$

$1$

$-1$

$2$

$-2$

Concepts:functional equation bounding to limit cases

Difficulty rating: 2020

Small Hint:

Set $a=b=0$ to pin down $f(0),$ then set $a=b$ to relate $f(2a)$ to $f(a)$

Big Hint:

From $a=b,$ $f(2a)=2f(a)^2-1\ge -1;$ taking $a=\tfrac12$ bounds $f(1)$ from below

Solution:

Setting $a=b=0$ gives $2f(0)=2f(0)^2,$ so $f(0)=0$ or $f(0)=1.$ If $f(0)=0,$ then setting $b=0$ forces $f\equiv 0,$ giving $f(1)=0.$ Otherwise $f(0)=1,$ and setting $a=b$ gives $f(2a)=2f(a)^2-1\ge -1$ for every $a.$ In particular, with $a=\tfrac12,$ $f(1)=2f\!\left(\tfrac12\right)^2-1\ge -1.$ So $f(1)\ge -1,$ and indeed every value in $[-1,\infty)$ is attainable (e.g. $f(x)=\cos(kx)$ or $\cosh(kx)$ ). Hence $-2$ is impossible.

Thus, the correct answer is E.

23.

When $n$ standard six-sided dice are rolled, the product of the numbers rolled can be any of $936$ possible values. What is $n?$

$11$

$6$

$8$

$10$

$9$

Concepts:prime factorization systematic listing

Difficulty rating: 2270

Small Hint:

A product is determined by the exponents of $2,3,5,$ so count the distinct exponent triples

Big Hint:

After fixing that exactly $k$ dice show $5,$ count the possible exponent pairs of $2$ and $3$ from the other $m=n-k$ dice

Solution:

Each die contributes an exponent vector in the primes $2,3,5$ (face $1\to(0,0,0),$ $2\to(1,0,0),$ $3\to(0,1,0),$ $4\to(2,0,0),$ $5\to(0,0,1),$ $6\to(1,1,0)$ ), and a product is determined by the sum of these vectors.

Fix the exponent of $5$ by requiring exactly $k$ dice to show $5,$ and put $m=n-k.$ For an exponent $b$ of $3,$ where $0\le b\le m,$ the possible exponents $a$ of $2$ are precisely $0,1,\ldots,2m-b.$ If $a\le b,$ use $a$ faces showing $6$ and $b-a$ showing $3.$ If $a\gt b,$ use $b$ faces showing $6$ and make the remaining $a-b$ factors of $2$ with faces $4$ and, if needed, one face $2.$ The bound $a\le 2m-b$ says this uses at most $m$ dice; fill unused dice with $1$ ’s. Thus the number of exponent pairs is $\begin{gathered} \sum_{b=0}^{m}(2m-b+1)\\ {}=\frac{(m+1)(3m+2)}2. \end{gathered}$

Summing over $k=0,1,\ldots,n$ , equivalently over $m=0,1,\ldots,n,$ the number of distinct products is $\begin{gathered} \sum_{m=0}^{n}\frac{(m+1)(3m+2)}2\\ {}=\frac{(n+1)^2(n+2)}2. \end{gathered}$ For $n=11$ this is $\dfrac{12^2\cdot13}{2}=936,$ so $n=11.$

Thus, the correct answer is A.

24.

Suppose that $a,$ $b,$ $c,$ and $d$ are positive integers satisfying all of the following relations.

$\begin{aligned} abcd &= 2^6\cdot 3^9\cdot 5^7\\ \operatorname{lcm}(a,b) &= 2^3\cdot 3^2\cdot 5^3\\ \operatorname{lcm}(a,c) &= 2^3\cdot 3^3\cdot 5^3\\ \operatorname{lcm}(a,d) &= 2^3\cdot 3^3\cdot 5^3\\ \operatorname{lcm}(b,c) &= 2^1\cdot 3^3\cdot 5^2\\ \operatorname{lcm}(b,d) &= 2^2\cdot 3^3\cdot 5^2\\ \operatorname{lcm}(c,d) &= 2^2\cdot 3^3\cdot 5^2 \end{aligned}$

What is $\gcd(a,b,c,d)?$

$30$

$45$

$3$

$15$

$6$

Concepts:least common multiple greatest common divisor prime factorization

Difficulty rating: 2270

Small Hint:

Work one prime at a time; $\gcd$ uses the minimum exponent and $\operatorname{lcm}$ the maximum

Big Hint:

For each prime, pin down the four exponents from the max (lcm) and total (product) conditions, then take the minimum

Solution:

Handle each prime separately using the exponents of $a,b,c,d.$

Prime $2$ (total $6$ ): $\max(b,c)=1$ forces $a=3;$ then $b+c+d=3$ with $\max(b,d)=\max(c,d)=2$ gives $d=2$ and $\{b,c\}=\{0,1\},$ so the minimum exponent is $0.$

Prime $3$ (total $9$ ): $\max(a,b)=2$ with the other lcms equal to $3$ forces $c=d=3;$ then $a+b=3$ with $\max(a,b)=2$ gives $\{a,b\}=\{1,2\},$ so the minimum is $1.$

Prime $5$ (total $7$ ): $\max(a,b)=3$ with $\max(b,c),$ $\max(b,d),\max(c,d)=2$ forces $a=3;$ then $b+c+d=4$ with each $\le 2$ and pairwise maxima $2$ gives two of them equal to $2$ and one equal to $0,$ so the minimum is $0.$

Therefore $\gcd(a,b,c,d)=2^0\cdot 3^1\cdot 5^0=3.$

Thus, the correct answer is C.

25.

A regular pentagon with area $\sqrt{5}+1$ is printed on paper and cut out. The five vertices of the pentagon are folded into the center of the pentagon, creating a smaller pentagon. What is the area of the new pentagon?

$4-\sqrt{5}$

$\sqrt{5}-1$

$8-3\sqrt{5}$

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

$\dfrac{2+\sqrt{5}}{3}$

Concepts:regular polygon paper folding area ratio

Difficulty rating: 2490

Small Hint:

Folding a vertex to the center creases along the perpendicular bisector of the segment from the center to that vertex

Big Hint:

The new pentagon is regular with apothem $\frac{R}{2};$ compare to the original apothem $R\cos 36^\circ$

Solution:

Let the original pentagon have circumradius $R.$ Folding a vertex to the center creases along the perpendicular bisector of the segment from the center to that vertex, a line at distance $\tfrac{R}{2}$ from the center. The five creases bound a regular pentagon with apothem $\tfrac{R}{2},$ whereas the original has apothem $R\cos 36^\circ.$ Areas scale as the square of the apothem, so the ratio is $\frac{(\frac{R}{2})^2}{(R\cos 36^\circ)^2}=\frac{1}{4\cos^2 36^\circ}.$ Since $\cos 36^\circ=\tfrac{1+\sqrt5}{4},$ this ratio is $\tfrac{4}{(1+\sqrt5)^2}$ $=\tfrac{4}{6+2\sqrt5}$ $=\tfrac{2}{3+\sqrt5}$ $=\tfrac{3-\sqrt5}{2}.$ Multiplying by the original area $\sqrt5+1$ gives $\tfrac{(3-\sqrt5)(\sqrt5+1)}{2}$ $=\tfrac{2\sqrt5-2}{2}$ $=\sqrt5-1.$

Thus, the correct answer is B.