2022 AMC 12A Solutions | LIVE by Po-Shen Loh

1.

What is the value of

$3+\cfrac{1}{3+\cfrac{1}{3+\frac13}}?$

$\dfrac{31}{10}$

$\dfrac{49}{15}$

$\dfrac{33}{10}$

$\dfrac{109}{33}$

$\dfrac{15}{4}$

Concepts:continued fraction fraction

Difficulty rating: 890

Small Hint:

Work from the innermost fraction outward

Big Hint:

$3+\dfrac13=\dfrac{10}{3},$ so the next layer up is $3+\dfrac{3}{10}$

Solution:

Simplify from the bottom. The innermost fraction is $3+\dfrac13=\dfrac{10}{3}.$

The next layer is $3+\dfrac{1}{\frac{10}{3}}=3+\dfrac{3}{10}=\dfrac{33}{10}.$

Finally, $3+\dfrac{1}{\frac{33}{10}}=3+\dfrac{10}{33}=\dfrac{109}{33}.$

Thus, the correct answer is D.

2.

The sum of three numbers is $96.$ The first number is $6$ times the third number, and the third number is $40$ less than the second number. What is the absolute value of the difference between the first and second numbers?

$1$

$2$

$3$

$4$

$5$

Concepts:system of equations substitution

Difficulty rating: 1020

Small Hint:

Let the third number be $t$ and write the others in terms of $t$

Big Hint:

The first is $6t,$ the second is $t+40,$ and they sum to $96$

Solution:

Let the third number be $t.$ Then the first is $6t$ and the second is $t+40.$ Their sum is $6t+(t+40)+t=8t+40=96,$ so $t=7.$

The first number is $42$ and the second is $47,$ so the difference has absolute value $5.$

Thus, the correct answer is E.

3.

Five rectangles, $A,$ $B,$ $C,$ $D,$ and $E,$ are arranged in a square as shown below. These rectangles have dimensions $1\times6,$ $2\times4,$ $5\times6,$ $2\times7,$ and $2\times3,$ respectively. (The figure is not drawn to scale.) Which of the five rectangles is the shaded one in the middle?

$A$

$B$

$C$

$D$

$E$

Concepts:area tiling rectangle

Difficulty rating: 1130

Small Hint:

The five areas add to the square’s area, which fixes the side length

Big Hint:

Anchor the large $5\times6$ and $2\times7$ pieces in corners, then fit the rest

Solution:

The five areas are $6,\ 8,\ 30,\ 14,$ and $6,$ which sum to $64.$ So the square is $8\times8.$

Placing $C$ ( $5\times6$ ) across the top left, $D$ ( $2\times7$ ) up the right side, $E$ ( $2\times3$ ) in the lower left, and $A$ ( $1\times6$ ) along the bottom leaves a central $2\times4$ gap, which is exactly rectangle $B.$

Thus, the correct answer is B.

4.

The least common multiple of a positive integer $n$ and $18$ is $180,$ and the greatest common divisor of $n$ and $45$ is $15.$ What is the sum of the digits of $n?$

$3$

$6$

$8$

$9$

$12$

Concepts:least common multiple greatest common divisor prime factorization

Difficulty rating: 1200

Small Hint:

Factor $180=2^2\cdot3^2\cdot5$ and see which powers $n$ must supply beyond $18=2\cdot3^2$

Big Hint:

$\gcd(n,45)=15$ forces exactly one factor of $3$ and at least one factor of $5$

Solution:

Since $180=2^2\cdot3^2\cdot5$ and $18=2\cdot3^2,$ the condition $\operatorname{lcm}(n,18)=180$ forces $n$ to contribute $2^2$ and $5,$ with its power of $3$ at most $2.$

From $\gcd(n,45)=\gcd(n,3^2\cdot5)$ $=15=3\cdot5,$ the power of $3$ in $n$ is exactly $1$ and the power of $5$ is at least $1.$

Therefore $n=2^2\cdot3\cdot5=60,$ whose digits sum to $6.$

Thus, the correct answer is B.

5.

Let the taxicab distance between points $(x_1,y_1)$ and $(x_2,y_2)$ in the coordinate plane be given by $|x_1-x_2|+|y_1-y_2|.$ For how many points $P$ with integer coordinates is the taxicab distance between $P$ and the origin less than or equal to $20?$

$441$

$761$

$841$

$921$

$924$

Concepts:lattice point arithmetic sequence

Difficulty rating: 1350

Small Hint:

Count the lattice points on each “diamond” $|x|+|y|=k$ separately

Big Hint:

There are $4k$ points with $|x|+|y|=k$ for $k\ge1,$ plus the origin

Solution:

For each $k\ge1,$ the set $|x|+|y|=k$ contains exactly $4k$ lattice points, and $k=0$ gives the single origin.

The total is $\begin{aligned} &1+\sum_{k=1}^{20}4k=1+4\cdot\frac{20\cdot21}{2} \\ &=1+840=841. \end{aligned}$

Thus, the correct answer is C.

6.

A data set consists of $6$ (not distinct) positive integers: $1,$ $7,$ $5,$ $2,$ $5,$ and $X.$ The average (arithmetic mean) of the $6$ numbers equals a value in the data set. What is the sum of all positive values of $X?$

$10$

$26$

$32$

$36$

$40$

Concepts:mean casework

Difficulty rating: 1270

Small Hint:

The five known numbers sum to $20,$ so the mean is $\dfrac{20+X}{6}$

Big Hint:

The mean must equal one of $1,2,5,7,$ or $X;$ only some give positive $X$

Solution:

The known numbers sum to $20,$ so the mean is $\dfrac{20+X}{6},$ which must equal an element of the set.

Setting it to $5$ gives $X=10;$ to $7$ gives $X=22;$ and to $X$ itself gives $20+X=6X,$ so $X=4.$ Values $1$ and $2$ give negative $X.$

The positive values are $10,22,4,$ summing to $36.$

Thus, the correct answer is D.

7.

A rectangle is partitioned into $5$ regions as shown. Each region is to be painted a solid color - red, orange, yellow, blue, or green - so that regions that touch are painted different colors, and colors can be used more than once. How many different colorings are possible?

$120$

$270$

$360$

$540$

$720$

Concepts:graph theory multiplication principle

Difficulty rating: 1380

Small Hint:

Color the region that borders the most others first

Big Hint:

One region touches all four others; each remaining region must avoid two already-used colors

Solution:

The bottom-middle region shares a border with all four other regions. Color it first in $5$ ways.

The top-left region borders it, giving $4$ choices. Each of the three remaining regions borders exactly two already-colored regions, which have different colors, leaving $3$ choices apiece.

The total is $5\cdot4\cdot3\cdot3\cdot3=540.$

Thus, the correct answer is D.

8.

The infinite product

$\sqrt[3]{10}\cdot\sqrt[3]{\sqrt[3]{10}}\cdot\sqrt[3]{\sqrt[3]{\sqrt[3]{10}}}\cdots$

evaluates to a real number. What is that number?

$\sqrt{10}$

$\sqrt[3]{100}$

$\sqrt[4]{1000}$

$10$

$10\sqrt[3]{10}$

Concepts:exponent geometric sequence

Difficulty rating: 1500

Small Hint:

Write every factor as a power of $10$

Big Hint:

The exponents are $\dfrac13,\dfrac19,\dfrac1{27},\ldots,$ a geometric series

Solution:

The $k$ th factor is $10$ raised to the $k$ -fold cube root, namely $10^{\frac{1}{3^k}}.$

The product is $10$ raised to $\begin{aligned} &\frac13+\frac19+\frac1{27}+\cdots=\frac{\frac{1}{3}}{1-\frac{1}{3}} \\ &=\frac12. \end{aligned}$

So the value is $10^{\frac{1}{2}}=\sqrt{10}.$

Thus, the correct answer is A.

9.

On Halloween $31$ children walked into the principal’s office asking for candy. They can be classified into three types: some always lie; some always tell the truth; and some alternately lie and tell the truth. The alternaters arbitrarily choose their first response, either a lie or the truth, but each subsequent statement has the opposite truth value from its predecessor. The principal asked everyone the same three questions in this order.

“Are you a truth-teller?” The principal gave a piece of candy to each of the $22$ children who answered yes.

“Are you an alternater?” The principal gave a piece of candy to each of the $15$ children who answered yes.

“Are you a liar?” The principal gave a piece of candy to each of the $9$ children who answered yes.

How many pieces of candy in all did the principal give to the children who always tell the truth?

$7$

$12$

$21$

$27$

$31$

Concepts:truth-tellers and liars logical deduction casework

Difficulty rating: 1530

Small Hint:

Work out how each of the three types answers each question

Big Hint:

Only alternaters who lie first answer the last question yes, which pins down their count

Solution:

To “Are you a truth-teller?” the truth-tellers and liars both answer yes, and only alternaters who lie on this question answer yes. To “Are you an alternater?” the liars answer yes, and among alternaters only those telling the truth on this question answer yes. To “Are you a liar?” only alternaters lying on this question answer yes.

Split the alternaters by first response. Those starting with a lie answer (lie, truth, lie), so they say yes to all three questions; those starting truthful answer (truth, lie, truth) and say yes to none of the three. The $9$ yeses on the last question are exactly the lie-first alternaters, so there are $9$ of them.

The second question’s $15$ yeses are the liars plus these $9,$ so there are $6$ liars. The first question’s $22$ yeses are truth-tellers plus liars plus the $9,$ so the truth-tellers number $22-6-9=7.$

Truth-tellers answer yes only to the first question, receiving one candy each, for $7\cdot1=7$ pieces.

Thus, the correct answer is A.

10.

What is the number of ways the numbers from $1$ to $14$ can be split into $7$ pairs such that for each pair, the greater number is at least $2$ times the smaller number?

$108$

$120$

$126$

$132$

$144$

Concepts:arrangements with restrictions multiplication principle

Difficulty rating: 1570

Small Hint:

A number $m$ can be the smaller of its pair only if $2m\le14$

Big Hint:

So $1$ – $7$ are the smaller elements and $8$ – $14$ the larger; assign greedily from the tightest smaller number

Solution:

Any number $8$ or larger cannot be a smaller element (its double exceeds $14$ ), so $8$ – $14$ are all larger elements and $1$ – $7$ are all smaller elements.

Match each smaller $s$ to a larger $g\ge2s.$ Processing from the most restrictive: $s=7$ forces $g=14$ ( $1$ way); then $s=6$ has $\{12,13\}$ left ( $2$ ); $s=5$ has $3;$ $s=4$ has $4;$ $s=3$ has $3;$ $s=2$ has $2;$ $s=1$ has $1.$

The number of matchings is $1\cdot2\cdot3\cdot4\cdot3\cdot2\cdot1=144.$

Thus, the correct answer is E.

11.

What is the product of all real numbers $x$ such that the distance on the number line between $\log_6 x$ and $\log_6 9$ is twice the distance on the number line between $\log_6 10$ and $1?$

$10$

$18$

$25$

$36$

$81$

Concepts:logarithm absolute value

Difficulty rating: 1530

Small Hint:

$\log_6 10-1=\log_6\dfrac{10}{6}=\log_6\dfrac53$

Big Hint:

The condition is $\left|\log_6\dfrac{x}{9}\right|=\log_6\dfrac{25}{9}$

Solution:

The right-hand distance is $|\log_6 10-1|=\log_6\dfrac53,$ so twice it is $2\log_6\dfrac53=\log_6\dfrac{25}{9}.$

Thus $\left|\log_6\dfrac{x}{9}\right|=\log_6\dfrac{25}{9},$ giving $\dfrac{x}{9}=\dfrac{25}{9}$ or $\dfrac{x}{9}=\dfrac{9}{25},$ so $x=25$ or $x=\dfrac{81}{25}.$

Their product is $25\cdot\dfrac{81}{25}=81.$

Thus, the correct answer is E.

12.

Let $M$ be the midpoint of $\overline{AB}$ in regular tetrahedron $ABCD.$ What is $\cos(\angle CMD)?$

$\dfrac14$

$\dfrac13$

$\dfrac25$

$\dfrac12$

$\dfrac{\sqrt3}{2}$

Concepts:3D geometry median (geometry)law of cosines

Difficulty rating: 1630

Small Hint:

$CM$ and $DM$ are medians of equilateral faces with edge length $1$

Big Hint:

$CM=DM=\dfrac{\sqrt3}{2}$ and $CD=1;$ apply the Law of Cosines in $\triangle CMD$

Solution:

Take edge length $1.$ Since $M$ is the midpoint of $\overline{AB},$ segments $CM$ and $DM$ are altitudes of the equilateral faces, each of length $\dfrac{\sqrt3}{2}.$ Also $CD=1.$

By the Law of Cosines in $\triangle CMD,$ $\begin{aligned} \cos(\angle CMD) &=\frac{\frac34+\frac34-1}{2\cdot\frac34} \\ &=\frac{\frac{1}{2}}{\frac{3}{2}}=\frac13. \end{aligned}$

Thus, the correct answer is B.

13.

Let $\mathcal{R}$ be the region in the complex plane consisting of all complex numbers $z$ that can be written as the sum of complex numbers $z_1$ and $z_2,$ where $z_1$ lies on the segment with endpoints $3$ and $4i,$ and $z_2$ has magnitude at most $1.$ What integer is closest to the area of $\mathcal{R}?$

$13$

$14$

$15$

$16$

$17$

Concepts:area distance formula

Difficulty rating: 1660

Small Hint:

$\mathcal{R}$ is the set of all points within distance $1$ of the segment

Big Hint:

The segment has length $5,$ so $\mathcal{R}$ is a $5\times2$ rectangle capped by two half-disks

Solution:

Adding a disk of radius $1$ to every point of the segment sweeps out all points within distance $1$ of it. The segment from $3$ to $4i$ has length $\sqrt{3^2+4^2}=5.$

This “stadium” is a $5\times2$ rectangle plus two half-disks of radius $1,$ with area $5\cdot2+\pi(1)^2=10+\pi\approx13.14.$

The closest integer is $13.$

Thus, the correct answer is A.

14.

What is the value of

$\begin{aligned} &(\log 5)^3+(\log 20)^3 \\ &\quad {}+(\log 8)(\log 0.25) \end{aligned}$

where $\log$ denotes the base-ten logarithm?

$\dfrac32$

$\dfrac74$

$2$

$\dfrac94$

$3$

Concepts:logarithm sum and difference of cubes

Difficulty rating: 1730

Small Hint:

Let $u=\log 2;$ then $\log 5=1-u$ and $\log 20=1+u$

Big Hint:

Also $\log 8=3u$ and $\log 0.25=-2u;$ use $a^3+b^3=(a+b)(a^2-ab+b^2)$

Solution:

Let $u=\log 2.$ Then $\log 5=1-u,$ $\log 20=1+u,$ $\log 8=3u,$ and $\log 0.25=-2u.$

With $a=1-u,\ b=1+u,$ we have $a+b=2$ and $ab=1-u^2,$ so $\begin{gathered} a^3+b^3 \\ =(a+b)\big((a+b)^2-3ab\big) \\ =2\big(4-3(1-u^2)\big) \\ =2+6u^2. \end{gathered}$

The last term is $(3u)(-2u)=-6u^2,$ so the total is $2+6u^2-6u^2=2.$

Thus, the correct answer is C.

15.

The roots of the polynomial $10x^3-39x^2+29x-6$ are the height, length, and width of a rectangular box (right rectangular prism). A new rectangular box is formed by lengthening each edge of the original box by $2$ units. What is the volume of the new box?

$\dfrac{24}{5}$

$\dfrac{42}{5}$

$\dfrac{81}{5}$

$30$

$48$

Concepts:Vieta’s Formulas volume

Difficulty rating: 1630

Small Hint:

By Vieta, $r+s+t=\dfrac{39}{10},$ $rs+rt+st=\dfrac{29}{10},$ $rst=\dfrac35$

Big Hint:

Expand $(r+2)(s+2)(t+2)$ using these symmetric sums

Solution:

Let the roots be $r,s,t.$ By Vieta’s formulas, $r+s+t=\dfrac{39}{10},$ $rs+rt+st=\dfrac{29}{10},$ and $rst=\dfrac{6}{10}=\dfrac35.$

The new volume is $\begin{gathered} (r+2)(s+2)(t+2) \\ =rst+2(rs+rt+st) \\ \quad {}+4(r+s+t)+8 \\ =\frac35+\frac{58}{10} \\ \quad {}+\frac{156}{10}+8 \\ =30. \end{gathered}$

Thus, the correct answer is D.

16.

A triangular number is a positive integer that can be expressed in the form $t_n=1+2+3+\cdots+n,$ for some positive integer $n.$ The three smallest triangular numbers that are also perfect squares are $t_1=1=1^2,$ $t_8=36=6^2,$ and $t_{49}=1225=35^2.$ What is the sum of the digits of the fourth smallest triangular number that is also a perfect square?

$6$

$9$

$12$

$18$

$27$

Concepts:triangular number perfect square recursion

Difficulty rating: 1800

Small Hint:

Rewrite $\frac{n(n+1)}{2}=y^2$ as the Pell equation $(2n+1)^2-8y^2=1$

Big Hint:

Generate successive positive solutions by multiplying $(2n+1)+y\sqrt8$ by $3+\sqrt8$

Solution:

If $t_n=y^2,$ then $\frac{n(n+1)}{2}=y^2,$ or $(2n+1)^2-8y^2=1.$ The positive Pell solutions occur successively by multiplying $(2n+1)+y\sqrt8$ by $3+\sqrt8.$

Starting from $(2n+1,y)=(3,1),$ this gives $(17,6),$ $(99,35),$ and then $(577,204).$ Thus the fourth value has $n=288$ and equals $204^2=41616.$

The sum of its digits is $4+1+6+1+6=18.$

Thus, the correct answer is D.

17.

Suppose $a$ is a real number such that the equation

$a\cdot(\sin x+\sin(2x))=\sin(3x)$

has more than one solution in the interval $(0,\pi).$ The set of all such $a$ can be written in the form $(p,q)\cup(q,r),$ where $p,$ $q,$ and $r$ are real numbers with $p\lt q\lt r.$ What is $p+q+r?$

$-4$

$-1$

$0$

$1$

$4$

Concepts:trigonometric identity casework

Difficulty rating: 1990

Small Hint:

Use $\sin 2x=2\sin x\cos x$ and $\sin 3x=\sin x(4\cos^2 x-1)$

Big Hint:

After dividing by $\sin x,$ note $x=\dfrac{2\pi}{3}$ (where $\cos x=-\tfrac12$ ) is always a solution

Solution:

Since $\sin x\ne0$ on $(0,\pi),$ divide by $\sin x:$ $\begin{gathered} a(1+2\cos x)=4\cos^2 x-1 \\ =(2\cos x-1)(2\cos x+1). \end{gathered}$

When $\cos x=-\tfrac12$ (that is, $x=\tfrac{2\pi}{3}$ ) both sides vanish, so this is a solution for every $a.$ Otherwise we may cancel $1+2\cos x$ to get $a=2\cos x-1,$ i.e. $\cos x=\dfrac{a+1}{2}.$

This yields a second solution in $(0,\pi)$ exactly when $-1\lt\dfrac{a+1}{2}\lt1,$ that is $a\in(-3,1),$ and it is distinct from $x=\tfrac{2\pi}{3}$ unless $a=-2.$

So more than one solution occurs for $a\in(-3,-2)\cup(-2,1),$ giving $p+q+r=-3-2+1=-4.$

Thus, the correct answer is A.

18.

Let $T_k$ be the transformation of the coordinate plane that first rotates the plane $k$ degrees counterclockwise around the origin and then reflects the plane across the $y$ -axis. What is the least positive integer $n$ such that performing the sequence of transformations $T_1,$ $T_2,$ $T_3,$ $\ldots,$ $T_n$ returns the point $(1,0)$ back to itself?

$359$

$360$

$719$

$720$

$721$

Concepts:transformation casework

Difficulty rating: 2010

Small Hint:

A point at angle $\theta$ is sent by $T_k$ to angle $(180-k)-\theta$

Big Hint:

Track the angle of $(1,0);$ also consider odd $n,$ where the net map is a reflection

Solution:

Rotating a point at angle $\theta$ by $k^\circ$ gives $\theta+k,$ and reflecting across the $y$ -axis sends angle $\phi$ to $180-\phi.$ So $T_k$ sends $\theta$ to $(180-k)-\theta.$

Starting from $(1,0)$ at angle $0,$ applying $T_1,T_2,\ldots$ gives angles $179,-1,178,-2,177,\ldots.$ After an even number $2m$ of steps the angle is $-m,$ and after an odd number $2m+1$ it is $179-m.$

For the point to return, the angle must be a multiple of $360^\circ.$ The even case needs $m=360,$ i.e. $n=720.$ The odd case needs $179-m=0,$ i.e. $m=179$ and $n=359,$ where the net reflection fixes $(1,0).$

The least such $n$ is $359.$

Thus, the correct answer is A.

19.

Suppose that $13$ cards numbered $1,$ $2,$ $3,$ $\ldots,$ $13$ are arranged in a row. The task is to pick them up in numerically increasing order, working repeatedly from left to right. In the example below, cards $1,$ $2,$ $3$ are picked up on the first pass, $4$ and $5$ on the second pass, $6$ on the third pass, $7,$ $8,$ $9,$ $10$ on the fourth pass, and $11,$ $12,$ $13$ on the fifth pass. For how many of the $13!$ possible orderings of the cards will the $13$ cards be picked up in exactly two passes?

$4082$

$4095$

$4096$

$8178$

$8191$

Concepts:permutations pattern recognition

Difficulty rating: 2010

Small Hint:

A new pass begins exactly when the next number to pick up lies to the left of the previous one

Big Hint:

Two passes means the sequence of positions $\text{pos}(1),\ldots,\text{pos}(13)$ has exactly one descent

Solution:

Let $\text{pos}(k)$ be the position of card $k.$ A fresh pass is needed exactly when $\text{pos}(k+1)\lt\text{pos}(k),$ so the number of passes is one more than the number of descents in the sequence $\text{pos}(1),\text{pos}(2),\ldots,\text{pos}(13).$

To build a permutation with at most one descent, choose the entries before the possible descent and write both chosen blocks in increasing order. There are $2^{13}$ subsets. The $14$ initial segments $\varnothing,\{1\},\ldots,\{1,\ldots,13\}$ produce no descent; every other subset produces a unique permutation with one descent. Hence the count is $2^{13}-14=8178.$

Thus, the correct answer is D.

20.

Isosceles trapezoid $ABCD$ has parallel sides $\overline{AD}$ and $\overline{BC},$ with $BC\lt AD$ and $AB=CD.$ There is a point $P$ in the plane such that $PA=1,$ $PB=2,$ $PC=3,$ and $PD=4.$ What is $\dfrac{BC}{AD}?$

$\dfrac14$

$\dfrac13$

$\dfrac12$

$\dfrac23$

$\dfrac34$

Concepts:coordinate geometry trapezoid symmetry

Difficulty rating: 2110

Small Hint:

Put $A=(-p,0),$ $D=(p,0),$ $B=(-q,h),$ $C=(q,h),$ symmetric about the $y$ -axis

Big Hint:

Both $PA^2-PD^2$ and $PB^2-PC^2$ equal a multiple of the $x$ -coordinate of $P$

Solution:

Place the trapezoid symmetric about the $y$ -axis: $A=(-p,0),$ $D=(p,0),$ $B=(-q,h),$ $C=(q,h),$ with $P=(x,y).$

Then $PA^2-PD^2=4px$ $=1-16=-15$ and $PB^2-PC^2=4qx$ $=4-9=-5.$ Dividing gives $\dfrac{p}{q}=3.$

Since $AD=2p$ and $BC=2q,$ we get $\dfrac{BC}{AD}=\dfrac{q}{p}=\dfrac13.$

Thus, the correct answer is B.

21.

Let $P(x)=x^{2022}+x^{1011}+1.$ Which of the following polynomials divides $P(x)?$

$x^2-x+1$

$x^2+x+1$

$x^4+1$

$x^6-x^3+1$

$x^6+x^3+1$

Concepts:roots of unity polynomial

Difficulty rating: 2170

Small Hint:

A divisor $D(x)$ works iff every root $\zeta$ of $D$ satisfies $\zeta^{2022}+\zeta^{1011}+1=0$

Big Hint:

For $x^6+x^3+1$ the roots are primitive $9$ th roots of unity, so $\zeta^3$ is a primitive cube root of unity

Solution:

If $\zeta$ is a root of a divisor, then $P(\zeta)=\zeta^{2022}+\zeta^{1011}+1=0$ is required.

The roots of $x^6+x^3+1$ are the primitive $9$ th roots of unity, so $\zeta^9=1.$ Reducing exponents modulo $9,$ $2022\equiv6$ and $1011\equiv3,$ giving $\zeta^6+\zeta^3+1.$ Here $\omega=\zeta^3$ is a primitive cube root of unity, so this equals $\omega^2+\omega+1=0.$

The other four options fail: substituting their roots yields nonzero values (for instance, the primitive cube roots of unity give $P=3$ ).

Thus, the correct answer is E.

22.

Let $c$ be a real number, and let $z_1,$ $z_2$ be the two complex numbers satisfying the quadratic $z^2-cz+10=0.$ Points $z_1,$ $z_2,$ $\dfrac{1}{z_1},$ and $\dfrac{1}{z_2}$ are the vertices of a (convex) quadrilateral $Q$ in the complex plane. When the area of $Q$ obtains its maximum value, $c$ is the closest to which of the following?

$4.5$

$5$

$5.5$

$6$

$6.5$

Concepts:complex number trapezoid optimization

Difficulty rating: 2270

Small Hint:

For real $c$ with complex roots, $z_2=\overline{z_1}$ and $|z_1|=|z_2|=\sqrt{10}$

Big Hint:

The four points form an isosceles trapezoid symmetric about the real axis; write its area in terms of the root’s angle $\theta$

Solution:

If the roots are non-real, then $z_1=\sqrt{10}\,e^{i\theta}$ and $z_2=\overline{z_1},$ since $z_1z_2=10.$ Then $\dfrac{1}{z_1}=\dfrac{1}{\sqrt{10}}e^{-i\theta}$ and $\dfrac{1}{z_2}=\dfrac{1}{\sqrt{10}}e^{i\theta}.$

The two vertical sides of this trapezoid have lengths $2\sqrt{10}\sin\theta$ and $\frac{2\sin\theta}{\sqrt{10}},$ and their horizontal separation is $(\sqrt{10}-\frac{1}{\sqrt{10}})\cos\theta.$ Hence its area is $\frac{99}{10}\sin\theta\cos\theta =\frac{99}{20}\sin2\theta,$ which is maximized at $\theta=45^\circ.$

Then $c=z_1+z_2$ $=2\sqrt{10}\cos45^\circ$ $=2\sqrt5\approx4.47,$ closest to $4.5.$

Thus, the correct answer is A.

23.

Let $h_n$ and $k_n$ be the unique relatively prime positive integers such that

$\frac11+\frac12+\frac13+\cdots+\frac1n=\frac{h_n}{k_n}.$

Let $L_n$ denote the least common multiple of the numbers $1,$ $2,$ $3,$ $\ldots,$ $n.$ For how many integers $n$ with $1\le n\le22$ is $k_n\lt L_n?$

$0$

$3$

$7$

$8$

$10$

Concepts:least common multiple prime factorization modular arithmetic

Difficulty rating: 2520

Small Hint:

Writing $H_n$ over $L_n,$ the reduced denominator is smaller exactly when a prime cancels from the numerator $\sum_k \frac{L_n}{k}$

Big Hint:

For a prime power $p^a$ with $p^a\le n\lt p^{a+1},$ check whether $p$ divides $\sum_{v_p(k)=a} \frac{L_n}{k}$

Solution:

$L_n$ is always divisible by $k_n,$ so $k_n\lt L_n$ exactly when some prime $p$ divides both $L_n$ and the numerator $N=\sum_{k=1}^n \tfrac{L_n}{k}$ (i.e. a prime cancels).

For a prime $p$ with maximal power $p^a\le n,$ only the terms with $v_p(k)=a$ keep $p$ out of $\frac{L_n}{k};$ all others are divisible by $p.$ So $p$ cancels iff $\sum_{v_p(k)=a}\tfrac{L_n}{k}\equiv0\pmod p.$

Applying this test recursively with $H_n=H_{n-1}+\frac{1}{n}$ gives $k_n=L_n$ for $1\le n\le5,$ $k_n=\frac{L_n}{3}$ for $6\le n\le8,$ and $k_n=L_n$ again for $9\le n\le17.$ Finally, the ratios $\frac{L_n}{k_n}$ for $n=18,19,20,21,22$ are $3,3,15,45,45,$ respectively. Thus cancellation occurs precisely for $n=6,7,8,18,19,20,21,22,$ which is $8$ values.

Thus, the correct answer is D.

24.

How many strings of length $5$ formed from the digits $0,$ $1,$ $2,$ $3,$ $4$ are there such that for each $j\in\{1,2,3,4\},$ at least $j$ of the digits are less than $j?$ (For example, $02214$ satisfies the condition because it contains at least $1$ digit less than $1,$ at least $2$ digits less than $2,$ at least $3$ digits less than $3,$ and at least $4$ digits less than $4.$ The string $23404$ does not satisfy the condition because it does not contain at least $2$ digits less than $2.$ )

$500$

$625$

$1089$

$1199$

$1296$

Concepts:arrangements with restrictions casework

Difficulty rating: 2380

Small Hint:

“At least $j$ digits are less than $j$ ” means the $j$ th smallest digit is at most $j-1$

Big Hint:

These are parking functions of length $5;$ use the circular-parking count with $6$ spaces

Solution:

Sort the five digits as $d_{(1)}\le d_{(2)}\le\cdots\le d_{(5)}.$ The requirement “at least $j$ digits less than $j$ ” is equivalent to $d_{(j)}\le j-1$ for $j=1,2,3,4,$ i.e. $d_{(1)}=0,\ d_{(2)}\le1,$ $\ d_{(3)}\le2,\ d_{(4)}\le3$ (with $d_{(5)}\le4$ automatic).

These strings are exactly the parking functions of length $5.$ To count them, arrange $6$ parking spaces in a circle and let $5$ labeled cars choose arbitrary preferred spaces. Each car moves forward to the first open space. Among the $6^5$ preference strings, rotating all preferences cycles the unique empty space through all $6$ positions. Therefore exactly $\frac{6^5}{6}=6^4=1296$ strings leave a specified space empty. Choosing that space as the extra sixth space gives precisely the sorted inequalities above.

Thus, the correct answer is E.

25.

A circle with integer radius $r$ is centered at $(r,r).$ Distinct line segments of length $c_i$ connect points $(0,a_i)$ to $(b_i,0)$ for $1\le i\le14$ and are tangent to the circle, where $a_i,$ $b_i,$ and $c_i$ are all positive integers and $c_1\le c_2\le\cdots\le c_{14}.$ What is the ratio $\dfrac{c_{14}}{c_1}$ for the least possible value of $r?$

$\dfrac{21}{5}$

$\dfrac{85}{13}$

$7$

$\dfrac{39}{5}$

$17$

Concepts:incircle, incenter, and inradius Pythagorean Triple factor counting

Difficulty rating: 2650

Small Hint:

A segment from $(0,a)$ to $(b,0)$ tangent to this circle makes $r$ the inradius or the semiperimeter of the right triangle with legs $a,b$

Big Hint:

For the inradius case, $(a-2r)(b-2r)=2r^2,$ so positive divisors of $2r^2$ count the oriented segments

Solution:

The circle centered $(r,r)$ with radius $r$ is tangent to both axes. A segment from $(0,a)$ to $(b,0)$ with $a^2+b^2=c^2$ is tangent to it when $r$ equals either the inradius $\tfrac{a+b-c}{2}$ or the semiperimeter $\tfrac{a+b+c}{2}$ of the right triangle with legs $a,b.$

In the inradius case, put $x=a-2r$ and $y=b-2r.$ Then $xy=2r^2,$ and every positive divisor $x$ determines one oriented segment, with $a=x+2r,$ $b=\frac{2r^2}{x}+2r,$ and $c=x+\frac{2r^2}{x}+2r.$ Thus there are exactly $d(2r^2)$ such segments.

For $r=1,2,3,4,5,$ these counts are $2,4,6,6,6,$ and no semiperimeter case is possible because the smallest integer right triangle has semiperimeter $6.$ At $r=6,$ $d(72)=12,$ and the $3$ - $4$ - $5$ triangle contributes two more oriented segments. Hence $6$ is the least possible radius and gives exactly $14$ segments.

The two semiperimeter segments have $c=5.$ In the inradius family, $c=x+\frac{72}{x}+12$ is largest at $x=1$ or $72,$ giving $c=85.$ Therefore $c_1=5,$ $c_{14}=85,$ and $\frac{c_{14}}{c_1}=17.$

Thus, the correct answer is E.