2022 AMC 10A — Problems & Online Timed Exam

Timed

1:15:00

1.

What is the value of $3+\dfrac{1}{3+\dfrac{1}{3+\dfrac{1}{3}}}?$

$\dfrac{31}{10}$

$\dfrac{49}{15}$

$\dfrac{33}{10}$

$\dfrac{109}{33}$

$\dfrac{15}{4}$

Answer: D

Concepts:continued fraction fraction

Difficulty rating: 770

Small Hint:

Start from the innermost denominator

Big Hint:

Simplify one layer of the continued fraction at a time

Solution:

We can simplify this expression as follows:

$\begin{aligned} 3 +& \dfrac{1}{3 + \dfrac{1}{3 + \dfrac{1}{3}}} \\ &= 3 + \dfrac{1}{3 + \dfrac{1}{\dfrac{10}{3}}} \\ &= 3 + \dfrac{1}{3 + \dfrac{3}{10}} \\ &= 3 + \dfrac{1}{\dfrac{33}{10}} \\ &= 3 + \dfrac{10}{33} \\ &= \dfrac{109}{33}. \end{aligned}$

Thus, D is the correct answer.

2.

Mike cycled $15$ laps in $57$ minutes. Assume he cycled at a constant speed throughout. Approximately how many laps did he complete in the first $27$ minutes?

$5$

$7$

$9$

$11$

$13$

Answer: B

Concepts:ratio and proportion estimation

Difficulty rating: 560

Small Hint:

Use a proportion because the speed is constant

Big Hint:

Estimate only after forming the exact fraction

Solution:

We can set up a proportion to solve this problem:

$\dfrac{15}{57} = \dfrac{x}{27}.$

Cross multiplying, we get $x = \dfrac{15}{57} \cdot 27 = \dfrac{135}{19} \approx 7.$

Thus, B is the correct answer.

3.

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?

$1$

$2$

$3$

$4$

$5$

Answer: E

Concepts:system of equations linear equation

Difficulty rating: 900

Small Hint:

Let the third number be the variable

Big Hint:

Write the other two numbers in terms of that variable

Solution:

Let $x, y,$ and $z$ be the three numbers. The conditions from the problem give us the following relations:

$\begin{aligned} x+y+z&=96 &&\text{(1)} \\ x&=6z &&\text{(2)} \\ z&=y-40 &&\text{(3)}. \end{aligned}$

Rearranging $(3),$ we get $y = z + 40.$ Plugging this new equation and $(2)$ into $(1),$ we get $6z + z + 40 + z = 96$ $8z + 40 = 96$ $8z = 56 \Rightarrow z = 7.$

From this, we get that $x = 6 \cdot z = 6 \cdot 7 = 42$ and $y = z + 40 = 7 + 40 = 47.$

Therefore, $y - x = 47 - 42 = 5.$

Thus, E is the correct answer.

4.

In some countries, automobile fuel efficiency is measured in liters per $100$ kilometers while other countries use miles per gallon. Suppose that $1$ kilometer equals $m$ miles, and $1$ gallon equals $l$ liters. Which of the following gives the fuel efficiency in liters per $100$ kilometers for a car that gets $x$ miles per gallon?

$\dfrac{x}{100lm}$

$\dfrac{xlm}{100}$

$\dfrac{lm}{100x}$

$\dfrac{100}{xlm}$

$\dfrac{100lm}{x}$

Answer: E

Concepts:unit conversion rate

Difficulty rating: 1020

Small Hint:

Convert miles per gallon into gallons per kilometer first

Big Hint:

Liters per $100$ kilometers is the reciprocal style of miles per gallon

Video solution:

Click to load, then click again to play

Written solution:

A car that gets $x$ miles per gallon uses $\frac{1}{x}$ gallons per mile.

Since $1$ mile is $\frac{1}{m}$ kilometers, this is $\frac{m}{x}$ gallons per kilometer.

Multiplying by $l$ liters per gallon gives $\frac{lm}{x}$ liters per kilometer, so for $100$ kilometers the fuel used is $\frac{100lm}{x}$ liters.

Thus, E is the correct answer.

5.

Square $ABCD$ has side length $1.$ Points $P,$ $Q,$ $R,$ and $S$ each lie on a side of $ABCD$ such that $APQCRS$ is an equilateral convex hexagon with side length $s.$ What is $s?$

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

$\dfrac{1}{2}$

$2 - \sqrt{2}$

$1 - \dfrac{\sqrt{2}}{4}$

$\dfrac{2}{3}$

Answer: C

Concepts:square (geometry)special right triangle rationalizing denominator

Difficulty rating: 1540

Small Hint:

Use the $45$ – $45$ – $90$ triangle at a corner of the square

Big Hint:

Write the side of the square as one hexagon side plus one corner leg

Solution:

Consider the diagram:

Since $AP = QC = s,$ we know that $PB = BQ.$ This shows that $\triangle PBQ$ is an isosceles right triangle with hypotenuse $PQ=s.$ Using the Pythagorean theorem, we get that $PB = \dfrac{s}{\sqrt{2}}.$

We also know that $1 = AB = AP + PB = s + \dfrac{s}{\sqrt{2}}.$

This equation simplifies to $1 = (1 + \dfrac{1}{\sqrt{2}})s$ Which implies that $s = \dfrac{1}{1 + \dfrac{1}{\sqrt{2}}} = \dfrac{\sqrt{2}}{\sqrt{2} + 1}.$

We can rationalize this fraction to get

$\dfrac{\sqrt{2}}{\sqrt{2} + 1} \cdot \dfrac{\sqrt{2} - 1}{\sqrt{2} - 1} = 2 - \sqrt{2}.$

Thus, C is the correct answer.

6.

Which expression is equal to $\left|a-2-\sqrt{(a-1)^2}\right|$ for $a < 0?$

$3 - 2a$

$1 - a$

$1$

$a + 1$

$3$

Answer: A

Concepts:absolute value radical

Difficulty rating: 900

Small Hint:

Replace the square root of a square by an absolute value

Big Hint:

Use the sign of each expression when a is negative

Solution:

By the definition of square root, we get that $\sqrt{(a - 1)^2} = |a - 1|.$

Since $a < 0,$ we get that $a - 1 < 0,$ which means that $|a - 1| = 1 - a.$

The whole expression therefore simplifies to $|a - 2 - (1 - a)| = |2a - 3|.$ Since $a < 0,$ we know that $2a - 3 < 0.$ This means that $|2a - 3| = 3 - 2a.$

Thus, A is the correct answer.

7.

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 $n?$

$3$

$6$

$8$

$9$

$12$

Answer: B

Concepts:least common multiple greatest common divisor prime factorization

Difficulty rating: 1140

Small Hint:

Compare prime exponents in the lcm and gcd conditions

Big Hint:

The gcd condition controls the exponent of $3$ differently from the lcm condition

Solution:

Prime factorize $18=2\cdot3^2,$ $45=3^2\cdot5,$ and $180=2^2\cdot3^2\cdot5.$

The lcm condition forces the exponent of $2$ in $n$ to be $2,$ the exponent of $5$ to be $1,$ the exponent of $3$ to be at most $2,$ and every other prime exponent to be $0.$ The gcd condition $\gcd(n,45)=15=3\cdot5$ further forces the exponent of $3$ in $n$ to be exactly $1.$

Therefore $n=2^2\cdot3\cdot5=60,$ and the sum of its digits is $6.$

Thus, B is the correct answer.

8.

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 possible values of $X?$

$10$

$26$

$32$

$36$

$40$

Answer: D

Concepts:mean casework

Difficulty rating: 1070

Small Hint:

The average must equal one of the listed data values or X

Big Hint:

Set the average equal to each possible value and keep only positive X values

Solution:

The average of the $6$ numbers is $\dfrac{1 + 7 + \cdots + X}{6} = \dfrac{20 + X}{6}.$

This value can equal any of the terms in the set, so we can case on what it equals.

$\dfrac{20 + X}{6} = 1 \iff X = -14$

$\dfrac{20 + X}{6} = 7 \iff X = 22$

$\dfrac{20 + X}{6} = 5 \iff X = 10$

$\dfrac{20 + X}{6} = 2 \iff X = -8$

$\dfrac{20 + X}{6} = X \iff X = 4$

Adding up all the positive values for $X,$ we get $36.$

Thus, D is the correct answer.

9.

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?

$120$

$270$

$360$

$540$

$720$

Answer: D

Concepts:arrangements with restrictions multiplication principle

Difficulty rating: 1220

Small Hint:

Color the regions in an order where each new region has known neighbors

Big Hint:

After the first two colors, each remaining region has three choices

Solution:

There are $5$ choices for the color of the bottom left rectangle. This forces there to be $4$ choices for the top left rectangle. The middle bottom rectangle touches both of the previous ones, so there are $3$ color options for this rectangle.

The rectangle in the top right is also limited to $3$ colors since it touches the two previous rectangles. Finally, the rectangle in the bottom right also has $3$ color options.

Multiplying these together, we get $5\cdot4\cdot3\cdot3\cdot3=540$ total colorings.

Thus, D is the correct answer.

10.

Daniel finds a rectangular index card and measures its diagonal to be $8$ centimeters. Daniel then cuts out equal squares of side $1$ cm at two opposite corners of the index card and measures the distance between the two closest vertices of these squares to be $4 \sqrt{2}$ centimeters, as shown below. What is the area of the original index card?

$14$

$10 \sqrt{2}$

$16$

$12 \sqrt{2}$

$18$

Answer: E

Concepts:rectangle Pythagorean Theorem algebraic manipulation

Difficulty rating: 1370

Small Hint:

Let the original side lengths be a and b

Big Hint:

Use the two given diagonal lengths to find a+b and ab

Solution:

We can label $a$ and $b$ as the width and height as in the diagram. Then we get that $a^2 + b^2 = 64$ and $(a - 2)^2 + (b - 2)^2 = 32.$

The latter expression simplifies to $a^2 + b^2 - 4a - 4b + 4 + 4 = 32,$ which is the same as $72 - 4(a + b) = 32.$ From this we get $a + b = 10.$

Squaring this, we get $a^2 + b^2 + 2ab = 100,$ which gets us that $2ab = 36,$ which means that the area $(ab)$ is $18.$

Thus, E is the correct answer.

11.

Ted mistakenly wrote $2^m\cdot\sqrt{\dfrac{1}{4096}}$ as $2\cdot\sqrt[m]{\dfrac{1}{4096}}.$

What is the sum of all real numbers $m$ for which these two expressions have the same value?

$5$

$6$

$7$

$8$

$9$

Answer: C

Concepts:exponent quadratic

Difficulty rating: 1280

Small Hint:

Rewrite $4096$ as a power of $2$

Big Hint:

After matching powers of $2,$ solve the resulting quadratic equation in m

Solution:

We can rewrite $4096$ as $2^{12},$ so $\dfrac{1}{4096} = 2^{-12}.$ Then if we equate the given expressions, we get $2^m \cdot 2^{-6} = 2 \cdot 2^{\frac{-12}{m}}.$ Equating the exponents, we get $m - 6 = 1 + \dfrac{-12}{m}.$

Multiplying by $m,$ we get $m^2 - 6m = m - 12$ and so $m^2 - 7m + 12 = 0$ $(m-4)(m-3)$ $m=4,~m=3$

Therefore, we can see that the sum of the solutions is $7.$

Thus, C is the correct answer.

12.

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?

$7$

$12$

$21$

$27$

$31$

Answer: A

Concepts:truth-tellers and liars logical deduction

Difficulty rating: 1950

Small Hint:

Separate alternaters by whether they start with truth or lie

Big Hint:

The first two yes-counts are enough to determine the number of truth-tellers

Solution:

For the first question, the truth-tellers will respond yes, the liars will respond yes, and the alternaters who decided to lie first will say yes. The alternaters who decide to tell the truth first will say no. Denote this as $22 = t + l + a_l.$

For the second questions, the liars will respond yes, and the alternaters who decided to lie first will say yes (they are forced to tell the truth for this question). The truth-tellers will respond no, and the alternaters who told the truth first would lie this round, responding no. Denote this as $15 = l + a_l.$

From this, we get that $t = 7.$ The principal only gives candy to children who always tell the truth in the first round, therefore only giving them $7$ candies total.

Thus, A is the correct answer.

13.

Let $\triangle ABC$ be a scalene triangle. Point $P$ lies on $\overline{BC}$ so that $\overline{AP}$ bisects $\angle BAC.$ The line through $B$ perpendicular to $\overline{AP}$ intersects the line through $A$ parallel to $\overline{BC}$ at point $D.$ Suppose $BP = 2$ and $PC = 3.$ What is $AD?$

$8$

$9$

$10$

$11$

$12$

Answer: C

Concepts:angle bisector theorem similarity isosceles triangle

Difficulty rating: 1540

Small Hint:

Use the angle bisector theorem on triangle ABC

Big Hint:

The perpendicular line creates an isosceles triangle before the final similarity

Solution:

Consider the following diagram:

Let $Y$ be the intersection of $\overline{BD}$ and $\overline{AC}.$ By the Angle Bisector Theorem, $AB:AC=BP:PC=2:3,$ so write $AB=2x$ and $AC=3x.$

Reflection across the angle bisector $\overline{AP}$ sends ray $AB$ to ray $AC.$ Because $BY\perp AP,$ it sends $B$ to $Y.$ Thus $AY=AB=2x,$ and hence $YC=AC-AY=x.$

Since $AD\parallel BC,$ with $B,Y,D$ collinear and $A,Y,C$ collinear, we have $\triangle BYC\sim\triangle DYA.$ Therefore $\frac{AD}{BC}=\frac{AY}{YC}=2.$

Finally, $BC=BP+PC=5,$ so $AD=2BC=10.$

Thus, C is the correct answer.

14.

How many ways are there to split the integers $1$ through $14$ into $7$ pairs such that in each pair, the greater number is at least $2$ times the lesser number?

$108$

$120$

$126$

$132$

$144$

Answer: E

Concepts:arrangements with restrictions pairing and grouping multiplication principle

Difficulty rating: 2390

Small Hint:

The largest small number has very few possible partners

Big Hint:

After $7$ is forced with $14,$ count partner choices from $8$ through $13$

Solution:

The numbers from $8$ through $14$ cannot be paired with one another, so they must be paired with the numbers from $1$ through $7.$ In particular, $7$ must be paired with $14,$ since no other available number is at least twice $7.$

Now let’s look at what the other numbers can pair with. $8$ and $9$ can pair with any number $1-4.$ $10$ and $11$ can pair with any number $1-5,$ and $12$ and $13$ can pair with any number $1-6.$

$8$ can pair with $4$ numbers, but then $9$ only has $3$ options since $8$ took one. $10$ then has $3$ options, since $2$ choices are taken, but it has one more to choose from $(5).$ $11$ then has $2$ options, $12$ has $2$ options, and $13$ only has $1.$

Multiplying these together yields $4 \cdot 3 \cdot 3 \cdot 2 \cdot 2 = 144.$

Thus, E is the correct answer.

15.

Quadrilateral $ABCD$ with side lengths $AB = 7,$ $BC = 24,$ $CD = 20,$ $DA = 15$ is inscribed in a circle. The area interior to the circle but exterior to the quadrilateral can be written in the form $\dfrac{a \pi - b}{c},$ where $a,$ $b,$ and $c$ are positive integers such that $a$ and $c$ have no common prime factor. What is $a + b + c?$

$260$

$855$

$1235$

$1565$

$1997$

Answer: D

Concepts:cyclic quadrilateral Pythagorean Triple circle area

Difficulty rating: 1950

Small Hint:

Check whether the two triangles formed by a diagonal are right triangles

Big Hint:

Use the common hypotenuse as the circle diameter

Solution:

Notice that $7^2 + 24^2$ and $15^2 + 20^2$ are both the same. This forces $AC = 25$ since otherwise $\angle B$ and $\angle D$ would both be acute or obtuse, violating the fact that their sum is $180^{\circ}.$

Also since $\angle B$ is right, we know that $AC$ is the diameter of the circle. The area of the circle is then $\dfrac{625}{4} \pi.$

To find the area of the quadrilateral, we can find the area of each of the triangles, which is $\dfrac{1}{2}(7 \cdot 24 + 20 \cdot 15) =$ $84 + 150 = 234.$

To find the area outside the quadrilateral, we subtract to get $\dfrac{625}{4} \pi - 234 = \dfrac{625 \pi - 936}{4}.$

Therefore, $a + b + c = 625 + 936 + 4$ $= 1565.$

Thus, D is the correct answer.

16.

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?

$\dfrac{24}{5}$

$\dfrac{42}{5}$

$\dfrac{81}{5}$

$30$

$48$

Answer: D

Concepts:Vieta’s Formulas polynomial volume

Difficulty rating: 1420

Small Hint:

Use Vieta formulas for the three original dimensions

Big Hint:

Expand the new volume before substituting symmetric sums

Solution:

Let $h, l,$ and $w$ be the dimensions of the old box. Then the volume of the new box is $(h + 2)(l + 2)(w + 2).$ Expanding, we get $hlw + 2(hl + hw + lw)$ $+ 4(l + h + w) + 8.$ We can use Vieta’s formulas to find the terms in this expression. We get that $hlw = -\dfrac{D}{A} = \dfrac{3}{5},$ $hl + hw + lw = \dfrac{C}{A} = \dfrac{29}{10},$ and $l + h + w = -\dfrac{B}{A} = \dfrac{39}{10}.$

Plugging these values into the expression, we get $\dfrac{3}{5} + 2 \cdot \dfrac{29}{10} + 4 \cdot \dfrac{39}{10} + 8 = 30.$

Thus, D is the correct answer.

17.

How many three-digit positive integers $\underline{a} \ \underline{b} \ \underline{c}$ are there whose nonzero digits $a,$ $b,$ and $c$ satisfy $0.\overline{\underline{a}~\underline{b}~\underline{c}} = \dfrac{1}{3} (0.\overline{a} + 0.\overline{b} + 0.\overline{c})?$ (The bar indicates digit repetition, thus $0.\overline{\underline{a}~\underline{b}~\underline{c}}$ is the infinite repeating decimal $0.\underline{a}~\underline{b}~\underline{c}~\underline{a}~\underline{b}~\underline{c}~\cdots$ )

$9$

$10$

$11$

$13$

$14$

Answer: D

Concepts:repeating decimal digits Diophantine Equation

Difficulty rating: 1660

Small Hint:

Convert the repeating decimals into fractions

Big Hint:

Reduce the condition to a linear equation in the three digits

Solution:

The repeating decimals satisfy $\begin{aligned} 0.\overline{\underline{a}\underline{b}\underline{c}}&=\frac{100a+10b+c}{999},\\ 0.\overline a&=\frac a9,\quad 0.\overline b=\frac b9,\\ 0.\overline c&=\frac c9. \end{aligned}$

Substitution and multiplication by $999$ give $100a+10b+c=37(a+b+c),$ or $7a=3b+4c.$

For each $a\in\{1,2,\ldots,9\},$ checking the nonzero digits $b,c$ in this linear equation gives $\begin{gathered} (1,1,1),(2,2,2),(3,3,3),\\ (4,4,4),(4,8,1),(5,1,8),\\ (5,5,5),(5,9,2),(6,2,9),\\ (6,6,6),(7,7,7),(8,8,8),\\ (9,9,9). \end{gathered}$ Thus there are $13$ integers.

Thus, D is the correct solution.

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 $T_1,$ $T_2,$ $T_3,$ $\cdots,$ $T_n$ returns the point $(1,0)$ back to itself?

$359$

$360$

$719$

$720$

$721$

Answer: A

Concepts:transformation pattern recognition

Difficulty rating: 1950

Small Hint:

Track only the angle in polar coordinates

Big Hint:

Two consecutive transformations simplify to a one-degree rotation

Solution:

Since we are working with angles and reflections, working with polar coordinates would make this problem easier to deal with.

Let $(r, \theta)$ be a polar coordinate. Rotating this by $k$ degrees counterclockwise maps the point to $(r, \theta + k^{\circ})$ and then reflecting it maps it to $(r, 180^{\circ} - \theta - k^{\circ}).$

Therefore, we have that $T_k(r, \theta) = (r, 180^{\circ} - \theta - k^{\circ}).$

From this, we can see that $T_{k + 1}(T_k(r, \theta)) =$ $T_{k + 1}(r, 180^{\circ} - \theta - k^{\circ}) =$ $(r, \theta - 1^{\circ}).$

Now, let’s analyze what happens to the point $(1, 0^{\circ}).$

After $T_1,$ we get $(1, 179^{\circ}).$

After $T_2,$ we get $(1, -1^{\circ}).$

After $T_3,$ we get $(1, 178^{\circ}).$

After $T_4,$ we get $(1, -2^{\circ}).$

$\vdots$

After $T_{2m - 1},$ we get $(1, 180^{\circ} - m^{\circ}).$

After $T_{2m},$ we get $(1, -m^{\circ}).$

From this, we can see that the first time the angle is back to $0^{\circ}$ is after $T_{2(180)-1}=T_{359}.$ Therefore $n=359.$

Thus, A is the correct answer.

19.

Let $L_n$ denote the least common multiple of the numbers $1,$ $2,$ $3,$ $\ldots,$ $n,$ and let $h$ be the unique positive integer such that $\dfrac{1}{1} + \dfrac{1}{2} + \dfrac{1}{3} + \cdots + \dfrac{1}{17} = \dfrac{h}{L_{17}}$ What is the remainder when $h$ is divided by $17?$

$1$

$3$

$5$

$7$

$9$

Answer: C

Concepts:modular arithmetic least common multiple prime factorization

Difficulty rating: 2150

Small Hint:

Work modulo $17$ after multiplying by the least common multiple

Big Hint:

All terms except the one from $\frac{1}{17}$ vanish modulo $17$

Solution:

Multiplying the harmonic sum by $L_{17},$ we get $h=\sum_{i=1}^{17}\frac{L_{17}}{i}.$

For $1\le i\le16,$ the term $\frac{L_{17}}{i}$ is still divisible by $17,$ so these terms contribute $0\pmod{17}.$

Thus $h\equiv \frac{L_{17}}{17}\pmod{17}.$ The least common multiple $L_{17}$ contains the prime-power factors $16,9,5,7,11,13,17,$ so $\begin{aligned} \frac{L_{17}}{17} &\equiv 16\cdot9\cdot5\cdot7\cdot11\cdot13 \\ &\pmod{17}. \end{aligned}$

Reducing modulo $17,$ this is $(-1)\cdot9\cdot5\cdot7\cdot11\cdot13$ $\equiv5\pmod{17}.$

Thus, C is the correct answer.

20.

A four-term sequence is formed by adding each term of a four-term arithmetic sequence of positive integers to the corresponding term of a four-term geometric sequence of positive integers. The first three terms of the resulting four-term sequence are $57,$ $60,$ and $91.$ What is the fourth term of this sequence?

$190$

$194$

$198$

$202$

$206$

Answer: E

Concepts:arithmetic sequence geometric sequence Diophantine Equation

Difficulty rating: 2230

Small Hint:

Subtract consecutive resulting terms

Big Hint:

Use the integer factorization of $b(r-1)^2$

Solution:

Let the arithmetic sequence be $a, a + d, a + 2d, a + 3d$ and the geometric sequence be $b, br, br^2, br^3.$

Then $a + b = 57 \tag*{(1)},$ $a + d + br = 60 \tag*{(2)},$ and $a + 2d + br^2 = 91 \tag*{(3)}.$

Subtracting $(1)$ from $(2)$ and $(2)$ from $(3),$ we get $d + b(r - 1) = 3$ and $d + br(r - 1) = 31.$

Subtracting these, we get $b(r - 1)^2 = 28.$

Let $t=b(r-1)=br-b,$ which is an integer. Then $t^2=28b,$ so $t=14u$ and $b=7u^2$ for some nonzero integer $u.$ Because $a=57-b>0,$ we must have $u\in\{-2,-1,1,2\}.$

Now $r=1+\frac{t}{b}=1+\frac{2}{u}.$ The cases $u=-1$ and $u=-2$ give $r=-1$ and $r=0,$ respectively, so they cannot produce a positive geometric sequence. If $u=2,$ then $b=28, r=2, a=29,$ and $d=3-t=-25,$ making the third arithmetic term negative.

Therefore $u=1,$ so $b = 7, r = 3, a = 50,$ and $d = -11.$ The arithmetic sequence is $50,39,28,17,$ and the geometric sequence is $7,21,63,189.$

The desired answer is $17 + 189 = 206.$

Thus, E is the correct answer.

21.

A bowl is formed by attaching four regular hexagons of side $1$ to a square of side $1.$ The edges of the adjacent hexagons coincide, as shown in the figure. What is the area of the octagon obtained by joining the top eight vertices of the four hexagons, situated on the rim of the bowl?

$6$

$7$

$5 + 2 \sqrt{2}$

$8$

$9$

Answer: B

Concepts:3D geometry regular polygon area decomposition

Difficulty rating: 2390

Small Hint:

Look at the rim from above

Big Hint:

The rim octagon can be viewed as a square with four corner triangles removed

Solution:

View the rim from directly above. Place the bottom square at $(\pm\tfrac12,\pm\tfrac12,0).$ In a regular hexagon, an edge adjacent to the attached side has components $\tfrac12$ parallel and $\tfrac{\sqrt3}{2}$ perpendicular to that side.

At a corner of the bottom square, the corresponding edges of two adjacent hexagons coincide. Comparing their horizontal components shows that the horizontal component of a unit vector perpendicular to an attached side within its hexagon is $\frac{1}{\sqrt3}.$ The opposite side of a regular hexagon is $\sqrt3$ units from the attached side, so its horizontal outward displacement is $1.$

Consequently, the four unit-length sides of the rim lie one unit beyond the four sides of the bottom square. Its top view is therefore a $3$ -by- $3$ square with four isosceles right corner triangles of leg $1$ removed:

Its area is $3^2-4\left(\frac12\cdot1^2\right)=7.$

Thus, B is the correct answer.

22.

Suppose that $13$ cards numbered $1,$ $2,$ $3,$ $\cdots,$ $13$ 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?

$4082$

$4095$

$4096$

$8178$

$8191$

Answer: D

Concepts:combinations complementary counting

Difficulty rating: 1660

Small Hint:

The cards picked up on the first pass must occupy an increasing subsequence

Big Hint:

Choose the positions of the first-pass cards and exclude one-pass arrangements

Solution:

Let $n$ be the number of cards picked up on the first pass, where $1 \leq n \leq 12.$

If we choose the spaces that the $n$ cards occupy, the positions of the remaining cards are determined since they must be placed in order.

There are $\binom{13}{n}$ ways to choose where the $n$ cards go, but if the $n$ cards are placed at the very beginning, then all the cards will be picked up on the first pass.

Therefore, for a given $n$ there are $\binom{13}{n} - 1$ ways to arrange the cards.

Summing over $1\le n\le12,$ we get $\begin{gathered} \sum_{n=1}^{12}\left(\binom{13}{n}-1\right) \\ = (2^{13}-2)-12 \\ = 8178. \end{gathered}$

Thus, D is the correct answer.

23.

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

$\dfrac{1}{4}$

$\dfrac{1}{3}$

$\dfrac{1}{2}$

$\dfrac{2}{3}$

$\dfrac{3}{4}$

Answer: B

Concepts:trapezoid Ptolemy’s Theorem symmetry

Difficulty rating: 2230

Small Hint:

Reflect P across the symmetry line of the isosceles trapezoid

Big Hint:

Apply Ptolemy to the two cyclic trapezoids formed by the reflection

Solution:

Let $P'$ be the reflection of $P$ across the perpendicular bisector of $\overline{BC}.$

This forms two new isosceles trapezoids: $CBPP'$ and $DAPP'.$

Therefore, we get $\begin{gathered} P'A = PD = 4 \\ P'D = PA = 1 \\ P'C = PB = 2 \\ P'B = PC = 3. \end{gathered}$

Using Ptolemy’s theorem, we know that the product of the diagonals is equal to the sum of the products of the opposite sides. Therefore: $\begin{gathered} PP' \cdot AD + 1 = 16 \\ PP' \cdot BC + 4 = 9. \end{gathered}$

This gets us $PP' \cdot AD = 15$ and $PP' \cdot BC = 5.$ Dividing these two equations yields $\dfrac{BC}{AD} = \dfrac{1}{3}.$

Thus, B is the correct answer.

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 this 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 $2.$ )

$500$

$625$

$1089$

$1199$

$1296$

Answer: E

Concepts:parking functions circular counting arrangements with restrictions

Difficulty rating: 2390

Small Hint:

Interpret each digit as a car’s preferred parking space

Big Hint:

Add a sixth parking space and arrange the spaces in a circle

Solution:

Regard the five digits, in order, as the preferred parking spaces of five cars. Spaces are numbered $0,1,2,3,4,$ and each car takes its preferred space if possible, or else the first empty space to its right. If the preferences sorted into nondecreasing order are $b_1\le b_2\le\cdots\le b_5,$ all cars park exactly when $b_i\le i-1\qquad(1\le i\le5).$ These inequalities are precisely the conditions in the problem.

To count such preference strings, add a sixth space and arrange spaces $0,1,\ldots,5$ in a circle. For any of the $6^5$ preference strings, all five cars park and exactly one space remains empty. Rotating every preference by one position rotates the empty space as well. Thus each orbit of six preference strings has each possible empty space exactly once.

Therefore exactly $\frac{6^5}{6}=6^4=1296$ circular preference strings leave space $5$ empty. No car in such a string prefers space $5,$ and cutting the circle immediately after that empty space gives exactly a successful parking sequence on spaces $0$ through $4.$ Hence the desired number of strings is $1296.$

Thus, E is the correct answer.

25.

Let $R,$ $S,$ and $T$ be squares that have vertices at lattice points (i.e., points whose coordinates are both integers) in the coordinate plane, together with their interiors.

The bottom edge of each square is on the $x$ -axis. The left edge of $R$ and the right edge of $S$ are on the $y$ -axis, and $R$ contains $\dfrac{9}{4}$ as many lattice points as does $S.$ The top two vertices of $T$ are in $R \cup S,$ and $T$ contains $\dfrac{1}{4}$ of the lattice points contained in $R \cup S.$ See the figure (not drawn to scale).

The fraction of lattice points in $S$ that are in $S \cap T$ is $27$ times the fraction of lattice points in $R$ that are in $R \cap T.$ What is the minimum possible value of the edge length of $R$ plus the edge length of $S$ plus the edge length of $T?$

$336$

$337$

$338$

$339$

$340$

Answer: B

Concepts:lattice point modular arithmetic optimization

Difficulty rating: 2600

Small Hint:

Use lattice-point counts along each side before converting to edge lengths

Big Hint:

The overlap condition forces a useful congruence modulo $13$

Video solution:

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Written solution:

Let $r$ be the number of lattice points on the side length of $R.$ Similarly define $s$ for $S$ and $t$ for $T.$ Note that the number of lattice points in a rectangle is the product of the number of lattice points along its width and the number of lattice points along its length.

The first condition gives us that $r^2 = \dfrac{9}{4} \cdot s^2$ $r = \dfrac{3}{2} \cdot s \tag*{(1)}$

The number of lattice points in $R \cup S$ is the sum of the lattice points in each of the regions, but there is overlap along the $y$ -axis where $S$ touches it.

The second condition, therefore, yields $t^2 = \dfrac{1}{4}(r^2 + s^2 - s)$ $t^2 = \dfrac{1}{4}(\dfrac{9}{4} \cdot s^2 + s^2 - s)$ $t^2 = \dfrac{1}{4} \cdot \dfrac{13s^2 - 4s}{4}$ $16t^2 = s(13s - 4).$ From $(1),$ we get that $s$ is a multiple of $2.$ We can substitute $s$ with $2j$ to get $16t^2 = 2j(26j - 4)$ $4t^2 = j(13j - 2).$ For the product to be divisible by $4,$ $j$ must be divisible by $2.$ We can again substitute $j$ with $2k$ to get $4t^2 = 2k(26k - 2)$ $t^2 = k(13k - 1) \tag*{(2)}$

Let $x$ be the number of lattice points along the bottom of the rectangle formed by $S \cap T$ and $y$ be the number of lattice points along the bottom of the rectangle formed by $R \cap T.$

Using these variables, we get that the number of lattice points in $S \cap T$ is $xt$ and in $R \cap T$ is $yt.$

The third condition gives us that $\dfrac{xt}{s^2} = 27 \cdot \dfrac{yt}{r^2}$ $\dfrac{x}{s^2} = 27 \cdot \dfrac{y}{\dfrac{9}{4} s^2}$ $x = 12y.$

We also know that $t = x + y - 1$ (accounting for overlap), and this yields $t = 13y - 1 \tag*{(3)}$

$(3)$ gives us that $\begin{gathered} t \equiv -1 \pmod{13}, \\ t^2 \equiv 1 \pmod{13}. \end{gathered}$

However, by $(2),$ we get that $t^2 \equiv -k \pmod{13}$ $k \equiv -1 \pmod{13}.$

By $(2),$ we also get that $k$ is a perfect square since it is relatively prime to $13k - 1,$ and they must multiply to a perfect square.

Thus $k$ must be a perfect square satisfying $k\equiv-1\pmod{13}.$ The smaller positive squares $1,4,9,16$ do not satisfy this congruence, while $k=25$ does, so $25$ is the least possible value.

From this value of $k,$ we get that $j = 2 \cdot 25 = 50,$ $s = 2 \cdot 50 = 100,$ and $r = \dfrac{3}{2} \cdot 100 = 150.$ We can also find that $t^2 = 25(13 \cdot 25 - 1) = 25 \cdot 324$ $t = 5 \cdot 18 = 90.$ Therefore, $r + s + t = 340.$ The question, however, asked for the sum of the side lengths. The side lengths of the squares are $1$ less than the number of lattice points on the side, so we have to subtract $3.$

This value is attainable: equation $(3)$ gives $y=7$ and hence $x=84.$ Since $x\le s=100$ and $y\le r=150,$ a square $T$ with $90$ lattice points per side can straddle the $y$ -axis with the required overlaps, and its top vertices lie in $R\cup S.$

Therefore, the desired answer is $340 - 3 = 337.$

Thus, B is the correct answer.