2023 AMC 10A — Problems & Online Timed Exam

Timed

1:15:00

1.

Cities $A$ and $B$ are $45$ miles apart. Alicia lives in $A$ and Beth lives in $B.$ Alicia bikes towards $B$ at $18$ miles per hour. Leaving at the same time, Beth bikes toward $A$ at $12$ miles per hour. How many miles from City $A$ will they be when they meet?

$20$

$24$

$25$

$26$

$27$

Answer: E

Concepts:relative speed distance rate and time

Difficulty rating: 890

Small Hint:

They ride toward each other, so their speeds add; find the time until they meet

Big Hint:

The meeting time is $\frac{45}{18 + 12}$ hours; multiply Alice’s speed by that time

Solution:

They ride toward each other, so their speeds add. That closes the $45$ -mile gap at $18 + 12 = 30$ mph, and they meet after $\frac{45}{30} = 1.5$ hours. Alice starts at $A,$ so by then she’s gone $18 \cdot 1.5 = 27$ miles. Thus, E is the correct answer.

2.

The weight of $\frac{1}{3}$ of a large pizza together with $3\frac{1}{2}$ cups of orange slices is the same as the weight of $\frac{3}{4}$ of a large pizza together with $\frac{1}{2}$ cup of orange slices. A cup of orange slices weighs $\frac{1}{4}$ of a pound. What is the weight, in pounds, of a large pizza?

$1\frac{4}{5}$

$2$

$2\frac{2}{5}$

$3$

$3\frac{3}{5}$

Answer: A

Concepts:linear equation fraction

Difficulty rating: 1080

Small Hint:

Let $p$ be the pizza’s weight and convert cups of orange slices to pounds using $\frac14$ pound per cup

Big Hint:

Set $\frac13 p + \frac72 \cdot \frac14 = \frac34 p + \frac12 \cdot \frac14$ and solve for $p$

Solution:

Let $p$ be the pizza’s weight. A cup of orange slices is $\frac14$ pound, so the two sides balance as $\frac13 p + \frac72 \cdot \frac14 = \frac34 p + \frac12 \cdot \frac14,$ that is $\frac13 p + \frac78 = \frac34 p + \frac18.$ Collect the pizza terms: $\frac68 = \left(\frac34 - \frac13\right)p = \frac{5}{12}p.$ So $p = \frac34 \cdot \frac{12}{5} = \frac95 = 1\frac45.$ Therefore, the answer is A.

3.

How many positive perfect squares less than $2023$ are divisible by $5?$

$8$

$9$

$10$

$11$

$12$

Answer: A

Concepts:perfect square divisibility

Difficulty rating: 1050

Small Hint:

A perfect square divisible by $5$ must be divisible by $25$

Big Hint:

Write the square as $(5k)^2 = 25k^2 \lt 2023$ and count the valid $k$

Solution:

If a perfect square is divisible by $5,$ it’s divisible by $25,$ so it looks like $(5k)^2 = 25k^2.$ We need $25k^2 \lt 2023,$ i.e. $k^2 \lt 80.9.$ That allows $k = 1, 2, \ldots, 8,$ which is $8$ squares. Thus, A is the correct answer.

4.

A quadrilateral has all integer side lengths, a perimeter of $26,$ and one side of length $4.$ What is the greatest possible length of one side of this quadrilateral?

$9$

$10$

$11$

$12$

$13$

Answer: D

Concepts:triangle inequality bounding to limit cases

Difficulty rating: 1130

Small Hint:

In any quadrilateral, each side is less than the sum of the other three

Big Hint:

If the longest side is $s,$ the other three sum to $26 - s,$ so $s \lt 26 - s$

Solution:

In any quadrilateral each side is shorter than the sum of the other three. Call the longest side $s.$ The rest sum to $26 - s,$ so $s \lt 26 - s,$ which gives $s \lt 13$ and hence $s \le 12.$ Can we hit $12?$ The sides $4, 12, 9, 1$ work, since $12 \lt 4 + 9 + 1.$ So the greatest length is $12.$ Therefore, the answer is D.

5.

How many digits are in the base-ten representation of $8^5 \cdot 5^{10} \cdot 15^5?$

$14$

$15$

$16$

$17$

$18$

Answer: E

Concepts:prime factorization digits

Difficulty rating: 1200

Small Hint:

Factor each base into primes: $8 = 2^3$ and $15 = 3 \cdot 5$

Big Hint:

Combine into $10^{15} \cdot 3^5,$ which is $3^5$ followed by $15$ zeros

Solution:

Factor everything into primes. $8^5 \cdot 5^{10} \cdot 15^5$ $= 2^{15} \cdot 5^{10} \cdot 3^5 \cdot 5^5$ $= 2^{15} \cdot 5^{15} \cdot 3^5$ $= 10^{15} \cdot 243.$ That’s $243$ followed by $15$ zeros, so it has $3 + 15 = 18$ digits. Thus, E is the correct answer.

6.

An integer is assigned to each vertex of a cube. The value of an edge is defined to be the sum of the values of the two vertices it touches, and the value of a face is defined to be the sum of the values of the four edges surrounding it. The value of the cube is defined as the sum of the values of its six faces. Suppose the sum of the integers assigned to the vertices is $21.$ What is the value of the cube?

$42$

$63$

$84$

$126$

$252$

Answer: D

Concepts:double counting cube geometry

Difficulty rating: 1270

Small Hint:

Each edge belongs to $2$ faces, so the total face value is $2$ times the total edge value

Big Hint:

Each vertex belongs to $3$ edges, so the total edge value is $3$ times the vertex total

Solution:

Count by incidences. Each edge lies on $2$ faces, so the six face values together are $2$ times the total of all edge values. Each vertex lies on $3$ edges, so the total edge value is $3$ times the vertex sum. Chaining these, the cube’s value is $2 \cdot 3 \cdot 21 = 126.$ Therefore, the answer is D.

7.

Janet rolls a standard $6$ -sided die $4$ times and keeps a running total of the numbers she rolls. What is the probability that at some point her running total will equal $3?$

$\dfrac{2}{9}$

$\dfrac{49}{216}$

$\dfrac{25}{108}$

$\dfrac{17}{72}$

$\dfrac{13}{54}$

Answer: B

Concepts:dice (probability)casework

Difficulty rating: 1340

Small Hint:

The running total reaches $3$ only through the first few rolls; list the disjoint ways to hit exactly $3$

Big Hint:

The cases are first roll $3;$ rolls $1,2;$ rolls $2,1;$ rolls $1,1,1;$ add their probabilities

Solution:

The total can only reach exactly $3$ through the opening rolls, and these ways are disjoint: $3$ alone (probability $\frac16$ ), then $1,2$ and $2,1$ (each $\frac1{36}$ ), and $1,1,1$ (probability $\frac1{216}$ ). Add them up: $\frac{36}{216} + \frac{6}{216} + \frac{6}{216} + \frac{1}{216} = \frac{49}{216}.$ Thus, B is the correct answer.

8.

Barb the baker creates a new temperature system for baking bread, Breadus, which is linearly based on Fahrenheit. Bread rises at $110\text{ F}^{\circ},$ which is $0$ on the Breadus scale. Bread bakes at $350\text{ F}^{\circ},$ which is $100$ on the Breadus scale. Bread is done when its internal temperature is $200\text{ F}^{\circ}.$ What is this temperature on the Breadus scale?

$33$

$34.5$

$36$

$37.5$

$39$

Answer: D

Concepts:linear equation slope

Difficulty rating: 1130

Small Hint:

The Breadus reading is linear in Fahrenheit through $(110, 0)$ and $(350, 100)$

Big Hint:

The slope is $\frac{100}{350 - 110};$ apply it to the $200 - 110$ degrees above the $0$ point

Solution:

The Breadus reading is linear in Fahrenheit through $(110, 0)$ and $(350, 100),$ so $B = \frac{100}{350 - 110}(F - 110)$ $= \frac{5}{12}(F - 110).$ Plug in $F = 200:$ $B = \frac{5}{12} \cdot 90 = 37.5.$ Therefore, the answer is D.

9.

A digital display shows the current date as an $8$ -digit integer consisting of a $4$ -digit year, followed by a $2$ -digit month, followed by a $2$ -digit date within the month. For example, Arbor Day this year is displayed as $20230428.$ For how many dates in $2023$ will each digit appear an even number of times in the $8$ -digit display for that date?

$5$

$6$

$7$

$8$

$9$

Answer: E

Concepts:digits parity casework

Difficulty rating: 1410

Small Hint:

The year $2023$ contributes two $2$ s (even), one $0,$ and one $3$

Big Hint:

The four month-and-day digits must supply one more $0$ and one more $3,$ with the remaining two digits equal to each other

Solution:

The year $2023$ already gives two $2$ s (even), one $0$ , and one $3$ . So to make every digit occur an even number of times, the four digits of $MM$ and $DD$ must supply one more $0$ , one more $3$ , and two equal digits, while keeping the number of $2$ s even. Checking the legal dates gives $01/13,$ $01/31,$ $02/23,$ $03/11,$ $03/22,$ $10/13,$ $10/31,$ $11/03,$ and $11/30,$ exactly $9$ dates. Thus, E is the correct answer.

10.

Maureen is keeping track of the mean of her quiz scores this semester. If Maureen scores an $11$ on the next quiz, her mean will increase by $1.$ If she scores an $11$ on each of the next three quizzes, her mean will increase by $2.$ What is the mean of her quiz scores currently?

$4$

$5$

$6$

$7$

$8$

Answer: D

Concepts:mean system of equations

Difficulty rating: 1270

Small Hint:

Let the current mean be $m$ over $n$ tests, so the current total is $mn$

Big Hint:

Adding one $11$ gives mean $m + 1$ and adding three $11$ s gives mean $m + 2;$ form two equations

Solution:

Let $m$ be the current mean over $n$ quizzes. One more $11$ makes the mean $m + 1:$ $\frac{mn + 11}{n + 1} = m + 1,$ which tidies up to $m + n = 10.$ Three more $11$ s make it $m + 2:$ $\frac{mn + 33}{n + 3} = m + 2,$ i.e. $3m + 2n = 27.$ Solve the pair and $m = 7.$ Therefore, the answer is D.

11.

A square of area $2$ is inscribed in a square of area $3,$ creating four congruent triangles, as shown below. What is the ratio of the shorter leg to the longer leg in the shaded right triangle?

$\dfrac{1}{5}$

$\dfrac{1}{4}$

$2 - \sqrt{3}$

$\sqrt{3} - \sqrt{2}$

$\sqrt{2} - 1$

Answer: C

Concepts:square (geometry)Pythagorean Theorem Vieta’s Formulas

Difficulty rating: 1500

Small Hint:

Each corner triangle has legs $a,b$ with $a+b=\sqrt3$ (a side of the big square) and $a^2+b^2=2$ (a side of the small square)

Big Hint:

$(a+b)^2-(a^2+b^2)$ gives $2ab;$ then $a$ and $b$ are the roots of $t^2-\sqrt3\,t+\tfrac12=0$

Solution:

Each corner right triangle has legs $a$ and $b.$ A side of the outer square gives $a+b=\sqrt3,$ and a side of the inscribed square gives $a^2+b^2=2.$ Subtract to find the product: $2ab=(a+b)^2-(a^2+b^2)$ $=3-2$ $=1,$ so $ab=\tfrac12.$ Then $a$ and $b$ are the roots of $t^2-\sqrt3\,t+\tfrac12=0,$ namely $\tfrac{\sqrt3\pm1}{2}.$ The ratio of the smaller leg to the larger is $\dfrac{\sqrt3-1}{\sqrt3+1}$ $=\dfrac{(\sqrt3-1)^2}{2}$ $=2-\sqrt3.$ Thus, C is the correct answer.

12.

How many three-digit positive integers $N$ satisfy the following properties?

• The number $N$ is divisible by $7.$

• The number formed by reversing the digits of $N$ is divisible by $5.$

$13$

$14$

$15$

$16$

$17$

Answer: B

Concepts:divisibility digits counting integers in a range

Difficulty rating: 1440

Small Hint:

The reversed number ends in the first digit of $N,$ and for divisibility by $5$ that digit must be $0$ or $5$

Big Hint:

Since $N$ is three digits it starts with $5;$ count the multiples of $7$ from $500$ to $599$

Solution:

When we reverse $N,$ its last digit is the first digit of $N.$ For the reversal to be divisible by $5,$ that digit is $0$ or $5.$ A three-digit number can’t start with $0,$ so $N$ starts with $5,$ meaning $500 \le N \le 599$ (and the reversal ends in $5,$ always fine). Now just count multiples of $7$ here: from $7 \cdot 72 = 504$ to $7 \cdot 85 = 595,$ that’s $14$ numbers. Therefore, the answer is B.

13.

Abdul and Chiang are standing $48$ feet apart in a field. Bharat is standing in the same field as far from Abdul as possible so that the angle formed by his lines of sight to Abdul and Chiang measures $60^\circ.$ What is the square of the distance (in feet) between Abdul and Bharat?

$1728$

$2601$

$3072$

$4608$

$6912$

Answer: C

Concepts:inscribed angle law of sines optimization

Difficulty rating: 1590

Small Hint:

Points that see segment $AC$ at a fixed $60^\circ$ lie on a circular arc through $A$ and $C$

Big Hint:

The farthest such point from Abdul is a diameter endpoint; find the circle’s diameter with the law of sines

Solution:

Let $A$ be Abdul, $C$ be Chiang with $AC = 48,$ and $B$ be Bharat with $\angle B = 60^\circ.$ Every point seeing $AC$ at $60^\circ$ lies on one circular arc, so all valid $B$ sit on a circle where chord $AC$ subtends $60^\circ.$ The law of sines gives its diameter, $\frac{AC}{\sin 60^\circ} = \frac{48}{\frac{\sqrt3}{2}} = 32\sqrt3.$ Now $AB$ is a chord, and a chord is longest when it’s a diameter. So $AB = 32\sqrt3$ and $AB^2 = 1024 \cdot 3 = 3072.$ Thus, C is the correct answer.

14.

A number is chosen at random from among the first $100$ positive integers, and a positive integer divisor of that number is then chosen at random. What is the probability that the chosen divisor is divisible by $11?$

$\dfrac{4}{100}$

$\dfrac{9}{200}$

$\dfrac{1}{20}$

$\dfrac{11}{200}$

$\dfrac{3}{50}$

Answer: B

Concepts:factor counting basic probability

Difficulty rating: 1630

Small Hint:

Only multiples of $11$ up to $100$ can have a divisor divisible by $11$

Big Hint:

For $n = 11m$ with $m \le 9,$ exactly half of $n$ ’s divisors are multiples of $11,$ since $m$ is not a multiple of $11$

Solution:

A number $n \le 100$ can only have a divisor divisible by $11$ when $n$ is divisible by $11,$ so $n \in \{11, 22, \ldots, 99\}.$ Write $n = 11m$ with $m \le 9.$ Here $m$ is not a multiple of $11,$ so $d(11m) = 2\,d(m),$ and the divisors that are multiples of $11$ are exactly the $d(m)$ numbers $11d.$ That makes the chance $\frac{d(m)}{2\,d(m)} = \frac12$ for each such $n.$ Averaging over all $100$ starting numbers, the probability is $\frac{1}{100}\sum_{m=1}^{9}\frac12 = \frac{9}{200}.$ Therefore, the answer is B.

15.

An even number of circles are nested, starting with a radius of $1$ and increasing by $1$ each time, all sharing a common point. The region between every other circle is shaded, starting with the region inside the circle of radius $2$ but outside the circle of radius $1.$ An example showing $8$ circles is displayed below. What is the least number of circles needed to make the total shaded area at least $2023\pi?$

$46$

$48$

$56$

$60$

$64$

Answer: E

Concepts:circle area arithmetic sequence summation

Difficulty rating: 1560

Small Hint:

The shaded region between the circles of radius $2k$ and $2k-1$ has area $\pi\big((2k)^2 - (2k-1)^2\big)$ $= (4k-1)\pi$

Big Hint:

With $2n$ circles the shaded area is $\pi\sum_{k=1}^{n}(4k-1) = \pi(2n^2 + n);$ solve $2n^2 + n \ge 2023$

Solution:

A circle of radius $r$ has area $\pi r^2.$ So the shaded ring between radius $2k$ and $2k-1$ has area $\pi\big((2k)^2 - (2k-1)^2\big)$ $= (4k-1)\pi.$ With $2n$ circles the shaded total is $\pi\sum_{k=1}^{n}(4k-1) = \pi(2n^2 + n).$ We want $2n^2 + n \ge 2023.$ At $n = 31$ it’s $1953,$ at $n = 32$ it’s $2080.$ So $n = 32,$ which means $2n = 64$ circles. Thus, E is the correct answer.

16.

In a table tennis tournament every participant played every other participant exactly once. Although there were twice as many right-handed players as left-handed players, the number of games won by left-handed players was $40\%$ more than the number of games won by right-handed players. (There were no ties and no ambidextrous players.) What is the total number of games played?

$15$

$36$

$45$

$48$

$66$

Answer: B

Concepts:combinations ratio and proportion

Difficulty rating: 1730

Small Hint:

With $L$ left-handed and $2L$ right-handed players, the number of games is $\binom{3L}{2}$

Big Hint:

Every game has one winner, and left wins to right wins is $1.4 : 1 = 7 : 5,$ so the total is a multiple of $12$

Solution:

Say there are $L$ left-handers and $2L$ right-handers, so $3L$ players and $\binom{3L}{2}$ games. Every game has one winner, and left wins are $1.4$ times right wins, so the wins split $7 : 5$ and the total must be a multiple of $12$ . Left-handers can win at most all games involving at least one left-hander, namely $\binom{L}{2} + 2L^2$ . Hence $\frac{7}{12}\binom{3L}{2} \leq \binom{L}{2} + 2L^2,$ which gives $L \leq 3$ . For $L = 1$ and $L = 2$ , the total is not divisible by $12$ . For $L = 3$ , there are $\binom{9}{2} = 36$ games. This is attainable if the left-handers win all $18$ cross-group games and all $3$ games among themselves, while the right-handers win their $15$ internal games. The win totals are $21$ and $15$ , so the answer is $36$ . Therefore, the answer is B.

17.

Let $ABCD$ be a rectangle with $AB = 30$ and $BC = 28.$ Points $P$ and $Q$ lie on $BC$ and $CD$ respectively so that all sides of $\triangle ABP,$ $\triangle PCQ,$ and $\triangle QDA$ have integer lengths. What is the perimeter of $\triangle APQ?$

$84$

$86$

$88$

$90$

$92$

Answer: A

Concepts:Pythagorean Triple rectangle systematic listing

Difficulty rating: 1840

Small Hint:

Each of the three triangles is right-angled at a corner, so $AP^2 = 30^2 + BP^2$ and $QA^2 = 28^2 + DQ^2$ must be perfect squares

Big Hint:

Use the triples $30$ - $16$ - $34$ and $21$ - $28$ - $35;$ then $PQ$ comes out an integer too

Solution:

Set $A = (0,0)$ , $B = (30,0)$ , $C = (30,28)$ , $D = (0,28)$ , with $P = (30, p)$ on $BC$ and $Q = (30 - q, 28)$ on $CD$ . The three right triangles give $AP = \sqrt{30^2 + p^2}$ , $QA = \sqrt{28^2 + (30 - q)^2}$ , and $PQ = \sqrt{(28 - p)^2 + q^2}$ . For $0 \leq p \leq 28$ , the equation $(AP-p)(AP+p)=900$ gives only $p=0$ and $p=16$ , with $AP=30$ and $AP=34$ . Similarly, setting $x=30-q$ , the equation $(QA-x)(QA+x)=784$ with $0 \leq x \leq 30$ gives $x=0$ or $x=21$ . Testing these four combinations in the formula for $PQ$ , only $p=16$ , $x=21$ works. Thus $q=9$ , $QA=35$ , and $PQ=\sqrt{12^2+9^2}=15$ . The perimeter of $\triangle APQ$ is $34+15+35=84$ . Thus, A is the correct answer.

18.

A rhombic dodecahedron is a solid with $12$ congruent rhombus faces. At every vertex, $3$ or $4$ edges meet, depending on the vertex. How many vertices have exactly $3$ edges meeting?

$5$

$6$

$7$

$8$

$9$

Answer: D

Concepts:Euler’s Polyhedron Formula polyhedron graph theory double counting

Difficulty rating: 1660

Small Hint:

The solid has $12$ faces and $24$ edges; use Euler’s formula to get the number of vertices

Big Hint:

If $x$ vertices have degree $3$ and the rest degree $4,$ the sum of all degrees is $2 \cdot 24$

Solution:

Each rhombus has $4$ edges, and every edge is shared by $2$ faces, so $E = \frac{12 \cdot 4}{2} = 24.$ With $F = 12,$ Euler’s formula gives $V = 2 - F + E = 14.$ Suppose $x$ vertices have $3$ edges and the other $14 - x$ have $4.$ The degrees sum to twice the edge count: $3x + 4(14 - x) = 2E = 48,$ so $x = 8.$ Therefore, the answer is D.

19.

The line segment formed by $A(1, 2)$ and $B(3, 3)$ is rotated to the line segment formed by $A'(3, 1)$ and $B'(4, 3)$ about the point $P(r, s).$ What is $|r - s|?$

$\dfrac{1}{4}$

$\dfrac{1}{2}$

$\dfrac{3}{4}$

$\dfrac{2}{3}$

$1$

Answer: E

Concepts:transformation perpendicular bisector coordinate geometry

Difficulty rating: 1730

Small Hint:

The rotation center is equidistant from each point and its image, so it lies on the perpendicular bisectors of $AA'$ and $BB'$

Big Hint:

Intersect the perpendicular bisector of $BB'$ (a vertical line) with that of $AA'$

Solution:

A rotation keeps its center equidistant from each point and its image. So $P$ is equidistant from $A$ and $A',$ and from $B$ and $B',$ which puts it at the intersection of two perpendicular bisectors. The bisector of $BB'$ from $(3,3)$ to $(4,3)$ is $x = 3.5.$ The bisector of $AA'$ from $(1,2)$ to $(3,1)$ is $2x - y = 2.5.$ Then $y = 2(3.5) - 2.5 = 4.5,$ so $P = (3.5, 4.5)$ and $|r - s| = |3.5 - 4.5| = 1.$ Thus, E is the correct answer.

20.

Each square in a $3 \times 3$ grid of squares is colored red, white, blue, or green so that every $2 \times 2$ square contains one square of each color. One such coloring is shown on the right below. How many different colorings are possible?

$24$

$48$

$60$

$72$

$96$

Answer: D

Concepts:arrangements with restrictions casework

Difficulty rating: 2080

Small Hint:

Label the cells row by row as $a,$ $b,$ $c;$ $d,$ $e,$ $f;$ $g,$ $h,$ $i;$ the top-left four are a permutation of all four colors

Big Hint:

Then $\{c, f\}$ and $\{g, h\}$ are each the two remaining colors in some order, and $i$ is forced but works only when $f \ne h$

Solution:

Label the cells row by row as $a,$ $b,$ $c;$ $d,$ $e,$ $f;$ $g,$ $h,$ $i.$ The top-left block $a, b, d, e$ is a permutation of the four colors, so $4! = 24$ ways. The block $\{b, c, e, f\}$ is also all four colors, and $b, e$ are fixed, so $\{c, f\}$ is the remaining two in some order: $2$ ways. Same story for $\{g, h\},$ the two colors apart from $d, e,$ another $2$ ways. That leaves $i,$ forced to whatever color is missing from $\{e, f, h\},$ and that only works when $f \ne h.$ Of the $2 \cdot 2 = 4$ order combinations, exactly one has $f = h,$ so $3$ survive. The total is $24 \cdot 3 = 72.$ Therefore, the answer is D.

21.

There is a unique polynomial $P(x)$ of least degree with leading coefficient $1$ satisfying all of the following:

$1$ is a root of $P(x) - 1,$ $2$ is a root of $P(x - 2),$ $3$ is a root of $P(3x),$ and $4$ is a root of $4P(x).$

All the roots of $P(x)$ except one are integers. If the one non-integer root can be written as $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers, what is $m + n?$

$41$

$43$

$45$

$47$

$49$

Answer: D

Concepts:polynomial substitution

Difficulty rating: 2120

Small Hint:

Translate each condition into a value: $P(1) = 1,$ $P(0) = 0,$ $P(9) = 0,$ $P(4) = 0$

Big Hint:

The integer roots $0, 4, 9$ force $P(x) = x(x - 4)(x - 9)(x - c);$ use $P(1) = 1$ to find $c$

Solution:

Translate each condition into a value: $P(1) = 1,$ $P(0) = 0,$ $P(9) = 0,$ and $P(4) = 0.$ So $0, 4, 9$ are roots. Could a cubic do it? A monic cubic with those roots has $P(1) = (1)(-3)(-8) = 24 \ne 1,$ so no. The least-degree monic polynomial is degree $4:$ $P(x) = x(x - 4)(x - 9)(x - c).$ Now $P(1) = (1)(-3)(-8)(1 - c)$ $= 24(1 - c)$ $= 1,$ so $1 - c = \frac{1}{24}$ and $c = \frac{23}{24}.$ That’s the lone non-integer root, so $m + n = 23 + 24 = 47.$ Thus, D is the correct answer.

22.

Circle $C_1$ and $C_2$ each have radius $1,$ and the distance between their centers is $\frac{1}{2}.$ Circle $C_3$ is the largest circle internally tangent to both $C_1$ and $C_2.$ Circle $C_4$ is internally tangent to both $C_1$ and $C_2$ and externally tangent to $C_3.$ What is the radius of $C_4?$

$\dfrac{1}{14}$

$\dfrac{1}{12}$

$\dfrac{1}{10}$

$\dfrac{3}{28}$

$\dfrac{1}{9}$

Answer: D

Concepts:tangent circles coordinate geometry symmetry

Difficulty rating: 2270

Small Hint:

Put the centers of $C_1, C_2$ at $\left(\pm\frac14, 0\right);$ $C_3$ is centered at the origin with radius from $1 - r_3 = \frac14$

Big Hint:

Let $C_4$ be centered at $(0, y)$ with radius $r;$ internal tangency to $C_1$ and external tangency to $C_3$ give two equations

Solution:

Put the centers of $C_1, C_2$ at $\left(\pm\frac14, 0\right).$ By symmetry the largest circle inside both sits at the origin with radius $r_3,$ where $1 - r_3 = \frac14,$ so $r_3 = \frac34.$ Let $C_4$ be centered at $(0, y)$ with radius $r.$ Internal tangency to $C_1$ gives $\sqrt{\frac1{16} + y^2} = 1 - r,$ and external tangency to $C_3$ gives $y = \frac34 + r.$ Substitute the second into the first: $\frac1{16} + \left(\frac34 + r\right)^2 = (1 - r)^2.$ This collapses to $\frac72 r = \frac38,$ so $r = \frac{3}{28}.$ Therefore, the answer is D.

23.

Positive integer divisors $a$ and $b$ of $N$ are called complementary if $ab = N.$ Given that $N$ has a pair of complementary divisors that differ by $20$ and a pair of complementary divisors that differ by $23,$ find the sum of the digits of $N.$

$11$

$13$

$15$

$17$

$19$

Answer: C

Concepts:difference of squares factor Diophantine Equation

Difficulty rating: 2380

Small Hint:

Complementary divisors with product $N$ differing by $20$ are $b$ and $b + 20,$ so $N + 100 = (b + 10)^2$

Big Hint:

Then $4N + 529$ is also a perfect square; combine the two to get a difference of squares equal to $129$

Solution:

Complementary divisors differing by $20$ are $b$ and $b + 20$ with product $N$ , so $N = b^2 + 20b$ and $N + 100 = (b + 10)^2$ . A pair differing by $23$ gives $4N + 529 = (2d + 23)^2$ . Set $N + 100 = k^2$ . Then $4k^2 + 129 = m^2$ , so $(m - 2k)(m + 2k)$ $= 129$ $= 3 \cdot 43$ . The positive factor pair $3,43$ gives $k=10$ , hence the inadmissible value $N=0$ . The pair $1,129$ gives $m = 65$ , $k = 32$ , hence $N = 32^2 - 100 = 924$ . Check it: $924 = 22 \cdot 42 = 21 \cdot 44$ , and the digit sum is $9 + 2 + 4 = 15$ . Thus, C is the correct answer.

24.

Six regular hexagonal blocks of side length $1$ unit are arranged inside a regular hexagonal frame. Each block lies along an inside edge of the frame and is aligned with two other blocks, as shown in the figure below. The distance from any corner of the frame to the nearest vertex of a block is $\frac{3}{7}$ unit. What is the area of the region inside the frame not occupied by the blocks?

$\dfrac{13\sqrt{3}}{3}$

$\dfrac{216\sqrt{3}}{49}$

$\dfrac{9\sqrt{3}}{2}$

$\dfrac{14\sqrt{3}}{3}$

$\dfrac{243\sqrt{3}}{49}$

Answer: C

Concepts:regular polygon area area decomposition

Difficulty rating: 2520

Small Hint:

The uncovered region is the frame’s area minus the six unit hexagons; a regular hexagon of side $t$ has area $\tfrac{3\sqrt3}{2}t^2$

Big Hint:

Extend the block edges to the frame: one frame side splits into lengths $\tfrac37, 1, 1,$ and $1-\tfrac37;$ then subtract the six block areas

Solution:

Let $d=\tfrac37.$ Extend the slanted edges of the blocks that meet a fixed side of the frame. Because all the relevant angles are $60^\circ,$ the extensions form an equilateral triangle of side $1$ at one end and an equilateral triangle of side $1-d$ at the other. Thus that frame side is partitioned into lengths $d,1,1,$ and $1-d,$ so its length is $d+1+1+(1-d)=3.$ A regular hexagon of side $t$ has area $\tfrac{3\sqrt3}{2}t^2.$ Therefore the uncovered area is the area of the side- $3$ frame minus the areas of the six unit blocks: $\tfrac{3\sqrt3}{2}\cdot 3^2$ $- 6\cdot\tfrac{3\sqrt3}{2}$ $= \tfrac{27\sqrt3}{2} - 9\sqrt3$ $= \tfrac{9\sqrt3}{2}.$ Therefore, the answer is C.

25.

If $A$ and $B$ are vertices of a polyhedron, define the distance $d(A, B)$ to be the minimum number of edges of the polyhedron one must traverse in order to connect $A$ and $B.$ For example, if $AB$ is an edge of the polyhedron, then $d(A, B) = 1,$ but if $AC$ and $CB$ are edges and $AB$ is not an edge, then $d(A, B) = 2.$ Let $Q,$ $R,$ and $S$ be randomly chosen distinct vertices of a regular icosahedron (a regular polyhedron made up of $20$ equilateral triangles). What is the probability that $d(Q, R) \gt d(R, S)?$

$\dfrac{7}{22}$

$\dfrac{1}{3}$

$\dfrac{3}{8}$

$\dfrac{5}{12}$

$\dfrac{1}{2}$

Answer: A

Concepts:graph theory symmetry basic probability

Difficulty rating: 2600

Small Hint:

From any vertex of an icosahedron, $5$ vertices are at distance $1,$ $5$ at distance $2,$ and $1$ (the opposite vertex) at distance $3$

Big Hint:

Fix $R;$ by symmetry $P(d(Q,R) \gt d(R,S))$ $= P(d(Q,R) \lt d(R,S)),$ so it equals $\frac{1 - P(\text{equal})}{2}$

Solution:

Fix $R.$ Of the other $11$ vertices, $5$ sit at distance $1,$ $5$ at distance $2,$ and $1$ (the opposite vertex) at distance $3.$ Pick ordered distinct $Q, S$ from these $11:$ that’s $11 \cdot 10 = 110$ pairs. The ones with $d(R,Q) = d(R,S)$ number $5\cdot4 + 5\cdot4 + 1\cdot0 = 40,$ so $P(\text{equal}) = \frac{40}{110} = \frac{4}{11}.$ By the symmetry between $Q$ and $S,$ the $\gt$ and $\lt$ cases split the rest evenly, so $P(d(Q,R) \gt d(R,S))$ $= \frac{1 - \frac4{11}}{2}$ $= \frac{7}{22}.$ Thus, A is the correct answer.