2013 AMC 12A Solutions | LIVE by Po-Shen Loh

1.

Square $ABCD$ has side length $10.$ Point $E$ is on $\overline{BC},$ and the area of $\triangle ABE$ is $40.$ What is $BE?$

$4$

$5$

$6$

$7$

$8$

Concepts:triangle area right triangle

Difficulty rating: 840

Small Hint:

The legs of right triangle $ABE$ are $AB$ and $BE$

Big Hint:

Set $\tfrac12\cdot AB\cdot BE = 40$ with $AB = 10$

Solution:

The legs of right triangle $ABE$ are $AB = 10$ and $BE.$ From $\tfrac12\cdot 10\cdot BE = 40,$ we get $BE = 8.$

Thus, the correct answer is E.

2.

A softball team played ten games, scoring $1,$ $2,$ $3,$ $4,$ $5,$ $6,$ $7,$ $8,$ $9,$ and $10$ runs. They lost by one run in exactly five games. In each of their other games, they scored twice as many runs as their opponent. How many total runs did their opponents score?

$35$

$40$

$45$

$50$

$55$

Concepts:parity casework

Difficulty rating: 1010

Small Hint:

Scoring twice the opponent is only possible when the team’s own score is even

Big Hint:

The five one-run losses are the odd-scored games; each opponent scored one more

Solution:

The team can only score twice as many runs as its opponent when its own score is even. Those games have scores $2, 4, 6, 8, 10,$ so their opponents scored $1 + 2 + 3 + 4 + 5 = 15.$

The other five games had scores $1, 3, 5, 7, 9$ and were one-run losses, so their opponents scored $2 + 4 + 6 + 8 + 10 = 30.$ The total is $15 + 30 = 45.$

Thus, the correct answer is C.

3.

A flower bouquet contains pink roses, red roses, pink carnations, and red carnations. One third of the pink flowers are roses, three fourths of the red flowers are carnations, and six tenths of the flowers are pink. What percent of the flowers are carnations?

$15$

$30$

$40$

$60$

$70$

Concepts:fraction percentage

Difficulty rating: 1100

Small Hint:

Six tenths of the flowers are pink and four tenths are red

Big Hint:

Two thirds of the pink and three fourths of the red are carnations; add the two amounts

Solution:

Six tenths of the flowers are pink and four tenths are red. Since two thirds of the pink flowers are carnations, pink carnations make up $\tfrac{2}{3}\cdot\tfrac{6}{10} = \tfrac{4}{10}$ of the flowers.

Since three fourths of the red flowers are carnations, red carnations make up $\tfrac{3}{4}\cdot\tfrac{4}{10} = \tfrac{3}{10}$ of the flowers. Together the carnations are $\tfrac{4}{10} + \tfrac{3}{10} = \tfrac{7}{10} = 70\%.$

Thus, the correct answer is E.

4.

What is the value of $\dfrac{2^{2014} + 2^{2012}}{2^{2014} - 2^{2012}}?$

$-1$

$1$

$\dfrac{5}{3}$

$2013$

$2^{4024}$

Concepts:exponent factoring

Difficulty rating: 1130

Small Hint:

Factor $2^{2012}$ out of both the numerator and the denominator

Big Hint:

The ratio reduces to $\dfrac{2^2 + 1}{2^2 - 1}$

Solution:

Factoring $2^{2012}$ from each term gives $\dfrac{2^{2012}(2^2 + 1)}{2^{2012}(2^2 - 1)} = \dfrac{4 + 1}{4 - 1} = \dfrac{5}{3}.$

Thus, the correct answer is C.

5.

Tom, Dorothy, and Sammy went on a vacation and agreed to split the costs evenly. During their trip Tom paid $\$105,$ Dorothy paid $\$125,$ and Sammy paid $\$175.$ In order to share the costs equally, Tom gave Sammy $t$ dollars, and Dorothy gave Sammy $d$ dollars. What is $t - d?$

$15$

$20$

$25$

$30$

$35$

Concepts:mean algebraic manipulation

Difficulty rating: 1130

Small Hint:

Each person’s fair share is one third of the total spent

Big Hint:

$t$ and $d$ measure how much Tom and Dorothy each fall short of their fair share

Solution:

The total spent was $105 + 125 + 175 = 405,$ so each fair share is $\tfrac13\cdot 405 = 135$ dollars.

Then $t = 135 - 105 = 30$ and $d = 135 - 125 = 10,$ so $t - d = 30 - 10 = 20.$

Thus, the correct answer is B.

6.

In a recent basketball game, Shenille attempted only three-point shots and two-point shots. She was successful on $20\%$ of her three-point shots and $30\%$ of her two-point shots. Shenille attempted $30$ shots. How many points did she score?

$12$

$18$

$24$

$30$

$36$

Concepts:percentage invariant

Difficulty rating: 1250

Small Hint:

Let $x$ be the number of three-point attempts, so $30 - x$ are two-point attempts

Big Hint:

Her score is $0.2\cdot 3\cdot x + 0.3\cdot 2\cdot(30 - x);$ the $x$ terms cancel

Solution:

If Shenille attempted $x$ three-point shots and $30 - x$ two-point shots, she scored $\begin{gathered} 0.2\cdot 3\cdot x + 0.3\cdot 2\cdot(30 - x) \\ = 0.6x + 0.6(30 - x) \\ = 0.6\cdot 30 = 18 \end{gathered}$ points.

Thus, the correct answer is B.

7.

The sequence $S_1,$ $S_2,$ $S_3,$ $\ldots,$ $S_{10}$ has the property that every term beginning with the third is the sum of the previous two. That is, $S_n = S_{n-2} + S_{n-1} \text{ for } n \ge 3.$ Suppose that $S_9 = 110$ and $S_7 = 42.$ What is $S_4?$

$4$

$6$

$10$

$12$

$16$

Concepts:recursion work backwards

Difficulty rating: 1270

Small Hint:

Rearrange the rule to $S_{n-2} = S_n - S_{n-1}$ and step downward

Big Hint:

From $S_9 = S_7 + S_8$ find $S_8,$ then work down to $S_4$

Solution:

Since $S_9 = S_7 + S_8,$ we get $S_8 = 110 - 42 = 68.$ Then $S_6 = S_8 - S_7 = 68 - 42 = 26,$ $S_5 = S_7 - S_6 = 42 - 26 = 16,$ and $S_4 = S_6 - S_5 = 26 - 16 = 10.$

Thus, the correct answer is C.

8.

Given that $x$ and $y$ are distinct nonzero real numbers such that $x + \dfrac{2}{x} = y + \dfrac{2}{y},$ what is $xy?$

$\dfrac{1}{4}$

$\dfrac{1}{2}$

$1$

$2$

$4$

Concepts:algebraic manipulation factoring

Difficulty rating: 1400

Small Hint:

Multiply both sides by $xy$ to clear the fractions

Big Hint:

The result factors as $(x - y)(xy - 2) = 0$

Solution:

Multiplying by $xy$ gives $x^2 y + 2y = xy^2 + 2x,$ so $\begin{gathered} x^2 y - xy^2 - 2x + 2y \\ = (x - y)(xy - 2) \\ = 0. \end{gathered}$

Since $x \ne y,$ it follows that $xy = 2.$

Thus, the correct answer is D.

9.

In $\triangle ABC,$ $AB = AC = 28$ and $BC = 20.$ Points $D,$ $E,$ and $F$ are on sides $\overline{AB},$ $\overline{BC},$ and $\overline{AC},$ respectively, such that $\overline{DE}$ and $\overline{EF}$ are parallel to $\overline{AC}$ and $\overline{AB},$ respectively. What is the perimeter of parallelogram $ADEF?$

$48$

$52$

$56$

$60$

$72$

Difficulty rating: 1460

Small Hint:

Since $EF \parallel AB,$ triangle $FEC$ is similar to $ABC$ and hence isosceles with $FE = FC$

Big Hint:

Half the perimeter equals $AF + FE = AF + FC = AC$

Solution:

Because $EF \parallel AB,$ triangle $FEC$ is similar to triangle $ABC,$ which is isosceles, so $FE = FC.$

Half the perimeter of parallelogram $ADEF$ is $AF + FE$ $= AF + FC$ $= AC = 28.$ The entire perimeter is $56.$

Thus, the correct answer is C.

10.

Let $S$ be the set of positive integers $n$ for which $\dfrac{1}{n}$ has the repeating decimal representation $0.\overline{ab} = 0.ababab\ldots,$ with $a$ and $b$ different digits. What is the sum of the elements of $S?$

$11$

$44$

$110$

$143$

$155$

Concepts:repeating decimal factor

Difficulty rating: 1510

Small Hint:

A two-digit repeating block means $\dfrac{1}{n} = \dfrac{\overline{ab}}{99}$

Big Hint:

So $\dfrac{99}{n}$ is the two-digit number $\overline{ab};$ check the divisors of $99$

Solution:

If $\dfrac{1}{n} = 0.\overline{ab},$ then $\dfrac{99}{n} = \overline{ab},$ a two-digit number. The positive divisors of $99$ are $1, 3, 9, 11, 33, 99.$

Only $n = 11, 33, 99$ make $\dfrac{99}{n}$ equal to $09, 03, 01,$ which have two different digits. The requested sum is $11 + 33 + 99 = 143.$

Thus, the correct answer is D.

11.

Triangle $ABC$ is equilateral with $AB = 1.$ Points $E$ and $G$ are on $\overline{AC}$ and points $D$ and $F$ are on $\overline{AB}$ such that both $\overline{DE}$ and $\overline{FG}$ are parallel to $\overline{BC}.$ Furthermore, triangle $ADE$ and trapezoids $DFGE$ and $FBCG$ all have the same perimeter. What is $DE + FG?$

$1$

$\dfrac{3}{2}$

$\dfrac{21}{13}$

$\dfrac{13}{8}$

$\dfrac{5}{3}$

Concepts:equilateral triangle parallel lines system of equations

Difficulty rating: 1610

Small Hint:

Let $x = DE$ and $y = FG,$ and write each of the three perimeters in terms of $x$ and $y$

Big Hint:

The three equal perimeters give $3x = 3y - x = 3 - y$

Solution:

Let $x = DE$ and $y = FG.$ The parallel cuts make the small regions equilateral or isosceles trapezoids, so the perimeters are $\begin{gathered} \triangle ADE: 3x, \\ \quad DFGE: 3y - x, \\ \quad FBCG: 3 - y. \end{gathered}$

Setting them equal, $3x = 3y - x$ gives $4x = 3y,$ and $3x = 3 - y.$ Solving yields $x = \tfrac{9}{13}$ and $y = \tfrac{12}{13},$ so $DE + FG = \tfrac{21}{13}.$

Thus, the correct answer is C.

12.

The angles in a particular triangle are in arithmetic progression, and the side lengths are $4,$ $5,$ and $x.$ The sum of the possible values of $x$ equals $a + \sqrt{b} + \sqrt{c},$ where $a,$ $b,$ and $c$ are positive integers. What is $a + b + c?$

$36$

$38$

$40$

$42$

$44$

Concepts:arithmetic sequence law of cosines casework

Difficulty rating: 1740

Small Hint:

Angles in arithmetic progression summing to $180^\circ$ force the middle angle to be $60^\circ$

Big Hint:

Apply the Law of Cosines in each case for which side is opposite the $60^\circ$ angle

Solution:

If the angles are $\alpha - \delta, \alpha, \alpha + \delta,$ their sum $3\alpha = 180^\circ$ gives $\alpha = 60^\circ,$ so one angle is $60^\circ.$

If $x$ is opposite the $60^\circ$ angle, the Law of Cosines gives $\begin{gathered} x^2 = 4^2 + 5^2 - 2\cdot 4\cdot 5\cos 60^\circ \\ = 21, \end{gathered}$ so $x = \sqrt{21}.$

If $5$ is opposite the $60^\circ$ angle, then $25 = x^2 - 4x + 16,$ whose positive solution is $x = 2 + \sqrt{13}.$ If $4$ is opposite, then $16 = x^2 - 5x + 25$ has no real solution.

The sum of the possible values is $2 + \sqrt{13} + \sqrt{21},$ so $a + b + c = 2 + 13 + 21 = 36.$

Thus, the correct answer is A.

13.

Let points $A = (0, 0),$ $B = (1, 2),$ $C = (3, 3),$ and $D = (4, 0).$ Quadrilateral $ABCD$ is cut into equal area pieces by a line passing through $A.$ This line intersects $\overline{CD}$ at point $\left(\dfrac{p}{q}, \dfrac{r}{s}\right),$ where these fractions are in lowest terms. What is $p + q + r + s?$

$54$

$58$

$62$

$70$

$75$

Concepts:shoelace formula coordinate geometry triangle area

Difficulty rating: 1740

Small Hint:

Find the area of $ABCD;$ the cutting line makes a triangle on base $AD$ with half that area

Big Hint:

With base $AD = 4,$ the triangle’s height is the $y$ -coordinate of the point on $CD$

Solution:

By the shoelace formula, the area of $ABCD$ is $\tfrac{15}{2}.$ Let the line meet $\overline{CD}$ at $G.$ Triangle $ADG$ must have area $\tfrac{15}{4}.$

Since $AD = 4$ lies on the $x$ -axis, $\tfrac12\cdot 4\cdot y_G = \tfrac{15}{4}$ gives $y_G = \tfrac{15}{8}.$ Line $CD$ is $y = -3(x - 4),$ so $x_G = \tfrac{27}{8}.$

Then $p + q + r + s$ $= 27 + 8 + 15 + 8$ $= 58.$

Thus, the correct answer is B.

14.

The sequence $\begin{gathered} \log_{12} 162, \ \log_{12} x, \ \log_{12} y, \\ \ \log_{12} z, \ \log_{12} 1250 \end{gathered}$ is an arithmetic progression. What is $x?$

$125\sqrt{3}$

$270$

$162\sqrt{5}$

$434$

$225\sqrt{6}$

Concepts:logarithm geometric sequence arithmetic sequence

Difficulty rating: 1800

Small Hint:

Equal differences of logs mean $162, x, y, z, 1250$ is a geometric sequence

Big Hint:

Find the common ratio $r$ from $162r^4 = 1250$

Solution:

Because the logarithms are in arithmetic progression, $162, x, y, z, 1250$ is a geometric sequence. Its common ratio $r$ satisfies $162 r^4 = 1250,$ so $r^4 = \tfrac{625}{81}$ and $r = \tfrac53.$

Therefore $x = 162\cdot\tfrac53 = 270.$

Thus, the correct answer is B.

15.

Rabbits Peter and Pauline have three offspring—Flopsie, Mopsie, and Cottontail. These five rabbits are to be distributed to four different pet stores so that no store gets both a parent and a child. It is not required that every store gets a rabbit. In how many different ways can this be done?

$96$

$108$

$156$

$204$

$372$

Concepts:casework multiplication principle

Difficulty rating: 1880

Small Hint:

Split into cases by whether the two parents go to the same store

Big Hint:

Same store: $4$ parent choices and $3^3$ child assignments; different stores: $4\cdot 3$ and $2^3$

Solution:

If the two parents share a store, there are $4$ choices for it, and each child must go to one of the other three stores: $4\cdot 3^3 = 108$ ways.

If the parents go to different stores, there are $4\cdot 3 = 12$ choices, and each child must go to one of the two remaining stores: $12\cdot 2^3 = 96$ ways.

The total is $108 + 96 = 204.$

Thus, the correct answer is D.

16.

$A,$ $B,$ and $C$ are three piles of rocks. The mean weight of the rocks in $A$ is $40$ pounds, the mean weight of the rocks in $B$ is $50$ pounds, the mean weight of the rocks in the combined piles $A$ and $B$ is $43$ pounds, and the mean weight of the rocks in the combined piles $A$ and $C$ is $44$ pounds. What is the greatest possible integer value for the mean in pounds of the rocks in the combined piles $B$ and $C?$

$55$

$56$

$57$

$58$

$59$

Concepts:mean optimization

Difficulty rating: 1980

Small Hint:

Let $a, b, c$ be the counts; the $A,B$ mean $43$ gives $7b = 3a$

Big Hint:

Write the $B,C$ mean in terms of $k$ and $c$ using $a = 7k,\ b = 3k$ and the $A,C$ mean $44$

Solution:

Let $a, b, c$ be the numbers of rocks in the piles. From $\dfrac{40a + 50b}{a + b} = 43,$ we get $7b = 3a,$ so $a = 7k$ and $b = 3k.$

Let $\mu_{BC}$ be the mean of $B$ and $C.$ Using the $A, C$ mean $44$ to express $\mu_C = \dfrac{28k + 44c}{c},$ we find $\mu_{BC} = \dfrac{178k + 44c}{3k + c},$ so $(\mu_{BC} - 44)c = (178 - 3\mu_{BC})k.$

Since $B$ is heavier than $A,$ the mean of $B$ and $C$ exceeds $44,$ forcing $178 - 3\mu_{BC} \gt 0,$ i.e. $\mu_{BC} \lt \tfrac{178}{3} = 59\tfrac13.$ The value $59$ is attainable by taking $k=15c;$ then pile $C$ has mean $464$ and the displayed formula gives $\mu_{BC}=59.$ Thus the greatest integer mean is $59.$

Thus, the correct answer is E.

17.

A group of $12$ pirates agree to divide a treasure chest of gold coins among themselves as follows. The $k$ th pirate to take a share takes $\dfrac{k}{12}$ of the coins that remain in the chest. The number of coins initially in the chest is the smallest number for which this arrangement will allow each pirate to receive a positive whole number of coins. How many coins does the $12$ th pirate receive?

$720$

$1296$

$1728$

$1925$

$3850$

Concepts:divisibility prime factorization factorial

Difficulty rating: 2050

Small Hint:

Before the $k$ th pirate takes a share, the chest holds $\dfrac{12}{12 - k}$ times what remains after

Big Hint:

If $n$ coins remain for the last pirate, the initial count is $\dfrac{12^{11}n}{11!};$ make it an integer with smallest $n$

Solution:

For $1 \le k \le 11,$ the number of coins before the $k$ th pirate takes a share is $\dfrac{12}{12 - k}$ times the number afterward. So if $n$ coins are left for the $12$ th pirate, the initial count is $\dfrac{12^{11}\, n}{11!} = \dfrac{2^{14}\cdot 3^{7}\, n}{5^2\cdot 7\cdot 11}.$

The smallest $n$ making this a positive integer is $5^2\cdot7\cdot11=1925.$ Before pirate $k,$ the remaining count is the initial count multiplied by $\frac{11!}{(12-k)!\,12^{k-1}};$ substituting this $n$ shows it is an integer for every $k.$ Hence all shares are integral, and the $12$ th pirate receives $1925$ coins.

Thus, the correct answer is D.

18.

Six spheres of radius $1$ are positioned so that their centers are at the vertices of a regular hexagon of side length $2.$ The six spheres are internally tangent to a larger sphere whose center is the center of the hexagon. An eighth sphere is externally tangent to the six smaller spheres and internally tangent to the larger sphere. What is the radius of this eighth sphere?

$\sqrt{2}$

$\dfrac{3}{2}$

$\dfrac{5}{3}$

$\sqrt{3}$

$2$

Concepts:sphere 3D geometry Pythagorean Theorem

Difficulty rating: 2100

Small Hint:

The large sphere has radius $3;$ let the eighth sphere have radius $r$ and center at distance $x$ from the center, with $x + r = 3$

Big Hint:

A center of a small sphere, the center $O,$ and the eighth center form a right triangle: $(r + 1)^2 = 2^2 + x^2$

Solution:

Each small center is $2$ from the center $O,$ and the small spheres have radius $1,$ so the large sphere has radius $3.$ Let the eighth sphere have radius $r$ and center $G$ at distance $x$ from $O;$ then $x + r = 3.$

Since $G$ is equidistant from two opposite hexagon vertices, $GO$ is perpendicular to the line to a vertex, and the Pythagorean Theorem gives $\begin{gathered} (r + 1)^2 = 2^2 + x^2 \\ = 4 + (3 - r)^2. \end{gathered}$

This simplifies to $2r + 1 = 13 - 6r,$ so $r = \tfrac32.$

Thus, the correct answer is B.

19.

In $\triangle ABC,$ $AB = 86,$ and $AC = 97.$ A circle with center $A$ and radius $AB$ intersects $\overline{BC}$ at points $B$ and $X.$ Moreover $\overline{BX}$ and $\overline{CX}$ have integer lengths. What is $BC?$

$11$

$28$

$33$

$61$

$72$

Concepts:power of a point prime factorization

Difficulty rating: 2200

Small Hint:

Power of a Point from $C$ gives $BC\cdot CX = AC^2 - AB^2$

Big Hint:

$AC^2 - AB^2 = 2013 = 3\cdot 11\cdot 61;$ factor with $CX \lt BC \lt AB + AC = 183$

Solution:

By the Power of a Point Theorem, $BC\cdot CX = AC^2 - AB^2$ where $AB$ is the radius. Thus $BC\cdot CX = 97^2 - 86^2 = 2013.$

Since $BC = BX + CX$ and $CX$ are integers, they are complementary factors of $2013 = 3\cdot 11\cdot 61.$ As $CX \lt BC \lt AB + AC = 183,$ the only possibility is $CX = 33$ and $BC = 61.$

Thus, the correct answer is D.

20.

Let $S$ be the set $\{1, 2, 3, \ldots, 19\}.$ For $a,$ $b \in S,$ define $a \succ b$ to mean that either $0 \lt a - b \le 9$ or $b - a \gt 9.$ How many ordered triples $(x, y, z)$ of elements of $S$ have the property that $x \succ y,$ $y \succ z,$ and $z \succ x?$

$810$

$855$

$900$

$950$

$988$

Concepts:modular arithmetic basic counting

Difficulty rating: 2220

Small Hint:

Work modulo $19;$ then $a \succ b$ means $0 \lt (a - b) \bmod 19 \le 9$

Big Hint:

Fix $x$ ( $19$ ways); let $y = x + i$ with $1 \le i \le 9,$ then $z$ has $i$ options

Solution:

Reading the elements modulo $19,$ the relation $a \succ b$ holds exactly when $0 \lt (a - b) \bmod 19 \le 9.$

There are $19$ choices for $x.$ Once $x$ is fixed, take $y = x + i$ for some $1 \le i \le 9.$ Then $z$ must satisfy $x + 10 \le z \le x + 9 + i,$ giving $i$ choices.

The total is $19(1 + 2 + \cdots + 9) = 19\cdot 45$ $= 855.$

Thus, the correct answer is B.

21.

Consider $\begin{gathered} A = \\ \tiny \log(2013 + \log(2012 + \log(2011 + \log(\cdots + \log(3 + \log 2)\cdots)))). \end{gathered}$

Which of the following intervals contains $A?$

$(\log 2016, \log 2017)$

$(\log 2017, \log 2018)$

$(\log 2018, \log 2019)$

$(\log 2019, \log 2020)$

$(\log 2020, \log 2021)$

Concepts:logarithm bounding to limit cases induction

Difficulty rating: 2210

Small Hint:

Track where the nested value sits: $0 \lt A_n \lt 1$ for small $n,$ then $1 \lt A_n \lt 2,$ etc.

Big Hint:

For $n = 2012$ the value satisfies $3 \lt A_{2012} \lt 4,$ so estimate $2013 + A_{2012}$

Solution:

Let $\tiny A_n = \log(n + \log((n-1) + \cdots + \log(3 + \log 2)\cdots)).$ One checks $0 \lt A_n \lt 1$ for $2 \le n \le 9,$ then $1 \lt A_n \lt 2$ for $10 \le n \le 98,$ then $2 \lt A_n \lt 3$ for $99 \le n \le 997,$ and $3 \lt A_n \lt 4$ for $998 \le n \le 9996.$

Hence $3 \lt A_{2012} \lt 4,$ so $2016 \lt 2013 + A_{2012} \lt 2017$ and therefore $\log 2016 \lt A \lt \log 2017.$

Thus, the correct answer is A.

22.

A palindrome is a nonnegative integer number that reads the same forwards and backwards when written in base $10$ with no leading zeros. A $6$ -digit palindrome $n$ is chosen uniformly at random. What is the probability that $\dfrac{n}{11}$ is also a palindrome?

$\dfrac{8}{25}$

$\dfrac{33}{100}$

$\dfrac{7}{20}$

$\dfrac{9}{25}$

$\dfrac{11}{30}$

Concepts:palindrome basic probability digits

Difficulty rating: 2440

Small Hint:

Show $m = \dfrac{n}{11}$ must be a $5$ -digit palindrome $\overline{abcba}$

Big Hint:

The product $11m$ is a palindrome exactly when $a + b \le 9$ and $b + c \le 9;$ count those

Solution:

Let $m= \frac{n}{11}.$ If $m$ had four digits, then $n<110000,$ so the first and last digits of the six-digit palindrome $n$ would both be $1.$ This forces the first and last digits of the palindromic $m$ to be $1,$ hence $m<2000$ and $n<22000,$ a contradiction. Therefore $m$ is a five-digit palindrome $\overline{abcba}.$

Writing $n=11m=\overline{abcba0}+\overline{abcba},$ no carries occur exactly when $a+b\le9$ and $b+c\le9;$ the resulting digits are $a,a+b,b+c,b+c,a+b,a.$ If $a+b\ge10,$ the leading and trailing digits differ; if only $b+c\ge10,$ the next pair differs. Thus the conditions are also necessary. The number of valid $m$ is $\sum_{b=0}^{9}(10 - b)(9 - b) = 330.$

There are $9\cdot 10^2 = 900$ six-digit palindromes, so the probability is $\dfrac{330}{900} = \dfrac{11}{30}.$

Thus, the correct answer is E.

23.

$ABCD$ is a square of side length $\sqrt{3} + 1.$ Point $P$ is on $\overline{AC}$ such that $AP = \sqrt{2}.$ The square region bounded by $ABCD$ is rotated $90^\circ$ counterclockwise with center $P,$ sweeping out a region whose area is $\dfrac{1}{c}(a\pi + b),$ where $a,$ $b,$ and $c$ are positive integers and $\gcd(a, b, c) = 1.$ What is $a + b + c?$

$15$

$17$

$19$

$21$

$23$

Concepts:transformation sector area decomposition

Difficulty rating: 2520

Small Hint:

Track the images $A', B', C', D';$ the swept region is four circular sectors plus four triangles

Big Hint:

$AP = \sqrt{2}$ and $PC = \sqrt{6}$ give the sector radii, and $\triangle BPH$ is a $30$ – $60$ – $90$ triangle

Solution:

Let $A', B', C', D'$ be the images of the vertices under the rotation. The swept region decomposes into four circular sectors and four triangles.

Since $AP = \sqrt{2}$ and $PC = AC - AP = \sqrt{6},$ the sectors at $A$ and $C$ have areas $\tfrac{\pi}{2}$ and $\tfrac{3\pi}{2}.$ If $H$ is the midpoint of $AA',$ then $PH=AH=1$ and $HB=\sqrt3,$ so $\triangle BPH$ is a $30$ - $60$ - $90^\circ$ triangle and $PB=2.$ Hence the two $60^\circ$ sectors along $BC$ each have area $\frac{2\pi}{3}.$ The two triangles with altitude $PH$ contribute $\sqrt3-1,$ and the other congruent pair has altitude $\sqrt3$ and contributes $3-\sqrt3.$ Thus the four triangles contribute $2.$

The total area is $\begin{gathered} \dfrac{\pi}{2} + \dfrac{3\pi}{2} + 2\cdot\dfrac{2\pi}{3} + 2 \\ = \dfrac{10\pi + 6}{3}, \end{gathered}$ so $a + b + c = 10 + 6 + 3 = 19.$

Thus, the correct answer is C.

24.

Three distinct segments are chosen at random among the segments whose endpoints are the vertices of a regular $12$ -gon. What is the probability that the lengths of these three segments are the three side lengths of a triangle with positive area?

$\dfrac{553}{715}$

$\dfrac{443}{572}$

$\dfrac{111}{143}$

$\dfrac{81}{104}$

$\dfrac{223}{286}$

Concepts:regular polygon triangle inequality complementary counting

Difficulty rating: 2650

Small Hint:

There are $6$ possible lengths $d_k = 2\sin(15k^\circ);$ count how many segments have each length

Big Hint:

Use complementary counting: subtract the triples whose longest length is at least the sum of the other two

Solution:

Inscribe the $12$ -gon in a unit circle. The segment lengths are $d_k = 2\sin(15k^\circ)$ for $1 \le k \le 6,$ with $12$ segments of each length $d_1, \ldots, d_5$ and $6$ of length $d_6.$

Comparing sums, the forbidden index triples $(a, b, c)$ with $d_a \le d_b \le d_c$ and $d_c \ge d_a + d_b$ are $\begin{gathered} (1,1,3),(1,1,4),(1,1,5), \\ (1,1,6),(1,2,4),(1,2,5), \\ (1,2,6),(1,3,5),(1,3,6), \\ (2,2,6). \end{gathered}$

The first three triples ending in $3,4,5$ contribute $3\binom{12}{2}12;$ the two repeated-length triples ending in $6$ contribute $2\binom{12}{2}6;$ the three triples of distinct non-diameter lengths contribute $3\cdot12^3;$ and the two remaining diameter triples contribute $2\cdot12^2\cdot6.$ Dividing their sum by $\binom{66}{3}$ gives failure probability $\frac{63}{286},$ so the answer is $1-\frac{63}{286}=\frac{223}{286}.$

Thus, the correct answer is E.

25.

Let $f : \mathbb{C} \to \mathbb{C}$ be defined by $f(z) = z^2 + iz + 1.$ How many complex numbers $z$ are there such that $\operatorname{Im}(z) \gt 0$ and both the real and the imaginary parts of $f(z)$ are integers with absolute value at most $10?$

$399$

$401$

$413$

$431$

$441$

Concepts:complex number lattice point

Difficulty rating: 2790

Small Hint:

Show $f$ is one-to-one on the upper half-plane, so counting $z$ equals counting valid $f(z)$

Big Hint:

The image is $\{w : \operatorname{Re}(w) \lt (\operatorname{Im}(w))^2 + 1\};$ count lattice points with $|\operatorname{Re}|, |\operatorname{Im}| \le 10$

Solution:

On the upper half-plane $H,$ if $f(z_1) = f(z_2)$ then $(z_1 - z_2)(z_1 + z_2 + i) = 0;$ since $\operatorname{Im}(z_1), \operatorname{Im}(z_2) \gt 0,$ the factor $z_1 + z_2 + i \ne 0,$ so $f$ is one-to-one on $H.$

For real $r,$ the boundary values $f(r)=r^2+1+ir$ trace the parabola $\operatorname{Re}(w)=(\operatorname{Im}(w))^2+1.$ Since $f(i)=-1$ lies to its left and $f$ is continuous and one-to-one on $H,$ its image consists precisely of the values $w$ satisfying $\operatorname{Re}(w)<(\operatorname{Im}(w))^2+1.$ Thus we count $w = a + ib$ with $a, b \in \mathbb{Z},$ $|a|, |b| \le 10,$ and $a \lt b^2 + 1:$ $\begin{gathered} |S| = 21^2 \\ {}- \sum_{b=-3}^{3}(10 - b^2) \\ = 441 - 42 = 399. \end{gathered}$

Thus, the correct answer is A.