2013 AMC 12A Exam Problems

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1.

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

$4$

$5$

$6$

$7$

$8$

Answer: E

Concepts:triangle area right triangle

Difficulty rating: 840

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?

$35$

$40$

$45$

$50$

$55$

Answer: C

Concepts:parity casework

Difficulty rating: 1010

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?

$15$

$30$

$40$

$60$

$70$

Answer: E

Concepts:fraction percentage

Difficulty rating: 1100

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 $\dfrac{2^{2014} + 2^{2012}}{2^{2014} - 2^{2012}}?$

$-1$

$1$

$\dfrac{5}{3}$

$2013$

$2^{4024}$

Answer: C

Concepts:exponent factoring

Difficulty rating: 1130

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 $t - d?$

$15$

$20$

$25$

$30$

$35$

Answer: B

Concepts:mean algebraic manipulation

Difficulty rating: 1130

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?

$12$

$18$

$24$

$30$

$36$

Answer: B

Concepts:percentage invariant

Difficulty rating: 1250

Solution:

If Shenille attempted $x$ three-point shots and $30 - x$ two-point shots, she scored $0.2\cdot 3\cdot x + 0.3\cdot 2\cdot(30 - x) = 0.6x + 0.6(30 - x) = 0.6\cdot 30 = 18$ 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 $S_4?$

$4$

$6$

$10$

$12$

$16$

Answer: C

Concepts:recursion work backwards

Difficulty rating: 1270

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 $xy?$

$\dfrac{1}{4}$

$\dfrac{1}{2}$

$1$

$2$

$4$

Answer: D

Concepts:algebraic manipulation factoring

Difficulty rating: 1400

Solution:

Multiplying by $xy$ gives $x^2 y + 2y = xy^2 + 2x,$ so $x^2 y - xy^2 - 2x + 2y = (x - y)(xy - 2) = 0.$

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 $ADEF?$

$48$

$52$

$56$

$60$

$72$

Answer: C

Difficulty rating: 1460

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 $S?$

$11$

$44$

$110$

$143$

$155$

Answer: D

Concepts:repeating decimal factor

Difficulty rating: 1510

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 $DE + FG?$

$1$

$\dfrac{3}{2}$

$\dfrac{21}{13}$

$\dfrac{13}{8}$

$\dfrac{5}{3}$

Answer: C

Concepts:equilateral triangle parallel lines system of equations

Difficulty rating: 1610

Solution:

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

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 $a + b + c?$

$36$

$38$

$40$

$42$

$44$

Answer: A

Concepts:arithmetic sequence law of cosines casework

Difficulty rating: 1740

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 $x^2 = 4^2 + 5^2 - 2\cdot 4\cdot 5\cos 60^\circ = 21,$ 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 $p + q + r + s?$

$54$

$58$

$62$

$70$

$75$

Answer: B

Concepts:shoelace formula coordinate geometry triangle area

Difficulty rating: 1740

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 $\log_{12} 162, \ \log_{12} x, \ \log_{12} y, \ \log_{12} z, \ \log_{12} 1250$ is an arithmetic progression. What is $x?$

$125\sqrt{3}$

$270$

$162\sqrt{5}$

$434$

$225\sqrt{6}$

Answer: B

Concepts:logarithm geometric sequence arithmetic sequence

Difficulty rating: 1800

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?

$96$

$108$

$156$

$204$

$372$

Answer: D

Concepts:casework multiplication principle

Difficulty rating: 1880

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 $C?$

$55$

$56$

$57$

$58$

$59$

Answer: E

Concepts:mean optimization

Difficulty rating: 1980

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, so 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?

$720$

$1296$

$1728$

$1925$

$3850$

Answer: D

Concepts:divisibility prime factorization factorial

Difficulty rating: 2050

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 $5^2\cdot 7\cdot 11 = 1925,$ and one checks each earlier pirate then receives a whole number of coins. 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?

$\sqrt{2}$

$\dfrac{3}{2}$

$\dfrac{5}{3}$

$\sqrt{3}$

$2$

Answer: B

Concepts:sphere 3D geometry Pythagorean Theorem

Difficulty rating: 2100

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 $(r + 1)^2 = 2^2 + x^2 = 4 + (3 - r)^2.$

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 $BC?$

$11$

$28$

$33$

$61$

$72$

Answer: D

Concepts:power of a point prime factorization

Difficulty rating: 2200

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 $z \succ x?$

$810$

$855$

$900$

$950$

$988$

Answer: B

Concepts:modular arithmetic basic counting

Difficulty rating: 2220

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 $A = \log(2013 + \log(2012 + \log(2011 + \log(\cdots + \log(3 + \log 2)\cdots)))).$

Which of the following intervals contains $A?$

$(\log 2016, \log 2017)$

$(\log 2017, \log 2018)$

$(\log 2018, \log 2019)$

$(\log 2019, \log 2020)$

$(\log 2020, \log 2021)$

Answer: A

Concepts:logarithm bounding to limit cases induction

Difficulty rating: 2210

Solution:

Let $A_n = \log(n + \log((n-1) + \cdots + \log(3 + \log 2)\cdots)).$ 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?

$\dfrac{8}{25}$

$\dfrac{33}{100}$

$\dfrac{7}{20}$

$\dfrac{9}{25}$

$\dfrac{11}{30}$

Answer: E

Concepts:palindrome basic probability digits

Difficulty rating: 2440

Solution:

Let $m = \dfrac{n}{11}.$ A $4$ -digit $m$ leads to a contradiction, so $m$ is a $5$ -digit palindrome $\overline{abcba}.$

Writing $n = 11m,$ there are no carries exactly when $a + b \le 9$ and $b + c \le 9,$ and only then is $n$ a palindrome. 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 $a + b + c?$

$15$

$17$

$19$

$21$

$23$

Answer: C

Concepts:transformation sector area decomposition

Difficulty rating: 2520

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 $\tfrac{3\pi}{2}.$ One finds $PB = 2,$ so the two $60^\circ$ sectors along $BC$ each have area $\tfrac{2\pi}{3}.$ The four triangles together contribute $(\sqrt{3} - 1) + (3 - \sqrt{3}) = 2.$

The total area is $\dfrac{\pi}{2} + \dfrac{3\pi}{2} + 2\cdot\dfrac{2\pi}{3} + 2 = \dfrac{10\pi + 6}{3},$ 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?

$\dfrac{553}{715}$

$\dfrac{443}{572}$

$\dfrac{111}{143}$

$\dfrac{81}{104}$

$\dfrac{223}{286}$

Answer: E

Concepts:regular polygon triangle inequality complementary counting

Difficulty rating: 2650

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 $(1,1,3),(1,1,4),(1,1,5),(1,1,6),(1,2,4),(1,2,5),(1,2,6),(1,3,5),(1,3,6),(2,2,6).$

Counting the corresponding segment selections and dividing by $\binom{66}{3}$ gives a failure probability of $\dfrac{63}{286},$ so the answer is $1 - \dfrac{63}{286} = \dfrac{223}{286}.$

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 $10?$

$399$

$401$

$413$

$431$

$441$

Answer: A

Concepts:complex number lattice point

Difficulty rating: 2790

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 $H.$

The image is $f(H) = \{w : \operatorname{Re}(w) \lt (\operatorname{Im}(w))^2 + 1\}.$ Thus we count $w = a + ib$ with $a, b \in \mathbb{Z},$ $|a|, |b| \le 10,$ and $a \lt b^2 + 1:$ $|S| = 21^2 - \sum_{b=-3}^{3}(10 - b^2) = 441 - 42 = 399.$

Thus, the correct answer is A.