Difference between revisions of "2012 AMC 12A Problems"

Latest revision as of 12:34, 12 August 2020

2012 AMC 12A (Answer Key) Printable versions: Wiki • AoPS Resources • PDF
Instructions This is a 25-question, multiple choice test. Each question is followed by answers marked A, B, C, D and E. Only one of these is correct. You will receive 6 points for each correct answer, 2.5 points for each problem left unanswered if the year is before 2006, 1.5 points for each problem left unanswered if the year is after 2006, and 0 points for each incorrect answer. No aids are permitted other than scratch paper, graph paper, ruler, compass, protractor and erasers (and calculators that are accepted for use on the test if before 2006. No problems on the test will require the use of a calculator). Figures are not necessarily drawn to scale. You will have 75 minutes working time to complete the test.
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25

Problem 1

A bug crawls along a number line, starting at $-2$ . It crawls to $-6$ , then turns around and crawls to $5$ . How many units does the bug crawl altogether?

$\textbf{(A)}\ 9\qquad\textbf{(B)}\ 11\qquad\textbf{(C)}\ 13\qquad\textbf{(D)}\ 14\qquad\textbf{(E)}\ 15$

Solution

Problem 2

Cagney can frost a cupcake every $20$ seconds and Lacey can frost a cupcake every $30$ seconds. Working together, how many cupcakes can they frost in $5$ minutes?

$\textbf{(A)}\ 10\qquad\textbf{(B)}\ 15\qquad\textbf{(C)}\ 20\qquad\textbf{(D)}\ 25\qquad\textbf{(E)}\ 30$

Solution

Problem 3

A box $2$ centimeters high, $3$ centimeters wide, and $5$ centimeters long can hold $40$ grams of clay. A second box with twice the height, three times the width, and the same length as the first box can hold $n$ grams of clay. What is $n$ ?

$\textbf{(A)}\ 120\qquad\textbf{(B)}\ 160\qquad\textbf{(C)}\ 200\qquad\textbf{(D)}\ 240\qquad\textbf{(E)}\ 280$

Solution

Problem 4

In a bag of marbles, $\tfrac{3}{5}$ of the marbles are blue and the rest are red. If the number of red marbles is doubled and the number of blue marbles stays the same, what fraction of the marbles will be red?

$\textbf{(A)}\ \dfrac{2}{5} \qquad\textbf{(B)}\ \dfrac{3}{7} \qquad\textbf{(C)}\ \dfrac{4}{7} \qquad\textbf{(D)}\ \dfrac{3}{5} \qquad\textbf{(E)}\ \dfrac{4}{5}$

Solution

Problem 5

A fruit salad consists of blueberries, raspberries, grapes, and cherries. The fruit salad has a total of $280$ pieces of fruit. There are twice as many raspberries as blueberries, three times as many grapes as cherries, and four times as many cherries as raspberries. How many cherries are there in the fruit salad?

$\textbf{(A)}\ 8\qquad\textbf{(B)}\ 16\qquad\textbf{(C)}\ 25\qquad\textbf{(D)}\ 64\qquad\textbf{(E)}\ 96$

Solution

Problem 6

The sums of three whole numbers taken in pairs are $12$ , $17$ , and $19$ . What is the middle number?

$\textbf{(A)}\ 4\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}\ 7\qquad\textbf{(E)}\ 8$

Solution

Problem 7

Mary divides a circle into $12$ sectors. The central angles of these sectors, measured in degrees, are all integers and they form an arithmetic sequence. What is the degree measure of the smallest possible sector angle?

$\textbf{(A)}\ 5\qquad\textbf{(B)}\ 6\qquad\textbf{(C)}\ 8\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 12$

Solution

Problem 8

An iterative average of the numbers $1$ , $2$ , $3$ , $4$ , and $5$ is computed in the following way. Arrange the five numbers in some order. Find the mean of the first two numbers, then find the mean of that with the third number, then the mean of that with the fourth number, and finally the mean of that with the fifth number. What is the difference between the largest and smallest possible values that can be obtained using this procedure?

$\textbf{(A)}\ \frac{31}{16}\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ \frac{17}{8}\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ \frac{65}{16}$

Solution

Problem 9

A year is a leap year if and only if the year number is divisible by $400$ (such as $2000$ ) or is divisible by $4$ but not by $100$ (such as $2012$ ). The $200\text{th}$ anniversary of the birth of novelist Charles Dickens was celebrated on February $7$ , $2012$ , a Tuesday. On what day of the week was Dickens born?

$\textbf{(A)}\ \text{Friday} \qquad\textbf{(B)}\ \text{Saturday} \qquad\textbf{(C)}\ \text{Sunday} \qquad\textbf{(D)}\ \text{Monday} \qquad\textbf{(E)}\ \text{Tuesday}$

Solution

Problem 10

A triangle has area $30$ , one side of length $10$ , and the median to that side of length $9$ . Let $\theta$ be the acute angle formed by that side and the median. What is $\sin{\theta}$ ?

$\textbf{(A)}\ \frac{3}{10}\qquad\textbf{(B)}\ \frac{1}{3}\qquad\textbf{(C)}\ \frac{9}{20}\qquad\textbf{(D)}\ \frac{2}{3}\qquad\textbf{(E)}\ \frac{9}{10}$

Solution

Problem 11

Alex, Mel, and Chelsea play a game that has $6$ rounds. In each round there is a single winner, and the outcomes of the rounds are independent. For each round the probability that Alex wins is $\frac{1}{2}$ , and Mel is twice as likely to win as Chelsea. What is the probability that Alex wins three rounds, Mel wins two rounds, and Chelsea wins one round?

$\textbf{(A)}\ \frac{5}{72}\qquad\textbf{(B)}\ \frac{5}{36}\qquad\textbf{(C)}\ \frac{1}{6}\qquad\textbf{(D)}\ \frac{1}{3}\qquad\textbf{(E)}\ 1$

Solution

Problem 12

A square region $ABCD$ is externally tangent to the circle with equation $x^2+y^2=1$ at the point $(0,1)$ on the side $CD$ . Vertices $A$ and $B$ are on the circle with equation $x^2+y^2=4$ . What is the side length of this square?

$\textbf{(A)}\ \frac{\sqrt{10}+5}{10}\qquad\textbf{(B)}\ \frac{2\sqrt{5}}{5}\qquad\textbf{(C)}\ \frac{2\sqrt{2}}{3}\qquad\textbf{(D)}\ \frac{2\sqrt{19}-4}{5}\qquad\textbf{(E)}\ \frac{9-\sqrt{17}}{5}$

Solution

Problem 13

Paula the painter and her two helpers each paint at constant, but different, rates. They always start at $\text{8:00 AM}$ , and all three always take the same amount of time to eat lunch. On Monday the three of them painted $50\%$ of a house, quitting at $\text{4:00 PM}$ . On Tuesday, when Paula wasn't there, the two helpers painted only $24\%$ of the house and quit at $\text{2:12 PM}$ . On Wednesday Paula worked by herself and finished the house by working until $\text{7:12 PM}$ . How long, in minutes, was each day's lunch break?

$\textbf{(A)}\ 30 \qquad\textbf{(B)}\ 36 \qquad\textbf{(C)}\ 42 \qquad\textbf{(D)}\ 48 \qquad\textbf{(E)}\ 60$

Solution

Problem 14

The closed curve in the figure is made up of $9$ congruent circular arcs each of length $\frac{2\pi}{3}$ , where each of the centers of the corresponding circles is among the vertices of a regular hexagon of side $2$ . What is the area enclosed by the curve?

$[asy] size(6cm); defaultpen(fontsize(6pt)); dotfactor=4; label("$\circ$",(0,1)); label("$\circ$",(0.865,0.5)); label("$\circ$",(-0.865,0.5)); label("$\circ$",(0.865,-0.5)); label("$\circ$",(-0.865,-0.5)); label("$\circ$",(0,-1)); dot((0,1.5)); dot((-0.4325,0.75)); dot((0.4325,0.75)); dot((-0.4325,-0.75)); dot((0.4325,-0.75)); dot((-0.865,0)); dot((0.865,0)); dot((-1.2975,-0.75)); dot((1.2975,-0.75)); draw(Arc((0,1),0.5,210,-30)); draw(Arc((0.865,0.5),0.5,150,270)); draw(Arc((0.865,-0.5),0.5,90,-150)); draw(Arc((0.865,-0.5),0.5,90,-150)); draw(Arc((0,-1),0.5,30,150)); draw(Arc((-0.865,-0.5),0.5,330,90)); draw(Arc((-0.865,0.5),0.5,-90,30)); [/asy]$

$\textbf{(A)}\ 2\pi+6\qquad\textbf{(B)}\ 2\pi+4\sqrt3 \qquad\textbf{(C)}\ 3\pi+4 \qquad\textbf{(D)}\ 2\pi+3\sqrt3+2 \qquad\textbf{(E)}\ \pi+6\sqrt3$

Solution

Problem 15

A $3\times3$ square is partitioned into $9$ unit squares. Each unit square is painted either white or black with each color being equally likely, chosen independently and at random. The square is then rotated $90^\circ$ clockwise about its center, and every white square in a position formerly occupied by a black square is painted black. The colors of all other squares are left unchanged. What is the probability that the grid is now entirely black?

$\textbf{(A)}\ \dfrac{49}{512} \qquad\textbf{(B)}\ \dfrac{7}{64} \qquad\textbf{(C)}\ \dfrac{121}{1024} \qquad\textbf{(D)}\ \dfrac{81}{512} \qquad\textbf{(E)}\ \dfrac{9}{32}$

Solution

Problem 16

Circle $C_1$ has its center $O$ lying on circle $C_2$ . The two circles meet at $X$ and $Y$ . Point $Z$ in the exterior of $C_1$ lies on circle $C_2$ and $XZ=13$ , $OZ=11$ , and $YZ=7$ . What is the radius of circle $C_1$ ?

$\textbf{(A)}\ 5\qquad\textbf{(B)}\ \sqrt{26}\qquad\textbf{(C)}\ 3\sqrt{3}\qquad\textbf{(D)}\ 2\sqrt{7}\qquad\textbf{(E)}\ \sqrt{30}$

Solution

Problem 17

Let $S$ be a subset of $\{1,2,3,\dots,30\}$ with the property that no pair of distinct elements in $S$ has a sum divisible by $5$ . What is the largest possible size of $S$ ?

$\textbf{(A)}\ 10\qquad\textbf{(B)}\ 13\qquad\textbf{(C)}\ 15\qquad\textbf{(D)}\ 16\qquad\textbf{(E)}\ 18$

Solution

Problem 18

Triangle $ABC$ has $AB=27$ , $BC=25$ , and $CA=26$ . Let $I$ denote the intersection of the internal angle bisectors of $\triangle ABC$ . What is $BI$ ?

$\textbf{(A)}\ 15\qquad\textbf{(B)}\ 5+\sqrt{26}+3\sqrt{3}\qquad\textbf{(C)}\ 3\sqrt{26}\qquad\textbf{(D)}\ \frac{2}{3}\sqrt{546}\qquad\textbf{(E)}\ 9\sqrt{3}$

Solution

Problem 19

Adam, Benin, Chiang, Deshawn, Esther, and Fiona have internet accounts. Some, but not all, of them are internet friends with each other, and none of them has an internet friend outside this group. Each of them has the same number of internet friends. In how many different ways can this happen?

$\textbf{(A)}\ 60 \qquad\textbf{(B)}\ 170 \qquad\textbf{(C)}\ 290 \qquad\textbf{(D)}\ 320 \qquad\textbf{(E)}\ 660$

Solution

Problem 20

Consider the polynomial

$P(x)=\prod_{k=0}^{10}(x^{2^k}+2^k)=(x+1)(x^2+2)(x^4+4)\cdots (x^{1024}+1024)$

The coefficient of $x^{2012}$ is equal to $2^a$ . What is $a$ ?

$\textbf{(A)}\ 5 \qquad\textbf{(B)}\ 6 \qquad\textbf{(C)}\ 7 \qquad\textbf{(D)}\ 10 \qquad\textbf{(E)}\ 24$

Solution

Problem 21

Let $a$ , $b$ , and $c$ be positive integers with $a\ge$ $b\ge$ $c$ such that $\begin{align*}a^{2}-b^{2}-c^{2}+ab&=2011\text{ and}\\ a^{2}+3b^{2}+3c^{2}-3ab-2ac-2bc&=-1997 \end{align*}$ What is $a$ ?

$\textbf{(A)}\ 249\qquad\textbf{(B)}\ 250\qquad\textbf{(C)}\ 251\qquad\textbf{(D)}\ 252\qquad\textbf{(E)}\ 253$

Solution

Problem 22

Distinct planes $p_1,p_2,....,p_k$ intersect the interior of a cube $Q$ . Let $S$ be the union of the faces of $Q$ and let $P =\bigcup_{j=1}^{k}p_{j}$ . The intersection of $P$ and $S$ consists of the union of all segments joining the midpoints of every pair of edges belonging to the same face of $Q$ . What is the difference between the maximum and minimum possible values of $k$ ?

$\textbf{(A)}\ 8\qquad\textbf{(B)}\ 12\qquad\textbf{(C)}\ 20\qquad\textbf{(D)}\ 23\qquad\textbf{(E)}\ 24$

Solution

Problem 23

Let $S$ be the square one of whose diagonals has endpoints $(0.1,0.7)$ and $(-0.1,-0.7)$ . A point $v=(x,y)$ is chosen uniformly at random over all pairs of real numbers $x$ and $y$ such that $0 \le x \le 2012$ and $0\le y\le 2012$ . Let $T(v)$ be a translated copy of $S$ centered at $v$ . What is the probability that the square region determined by $T(v)$ contains exactly two points with integer coordinates in its interior?

$\textbf{(A)}\ \frac{1}{8}\qquad\textbf{(B) }\frac{7}{50}\qquad\textbf{(C) }\frac{4}{25}\qquad\textbf{(D) }\frac{1}{4}\qquad\textbf{(E) }\frac{8}{25}$

Solution

Problem 24

Let $\{a_k\}_{k=1}^{2011}$ be the sequence of real numbers defined by $a_1=0.201,$ $a_2=(0.2011)^{a_1},$ $a_3=(0.20101)^{a_2},$ $a_4=(0.201011)^{a_3}$ , and in general,

$a_k=\begin{cases} (0.\underbrace{20101\cdots 0101}_{k+2\text{ digits}})^{a_{k-1}} & \text{if }k\text{ is odd,}\\ (0.\underbrace{20101\cdots 01011}_{k+2\text{ digits}})^{a_{k-1}}& \text{if }k\text{ is even.} \end{cases}$

Rearranging the numbers in the sequence $\{a_k\}_{k=1}^{2011}$ in decreasing order produces a new sequence $\{b_k\}_{k=1}^{2011}$ . What is the sum of all integers $k$ , $1\le k \le 2011$ , such that $a_k=b_k?$

$\textbf{(A)}\ 671\qquad\textbf{(B)}\ 1006\qquad\textbf{(C)}\ 1341\qquad\textbf{(D)}\ 2011\qquad\textbf{(E)}\ 2012$

Solution

Problem 25

Let $f(x)=|2\{x\}-1|$ where $\{x\}$ denotes the fractional part of $x$ . The number $n$ is the smallest positive integer such that the equation $nf(xf(x))=x$ has at least $2012$ real solutions. What is $n$ ? Note: the fractional part of $x$ is a real number $y=\{x\}$ such that $0\le y<1$ and $x-y$ is an integer.

$\textbf{(A)}\ 30\qquad\textbf{(B)}\ 31\qquad\textbf{(C)}\ 32\qquad\textbf{(D)}\ 62\qquad\textbf{(E)}\ 64$

Solution

2012 AMC 12A (Problems • Answer Key • Resources)
Preceded by 2011 AMC 12B Problems	Followed by 2012 AMC 12B Problems
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All AMC 12 Problems and Solutions

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Difference between revisions of "2012 AMC 12A Problems"

Latest revision as of 12:34, 12 August 2020

Contents

Problem 1

Problem 2

Problem 3

Problem 4

Problem 5

Problem 6

Problem 7

Problem 8

Problem 9

Problem 10

Problem 11

Problem 12

Problem 13

Problem 14

Problem 15

Problem 16

Problem 17

Problem 18

Problem 19

Problem 20

Problem 21

Problem 22

Problem 23

Problem 24

Problem 25

See also

@@ Line 1: / Line 1: @@
+{{AMC12 Problems|year=2012|ab=A}}
 == Problem 1 ==
-A bug crawls along a number line, starting at -2. It crawls to -6, then turns around and crawls to 5. How many units does the bug crawl altogether?
+A bug crawls along a number line, starting at <math>-2</math>. It crawls to <math>-6</math>, then turns around and crawls to <math>5</math>. How many units does the bug crawl altogether?
 <math> \textbf{(A)}\ 9\qquad\textbf{(B)}\ 11\qquad\textbf{(C)}\ 13\qquad\textbf{(D)}\ 14\qquad\textbf{(E)}\ 15 </math>
-[[2012 AMC 10A Problems/Problem 3|Solution]]
+[[2012 AMC 12A Problems/Problem 1|Solution]]
 == Problem 2 ==
-Cagney can frost a cupcake every 20 seconds and Lacey can frost a cupcake every 30 seconds. Working together, how many cupcakes can they frost in 5 minutes?
+Cagney can frost a cupcake every <math>20</math> seconds and Lacey can frost a cupcake every <math>30</math> seconds. Working together, how many cupcakes can they frost in <math>5</math> minutes?
 <math> \textbf{(A)}\ 10\qquad\textbf{(B)}\ 15\qquad\textbf{(C)}\ 20\qquad\textbf{(D)}\ 25\qquad\textbf{(E)}\ 30 </math>
-[[2012 AMC 10A Problems/Problem 1|Solution]]
+[[2012 AMC 12A Problems/Problem 2|Solution]]
 == Problem 3 ==
@@ Line 19: / Line 21: @@
 <math>\textbf{(A)}\ 120\qquad\textbf{(B)}\ 160\qquad\textbf{(C)}\ 200\qquad\textbf{(D)}\ 240\qquad\textbf{(E)}\ 280</math>
+[[2012 AMC 12A Problems/Problem 3|Solution]]
 == Problem 4 ==
@@ Line 30: / Line 34: @@
 \qquad\textbf{(E)}\ \dfrac{4}{5}
   </math>
+[[2012 AMC 12A Problems/Problem 4|Solution]]
 == Problem 5 ==
@@ Line 36: / Line 42: @@
 <math> \textbf{(A)}\ 8\qquad\textbf{(B)}\ 16\qquad\textbf{(C)}\ 25\qquad\textbf{(D)}\ 64\qquad\textbf{(E)}\ 96 </math>
+[[2012 AMC 12A Problems/Problem 5|Solution]]
 == Problem 6 ==
@@ Line 42: / Line 50: @@
 <math> \textbf{(A)}\ 4\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}\ 7\qquad\textbf{(E)}\ 8 </math>
+[[2012 AMC 12A Problems/Problem 6|Solution]]
 == Problem 7 ==
@@ Line 48: / Line 58: @@
 <math> \textbf{(A)}\ 5\qquad\textbf{(B)}\ 6\qquad\textbf{(C)}\ 8\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 12 </math>
+[[2012 AMC 12A Problems/Problem 7|Solution]]
 == Problem 8 ==
@@ Line 54: / Line 66: @@
 <math> \textbf{(A)}\ \frac{31}{16}\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ \frac{17}{8}\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ \frac{65}{16} </math>
+[[2012 AMC 12A Problems/Problem 8|Solution]]
 == Problem 9 ==
@@ Line 65: / Line 79: @@
 \qquad\textbf{(E)}\ \text{Tuesday}
   </math>
+[[2012 AMC 12A Problems/Problem 9|Solution]]
 == Problem 10 ==
@@ Line 71: / Line 87: @@
 <math> \textbf{(A)}\ \frac{3}{10}\qquad\textbf{(B)}\ \frac{1}{3}\qquad\textbf{(C)}\ \frac{9}{20}\qquad\textbf{(D)}\ \frac{2}{3}\qquad\textbf{(E)}\ \frac{9}{10} </math>
+[[2012 AMC 12A Problems/Problem 10|Solution]]
 == Problem 11 ==
@@ Line 77: / Line 95: @@
 <math> \textbf{(A)}\ \frac{5}{72}\qquad\textbf{(B)}\ \frac{5}{36}\qquad\textbf{(C)}\ \frac{1}{6}\qquad\textbf{(D)}\ \frac{1}{3}\qquad\textbf{(E)}\ 1 </math>
+[[2012 AMC 12A Problems/Problem 11|Solution]]
 == Problem 12 ==
@@ Line 83: / Line 103: @@
 <math> \textbf{(A)}\ \frac{\sqrt{10}+5}{10}\qquad\textbf{(B)}\ \frac{2\sqrt{5}}{5}\qquad\textbf{(C)}\ \frac{2\sqrt{2}}{3}\qquad\textbf{(D)}\ \frac{2\sqrt{19}-4}{5}\qquad\textbf{(E)}\ \frac{9-\sqrt{17}}{5} </math>
+[[2012 AMC 12A Problems/Problem 12|Solution]]
 == Problem 13 ==
@@ Line 94: / Line 116: @@
 \qquad\textbf{(E)}\ 60
   </math>
+[[2012 AMC 12A Problems/Problem 13|Solution]]
 == Problem 14 ==
@@ Line 100: / Line 124: @@
 <asy>
+size(6cm);
 defaultpen(fontsize(6pt));
 dotfactor=4;
@@ Line 127: / Line 152: @@
 <math> \textbf{(A)}\ 2\pi+6\qquad\textbf{(B)}\ 2\pi+4\sqrt3 \qquad\textbf{(C)}\ 3\pi+4 \qquad\textbf{(D)}\ 2\pi+3\sqrt3+2 \qquad\textbf{(E)}\ \pi+6\sqrt3 </math>
+[[2012 AMC 12A Problems/Problem 14|Solution]]
 == Problem 15 ==
-A <math>3\times3</math> square is partitioned into <math>9</math> unit squares.  Each unit square is painted either white or black with each color being equally likely, chosen independently and at random.  The square is the rotated <math>90^\circ</math> clockwise about its center, and every white square in a position formerly occupied by a black square is painted black.  The colors of all other squares are left unchanged.  What is the probability that the grid is now entirely black?
+A <math>3\times3</math> square is partitioned into <math>9</math> unit squares.  Each unit square is painted either white or black with each color being equally likely, chosen independently and at random.  The square is then rotated <math>90^\circ</math> clockwise about its center, and every white square in a position formerly occupied by a black square is painted black.  The colors of all other squares are left unchanged.  What is the probability that the grid is now entirely black?
 <math> \textbf{(A)}\ \dfrac{49}{512}
@@ Line 138: / Line 165: @@
 \qquad\textbf{(E)}\ \dfrac{9}{32}
   </math>
+[[2012 AMC 12A Problems/Problem 15|Solution]]
 == Problem 16 ==
@@ Line 144: / Line 173: @@
 <math> \textbf{(A)}\ 5\qquad\textbf{(B)}\ \sqrt{26}\qquad\textbf{(C)}\ 3\sqrt{3}\qquad\textbf{(D)}\ 2\sqrt{7}\qquad\textbf{(E)}\ \sqrt{30} </math>
+[[2012 AMC 12A Problems/Problem 16|Solution]]
 == Problem 17 ==
@@ Line 150: / Line 181: @@
 <math> \textbf{(A)}\ 10\qquad\textbf{(B)}\ 13\qquad\textbf{(C)}\ 15\qquad\textbf{(D)}\ 16\qquad\textbf{(E)}\ 18 </math>
+[[2012 AMC 12A Problems/Problem 17|Solution]]
 == Problem 18 ==
-Triangle <math>ABC</math> has <math>AB=27</math>, <math>AC=26</math>, and <math>BC=25</math>.  Let <math>I</math> denote the intersection of the internal angle bisectors of <math>\triangle ABC</math>.  What is <math>BI</math>?
+Triangle <math>ABC</math> has <math>AB=27</math>, <math>BC=25</math>, and <math>CA=26</math>.  Let <math>I</math> denote the intersection of the internal angle bisectors of <math>\triangle ABC</math>.  What is <math>BI</math>?
 <math> \textbf{(A)}\ 15\qquad\textbf{(B)}\ 5+\sqrt{26}+3\sqrt{3}\qquad\textbf{(C)}\ 3\sqrt{26}\qquad\textbf{(D)}\ \frac{2}{3}\sqrt{546}\qquad\textbf{(E)}\ 9\sqrt{3} </math>
+[[2012 AMC 12A Problems/Problem 18|Solution]]
 == Problem 19 ==
@@ Line 167: / Line 202: @@
 \qquad\textbf{(E)}\ 660
   </math>
+[[2012 AMC 12A Problems/Problem 19|Solution]]
 == Problem 20 ==
@@ Line 172: / Line 209: @@
 Consider the polynomial
-<cmath>P(x)=\prod_{k=0}^{10}=(x^2^+2^k)=(x+1)(x^2+2)(x^4+4)\cdots (x^{1024}+1024)</cmath>
+<cmath>P(x)=\prod_{k=0}^{10}(x^{2^k}+2^k)=(x+1)(x^2+2)(x^4+4)\cdots (x^{1024}+1024)</cmath>
 The coefficient of <math>x^{2012}</math> is equal to <math>2^a</math>.  What is <math>a</math>?
@@ Line 183: / Line 220: @@
   </math>
+[[2012 AMC 12A Problems/Problem 20|Solution]]
 == Problem 21 ==
 Let <math>a</math>, <math>b</math>, and <math>c</math> be positive integers with <math>a\ge</math> <math>b\ge</math> <math>c</math> such that
-<cmath> \begin{align*}a^{2}-b^{2}-c^{2}+ab&=2011\text{ and}\\ a^{2}+3b^{2}+3c^{2}-3ab-2ac-2bc&=-1997\end{align*} </cmath>
+<cmath>
+\begin{align*}a^{2}-b^{2}-c^{2}+ab&=2011\text{ and}\\
+a^{2}+3b^{2}+3c^{2}-3ab-2ac-2bc&=-1997
+\end{align*}
+</cmath>
 What is <math>a</math>?
 <math> \textbf{(A)}\ 249\qquad\textbf{(B)}\ 250\qquad\textbf{(C)}\ 251\qquad\textbf{(D)}\ 252\qquad\textbf{(E)}\ 253 </math>
-[[2012 AMC 10A Problems/Problem 24|Solution]]
+[[2012 AMC 12A Problems/Problem 21|Solution]]
 == Problem 22 ==
@@ Line 199: / Line 242: @@
 <math> \textbf{(A)}\ 8\qquad\textbf{(B)}\ 12\qquad\textbf{(C)}\ 20\qquad\textbf{(D)}\ 23\qquad\textbf{(E)}\ 24 </math>
+[[2012 AMC 12A Problems/Problem 22|Solution]]
 == Problem 23 ==
-Let <math>S</math> be the square one of whose diagonals has endpoints <math>(0.1,0.7)</math> and <math>(-0.1,-0.7)</math>.  A point <math>v=(x,y)</math> is chosen uniformly at random over all pairs of real numbers <math>x</math> and <math>y</math> such that <math>0 \le x \le 2012</math> and <math>0\le y\le 2012</math>.  Let <math>T(v)</math> be a translated copy of <math>S</math> centered at <math>v</math>.  What is the probability that the square region determined by <math>T(v)</math> contains exactly two points with integer coefficients in its interior?
+Let <math>S</math> be the square one of whose diagonals has endpoints <math>(0.1,0.7)</math> and <math>(-0.1,-0.7)</math>.  A point <math>v=(x,y)</math> is chosen uniformly at random over all pairs of real numbers <math>x</math> and <math>y</math> such that <math>0 \le x \le 2012</math> and <math>0\le y\le 2012</math>.  Let <math>T(v)</math> be a translated copy of <math>S</math> centered at <math>v</math>.  What is the probability that the square region determined by <math>T(v)</math> contains exactly two points with integer coordinates in its interior?
-<math> \textbf{(A)}\ 0.125\qquad\textbf{(B)}\ 0.14\qquad\textbf{(C)}\ 0.16\qquad\textbf{(D)}\ 0.25 \qquad\textbf{(E)}\ 0.32 </math>
+<math> \textbf{(A)}\ \frac{1}{8}\qquad\textbf{(B) }\frac{7}{50}\qquad\textbf{(C) }\frac{4}{25}\qquad\textbf{(D) }\frac{1}{4}\qquad\textbf{(E) }\frac{8}{25} </math>
-[[2012 AMC 10A Problems/Problem 24|Solution]]
+[[2012 AMC 12A Problems/Problem 23|Solution]]
 == Problem 24 ==
@@ Line 212: / Line 257: @@
 Let <math>\{a_k\}_{k=1}^{2011}</math> be the sequence of real numbers defined by <math>a_1=0.201,</math> <math>a_2=(0.2011)^{a_1},</math> <math>a_3=(0.20101)^{a_2},</math> <math>a_4=(0.201011)^{a_3}</math>, and in general,
-<cmath>a_k=\begin{cases}(0.\underbrace{20101\cdots 0101}_{k+2\text{ digits}})^{a_{k-1}}\qquad\text{if }k\text{ is odd,}\\(0.\underbrace{20101\cdots 01011}_{k+2\text{ digits}})^{a_{k-1}}\qquad\text{if }k\text{ is even.}\end{cases}</cmath>
+<cmath>
+a_k=\begin{cases}
+(0.\underbrace{20101\cdots 0101}_{k+2\text{ digits}})^{a_{k-1}} & \text{if }k\text{ is odd,}\\
+(0.\underbrace{20101\cdots 01011}_{k+2\text{ digits}})^{a_{k-1}}& \text{if }k\text{ is even.}
+\end{cases}
+</cmath>
 Rearranging the numbers in the sequence  <math>\{a_k\}_{k=1}^{2011}</math> in decreasing order produces a new sequence  <math>\{b_k\}_{k=1}^{2011}</math>.  What is the sum of all integers <math>k</math>, <math>1\le k \le 2011</math>, such that <math>a_k=b_k?</math>
-<math> \textbf{(A)}\ 671\qquad\textbf{(B)}\ 1006\qquad\textbf{(C)}\ 1341\qquad\textbf{(D)}\ 2011\qquad\textbf{(E)}\2012 </math>
+<math> \textbf{(A)}\ 671\qquad\textbf{(B)}\ 1006\qquad\textbf{(C)}\ 1341\qquad\textbf{(D)}\ 2011\qquad\textbf{(E)}\ 2012 </math>
+[[2012 AMC 12A Problems/Problem 24|Solution]]
 == Problem 25 ==
@@ Line 223: / Line 275: @@
 Let <math>f(x)=|2\{x\}-1|</math> where <math>\{x\}</math> denotes the fractional part of <math>x</math>.  The number <math>n</math> is the smallest positive integer such that the equation <cmath>nf(xf(x))=x</cmath> has at least <math>2012</math> real solutions.  What is <math>n</math>?  '''Note:''' the fractional part of <math>x</math> is a real number <math>y=\{x\}</math> such that <math>0\le y<1</math> and <math>x-y</math> is an integer.
-<math> \textbf{(A)}\ 30\qquad\textbf{(B)}\ 31\qquad\textbf{(C)}\ 32\qquad\textbf{(D)}\ 62\qquad\textbf{(E)}\64 </math>
+<math> \textbf{(A)}\ 30\qquad\textbf{(B)}\ 31\qquad\textbf{(C)}\ 32\qquad\textbf{(D)}\ 62\qquad\textbf{(E)}\ 64 </math>
+[[2012 AMC 12A Problems/Problem 25|Solution]]
+==See also==
+{{AMC12 box|year=2012|ab=A|before=[[2011 AMC 12B Problems]]|after=[[2012 AMC 12B Problems]]}}
+{{MAA Notice}}