2012AMC12A真题及答案解析

2012 AMC12A 真题

答案解析请参考文末

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?

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

 

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?

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

 

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 $n$?

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

 

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?

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

 

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?

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

 

Problem 6

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

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

 

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?

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

 

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?

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

 

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?

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

 

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 $\sin{\theta}$?

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

 

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?

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

 

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?

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

 

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?

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

 

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] 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]

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

 

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?

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

 

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 $C_1$?

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

 

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 $S$?

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

 

Problem 18

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

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

 

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?

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

 

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 $a$?

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

 

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 $a$?

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

 

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 $k$?

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

 

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?

$\textbf{(A)}\ 0.125\qquad\textbf{(B)}\ 0.14\qquad\textbf{(C)}\ 0.16\qquad\textbf{(D)}\ 0.25 \qquad\textbf{(E)}\ 0.32$

 

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 $a_k=b_k?$

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

 

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.

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

  1. E
  2. D
  3. D
  4. C
  5. D
  6. D
  7. C
  8. C
  9. A
  10. D
  11. B
  12. D
  13. D
  14. E
  15. A
  16. E
  17. B
  18. A
  19. B
  20. B
  21. E
  22. C
  23. C
  24. C
  25. C

Solution

1.

[asy] draw((-2,1)--(-6,1),red+dashed,EndArrow); draw((-6,2)--(5,2),blue+dashed,EndArrow); dot((-2,0)); dot((-6,0)); dot((5,0)); label("$-2$",(-2,0),dir(270)); label("$-6$",(-6,0),dir(270)); label("$5$",(5,0),dir(270)); label("$4$",(-4,0.9),dir(270)); label("$11$",(-1.5,2.5),dir(90)); [/asy]

Crawling from $-2$ to $-6$ takes it a distance of $4$ units. Crawling from $-6$ to $5$ takes it a distance of $11$ units. Add $4$ and $11$ to get $\boxed{\textbf{(E)}\ 15}$

 

2.

Cagney can frost one in $20$ seconds, and Lacey can frost one in $30$ seconds. Working together, they can frost one in $\frac{20\cdot30}{20+30} = \frac{600}{50} = 12$ seconds. In $300$ seconds ($5$ minutes), they can frost $\boxed{\textbf{(D)}\ 25}$ cupcakes.

 

3.

The first box has volume $2\times3\times5=30\text{ cm}^3$, and the second has volume $(2\times2)\times(3\times3)\times(5)=180\text{ cm}^3$. The second has a volume that is $6$ times greater, so it holds $6\times40=\boxed{\textbf{(D)}\ 240}$ grams.

 

4.

Assume that there are 5 total marbles in the bag. The actual number does not matter, since all we care about is the ratios, and the only operation performed on the marbles in the bag is doubling.

There are 3 blue marbles in the bag and 2 red marbles. If you double the amount of red marbles, there will still be 3 blue marbles but now there will be 4 red marbles. Thus, the answer is $\boxed{\textbf{(C)}\ \frac{4}{7}}$.

 

5.

So let the number of blueberries be $b,$ the number of raspberries be $r,$ the number of grapes be $g,$ and finally the number of cherries be $c.$

Observe that since there are $280$ pieces of fruit,\[b+r+g+c=280.\]

Since there are twice as many raspberries as blueberries,\[2b=r.\]

The fact that there are three times as many grapes as cherries implies,\[3c=g.\]

Because there are four times as many cherries as raspberries, we deduce the following:\[4r=c.\]

Note that we are looking for $c.$ So, we try to rewrite all of the other variables in terms of $c.$ The third equation gives us the value of $g$ in terms of $c$ already. We divide the fourth equation by $4$ to get that $r=\frac{c}{4}.$ Finally, substituting this value of $r$ into the first equation provides us with the equation $b=\frac{c}{8}$ and substituting yields:\[\frac{c}{4}+\frac{c}{8}+3c+c=280\]Multiply this equation by $8$ to get:\[2c+c+24c+8c=8\cdot 280,\]\[35c=8\cdot 280,\]\[c=64.\]\[\boxed{D}\]

 

6.

Let the three numbers be equal to $a$$b$, and $c$. We can now write three equations:

$a+b=12$

$b+c=17$

$a+c=19$

Adding these equations together, we get that

$2(a+b+c)=48$ and

$a+b+c=24$

Substituting the original equations into this one, we find

$c+12=24$

$a+17=24$

$b+19=24$

Therefore, our numbers are 12, 7, and 5. The middle number is $\boxed{\textbf{(D)}\ 7}$

 

7.

If we let $a$ be the smallest sector angle and $r$ be the difference between consecutive sector angles, then we have the angles $a, a+r, a+2r, \cdots. a+11r$. Use the formula for the sum of an arithmetic sequence and set it equal to 360, the number of degrees in a circle.

\begin{align*} \frac{a+a+11r}{2}\cdot 12 &= 360\\ 2a+11r &= 60\\ a &= \frac{60-11r}{2} \end{align*}

All sector angles are integers so $r$ must be a multiple of 2. Plug in even integers for $r$ starting from 2 to minimize $a.$ We find this value to be 4 and the minimum value of $a$ to be $\frac{60-11(4)}{2} = \boxed{\textbf{(C)}\ 8}$

 

8.

The minimum and maximum can be achieved with the orders $5, 4, 3, 2, 1$ and $1, 2, 3, 4, 5$.

$5,4,3,2,1 \Rightarrow \frac92,3,2,1 \Rightarrow \frac{15}{4},2,1 \Rightarrow \frac{23}{8},1 \Rightarrow \frac{31}{16}$

$1,2,3,4,5 \Rightarrow \frac32,3,4,5 \Rightarrow \frac94,4,5 \Rightarrow \frac{25}{8},5 \Rightarrow \frac{65}{16}$

The difference between the two is $\frac{65}{16}-\frac{31}{16}=\frac{34}{16}=\boxed{\textbf{(C)}\ \frac{17}{8}}$.

 

9.

Every year we go back is moving one day back, because $365 = 1\ (\text{mod}\ 7)$. Every leap year we go back two days, since $366 = 2\ (\text{mod}\ 7)$. A leap year is usually every four years, so 200 years would have $\frac{200}{4}$ = $50$ leap years, but the problem says that 1900 does not count as a leap year.

Therefore there would be 151 regular years and 49 leap years, so $1(151)+2(49)$ = $249$ days back. Since $249 = 4\ (\text{mod}\ 7)$, four days back from Tuesday would be $\boxed{\textbf{(A)}\ \text{Friday}}$.

 

10.

[asy] unitsize(5mm); defaultpen(linewidth(.8pt)+fontsize(10pt)); dotfactor=4; pair A=(-5,0), B=(5,0), C=(sqrt(45),6), D=(0,0), E=(sqrt(45),0); draw(A--B--C--cycle); draw(D--C); draw(E--C); pair[] ps={A,B,C,D,E}; dot(ps); label("$A$",A,SW); label("$B$",B,SE); label("$C$",C,N); label("$D$",D,SW); label("$E$",E,S); label("$9$",(D--C),NW); label("$6$",(E--C)); label("$10$",(A--B),SE); [/asy]$AB$ is the side of length $10$, and $CD$ is the median of length $9$. The altitude of $C$ to $AB$ is $6$ because the 0.5(altitude)(base)=Area of the triangle. $\theta$ is $\angle CDE$. To find $\sin{\theta}$, just use opposite over hypotenuse with the right triangle $\triangle DCE$. This is equal to $\frac69=\boxed{\textbf{(D)}\ \frac23}$.

 

11.

If $m$ is the probability Mel wins and $c$ is the probability Chelsea wins, $m=2c$ and $m+c=\frac12$. From this we get $m=\frac13$ and $c=\frac16$. For Alex to win three, Mel to win two, and Chelsea to win one, in that order, is $\frac{1}{2^3\cdot3^2\cdot6}=\frac{1}{432}$. Multiply this by the number of permutations (orders they can win) which is $\frac{6!}{3!2!1!}=60.$

\[\frac{1}{432}\cdot60=\frac{60}{432}=\boxed{\textbf{(B)}\ \frac{5}{36}}\]

 

12.

[asy] unitsize(15mm); defaultpen(linewidth(.8pt)+fontsize(10pt)); dotfactor=4; real a=1; real b=2; pair O=(0,0); pair A=(-(sqrt(19)-2)/5,1); pair B=((sqrt(19)-2)/5,1); pair C=((sqrt(19)-2)/5,1+2(sqrt(19)-2)/5); pair D=(-(sqrt(19)-2)/5,1+2(sqrt(19)-2)/5); pair E=(-(sqrt(19)-2)/5,0); path inner=Circle(O,a); path outer=Circle(O,b); draw(outer); draw(inner); draw(A--B--C--D--cycle); draw(O--D--E--cycle); label("$A$",D,NW); label("$E$",E,SW); label("$O$",O,SE); label("$s+1$",(D--E),W); label("$\frac{s}{2}$",(E--O),S); pair[] ps={A,B,C,D,E,O}; dot(ps); [/asy]The circles have radii of $1$ and $2$. Draw the triangle shown in the figure above and write expressions in terms of $s$ (length of the side of the square) for the sides of the triangle. Because $AO$ is the radius of the larger circle, which is equal to $2$, we can write the Pythagorean Theorem.

\begin{align*} \left( \frac{s}{2} \right) ^2 + (s+1)^2 &= 2^2\\ \frac14 s^2 + s^2 + 2s + 1 &= 4\\ \frac54 s^2 +2s - 3 &= 0\\ 5s^2 + 8s - 12 &=0 \end{align*}

Use the quadratic formula.

\[s = \frac{-8+\sqrt{8^2-4(5)(-12)}}{10} = \frac{-8+\sqrt{304}}{10} = \frac{-8+4\sqrt{19}}{10} = \boxed{\textbf{(D)}\ \frac{2\sqrt{19}-4}{5}}\]

 

13.

Let Paula work at a rate of $p$, the two helpers work at a combined rate of $h$, and the time it takes to eat lunch be $L$, where $p$ and $h$ are in house/hours and L is in hours. Then the labor on Monday, Tuesday, and Wednesday can be represented by the three following equations:

\[(8-L)(p+h)=50\]

\[(6.2-L)h=24\]

\[(11.2-L)p=26\]

With three equations and three variables, we need to find the value of $L$. Adding the second and third equations together gives us $6.2h+11.2p-L(p+h)=50$. Subtracting the first equation from this new one gives us $-1.8h+3.2p=0$, so we get $h=\frac{16}{9}p$. Plugging into the second equation:

\[(6.2-L)\frac{16}{9}p=24\]\[(6.2-L)p=\frac{27}{2}\]

We can then subtract this from the third equation:

\[5p=26-\frac{27}{2}\]\[p=\frac{5}{2}\]Plugging $p$ into our third equation gives:\[L=\frac{4}{5}\]

Converting $L$ from hours to minutes gives us $L=48$ minutes, which is $\boxed{\textbf{(D)}\ 48}$.

 

14.

Draw the hexagon between the centers of the circles, and compute its area $(6)(0.5)(2\sqrt{3})=6\sqrt{3}$. Then add the areas of the three sectors outside the hexagon ($2\pi$) and subtract the areas of the three sectors inside the hexagon but outside the figure($\pi$) to get the area enclosed in the curved figure $(\pi+6\sqrt{3})$, which is $\boxed{\textbf{(E)}\ \pi+6\sqrt{3}}$.

 

15.

First, there is only one way for the middle square to be black because it is not affected by the rotation. Then we can consider the corners and edges separately. Let's first just consider the number of ways we can color the corners. There is $1$ case with all black squares. There are four cases with one white square and all $4$ work. There are six cases with two white squares, but only the $2$ with the white squares diagonal from each other work. There are no cases with three white squares or four white squares. Then the total number of ways to color the corners is $1+4+2=7$. In essence, the edges work the same way, so there are also $7$ ways to color them. The number of ways to fit the conditions over the number of ways to color the squares is

\[\frac{7\times7}{2^9}=\boxed{\textbf{(A)}\ \frac{49}{512}}\]

 

16.

Let $r$ denote the radius of circle $C_1$. Note that quadrilateral $ZYOX$ is cyclic. By Ptolemy's Theorem, we have $11XY=13r+7r$ and $XY=20r/11$. Let t be the measure of angle $YOX$. Since $YO=OX=r$, the law of cosines on triangle $YOX$ gives us $\cos t =-79/121$. Again since $ZYOX$ is cyclic, the measure of angle $YZX=180-t$. We apply the law of cosines to triangle $ZYX$so that $XY^2=7^2+13^2-2(7)(13)\cos(180-t)$. Since $\cos(180-t)=-\cos t=79/121$ we obtain $XY^2=12000/121$. But$XY^2=400r^2/121$ so that $r=\sqrt{30}$$\boxed{E}$.

 

17.

Of the integers from $1$ to $30$, there are six each of $0,1,2,3,4\ (\text{mod}\ 5)$. We can create several rules to follow for the elements in subset $S$. No element can be $1\ (\text{mod}\ 5)$ if there is an element that is $4\ (\text{mod}\ 5)$. No element can be $2\ (\text{mod}\ 5)$ if there is an element that is $3\ (\text{mod}\ 5)$. Thus we can pick 6 elements from either $1\ (\text{mod}\ 5)$ or $4\ (\text{mod}\ 5)$ and 6 elements from either $2\ (\text{mod}\ 5)$ or $3\ (\text{mod}\ 5)$ for a total of $6+6=12$ elements. Considering $0\ (\text{mod}\ 5)$, there can be one element that is so because it will only be divisible by $5$ if paired with another element that is $0\ (\text{mod}\ 5)$. The final answer is $\boxed{\textbf{(B)}\ 13}$.

 

18.

Inscribe circle $C$ of radius $r$ inside triangle $ABC$ so that it meets $AB$ at $Q$$BC$ at $R$, and $AC$ at $S$. Note that angle bisectors of triangle $ABC$ are concurrent at the center $O$(also $I$) of circle $C$. Let $x=QB$$y=RC$ and $z=AS$. Note that $BR=x$$SC=y$ and $AQ=z$. Hence $x+z=27$$x+y=25$, and $z+y=26$. Subtracting the last 2 equations we have $x-z=-1$ and adding this to the first equation we have $x=13$.

By Heron's formula for the area of a triangle we have that the area of triangle $ABC$ is $\sqrt{39(14)(13)(12)}$. On the other hand the area is given by $(1/2)25r+(1/2)26r+(1/2)27r$. Then $39r=\sqrt{39(14)(13)(12)}$ so that $r^2=56$.

Since the radius of circle $O$ is perpendicular to $BC$ at $R$, we have by the pythagorean theorem $BO^2=BI^2=r^2+x^2=56+169=225$ so that $BI=15$.

 

19.

Note that if $n$ is the number of friends each person has, then $n$ can be any integer from $1$ to $4$, inclusive.

Also note that one person can only have at most 5 friends. That's right, 5 friends.

Also note that the cases of $n=1$ and $n=4$ are the same, since a map showing a solution for $n=1$ can correspond one-to-one with a map of a solution for $n=4$ by simply making every pair of friends non-friends and vice versa. The same can be said of configurations with $n=2$ when compared to configurations of $n=3$. Thus, we have two cases to examine, $n=1$ and $n=2$, and we count each of these combinations twice.

For $n=1$, if everyone has exactly one friend, that means there must be $3$ pairs of friends, with no other interconnections. The first person has $5$ choices for a friend. There are $4$ people left. The next person has $3$ choices for a friend. There are two people left, and these remaining two must be friends. Thus, there are $15$ configurations with $n=1$.

For $n=2$, there are two possibilities. The group of $6$ can be split into two groups of $3$, with each group creating a friendship triangle. The first person has $\binom{5}{2} = 10$ ways to pick two friends from the other five, while the other three are forced together. Thus, there are $10$ triangular configurations.

However, the group can also form a friendship hexagon, with each person sitting on a vertex, and each side representing the two friends that person has. The first person may be seated anywhere on the hexagon without loss of generality. This person has $\binom{5}{2} = 10$choices for the two friends on the adjoining vertices. Each of the three remaining people can be seated "across" from one of the original three people, forming a different configuration. Thus, there are $10 \cdot 3! = 60$ hexagonal configurations, and in total $70$ configurations for $n=2$.

As stated before, $n=3$ has $70$ configurations, and $n=4$ has $15$ configurations. This gives a total of $(70 + 15)\cdot 2 = 170$configurations, which is option $\boxed{\textbf{(B)}\ 170}$.

 

20.

Every term in the expansion of the product is formed by taking one term from each factor and multiplying them all together. Therefore, we pick a power of $x$ or a power of $2$ from each factor.

Every number, including $2012$, has a unique representation by the sum of powers of two, and that representation can be found by converting a number to its binary form. $2012 = 11111011100_2$, meaning $2012 = 1024 + 512 + 256 + 128 + 64 + 16 + 8 + 4$.

Thus, the $x^{2012}$ term was made by multiplying $x^{1024}$ from the $(x^{1024} + 1024)$ factor, $x^{512}$ from the $(x^{512} + 512)$ factor, and so on. The only numbers not used are $32$$2$, and $1$.

Thus, from the $(x^{32} + 32), (x^2+2), (x+1)$ factors, $32$$2$, and $1$ were chosen as opposed to $x^{32}, x^2$, and $x$.

Thus, the coefficient of the $x^{2012}$ term is $32 \times 2 \times 1 = 64 = 2^6$. So the answer is $6 \rightarrow \boxed{B}$

 

21.

Add the two equations.

$2a^2 + 2b^2 + 2c^2 - 2ab - 2ac - 2bc = 14$.

Now, this can be rearranged:

$(a^2 - 2ab + b^2) + (a^2 - 2ac + c^2) + (b^2 - 2bc + c^2) = 14$

and factored:

$(a - b)^2 + (a - c)^2 + (b - c)^2 = 14$

$a$$b$, and $c$ are all integers, so the three terms on the left side of the equation must all be perfect squares. Recognize that $14 = 9 + 4 + 1$.

$(a-c)^2 = 9 \rightarrow a-c = 3$, since $a-c$ is the biggest difference. It is impossible to determine by inspection whether $a-b = 4$ or $1$, or whether $b-c = 1$ or $4$.

We want to solve for $a$, so take the two cases and solve them each for an expression in terms of $a$. Our two cases are $(a, b, c) = (a, a-1, a-3)$ or $(a, a-2, a-3)$. Plug these values into one of the original equations to see if we can get an integer for $a$.

$a^2 - (a-1)^2 - (a-3)^2 + a(a-1) = 2011$, after some algebra, simplifies to $7a = 2021$$2021$ is not divisible by $7$, so $a$ is not an integer.

The other case gives $a^2 - (a-2)^2 - (a-3)^2 + a(a-2) = 2011$, which simplifies to $8a = 2024$. Thus, $a = 253$ and the answer is $\boxed{\textbf{(E)}\ 253}$.

 

22.

[asy] pair A=(0,0), B=(2,0), C=(2,2), D=(0,2), E=(-1,3), F=(1,3), G=(1,1), H=(-1,1), I=(1,0), J=(2,1), K=(1,2), L=(0,1), M=(-0.5,0.5), N=(-1,2), O=(-0.5,2.5), P=(0,3), Q=(1.5,2.5), R=(1,2), S=(1.5,0.5), T=(0,1); draw(A--B--C--D--E--F); draw(H--A); draw(A--D); draw(H--E); draw(F--C); draw(H--G--F); draw(G--B); label("\(A\)",A,SW); label("\(B\)",B,SE); label("\(C\)",C,NE); label("\(D\)",D,SW); label("\(E\)",E,NW); label("\(F\)",F,NE); label("\(G\)",G,NE); label("\(H\)",H,SW); [/asy][asy] pair A=(0,0), B=(2,0), C=(2,2), D=(0,2), E=(-1,3), F=(1,3), G=(1,1), H=(-1,1), I=(1,0), J=(2,1), K=(1,2), L=(0,1), M=(-0.5,0.5), N=(-1,2), O=(-0.5,2.5), P=(0,3), Q=(1.5,2.5), R=(1,2), S=(1.5,0.5), T=(0,1); draw(A--B--C--D--E--F); draw(H--A); draw(A--D); draw(H--E); draw(F--C); draw(I--J--K--L--I); draw(I--K); draw(J--L); draw(K--Q--P--O--K); draw(K--P); draw(Q--O); draw(O--L--M--N--O); draw(O--M); draw(L--N); label("\(A\)",A,SW); label("\(B\)",B,SE); label("\(C\)",C,NE); label("\(D\)",D,SW); label("\(E\)",E,NW); label("\(F\)",F,NE); label("\(H\)",H,SW); [/asy]
We need two different kinds of planes that only intersect $Q$ at the mentioned segments (we call them traces in this solution). These will be all the possible $p_j$'s.

First, there are two kinds of segments joining the midpoints of every pair of edges belonging to the same face of $Q$long traces are those connecting the midpoint of opposite sides of the same face of $Q$, and short traces are those connecting the midpoint of adjacent sides of the same face of $Q$.

Suppose $p_j$ contains a short trace $t_1$ of a face of $Q$. Then it must also contain some trace $t_2$ of an adjacent face of $Q$, where $t_2$ share a common endpoint with $t_1$. So, there are three possibilities for $t_2$, each of which determines a plane $p_j$ containing both $t_1$ and $t_2$.

Case 1: $t_2$ makes an acute angle with $t_1$. In this case, $p_j \cap Q$ is an equilateral triangle made by three short traces. There are $8$ of them, corresponding to the $8$ vertices.

Case 2: $t_2$ is a long trace. $p \cap Q$ is a rectangle. Each pair of parallel faces of $Q$ contributes $4$ of these rectangles so there are $12$ such rectangles.

Case 3: $t_2$ is the short trace other than the one described in case 1, i.e. $t_2$ makes an obtuse angle with $t_1$. It is possible to prove that $p \cap Q$ is a regular hexagon (See note #1 for a proof) and there are $4$ of them.

Case 4: $p_j$ contains no short traces. This can only make $p_j \cap Q$ be a square enclosed by long traces. There are $3$ such squares.

In total, there are $8+12+4+3=27$ possible planes in $P$. So the maximum of $k$ is $27$.

On the other hand, the most economic way to generate these long and short traces is to take all the planes in case 3 and case 4. Overall, they intersect at each trace exactly once (there is a quick way to prove this. See note #2 below.) and also covered all the $6\times 4 + 4\times 3 = 36$ traces. So the minimum of $k$ is $7$. The answer to this problem is then $27-7=20$ ... $\framebox{C}$.

 

23.

[asy] pair A=(0.1,0.7), C=(-0.1,-0.7), B=(-0.7,0.1), D=(0.7,-0.1), X=(1,0), W=(-1,0), Y=(0,1), Z=(0,-1); draw (A--B--C--D--A); draw(A--C); draw(B--D); draw(W--X); draw(Y--Z); label("\((0.1,0.7)\)",A,NE); label("\((-0.1,-0.7)\)",C,SW); label("\(x\)",X,NW); label("\(y\)",Y,NE); [/asy]The unit square's diagonal has a length of $\sqrt{0.2^2 + 1.4^2} = \sqrt{2}$. Because $S$ square is not parallel to the axis, the two points must be adjacent.

Now consider the unit square $U$ with vertices $(0,0), (1,0), (1,1)$ and $(0,1)$. Let us first consider only two vertices, $(0,0)$ and $(1,0)$. We want to find the area of the region within $U$ that the point $v=(x,y)$ will create the translation of $S$$T(v)$ such that it covers both $(0,0)$and $(1,0)$. By symmetry, there will be three equal regions that cover the other pairs of adjacent vertices.

For $T(v)$ to contain the point $(0,0)$$v$ must be inside square $S$. Similarly, for $T(v)$ to contain the point $(1,0)$$v$ must be inside a translated square $S$ with center at $(1,0)$, which we will call $S'$. Therefore, the area we seek is Area$(U \cap S \cap S')$.

To calculate the area, we notice that Area$(U \cap S \cap S') = \frac{1}{2} \cdot$ Area$(S \cap S')$ by symmetry. Let $S_1 = (0.1, 0.7), S_2 = (0.7, -0.1), S'_1 = (1.1, 0.7), S'_2 = (0.3, 0.1)$. Let $M = (0.7, 0.4)$ be the midpoint of $S'_1S'_2$, and $N = (0.7, 0.7)$along the line $S_1S'_1$. Let $I$ be the intersection of $S$ and $S'$ within $U$, and $J$ be the intersection of $S$ and $S'$ outside $U$. Therefore, the area we seek is $\frac{1}{2} \cdot$ Area$(S \cap S') = \frac{1}{2} [IS'_2JS_2]$. Because $S_2, M, N$ all have $x$ coordinate $0.7$, they are collinear. Noting that the side length of $S$ and $S'$ is $1$ (as shown above), we also see that $S_2M = MS'_1 = 0.5$, so $\triangle{S'_1NM} \cong \triangle{S_2IM}$. If follows that $IS_2 = NS'_1 = 1.1 - 0.7 = 0.4$ and $IS'2 = MS'_2 - MI = MS'_2 - MN = 0.5 - 0.3 = 0.2$. Therefore, the area is $\frac{1}{2} \cdot$ Area$(S \cap S') = \frac{1}{2} [IS'_2JS_2] = \frac{1}{2} \cdot 0.2 \cdot 0.4 = 0.04$.

Because there are three other regions in the unit square $U$ that we need to count, the total area of $v$ within $U$ such that $T(v)$ contains two adjacent lattice points is $0.04 \cdot 4 = 0.16$.

By periodicity, this probability is the same for all $0 \le x \le 2012$ and $0 \le y \le 2012$. Therefore, the answer is $\boxed{\textbf{(C)}\ 0.16}$.

 

24.

First, we must understand two important functions: $f(x) = b^x$ for $0 < b < 1$(decreasing exponential function), and $g(x) = x^k$ for $k > 0$(increasing exponential function for positive $x$). $f(x)$ is used to establish inequalities when we change the exponent and keep the baseconstant. $g(x)$ is used to establish inequalities when we change the base and keep the exponent constant.

We will now examine the first few terms.

Comparing $a_1$ and $a_2$$0 < a_1 = (0.201)^1 < (0.201)^{a_1} < (0.2011)^{a_1} = a_2 < 1 \Rightarrow 0 < a_1 < a_2 < 1$.

Therefore, $0 < a_1 < a_2 < 1$.

Comparing $a_2$ and $a_3$$0 < a_3 = (0.20101)^{a_2} < (0.20101)^{a_1} < (0.2011)^{a_1} = a_2 < 1 \Rightarrow 0 < a_3 < a_2 < 1$.

Comparing $a_1$ and $a_3$$0 < a_1 = (0.201)^1 < (0.201)^{a_2} < (0.20101)^{a_2} = a_3 < 1 \Rightarrow 0 < a_1 < a_3 < 1$.

Therefore, $0 < a_1 < a_3 < a_2 < 1$.

Comparing $a_3$ and $a_4$$0 < a_3 = (0.20101)^{a_2} < (0.20101)^{a_3} < (0.201011)^{a_3} = a_4 < 1 \Rightarrow 0 < a_3 < a_4 < 1$.

Comparing $a_2$ and $a_4$$0 < a_4 = (0.201011)^{a_3} < (0.201011)^{a_1} < (0.2011)^{a_1} = a_2 < 1 \Rightarrow 0 < a_4 < a_2 < 1$.

Therefore, $0 < a_1 < a_3 < a_4 < a_2 < 1$.

Continuing in this manner, it is easy to see a pattern(see Note 1).

Therefore, the only $k$ when $a_k = b_k$ is when $2(k-1006) = 2011 - k$. Solving gives $\boxed{\textbf{(C)} 1341}$.

 

25.

Our goal is to determine how many times the graph of $nf(xf(x))=x$ intersects the graph of $y=x$.

We begin by analyzing the behavior of $\{x\}$. It increases linearly with a slope of one, then when it reaches the next integer, it repeats itself. We can deduce that the function is like a sawtooth wave, with a period of one. We then analyze the function $f(x)=|2\{x\}-1|$. The slope of the teeth is multiplied by 2 to get 2, and the function is moved one unit downward. The function can then be described as starting at -1, moving upward with a slope of 2 to get to 1, and then repeating itself, still with a period of 1. The absolute value of the function is then taken. This results in all the negative segments becoming flipped in the Y direction. The positive slope starting at -1 of the function ranging from $u$ to $u.5$, where u is any arbitrary integer, is now a negative slope starting at positive 1. The function now looks like the letter V repeated within every square in the first row.

It is now that we address the goal of this, which is to determine how many times the function intersects the line $y=x$. Since there are two line segments per box, the function has two chances to intersect the line $y=x$ for every integer. If the height of the function is higher than $y=x$ for every integer on an interval, then every chance within that interval intersects the line.

Returning to analyzing the function, we note that it is multiplied by $x$, and then fed into $f(x)$. Since $f(x)$ is a periodic function, we can model it as multiplying the function's frequency by $x$. This gives us $2x$ chances for every integer, which is then multiplied by 2 once more to get $4x$ chances for every integer. The amplitude of this function is initially 1, and then it is multiplied by $n$, to give an amplitude of $n$. The function intersects the line $y=x$ for every chance in the interval of $0\leq x \leq n$, since the function is n units high. The function ceases to intersect $y=x$ when $n < x$, since the height of the function is lower than $y=x$.

The number of times the function intersects $y=x$ is then therefore equal to $4+8+12...+4x$. We want this sum to be greater than 2012 which occurs when $x=32$.

 

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