AMC12 2012 A

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 $\boxed{\textbf{(E)}\ 15}$

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.

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.

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 $\boxed{\textbf{(C)}\ \frac{4}{7}}$.

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: \[2c+c+24c+8c=8\cdot 280,\] \[35c=8\cdot 280,\] \[c=64.\] \[\boxed{D}\]

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 $\boxed{\textbf{(D)}\ 7}$

Let $a_1$ be the first term of the arithmetic progression and $a_{12}$ be the last term of the arithmetic progression. From the formula of the sum of an arithmetic progression (or arithmetic series), we have $12*\frac{a_1+a_{12}}{2}=360$, which leads us to $a_1 + a_{12} = 60$. $a_{12}$, the largest term of the progression, can also be expressed as $a_1+11d$, where $d$ is the common difference. Since each angle measure must be an integer, $d$ must also be an integer. We can isolate $d$ by subtracting $a_1$ from $a_{12}$ like so: $a_{12}-a_1=a_1+11d-a_1=11d$. Since $d$ is an integer, the difference between the first and last terms, $11d$, must be divisible by $11.$ Since the total difference must be less than $60$, we can start checking multiples of $11$ less than $60$ for the total difference between $a_1$ and $a_{12}$. We start with the largest multiple, because the maximum difference will result in the minimum value of the first term. If the difference is $55$, $a_1=\frac{60-55}{2}=2.5$, which is not an integer, nor is it one of the five options given. If the difference is $44$, $a_1=\frac{60-44}{2}$, or $\boxed{\textbf{(C)}\ 8}$

The iterative average of any 5 integers $a,b,c,d,e$ is defined as: \[\frac{\frac{\frac{\frac{a+b} 2+c} 2+d} 2+e} 2=\frac{a+b+2c+4d+8e}{16}\] Plugging in $1,2,3,4,5$ for $a,b,c,d,e$, we see that in order to maximize the fraction, $a=1,b=2,c=3,d=4,e=5$, and in order to minimize the fraction, $a=5,b=4,c=3,d=2,e=1$. After plugging in these values and finding the positive difference of the two fractions, we arrive with $\frac{34}{16} \Rightarrow \frac{17}{8}$, which is our answer of $\boxed{\textbf{(C)}}$

In this solution we refer to moving to the left as decreasing the year or date number and moving to the right as increasing the year or date number. Every non-leap year we move to the right results in moving one day to the right because $365\equiv 1\pmod 7$. Every leap year we move to the right results in moving $2$ days to the right since $366\equiv 2\pmod 7$. 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 $249 \equiv 4\ (\text{mod}\ 7)$, four days back from Tuesday would be $\boxed{\textbf{(A)}\ \text{Friday}}$.

$AB$ 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 $\frac69=\boxed{\textbf{(D)}\ \frac23}$.

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 $\frac{6!}{3!2!1!}=60.$ \[\frac{1}{432}\cdot60=\frac{60}{432}=\boxed{\textbf{(B)}\ \frac{5}{36}}\]

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

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 $\boxed{\textbf{(D)}\ 48}$.

We can draw the hexagon between the centers of the circles, and compute its area. The hexagon is made of $6$ equilateral triangles each with length $2$, so the area is: \[\frac{\sqrt{3}}{4} \cdot 2^2 \cdot 6=6 \sqrt{3}.\] Then, we add the areas of the three sectors outside the hexagon: \[\frac 23 \pi \cdot 3=2\pi.\] We now subtract the areas of the three sectors inside the hexagon but outside the figure (which is $\pi$) to get the area enclosed in the curved figure: \[6 \sqrt{3}+2\pi-\pi=\pi+6\sqrt{3}.\] Hence, our answer is $\boxed{\textbf{(E)}\ \pi+6\sqrt{3}},$ and we are done. \[\] (Minor edits, made by dbnl.)

First, look for invariants. The center, unaffected by rotation, must be black. So automatically, the chance is less than $\frac{1}{2}.$ Note that a $90^{\circ}$ rotation requires that black squares be across from each other across a vertical or horizontal axis. As such, $2$ squares directly across from each other must be black in the $4$ edge squares. Since there are $2$ configurations for this to be possible (top and bottom, right and left), this is a chance of \[\left (\frac{1}{2} \cdot \frac{1}{2} \right )+ \left (\frac{1}{2} \cdot \frac{1}{2} \right )=\frac{1}{2}\] However, by PIE, subtract the chance all 4 are black and both configurations are met: $\frac{1}{2}- \left (\frac{1}{2} \cdot \frac{1}{2} \right ) \cdot \left (\frac{1}{2} \cdot \frac{1}{2} \right )=\frac{7}{16}$. Through symmetrical reasoning, the corners also have a $\frac{7}{16}$ chance of having a configuration that yields all black corners. Then, the chance that all squares black is the intersection of all these probabilities: $\frac{1}{2} \left (\frac{7}{16} \right ) \left (\frac{7}{16} \right ) = \boxed{\textbf{(A)}\ \frac{49}{512}}$ Also, if you have little to no time and are guessing, notice there are a total of $2^9 = 512$ ways to permutate the colors on the square (Each square can be white or black, so there are 2 possibilities for each of the 9 squares). Thus, the answer must be in some form of $\frac{\text{the number of good cases}}{\text{the total number of cases}}$, so E is not possible. Also, since the number of good cases must be an integer, C is not possible. From there, your chances of guessing the right answer are slightly higher. ~Extremelysupercooldude

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 $r=\boxed{(E)\sqrt{30}}$.

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 $\boxed{\textbf{(B)}\ 13}$.

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 $BI=\boxed{\textbf{(A) } 15}$.

Note that if $n$ is the number of friends each person has, then $n$ can be any integer from $1$ to $4$, inclusive, truly. One person can have at most 4 friends since they cannot be all friends (stated in the problem). 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. (Note: If you aren’t familiar with one-to-one correspondences, think of it like this: the number of ways to choose 4 friends is equal to number of ways to exclude one friend from your friend group. Hence, since the number of ways to choose 1 friend is the same thing as choosing 1 to not be friends with, $n=1$ and $n=4$ have the same number of ways. Similarly, $n=2$ and $n=3$ have the same number of ways as well. ~peelybonehead) 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 $\boxed{\textbf{(B)}\ 170}$. I thought the answer was 60??? MisW (talk) Reply: 60 is just for the subcase with one large cycle. We can also calculate the triangular configurations by applying $\frac{\binom{6}{3}}{2} = \frac{20}{2}=10$ (Because choosing $A$,$B$ and $C$ is the same as choosing $D$,$E$ and $F$. For the hexagonal configurations, we know that the total amount of combinations is $6!=720$. However, we must correct for our overcounting because of rotation and reflection. We have that there are ${720 \over 6 \cdot 2} = 60$ because there $6$ rotations of the hexagon and 2 reflections (clockwise and counterclockwise) for each valid way. We can also calculate the hexagonal configurations by placing each person at a random vertex (6!) and dividing for rotations (by 6) and reflections (by 2): For the hexagonal configuration, it is essentially congruent to counting the number of paths that start and end with A. From $A$, we have $5$ options, then the next one has $4$, and so on, so we have: $5\cdot 4 \hdots 1 \cdot 1$, where the final $1$ is because the final point needs to reconnect to $A$. Thus, there are $5! = 120$ ways. Dividing by $2$ since we overcounted by accounting for both directions, we have $120/2 = \fbox{60}$ You could ALSO count the hexagonal arrangements by considering the rotation and then reflection. Each arrangement of the six friends maps to a sequence _ _ _ _ _ _, each person is friends with their two neighbors, and the rightmost and leftmost people are friends with each other. Notice that any configuration can be rotated to begin with A, so we have the sequence: A _ _ _ _ _. Now there are $5!$ ways to fill these blanks. For example, ABCDEF. However, since A is friends with both B and F, this sequence can be 'flipped' to yield AFEDCB, which corresponds to the same set of friendships. Since each sequence can be flipped to yield another valid sequence, we have overcounted by a factor of two. Thus the final number of hexagonal configurations = $\frac{5!}{2} = \fbox{60}$

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 $6 \rightarrow \boxed{B}$.

Add the two equations. $2a^2 + 2b^2 + 2c^2 - 2ab - 2ac - 2bc = 14$. Now, this can be rearranged and factored. $(a^2 - 2ab + b^2) + (a^2 - 2ac + c^2) + (b^2 - 2bc + c^2) = 14$ $(a - b)^2 + (a - c)^2 + (b - c)^2 = 14$ $a$, $b$, and $c$ are all integers, so the three terms on the left side of the equation must all be perfect squares. We see that the only is possibility is $14 = 9 + 4 + 1$. $(a-c)^2 = 9 \Rightarrow a-c = 3$, since $a-c$ is the biggest difference. It is impossible to determine by inspection whether $a-b = 1$ or $2$, or whether $b-c = 1$ or $2$. 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 $\boxed{\textbf{(E)}\ 253}$.

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 $27-7=20$ ... $\framebox{C}$. Note 1: Indeed, let $t_1=AB$ where $B=t_1\cap t_2$, and $C$ be the other endpoint of $t_2$ that is not $B$. Draw a line through $C$ parallel to $t_1$. This line passes through the center $O$ of the cube and therefore we see that the reflection of $A,B,C$, denoted by $A', B', C'$, respectively, lie on the same plane containing $A,B,C$. Thus $p_j \cap Q$ is the regular hexagon $ABCA'B'C'$. To count the number of these hexagons, just notice that each short trace uniquely determine a hexagon (by drawing the plane through this trace and the center), and that each face has $4$ short traces. Therefore, there are $4$ such hexagons. Note 2: The quick way to prove the fact that none of the planes described in case 3 and case 4 share the same trace is as follows: each of these plane contains the center and therefore the intersection of each pair of them is a line through the center, which obviously does not contain any traces.

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 $0.16 = \boxed{\frac{4}{25} \textbf{(C)} }$

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 power function for positive $x$). $f(x)$ is used to establish inequalities when we change the exponent and keep the base constant. $g(x)$ 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 $\boxed{\textbf{(C)} 1341}$. Note 1: We claim that $0 < a_1 < a_3 < ... < a_{2011} < a_{2010} < ... < a_4 < a_2 < 1$. We can use induction to prove this statement. (not necessary for AMC): Base Case: We have already shown the base case above, where $0 < a_1 < a_2 < 1$. Inductive Step: Rearranging in decreasing order gives $1 > b_1 = a_2 > b_2 = a_4 > ... > b_{1005} = a_{2010} > b_{1006} = a_{2011} > ... > b_{2010} = a_3 > b_{2011} = a_1 > 0$.

Our goal is to determine how many times the graph of $nf(xf(x))=x$ intersects the graph of $y=x$. (Conversely, we can also divide the equation by $n$ to get $f(xf(x))=\frac{x}{n}$ and look at the graph $y=\frac{x}{n}$) 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 $4+8+12...+4n$. We want this sum to be greater than 2012 which occurs when $n=32 \Rightarrow \boxed{(C)}$ .