AMC12 2012 A

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 $n=32 \Rightarrow \boxed{(C)}$ .