2012 AMC 12A Problems/Problem 25

Problem

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

Our goal is to determine how many times intersects the line $y=x$ .

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 $\int_0^n \! 4x \, \mathrm{d}x$ . It is easy to see that this is equal to $2n^2$ . The problem then simplifies to the algebraic expression $2n^2=2012$ , which simplifies to, $n^2=1006$ , and then to $n=\sqrt{1006}$ , which rounds up to 32. $\boxed{\text{C}}$ .

2012 AMC 12A (Problems • Answer Key • Resources)
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2012 AMC 12A Problems/Problem 25

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