2010 AMC 12B Problems/Problem 24

Problem 24

The set of real numbers $x$ for which

$\dfrac{1}{x-2009}+\dfrac{1}{x-2010}+\dfrac{1}{x-2011}\ge1$

is the union of intervals of the form $a<x\le b$ . What is the sum of the lengths of these intervals?

$\textbf{(A)}\ \dfrac{1003}{335} \qquad \textbf{(B)}\ \dfrac{1004}{335} \qquad \textbf{(C)}\ 3 \qquad \textbf{(D)}\ \dfrac{403}{134} \qquad \textbf{(E)}\ \dfrac{202}{67}$

Solution

Because the right side of the inequality is a horizontal line, the left side can be translated horizontally by any value and the intervals will remain the same. For simplicity of calculation, we will find the intervals where $\frac{1}{x+1}+\frac{1}{x}+\frac{1}{x-1}\ge1$ The left side of the inequality has three vertical asymptotes at $x=\{-1,0,1\}$ . Values immediately to the left of each asymptote are hugely negative, and values immediately to the right of each asymptote are hugely positive. In addition, the function has a horizontal asymptote at $y=0$ . The function intersects $1$ at some point from $x=-1$ to $x=0$ , and from $x=0$ to $x=1$ , and at some point to the right of $x=1$ . The intervals where the function is greater than 1 are between the points where $x=1$ and the vertical asymptotes.

If $p$ , $q$ , and $r$ are values of x where $\frac{1}{x+1}+\frac{1}{x}+\frac{1}{x-1}=1$ , then the sum of the lengths of the intervals is $(p-(-1))+(q-0)+(r-1)=p+q+r$ .