2010 AMC 12B Problems/Problem 24
Problem 24
The set of real numbers 
for which
![\[\dfrac{1}{x-2009}+\dfrac{1}{x-2010}+\dfrac{1}{x-2011}\ge1\]](http://latex.artofproblemsolving.com/f/9/d/f9d51dcb945e183da3554b60e54e9f18e92536e0.png)
is the union of intervals of the form 
. What is the sum of the lengths of these intervals?
Solution
First, we shift the graph so it will be easier to manipulate
![\[\dfrac{1}{x - 1} + \dfrac{1}{x} + \dfrac{1}{x+1} \ge 1\]](http://latex.artofproblemsolving.com/5/2/f/52f5f635e5d6f2864d9415163aacb0526724f91c.png)
We can see that this is equivalent to the original problem. We now want to solve this equation.
![\[\frac{(x)(x + 1) + (x-1)(x+1) + (x)(x-1)}{(x-1)(x)(x+1)} \ge 1\]](http://latex.artofproblemsolving.com/c/a/1/ca1c23f016acdfa366c5014f635cc3bddf0545f1.png)
![\[\frac{(x^2+x+x^2-1+x^2-x)}{x^3-x} \ge 1\]](http://latex.artofproblemsolving.com/a/4/4/a4435569c680b8d14fbb1a5ed9e1f17a5b11dd2a.png)
![\[\frac{3x^2-1}{x^3-x}\ge 1\]](http://latex.artofproblemsolving.com/f/6/5/f6528124edcb0ea512c081d33f6d41e0f5755695.png)
We first find where this will lead to equality. We have 
.
Our answer is just the sum of the roots, and using Vieta's, we get that to be 
.
