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Difference between revisions of "2010 AMC 12B Problems/Problem 24"
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<cmath>x^3-3x^2-x+1=0</cmath>
 
<cmath>x^3-3x^2-x+1=0</cmath>
  
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And now our job is simply to find the sum of the roots of <math>x^3-3x^2-x+1=0</math>.
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And now our job is simply to find the sum of the roots of <math>x^3-3x^2-x+1</math>.
  
 
<math>x^3-3x^2-x+1=(x-p)(x-q)(x-r)</math>
 
<math>x^3-3x^2-x+1=(x-p)(x-q)(x-r)</math>

Revision as of 23:41, 1 February 2011

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 $(p-(-1))+(q-0)+(r-1)=p+q+r$.

\[\frac{1}{x+1}+\frac{1}{x}+\frac{1}{x-1}=1\] \[x(x-1)+(x-1)(x+1)+x(x+1)=x(x-1)(x+1)\] \[x^3-3x^2-x+1=0\]

And now our job is simply to find the sum of the roots of $x^3-3x^2-x+1$.

$x^3-3x^2-x+1=(x-p)(x-q)(x-r)$ $x^3-3x^2-x+1=x^3-(q+p+r)x^2+(qp+qr+pr)x-qpr$

and the sum $q+p+r$ is the negative of the coefficient on the $x^2$ term, which is $3\Rightarrow\boxed{C}$

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