Difference between revisions of "2010 AMC 12B Problems/Problem 24"
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<math>\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}</math> | <math>\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}</math> | ||
| + | |||
| + | == 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 <cmath>\frac{1}{x+1}+\frac{1}{x}+\frac{1}{x-1}\ge1</cmath> | ||
| + | The left side of the inequality has three vertical asymptotes at <math>x=\{-1,0,1\}</math>. 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 <math>y=0</math>. The function intersects <math>1</math> at some point from <math>x=-1</math> to <math>x=0</math>, and from <math>x=0</math> to <math>x=1</math>, and at some point to the right of <math>x=1</math>. The intervals where the function is greater than 1 are between the points where <math>x=1</math> and the vertical asymptotes. | ||
| + | |||
| + | If <math>p</math>, <math>q</math>, and <math>r</math> are values of x where <math>\frac{1}{x+1}+\frac{1}{x}+\frac{1}{x-1}=1</math>, then the sum of the lengths of the intervals is <math>(p-(-1))+(q-0)+(r-1)=p+q+r</math>. | ||
| + | |||
| + | <cmath>\frac{1}{x+1}+\frac{1}{x}+\frac{1}{x-1}=1</cmath> | ||
| + | <cmath>x(x-1)+(x-1)(x+1)+x(x+1)=x(x-1)(x+1)</cmath> | ||
| + | <cmath>x^3-3x^2-x+1=0</cmath> | ||
| + | |||
| + | And now our job is simply to find the sum of the roots of <math>x^3-3x^2-x+1=0</math>. | ||
| + | |||
| + | <math>x^3-3x^2-x+1=(x-p)(x-q)(x-r)</math> | ||
| + | <math>x^3-3x^2-x+1=x^3-(q+p+r)x^2+(qp+qr+pr)x-qpr</math> | ||
| + | |||
| + | and the sum <math>q+p+r</math> is the negative of the coefficient on the <math>x^2</math> term, which is <math>3\Rightarrow\boxed{C}</math> | ||
Revision as of 23:39, 1 February 2011
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
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 ![\[\frac{1}{x+1}+\frac{1}{x}+\frac{1}{x-1}\ge1\]](http://latex.artofproblemsolving.com/3/2/4/324910843f199b5977e90e0a097a86d713c7c1e6.png)
The left side of the inequality has three vertical asymptotes at 
. 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 
. The function intersects 
at some point from 
to 
, and from to 
, and at some point to the right of . The intervals where the function is greater than 1 are between the points where and the vertical asymptotes.
If 
, 
, and 
are values of x where 
, then the sum of the lengths of the intervals is 
.
![\[\frac{1}{x+1}+\frac{1}{x}+\frac{1}{x-1}=1\]](http://latex.artofproblemsolving.com/e/e/7/ee708104d86af6b6b8ed65c2e151d9e270bb9073.png)
![\[x(x-1)+(x-1)(x+1)+x(x+1)=x(x-1)(x+1)\]](http://latex.artofproblemsolving.com/4/1/9/4190928593194dd20e2da72741eb2162c926dcaa.png)
![\[x^3-3x^2-x+1=0\]](http://latex.artofproblemsolving.com/0/a/6/0a6c95279cd8f4308f9441a7e5beaf5219fa54cf.png)
And now our job is simply to find the sum of the roots of 
.
and the sum 
is the negative of the coefficient on the 
term, which is 
