2010 AMC 12B Problem 24

Question

The set of real numbers

x

for which

\begin{aligned} &\frac{1}{x-2009}+\frac{1}{x-2010} \\ &\quad {}+\frac{1}{x-2011}\ge1 \end{aligned}

is the union of intervals of the form

a\lt x\le b.

What is the sum of the lengths of these intervals?

\dfrac{1003}{335}

\dfrac{1004}{335}

3

\dfrac{403}{134}

\dfrac{202}{67}

The left side is decreasing on each interval between its vertical asymptotes, so each solution interval ends at a root of the equation

=1

The three right endpoints are the roots of a cubic; sum them with Vieta

Let

f(x)

be the left-hand side. On each interval between consecutive asymptotes

2009, 2010, 2011,

the function

f

is decreasing, and

f\lt1

for all

x\lt2009.

On each of

(2009,2010),

(2010,2011),

and

(2011,\infty),

the solution is the part from the left asymptote up to a value

x_i

where

f(x_i)=1.

So the solution set consists of three intervals with left endpoints

2009, 2010, 2011

and right endpoints

x_1, x_2, x_3.

The total length is

(x_1-2009)

+(x_2-2010)

+(x_3-2011)

=x_1+x_2+x_3-6030.

Clearing denominators in

f(x)=1

gives

\begin{aligned} &x^3 \\ &\quad \small{}-(2009+2010+2011+3)x^2 \\ &\quad {}+\cdots=0, \end{aligned}

whose roots are

x_1, x_2, x_3.

By Vieta,

x_1+x_2+x_3=6033,

so the sum of lengths is

6033-6030=3.

Thus, the correct answer is C.

Problem 24 in other years