Problem:
The set of real numbers x for which
x−20091+x−20101+x−20111≥1
is the union of intervals of the form a<x≤b. What is the sum of the lengths of these intervals?
Answer Choices:
A. 3351003
B. 3351004
C. 3
D. 134403
E. 67202
Solution:
Let f(x)=x−20091+x−20101+x−20111. Note that
f(x)−f(y)=(y−x)((x−2009)(y−2009)1+(x−2010)(y−2010)1+(x−2011)(y−2011)1)
If x<y<2009, then y−x>0,
(x−2009)(y−2009)1>0,(x−2010)(y−2010)1>0 and (x−2011)(y−2011)1>0
Thus f is decreasing on the interval x<2009, and because f(x)<0 for x<0, it follows that no values x<2009 satisfy f(x)≥1.
If 2009<x<y<2010, then f(x)−f(y)>0 as before. Thus f is decreasing in the interval 2009<x<2010. Moreover, f(2009+101)=10−910−1910>1 and f(2010−101)=910−10−1110<1. Thus there is a number 2009<x1<2010 such that f(x)≥1 for 2009<x≤x1 and f(x)<1 for x1<x<2010.
Similarly, f is decreasing on the interval 2010<x<2011,f(2010+101)>1, and f(2011−101)<1. Thus there is a number 2010<x2<2011 such that f(x)≥1 for 2010<x≤x2 and f(x)<1 for x2<x<2011.
Finally, f is decreasing on the interval x>2011,f(2011+101)>1, and f(2014)=51+41+31<1. Thus there is a number x3>2011 such that f(x)≥1 for 2011<x≤x3 and f(x)<1 for x>x3.
The required sum of the lengths of these three intervals is
x1−2009+x2−2010+x3−2011=x1+x2+x3−6020
Multiplying both sides of the equation
x−20091+x−20101+x−20111=1
by (x−2009)(x−2010)(x−2011) and collecting terms on one side of the equation gives
x3−x2(2009+2010+2011+1+1+1)+ax+b=0
where a and b are real numbers. The three roots of this equation are x1,x2, and x3. Thus x1+x2+x3=6020+3, and consequently the required sum equals 3 .
The problems on this page are the property of the MAA's American Mathematics Competitions