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