1988 IMO Problems/Problem 4

Show that the solution set of the inequality

$\sum_{k=1}^{70}\frac{k}{x-k}\ge\frac{5}{4}$

is a union of disjoint intervals, the sum of whose length is $1988$ .

Consider the graph of $f(x)=\sum_{k=1}^{70}\frac{k}{x-k}\ge\frac{5}{4}$ . On the values of $x$ between $n$ and $n+1$ for $n\in\mathbb{N}$ $1\le n\le 69$ , the terms of the form $\frac{k}{x-k}$ for $k\ne n,n+1$ have a finite range. In contrast, the term $\frac{n}{x-n}$ has an infinite range, from $+\infty$ to $n$ . Similarly, the term $\frac{n+1}{x-n-1}$ has infinite range from $-n-1$ to $-\infty$ . Thus, since the two undefined values occur at the distinct endpoints, we can deduce that $f(x)$ takes on all values between $+\infty$ and $-\infty$ for $x\in(n,n+1)$ . Thus, by the Intermediate Value Theorem, we are garunteed a $n<r_n<n+1$ such that $f(r_n)=\frac{5}{4}$ . Additionally, we have that for $x>70$ , the value of $f(x)$ goes from $+\infty$ to $0$ , since as $x$ increases, all the terms go to $0$ . Thus, there exists some $r_{70}>70$ such that $f(r_{70})=\frac{5}{4}$ and so $f(x)\ge\frac{5}{4}$ for $x\in(70,r_{70})$ .

So, we have $70$ $r_i$ such that $f(r_i)=\frac{5}{4}$ . There are obviously no other such $r_i$ since $f(x)=\frac{5}{4}$ yields a polynomial of degree $70$ when combining fractions. Thus, we have that the solution set to the inequality $f(x)\ge\frac{5}{4}$ is the union of the intervals $(n,r_n]$ (since if $f(x)<\frac{5}{4}$ for $x\in(n,r_n)$ then there would exist another solution to the equation $f(x)=\frac{5}{4}$ .

Thus we have proven that the solution set is the union of disjoint intervals. Now we are to prove that the sum of their lengths is $1988$ .

The sum of their lengths is $r_1+r_2+\cdots+r_{70}-(1+2+\cdots+70)=r_1+r_2+\cdots+r_{70}-35\cdot71$ . We have that the equation $f(x)=\frac{5}{4}$ yields a polynomial with roots $r_i$ . Thus, opposite of the coeficient of $x^{69}$ divided by the leading coefficient is the sum of the $r_i$ . It is easy to see that the coefficient of $x^{69}$ is $-5(1+2+\cdots+70)-4(1+2+\cdots+70)=-9\cdot35\cdot 71$ . Thus, since the leading coefficient is $5$ we have $r_1+r_2+\cdots+r_{70}=9\cdot7\cdot71$ . Thus, the sum of the lengths of the intervals is $63\cdot71-35\cdot71=28\cdot71=1988$ as desired.

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1988 IMO Problems/Problem 4