IMO 1968 SL 6

Origin: HUN

Problem

If $a_i$ $(i = 1, 2, \ldots, n)$ are distinct non-zero real numbers, prove that the equation

$$\frac{a_1}{a_1 - x} + \frac{a_2}{a_2 - x} + \cdots + \frac{a_n}{a_n - x} = n$$

has at least $n - 1$ real roots.

Solution

We define

$$f(x) = \frac{a_1}{a_1 - x} + \frac{a_2}{a_2 - x} + \cdots + \frac{a_n}{a_n - x}.$$

Let us assume w.l.o.g.

$$a_1 < a_2 < \cdots < a_n.$$

We note that for all $1 \leq i < n$ the function $f$ is continuous in the interval $(a_i, a_{i+1})$ and satisfies

$$\lim_{x \to a_i} f(x) = -\infty$$

and

$$\lim_{x \to a_{i+1}} f(x) = \infty.$$

Hence the equation

$$f(x) = n$$

will have a real solution in each of the $n - 1$ intervals $(a_i, a_{i+1})$.

Remark.

In fact, this equation has exactly $n$ solutions, and hence they are all real. Moreover, the solutions are distinct if all $a_i$ are of the same sign, since $x = 0$ is an evident solution.