IMO 1988 SL 16

Origin: IRE

Problem

Show that the solution set of the inequality  k=1 k x −k \geq5 is a union of disjoint intervals the sum of whose lengths is 1988.

Solution

Let f(x) = 70 k=1 k x−k. For all integers i = 1, . . . , 70 we have that f(x) tends to plus infinity as x tends downward to i, and f(x) tends to minus infinity as x tends upward to i. As x tends to infinity, f(x) tends to 0. Hence it follows that there exist x1, x2, . . . , x70 such that 1 < x1 < 2 < x2 < 3 < \cdot \cdot \cdot < x69 < 70 < x70 and f(xi) = 5 4 for all i = 1, . . . , 70. Then the solution to the inequality is given by S = %70 i=1(i, xi]. For numbers x for which f(x) is well-defined, the equality f(x) = 5 4 is equivalent to

p(x) =

j=1 (x −j) −4  k=1 k

j=1 j̸=k (x −j) = 0. The numbers x1, x2, . . . , x70 are then the zeros of this polynomial. The sum 70 i=1 xi is then equal to minus the coefficient of x69 in p, which equals 70 i=1
i + 4 5i  . Finally, |S| =  i=1 (xi −i) = 4 5 \cdot  i=1 i = 4 5 \cdot 70 \cdot 71 = 1988 .