IMO 1987 SL 7

Origin: NET

Problem

Given five real numbers u0, u1, u2, u3, u4, prove that it is always possible to find five real numbers v0, v1, v2, v3, v4 that satisfy the following conditions: (i) ui −vi \inN. (ii)  0\leqi<j\leq4(vi −vj)2 < 4.

Solution

For all real numbers v the following inequality holds:  0\leqi<j\leq4 (vi −vj)2 \leq5  i=0 (vi −v)2. (1) Indeed,  0\leqi<j\leq4 (vi −vj)2 =  0\leqi<j\leq4 [(vi −v) −(vj −v)]2 = 5  i=0 (vi −v)2 −

 i=0 (vi −v) \leq5  i=0 (vi −v)2. Let us first take vi’s, satisfying condition (1), so that w.l.o.g. v0 \leqv1 \leq v2 \leqv3 \leqv4 \leq1 + v0. Defining v5 = 1 + v0, we see that one of the differences vj+1 −vj, j = 0, . . . , 4, is at most 1/5. Take v = (vj+1 + vj)/2, and then place the other three vj’s in the segment [v −1/2, v + 1/2]. Now we have |v −vj| \leq1/10, |v −vj+1| \leq1/10, and |v −vk| \leq1/2, for any k different from j, j + 1. The vi’s thus obtained have the required property. In fact, using the inequality (1), we obtain  0\leqi<j\leq4 (vi −vj)2 \leq5

 1 2

• 3 1 2 = 3.85 < 4. Remark. The best possible estimate for the right-hand side is 2.