IMO 1984 SL 4

Origin: MON

Problem

Let d be the sum of the lengths of all diagonals of a convex polygon of n (n > 3) vertices and let p be its perimeter. Prove that n −3 < d p < 1 n  n + 1 −2  .

Solution

Consider the convex n-gon A1A2 . . . An (the indices are considered modulo n). For any diagonal AiAj we have AiAj +Ai+1Aj+1 > AiAi+1 +AjAj+1. Summing all such n(n −3)/2 inequalities, we obtain 2d > (n−3)p, proving the first inequality. Let us now prove the second inequality. We notice that for each diagonal AiAi+j (we may assume w.l.o.g. that j \leq[n/2]) the following relation holds: AiAi+j < AiAi+1 + \cdot \cdot \cdot + Ai+j−1Ai+j. (1)

If n = 2k + 1, then summing the inequalities (1) for j = 2, 3, . . . , k and i = 1, 2, . . . , n yields d < (2 + 3 + \cdot \cdot \cdot + k)p = ([n/2] [n + 1/2] −2) p/2. If n = 2k, then summing the inequalities (1) for j = 2, 3, . . . , k −1, i = 1, 2, . . . , n and for j = k, i = 1, 2, . . ., k again yields d < (2 + 3 + \cdot \cdot \cdot + (k −1) + k/2)p = 1 2 ([n/2] [n + 1/2] −2) p.