Let d be the sum of the lengths of all the diagonals of a plane convex polygon with n > 3 vertices. Let p be its perimeter. Prove that:
n - 3 < 2d/p < [n/2] [(n+1)/2] - 2, where [x] denotes the greatest integer not exceeding x.
Solution
Given any diagonal AX, let B be the next vertex counterclockwise from A, and Y the next vertex counterclockwise from X. Then the diagonals AX and BY intersect at K. AK + KB > AB and XK + KY > XY, so AX + BY > AB + XY. Keeping A fixed and summing over X gives n - 3 cases. So if we then sum over A we get every diagonal appearing four times on the lhs and every side appearing 2(n-3) times on the rhs, giving 4d > 2(n-3)p.
Denote the vertices as A0, ... , An-1 and take subscripts mod n. The ends of a diagonal AX are connected by r sides and n-r sides. The idea of the upper limit is that its length is less than the sum of the shorter number of sides. Evaluating it is slightly awkward.
We consider n odd and n even separately. Let n = 2m+1. For the diagonal AiAi+r with r ≤ m, we have AiAi+r ≤ AiAi+2 + ... + AiAi+r. Summing from r = 2 to m gives for the rhs (m-1)AiAi+1 + (m-1)Ai+1Ai+2 + (m-2)Ai+2Ai+3 + (m-3)Ai+3Ai+4 + ... + 1.Ai+m-1Ai+m. Now summing over i gives d for the lhs and p( (m-1) + (1 + 2 + ... + m-1) ) = p( (m2 + m - 2)/2 ) for the rhs. So we get 2d/p ≤ m2 + m - 2 = [n/2] [(n+1)/2] - 2.
Let n = 2m. As before we have AiAi+r <= AiAi+2 + ... + AiAi+r for 2 ≤ r ≤ m-1. We may also take AiAi+m ≤ p/2. Summing as in the even case we get 2d/p = m2 - 2 = [n/2] [(n+1)/2] - 2.
Solutions are also available in Murray S Klamkin, International Mathematical Olympiads 1978-1985, MAA 1986, and in István Reiman, International Mathematical Olympiad 1959-1999, ISBN 189-8855-48-X.
© John Scholes
jscholes@kalva.demon.co.uk
19 Oct 1998