IMO 1984 · Problem 5

Let \(d\) be the sum of the lengths of all the diagonals of a plane convex polygon with \(n\) vertices (\(n>3\)), and let \(p\) be its perimeter. Prove that \[ n-3<\frac{2d}{p}<\left[\frac n2\right]\left[\frac{n+1}{2}\right]-2, \] where \([x]\) denotes the greatest integer not exceeding \(x\).

Official text of the problem. Official problem paper (PDF)

Solution

Let the vertices, in boundary order, be \(A_1,\ldots,A_n\), with indices interpreted modulo \(n\). Write \(e_i=|A_iA_{i+1}|\), so that \(\sum_i e_i=p\). The argument also allows collinear consecutive vertices, provided the polygon has positive area.

For the lower bound, consider any two nonadjacent sides \(A_iA_{i+1}\) and \(A_jA_{j+1}\). By convexity, the segments \(A_iA_j\) and \(A_{i+1}A_{j+1}\), which join alternating vertices, intersect at a point \(X\). The triangle inequality gives \[ \begin{aligned} e_i+e_j &\le |A_iX|+|XA_{i+1}|+|A_jX|+|XA_{j+1}|\\ &=|A_iA_j|+|A_{i+1}A_{j+1}|. \end{aligned} \] Equality would require \(X\) to lie on both sides \(A_iA_{i+1}\) and \(A_jA_{j+1}\). These nonadjacent sides are disjoint, so in fact \[ e_i+e_j<|A_iA_j|+|A_{i+1}A_{j+1}|. \]

Sum over all unordered pairs of nonadjacent sides. Each side is paired with exactly \(n-3\) others, so the left-hand sum is \((n-3)p\). The terms \(|A_iA_j|\) run through every diagonal once. Shifting both indices by one permutes the diagonals, so the terms \(|A_{i+1}A_{j+1}|\) also sum to \(d\). Hence \[ (n-3)p<2d. \]

For the upper bound, define \[ T_k=\sum_{i=1}^{n}|A_iA_{i+k}|, \qquad 1\le k\le n-1. \] Following the boundary from \(A_i\) to \(A_{i+k}\) and applying the triangle inequality yields \[ |A_iA_{i+k}|\le e_i+e_{i+1}+\cdots+e_{i+k-1}. \] On summing over \(i\), each side occurs \(k\) times, giving \(T_k\le kp\). Also \(T_k=T_{n-k}\), so \[ T_k\le \min(k,n-k)\,p. \]

At least one triple of consecutive vertices is noncollinear: otherwise, since consecutive vertices are distinct, all vertices would lie on one line. Thus at least one triangle inequality contributing to \(T_2\le2p\) is strict. Since \(n\ge4\), we therefore have \[ T_2<2p=\min(2,n-2)\,p. \]

Every unordered pair of distinct vertices is counted twice in \(\sum_{k=1}^{n-1}T_k\). Consequently, \[ 2(p+d) =\sum_{k=1}^{n-1}T_k <p\sum_{k=1}^{n-1}\min(k,n-k). \] If \(n=2m\), the last sum is \[ 2(1+\cdots+(m-1))+m=m^2; \] if \(n=2m+1\), it is \[ 2(1+\cdots+m)=m(m+1). \] In either case it equals \[ \left[\frac n2\right]\left[\frac{n+1}{2}\right]. \] Subtracting \(2p\) and dividing by \(p>0\), then combining with the lower bound, proves \[ \boxed{\displaystyle n-3<\frac{2d}{p} <\left[\frac n2\right]\left[\frac{n+1}{2}\right]-2}. \]

The solution was checked against John Scholes' IMO solution archive (Kalva).