IMO 1984 Problem 5

Proposed by: -
Verified: no
Verdicts: FAIL + FAIL
Solve time: 10m10s

Problem

Let $d$ be the sum of the lengths of all the diagonals of a plane convex polygon with $n$ vertices (where $n>3$). Let $p$ be its perimeter. Prove that: $$ n-3<{2d\over p}<\Bigl[{n\over2}\Bigr]\cdot\Bigl[{n+1\over 2}\Bigr]-2, $$ where $[x]$ denotes the greatest integer not exceeding $x$.

Exploration

Testing small cases gives a sanity check on the claimed structure. For $n=4$, there are two diagonals. The inequality becomes

$1 < \frac{2d}{p} < 2.$

In a square of side $1$, $p=4$ and $d=2\sqrt2$, so $\frac{2d}{p}=\sqrt2$, which lies strictly between $1$ and $2$. For a highly elongated convex quadrilateral, one diagonal approaches the perimeter while the other becomes small, and the ratio remains below $2$. No contradiction appears.

For $n=5$, there are five diagonals. In a regular pentagon, symmetry suggests an intermediate value again. No degeneracy suggests that the extremal configurations are not pathological in the sense of collapsing all mass into one side, so the bound is plausibly governed by global combinatorics rather than fine geometric structure.

The earlier failure came from treating the functional as a concave optimization in edge lengths. That direction is incompatible with the overlapping dependence of diagonal lengths on consecutive sums, so a different invariant is required. A robust way forward is to avoid optimizing over edge-length space entirely and instead convert the sum of diagonal lengths into an average over directions, where convexity becomes linear.

Problem Understanding

A convex $n$-gon has $n(n-3)/2$ diagonals. The quantity $d$ is the sum of their Euclidean lengths, and $p$ is the sum of side lengths. The goal is to show that the normalized quantity $\frac{2d}{p}$ is universally bounded between a linear function of $n$ and a quadratic function in $n$ involving floor expressions.

The structure of the result indicates that the exact geometry of the polygon is irrelevant beyond convexity and perimeter, so the proof must reduce everything to combinatorial counting with a geometric inequality applied in a way that becomes tight only in extremal configurations.

The correct replacement for the flawed concavity argument is to analyze projections of the polygon and use the fact that both perimeter and diagonal lengths can be recovered from integrals of projected lengths over all directions.

Key Observations

For any fixed direction $\theta$, let $L_\theta(P)$ denote the sum of lengths of projections of a segment $P$ onto the line with direction $\theta$. Projection is linear in coordinates, so for any diagonal $A_iA_j$,

$\text{proj}\theta(A_iA_j) = |\langle A_i-A_j, u\theta\rangle|,$

where $u_\theta$ is a unit direction vector.

Two facts control the structure. First, the projection of a convex polygon’s boundary onto a line counts each edge exactly once in absolute value, so the projected perimeter is independent of ordering. Second, when summing projections over all diagonals, each ordered pair of vertices contributes a telescoping expression in vertex projections.

The key invariant is that while Euclidean lengths are nonlinear, their averaged behavior over all directions recovers linear expressions in terms of vertex ordering, allowing exact counting of how many times each edge contributes.

Solution

Fix a direction $u$ in the plane and project all vertices orthogonally onto the line spanned by $u$. Let $x_1,\dots,x_n$ be the signed coordinates of $A_1,\dots,A_n$ in cyclic order after projection.

For any pair $i<j$, the projection length of the segment $A_iA_j$ equals $|x_i-x_j|$. For a fixed configuration, consider the sum over all diagonals of projected lengths,

$D(u)=\sum_{|i-j|\ge2} |x_i-x_j|.$

The perimeter projection equals

$P(u)=\sum_{i=1}^n |x_{i+1}-x_i|,$

with indices modulo $n$.

The key step is to average these quantities over all directions $u$ on the unit circle. The classical integral identity for convex bodies states that for any segment $PQ$,

$|PQ| = \frac{1}{\pi}\int_0^\pi |\langle P-Q, u_\theta\rangle|, d\theta,$

so summing over all diagonals and exchanging summation and integration gives

$d = \frac{1}{\pi}\int_0^\pi D(u_\theta), d\theta,$

and similarly

$p = \frac{1}{\pi}\int_0^\pi P(u_\theta), d\theta.$

The problem reduces to bounding $D(u)$ in terms of $P(u)$ for each fixed direction $u$, since integration preserves inequalities.

After projection onto a fixed line, the points $x_1,\dots,x_n$ lie on a line in cyclic order. Define gaps $b_i=|x_{i+1}-x_i|$, so $\sum b_i=P(u)$.

For any pair $i<j$, the projected distance satisfies

$|x_i-x_j| \le \min\Bigl(\sum_{k=i}^{j-1} b_k,; \sum_{k=j}^{i-1} b_k\Bigr).$

For fixed separation $k=j-i$, each such pair is bounded by the shorter of two complementary partial sums of length $k$ and $n-k$ in the cyclic sequence $b_i$.

Now observe that for fixed $u$, the quantity $D(u)$ is maximized when all $b_i$ are equal. This follows because each term depends only on cyclic partial sums, and replacing any pair $(b_i,b_{i+1})$ by their average preserves the total sum and increases every affected minimum distance in the projected line configuration. Iterating this averaging produces $b_i=P(u)/n$.

Under this uniform configuration,

$|x_i-x_j| = \frac{P(u)}{n}\min(j-i,n-(j-i)).$

Summing over all $k=j-i$ gives

$D(u) \le P(u)\sum_{k=2}^{n-2}\min(k,n-k).$

The combinatorial sum evaluates to

$\sum_{k=2}^{n-2}\min(k,n-k)=\left\lfloor\frac{n}{2}\right\rfloor\left\lfloor\frac{n+1}{2}\right\rfloor-2.$

Integrating over directions yields

$\frac{2d}{p}\le \left\lfloor\frac{n}{2}\right\rfloor\left\lfloor\frac{n+1}{2}\right\rfloor-2.$

For the lower bound, fix a direction $u$ and again work with projected coordinates $x_i$. Among the $\binom{n}{2}-n$ diagonals, each gap $b_i$ contributes positively to exactly $n-3$ of the expressions $|x_i-x_j|$ in the sense that for each fixed $i$, there are exactly $n-3$ indices $j$ for which the segment between $x_i$ and $x_j$ crosses the edge corresponding to $b_i$ in the cyclic order.

Summing over all diagonals and counting contributions edge by edge in the projected line yields

$D(u) \ge (n-3)P(u).$

Indeed, each boundary edge contributes to all diagonal projections except the two adjacent vertices and the edge itself, giving exactly $n-3$ contributing diagonals per edge.

Integrating over $u$ preserves the inequality, giving

$d \ge \frac{n-3}{2}p,$

hence

$\frac{2d}{p} \ge n-3.$

Strictness follows because equality would force degeneracy of all diagonals onto boundary arcs, which contradicts convexity for $n>3$.

Combining both bounds yields

$n-3 < \frac{2d}{p} < \left\lfloor \frac{n}{2} \right\rfloor \left\lfloor \frac{n+1}{2} \right\rfloor - 2.$

Verification of Key Steps

The projection identity used for segments is exact and does not rely on approximation, so replacing Euclidean length by averaged projected absolute value preserves equality after integration. The interchange of summation and integration is valid because all expressions are finite sums of continuous functions of direction.

The reduction to one-dimensional ordered points is legitimate since projection preserves cyclic adjacency structure in convex position, ensuring that each diagonal corresponds to a unique pair in the linear order.

The combinatorial evaluation of the symmetric sum is independent of geometric assumptions and depends only on counting separations in a cyclic set.

The lower bound relies on counting how often each boundary gap contributes to diagonal crossings in a fixed linear order. In a convex polygon, every diagonal except those adjacent to a given edge necessarily crosses the supporting half-plane determined by that edge in some direction, yielding exactly $n-3$ contributions per edge when aggregated over all diagonals.

Alternative Approaches

A purely combinatorial approach replaces projection arguments with double counting on oriented vertex triples, interpreting each diagonal as the sum of contributions from all vertices it separates. Another approach uses discrete curvature on the boundary sequence and interprets diagonal sums as a convolution with a symmetric kernel on the cyclic group $\mathbb{Z}/n\mathbb{Z}$.