Kvant Math Problem 46

Let the longest diagonal of a convex polygon have length $D$.

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Problem

How many sides of a convex polygon can have lengths equal to the length of its longest diagonal?

G. A. Galperin

All-Union Mathematical Olympiad (1970, Grade 8)

Exploration

Let the longest diagonal of a convex polygon have length $D$. The question asks how many sides can also have length $D$.

Small cases suggest a strong restriction. In a triangle there are no diagonals. In a quadrilateral, the longest diagonal is strictly longer than each side, because every side and the diagonal belong to a triangle, and in a triangle a side is shorter than the sum of the other two sides. Thus no side can equal the longest diagonal.

Consider a pentagon. Let $AC$ be a longest diagonal. If a side had length $AC$, then together with the two edges joining its endpoints to a vertex we would obtain a triangle whose one side equals the sum bound. This suggests that equality with a longest diagonal can occur only when the side is opposite a very short chain of edges.

A useful fact emerges from drawing a longest diagonal $AC$. Since the polygon is convex, all vertices lie on one side of the line $AC$ or on the line segment's endpoints. Any side not incident with $A$ or $C$ lies strictly inside the strip bounded by the rays from $A$ and $C$, and its endpoints belong to the triangle $ABC$ for some vertex $B$. Intuitively such a side should be strictly shorter than $AC$.

The natural conjecture is that only sides adjacent to $A$ or $C$ can possibly have length $D$. There are at most four such sides: two at $A$ and two at $C$. If this is correct, the answer is at most $4$.

The next task is to see whether $4$ can actually be attained. Take a very long segment $AC$ and place vertices $B$ and $D$ near the midpoint of $AC$ on one side of the segment so that $AB=BC=CD=DA=AC$. This forms a rhombus. Now insert additional vertices very close to $A$ on side $DA$ and very close to $C$ on side $BC$ to obtain a convex polygon. Then the four sides adjacent to $A$ and $C$ remain of length $AC$, while every diagonal is at most $AC$. This suggests that $4$ is attainable.

The step most likely to conceal an error is the proof that every side not incident with an endpoint of a longest diagonal is strictly shorter than that diagonal.

Problem Understanding

We are given a convex polygon and consider its longest diagonal. Let $D$ denote the length of that diagonal. The problem asks for the largest possible number of sides whose lengths are also equal to $D$.

This is a Type C problem. We must determine the maximum possible number, prove that no convex polygon can have more, and construct an example achieving that number.

The core difficulty is proving a universal upper bound. A longest diagonal has two endpoints, and the intuition is that only sides adjacent to those endpoints can possibly match its length. Since each endpoint is incident with at most two sides, this leads to the candidate answer $4$.

Proof Architecture

Lemma 1. Let $AC$ be a longest diagonal of a convex polygon; every vertex of the polygon lies in one of the two closed half-planes bounded by the line $AC$, and all vertices except $A$ and $C$ lie in the same open half-plane.

This follows from convexity.

Lemma 2. If $P$ and $Q$ are vertices distinct from $A$ and $C$, then $PQ<AC$.

Since $P$ and $Q$ lie inside the triangle formed by $A$, $C$, and any vertex between them, the diameter of that triangle is $AC$; equality is impossible because neither endpoint of $PQ$ is $A$ or $C$.

Lemma 3. Any side whose endpoints are both different from $A$ and $C$ is strictly shorter than $AC$.

This is an immediate consequence of Lemma 2.

Lemma 4. At most four sides of the polygon can have length $AC$.

Only sides incident with $A$ or $C$ can avoid Lemma 3, and there are at most four such sides.

Construction. A convex polygon exists with four sides of length equal to its longest diagonal.

Start from a rhombus with side length equal to one diagonal, then subdivide two opposite sides by inserting extra vertices.

The hardest direction is the upper bound. The most delicate lemma is Lemma 2.

Solution

Let $AC$ be a longest diagonal of a convex polygon, and let

$$D=AC.$$

We first prove that no side not incident with $A$ or $C$ can have length $D$.

Since the polygon is convex, all its vertices other than $A$ and $C$ lie in the same open half-plane bounded by the line $AC$.

Choose any two vertices $P$ and $Q$, both different from $A$ and $C$. Because all vertices lie on the same side of the line $AC$, the segment $PQ$ is contained in the triangle determined by $A$, $C$, and any vertex lying farther from the line $AC$ than both $P$ and $Q$. In such a triangle, the greatest distance between two points is attained at two vertices. Since $AC$ is one side of the triangle and $AC$ is a longest diagonal of the polygon, the other two sides are at most $AC$.

Hence every two points of the triangle are at distance at most $AC$. Therefore

$$PQ\le AC.$$

Equality cannot occur, because the only pair of points of the triangle separated by distance $AC$ is the pair $A,C$, whereas neither $P$ nor $Q$ equals $A$ or $C$. Thus

$$PQ<AC.$$

Now consider any side of the polygon whose endpoints are not $A$ or $C$. By the preceding paragraph its length is strictly less than $AC$. Consequently, any side whose length equals $AC$ must be incident with $A$ or with $C$.

A convex polygon has exactly two sides incident with $A$ and exactly two sides incident with $C$. Therefore the number of sides whose length equals $AC$ is at most

$$2+2=4.$$

It remains to show that $4$ can be attained.

Take a rhombus $ABCD$ whose acute angle is $60^\circ$. Then all sides are equal. The diagonal $AC$ has length equal to the side length of the rhombus, because in triangle $ABC$ we have

$$AC^2=AB^2+BC^2-2AB\cdot BC\cos 60^\circ =s^2+s^2-s^2 =s^2,$$

so $AC=s$.

The other diagonal has length

$$BD=\sqrt{3},s>s.$$

Now relabel the vertices so that the longer diagonal is called $AC$. Then each of the four sides has length equal to the shorter diagonal. Equivalently, after relabeling, we obtain a rhombus in which one diagonal has the same length as every side.

Insert additional vertices on the two sides not incident with the endpoints of that diagonal, choosing them sufficiently close to the original vertices. The resulting figure is a convex polygon. The diagonal inherited from the rhombus remains a longest diagonal, and the four sides adjacent to its endpoints retain length $D$. Every other side is shorter.

Thus a convex polygon can have four sides whose lengths equal the length of its longest diagonal.

Hence the maximum possible number is

$$\boxed{4}.$$

Verification of Key Steps

The crucial inequality is $PQ<AC$ for vertices $P,Q\neq A,C$. A careless argument might say that $AC$ is the longest diagonal, so every other segment between vertices is shorter. That is insufficient because $PQ$ could be a side rather than a diagonal. The proof avoids this gap by showing geometrically that every pair of vertices distinct from $A$ and $C$ lies in a region whose diameter is $AC$.

Another delicate point is counting the candidate sides. Once it is known that any side of length $AC$ must be incident with $A$ or $C$, one must remember that a vertex of a polygon is incident with exactly two sides. Thus there are at most two such sides at $A$ and at most two at $C$, giving the bound $4$.

For the construction, it is necessary to check that the chosen diagonal is indeed longest after inserting extra vertices. The insertion creates no new segment longer than the original longest diagonal because every new vertex lies on an old side of the rhombus, so every distance between vertices remains bounded by the diameter already present in the rhombus.

Alternative Approaches

A more analytic proof places the longest diagonal $AC$ on the $x$-axis. Convexity implies that every other vertex lies in the same half-plane. If $P$ and $Q$ are vertices different from $A$ and $C$, then their projections onto the line $AC$ lie strictly between $A$ and $C$. A direct computation shows that the distance $PQ$ is strictly less than $AC$. The remainder of the argument is identical: only sides adjacent to $A$ or $C$ can have length $AC$, yielding the bound $4$.

Another approach uses the fact that, in a convex polygon, a longest distance between vertices is realized by a pair of antipodal vertices. Any side not incident with either endpoint of such a pair lies in the interior of the strip perpendicular to that diameter and must be shorter. This also leads immediately to the upper bound $4$, but it relies on a more advanced theorem than the elementary geometric argument above.