A polygon has sides of fixed length, but is flexible about its vertices

How should it be arranged for maximum area?

For maximum area, all vertices should lie on a circle. The polygon is then called a "cyclic polygon". It does not matter in which order the polygon sides occur.

There are 2 elegant proofs that the polygon must be cyclic, which appeared in the newsgroup sci.math in October 2002, under the subject header: "Maximum area of a flexible polygon" (you can search google groups for this).

First proof (from David Eppstein): arrange the vertices so they lie on a circle, and 'glue' the arc-shaped (red) caps onto each side of the polygon:

We can still change the shape of the polygon, with the (red) caps on each of its sides, as shown below. The total outside perimeter (round the outside of all the red bits) does not change.

The area within the outside perimeter comprises the area of the polygon plus the area of all the red caps. The area of the red caps does not change as we alter the polygon. Therefore if we maximise the area within the outside perimeter we will maximise the area of the polygon.

The maximum area within the outside perimeter occurs when the outside perimeter is a circle. So the maximum area for the polygon occurs when the outside perimeter is a circle. So the polygon vertices lie on a circle. QED.

Second proof (from Don Reble): we know the maximum area polygon will be convex (or else a simple reflection of concave vertices would increase the area). Take four consecutive vertices, calling them the 1st, 2nd, 3rd and 4th. Join the 1st and 4th with a straight line, so that we have a quadrilateral.

It is known that a quadrilateral has maximum area when its vertices lie on a circle (so it is a so-called cyclic quadrilateral). An outline proof of this:- the area of a general quadrilateral is given by Bretschneider's formula (using s for the semi-perimeter and a,b,c,d as the four side lengths):

This is minimised when opposite interior angles A and C sum to 180 degrees, which implies the quadrilateral is cyclic.

The area of our polygon includes the area of the quadrilateral. If the polygon has maximum area, so does the quadrilateral (or you could increase the area of the polygon by increasing the area of the quadrilateral).

The quadrilateral is therefore cyclic - its vertices lie on a circle.

Any 3 points define a circle:

We now look at the quadrilateral formed by the 2nd, 3rd, 4th and 5th vertices:

By the same argument, this must be cyclic, and the circle must be the same, as 3 vertices are common to both quadrilaterals.

The argument extends to all vertices, as we go around the polygon. All vertices lie on a circle. QED.

The order of the side lengths around the polygon does not matter. Proof: form a triangle from two adjacent sides by joining their unjoined ends. You can swap the order of the two sides without altering the area of the triangle - and hence without altering the area of the polygon. By swapping sides enough, you can arrive at any order of sides for the polygon. QED.

1. See this note, which suggests this result is known as Cramer's Theorem, and gives some references to proofs. Elsewhere, Cramer's Theorem seems to relate to statistics - so there is some confusion here.
It's possible another mathematician of that time called Castillon did some work on this problem.
2. I'm reliably informed that a proof is on page 236 of Maxima and Minima without Calculus; Ivan Niven; MAA Press, 1981.

page date: 11Nov05, revised 17Jan18.      I enjoy correspondence stimulated by this site. You can contact me here.