Optimization – Proving the Regular n-gon Maximizes Area for Fixed Perimeter

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It is often assumed that, given $n$, the regular $n$-gon will make the most efficient use of perimeter for area. I have never seen this proven. Anyone have something slick?

(That is, how can we prove that, given some number of sides, the regular polygons maximize area for a fixed perimeter? Or minimize perimeter for a fixed area, or maximize the ratio, $\frac{A}{p}$, or minimize the reciprocal… You get the point.)

This is easy enough to show for rectangles, and I've done it for triangles as well. I'm struggling find a generalizable method, since the space of $n$-gons with fixed perimeter (up to congruence) is $2n-4$-dimensional, I believe. (Another problem I'm thinking on.)

Any thoughts on a general proof of this claim?

Best Answer

There must be a simple proof that doesn't involve analysis, but I'm too lazy to think of that now. If we allow analytic arguments, there is a simple plan. It is kind of rough.

  1. Argue from compactness that for a fixed perimeter, there exists a polygon with $n$ vertices and given perimeter that has the maximal area. Maybe in order for the compactness argument to work we will have to allow vertices to coincide, or for the polygon to be non-convex or even self-intersecting. Doesn't really matter.
  2. Look at that polygon with maximum area. I'll call it the optimal polygon. If it is not convex, then two adjacent edges $AB$ and $BC$ can be reflected against line $AC$, thus increasing the area. So it is convex.
  3. All the edges of the optimal polygon must have equal length. If two adjacent edges $AB$ and $BC$ have different length, we can move point $B$ in such a way that one of the edges loses some length, the other gains the same length, and the area of triangle $ABC$ increases. This contradicts the optimality of our polygon. So all edges have the same length.
  4. All the angles of the optimal polygon must be equal. Suppose two adjacent angles aren't equal: $\angle ABC \neq \angle BCD$, where $A,B,C,D$ are consecutive vertices. Then it must be possible to move points $B$ and $C$ around in such a way that the area of $ABCD$ increases and $|AB|+|BC|+|CD|$ stays the same. The details here can get a bit messy, but this is just a plan. So, a contradiction, therefore all the angles are equal.
  5. That's it, the optimal polygon must be regular.

This plan is kind of sketchy, but it looks perfectly doable. The main advantage is that you don't have to invent anything smart about the polygon as a whole, you just need to look at 3 or 4 consecutive vertices and come up with ways to increase the area if the polygon isn't regular.

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