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Regular polygons have largest area for a fixed perimeter
To be archived: How do we prove that a regular polygon will have a greater area than an irregular polygon with the same perimeter?
By anonymous on Thursday, February 28, 2002 - 07:26 pm:

How can I prove that a regular polygon will always have a larger area than an irregular one with the same perimeter? Having proven that, how can I prove that the more sides a polygon has (keeping the same perimeter), the larger the area will be?

By Arvan Pritchard on Friday, March 01, 2002 - 10:00 am:

One approach is:

1) Prove that given a non-convex polygon you can find a larger convex polygon with the same perimeter - I think moving concave vertices out keeping their edges the same, but you need to be careful about self-intersecting polygons.

2) Now you can take a convex irregular polygon and look at a vertex whose adjacent edges are unequal. If you replace just that vertex with one equi-distant from the neighbours then the resulting area is increased - you can prove this by looking at the triangles formed by the line joining the two neighbours and the two pairs of edges.

3) Once you have proved that the regular polygons have maximum area, you can derive a formula for the area of a regular polygon and that will give the last result.

By Dan Goodman on Sunday, March 03, 2002 - 09:15 pm:

Anon, here is a quick summary of how to show that the regular polygon has largest area, let us know which bits need more explaining.

First, a regular polygon is one in which all the angles are the same and all the edges have the same length. The first step in showing that this has the largest area (for a fixed perimeter and number of vertices - a vertex is a point where two edges meet and the plural of vertex is vertices) is to show that it is convex.

Do you know what convex means? Basically, a shape is convex if you can "see" every point inside the shape from any other point. An "L" isn't convex, because you can't see the bottom right hand corner of the L from the top left hand corner (imagine being in an L-shaped room), but a square or circle is convex. All of the regular polygons are convex.

A polygon of maximum area will be convex because you can take a polygon that isn't convex and make the area larger without making the perimeter larger (sometimes even making it smaller), this is the point of my diagram 3 on the other thread.

Then you show that you can increase the area, keeping the perimeter the same, by making all the edges the same length - by going round all of the pairs of edges on the polygon and changing them to make them the same length. (This is the point of diagram 1.)

Then you show that you can increase the area by making all the angles the same, by going round all of the groups of three adjacent edges and making the angles the same. (Diagram 2 and the other thread about "cyclic quadrilaterals".)

Then you end up with a regular polygon, because you've made all the edges and angles the same, and the final shape has the same perimeter as the first shape, but a larger area. So the regular polygon has maximum area for fixed perimeter.

Is that OK? The details of showing that the edges and angles must be the same length are a bit tricky. I'm not sure if there is a simpler way of explaining it than on the other threads. Anyone want to have a go? :-)

You can use the fact that a regular polygon has the biggest area for a fixed perimeter (and number of vertices) to show that a circle is the shape with the largest area for a fixed perimeter. For any shape, you can draw a polygon (of the same perimeter) that looks very similar to it. The more edges (or vertices) you use, the more similar the polygon and the shape will look. For example:

Polygon Approximations

As you increase the number of edges, the polygon looks more and more like the original shape.

So any shape which has maximum area for fixed perimeter will have to look more and more like a regular polygon as the number of edges gets bigger and bigger. But the only shape that looks more and more like a regular polygon as the number of edges gets bigger is a circle, as you can probably guess by drawing a regular polygon with a thousand edges (don't actually bother doing this, it looks exactly like a circle). So the shape with maximum area for a fixed perimeter must be a circle.

That argument is, as some mathematicians like to say, a bit wishy-washy. Properly proving that the circle has the maximum area is something that would take much too long, it's first or second year at university difficulty.

(Technically, for advanced readers only, I think what we've shown above is that the circle is the shape of maximum area for shapes bounded by a Jordan curve, which IIRC is a limit curve of piecewise linear curves. Does anyone know if this concept be extended in some sense to measurable shapes (e.g. to find the measurable subset of R2 with maximal 2D Lebesque measure for fixed 1D Lebesque measure of the boundary set)? If so, either start another thread or email me rather than posting here, I'd be interested to know.)

By Dan Goodman on Tuesday, March 05, 2002 - 10:53 pm:

(Another technical note, what I wrote above is wrong. A Jordan curve is not a limit of piecewise linear curves because that would allow the possibility of space filling curves.)

By anonymous on Monday, March 04, 2002 - 05:08 pm:

wow thanks that was really helpful!!