Given a polygon of n-sides, why does the regular one (i.e. all sides equal) enclose the greatest area given a constant perimeter?

Question

This doesn't require much more than the title. I just need an explanation, but an algebraic proof would be a bonus.

We can demonstrate this for quadrilaterals, a square is best as shown by this graph- the area peaks when both sides are equal at 250.

Answer 1

Here is an elegant proof, but it relies on a bit of real analysis and not just euclidean geometry. I assume you are concerned with only non-self-intersecting polygons since it is not clear what should be defined as the area of a self-intersecting polygon.

[Edit: I don't know how I misread the question to be about polygons inscribed in a given circle with the maximum area. The first proof solves that and the second proof solves the original problem.]

Cyclic Polygon Maximum Area

Let $D$ be the set of vectors in $\mathbb{R}^n$ that describe the angles subtended by the sides of a non-self-intersecting cyclic polygon (which is allowed to have sides of length $0$).

Then $D$ is clearly a closed bounded set and hence compact.

Let $f(v)$ be the area of the polygon $P$ that is described by the vector $v$.

Then $f$ is continuous on $D$.

Therefore $f$ attains a maximum on $D$.

Let $a \in D$ such that $f(a) = \max_{v \in D} f(v)$ and let R be the polygon described by $a$.

If $R$ is not regular:

  $R$ has two sides $AB,BC$ of unequal length.

  Move $B$ on the arc $AC$ to make $AB,BC$ of equal length.

  Then $B$ is now further away from $AC$ and hence $Area(\triangle ABC)$ increases.

  Thus $Area(R)$ increases.

  Contradiction.

Therefore $R$ is regular.

Isoperimetric Polygon Maximum Area

Let $p > 0$ be the given perimeter.

Let $D$ be the set of points in $(\mathbb{R}^2)^n$ that describe the vertices of a directed polygon of perimeter $p$ (which is allowed to have sides of length $0$ and self intersections) such that one vertex is $(0,0)$.

Then $D$ is a closed (since it can be expressed as non-strict inequalities) bounded set and hence compact.

Let $f(v)$ be the signed area of the polygon $P$ that is described by the vector $v$.

Then $f$ is continuous on $D$.

Therefore $f$ attains a maximum on $D$.

Let $a \in D$ such that $f(a) = \max_{v \in D} f(v)$ and let R be the polygon described by $a$.

If $R$ has two sides $AB,BC$ such that $|AB| \ne |BC|$ or $Area(\triangle ABC) < 0$:

  Move $B$ to preserve $|AB|+|BC|$ but make $|AB| = |BC|$ and $Area(\triangle ABC) \ge 0$.

  The locus of $B$ that preserves $|AB|+|BC|$ is an ellipse with $AC$ as a diameter.

  Thus $Area(\triangle ABC)$ increases and hence $Area(R)$ increases.

  Contradiction.

Therefore $R$ has all sides of equal length and each internal angle being at most $180^\circ$.

If $R$ is not cyclic:

  Let $X,Y,Z,W$ be four vertices of $R$ in order that do not lie on a circle.

  These four vertices divide the sides of $R$ into four sections.

  Move those four sections rigidly so that $R$ remains a polygon but $XYZW$ becomes cyclic.

  Then $Area(R)$ increases by Bretschneider's formula.

  Contradiction.

Therefore $R$ is cyclic.

Therefore $R$ is a regular polygon (possibly a star).

Thus $R$ has area $n·\tan(α/2)$ where $α$ is the internal angle between sides of $R$.

Thus $R$ is a regular polygon since it alone maximizes $α$.

Notes

The above proofs also show with essentially no change that the regular polygon is the only case with the maximum area. My second proof uses signed area since it was easier than requiring the polygon to be non-self-intersecting, but I'll leave my first proof the way it is.