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

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.

• How did you prove it for quadrilaterals? – Jorge Fernández Hidalgo Jul 15 '14 at 15:01
• I think this might help physicsforums.com/showthread.php?t=608769 – Jam Jul 15 '14 at 15:02
• @Bananarama I didn't prove it algebraically, but it was the only one that I could figure out with a graph. Polygons are more complex, and as I said ideally I will have an indisputable proof. – Alex Thornton Jul 15 '14 at 15:03
• but did you prove it for all quadrilaterals or only rectangles? – Jorge Fernández Hidalgo Jul 15 '14 at 15:04
• @Bananarama Oh, it appears I misunderstood you. I haven't proven that 'regular' (if you will) quadrilaterals are best in that sense, but I guess that would still work under the proof as they are still technically polygons. – Alex Thornton Jul 15 '14 at 15:06

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.

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.

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.

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.

• If you accept that a regular polygon of $n$ sides exists for any $n$, and you can show that any other arbitrary polygon of $n$ sides has less area, do you really need Bolzano-Weierstrauss in order to establish that the regular polygon has the maximum area? – David K Jul 15 '14 at 16:23
• @DavidK: This method does not directly compare an arbitrary polygon with the regular one, which is why we needed compactness to show that there is a maximum area polygon and then prove that it is regular. (Anyway I'm in the process of fixing my answer for the non-cyclic version. Please give me a little while more..) – user21820 Jul 15 '14 at 16:26
• True, you can't use that method to make a one-step comparison between the regular polygon and the arbitrary polygon. I was thinking you could build a chain of inequalities based on numbers of equal sides and angles. Now I'm having some doubts about it, and in any case it looks like B-W is a faster way to the result after all. – David K Jul 15 '14 at 16:35
• @DavidK: It's possible if we can construct a sequence of polygons with increasing area, but we will still need to use sequential compactness to extract a convergent subsequence of polygons and even then we will have a problem proving that the limit polygon is regular. At least that's what I think. – user21820 Jul 15 '14 at 16:38
• @bigant146: Hmm your last comment works if we aim to prove that the maximum-area non-self-intersecting polygon is convex. But if we want to prove that the maximum-signed-area polygon is convex, I'm afraid I don't see how it works, which is also why I ended up with the above argument, which I've just edited in. – user21820 Aug 13 '17 at 15:22