Proving the regular n-gon maximizes area for fixed perimeter. 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?
 A: 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.


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*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.

*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.

*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.

*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.

*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.
A: (a) Good to see you here, Paul.
(b) To address the missing link in Dan's solution (step 1):
Let $P$ be the set of all plane polygons with $n$ vertices, modulo congruence. We need a topological structure on $P$ such that the map $A:P\rightarrow\mathbb{R}$ that takes a polygon to its area is continuous, and such that the subset $P_\ell$ of $P$ consisting of polygons of a fixed perimeter $\ell$ is compact. Then, compactness of $P_\ell$ and continuity of $A$ together imply that $A$ attains a maximum value on $P_\ell$, and Dan's argument can get started.
I am sure there is a completely standard way to do this, but I don't know it, so I'll take it from scratch:
I feel that there is only one sensible choice for the topological structure on $P$, so we had better hope it has these properties.
Start with $(\mathbb{R}^2)^n$, viewed as the set of $n$-tuples of points in the plane $(p_1,\dots,p_n)$. (From here on, "coordinate" means a point in the plane, not a real number.) Each $n$-tuple characterizes some polygon uniquely: Start at $p_1$ and draw a line to $p_2$, etc., returning to $p_1$ at the end. However, the same polygon is characterized by different $n$-tuples: $(p_2,\dots,p_n,p_1)$ and $(p_n,p_{n-1},\dots,p_1)$ both give us the same polygon as $(p_1,\dots,p_n)$ for example. The amount of ambiguity is given precisely by the dihedral group $D_n$, acting on the coordinates, since we could start at any of the vertices and go in either direction to turn a polygon into an $n$-tuple of points. So we need to quotient $(\mathbb{R}^2)^n$ by the action of $D_n$ on the coordinates. (Here $D_n$ is being realized as the subgroup of $S_n$ generated by the long cycle $(1\dots n)$, and the permutation $(1n)(2(n-1))\dots$ that reverses the order of the coordinates.)
Thus the quotient $(\mathbb{R}^2)^n / D_n$ is the set of distinct polygons. To reduce to $P$ we need to mod out by congruence, thus to quotient by the group $E(2)$ of Euclidean motions of the plane. Since $D_n$ is permuting the coordinates and $E(2)$ is acting separately on each coordinate, their actions on $(\mathbb{R}^2)^n$ commute with each other, thus $P=(\mathbb{R}^2)^n/(D_n\times E(2))$.
The only sensible topology is the quotient topology.
Now, for sure, this topology does make the area map $A:P\rightarrow \mathbb{R}$ continuous. We can define an area map starting with $(\mathbb{R}^2)^n$ by drawing the polygon given by the $n$-tuple and then calculating its area; I don't feel the need to argue that this is continuous. But, this map factors through $P$ because the group $D_n\times E(2)$ acting on the $n$-tuples will not affect the area. The whole point of the quotient topology is that continuous maps from the original space that factor through the quotient are automatically continuous, so $A:P\rightarrow\mathbb{R}$ is continuous.
By the same logic, the perimeter map $p:P\rightarrow\mathbb{R}$ is also continuous, and the set $P_\ell$ of polygons of fixed perimeter $\ell$ is precisely $p^{-1}(\{\ell\})$. The set $\{\ell\}\subset \mathbb{R}$ is closed, so its preimage in $P$ under the continuous map $p$ is also closed; thus $P_\ell$ is a closed subset of $P$.
My plan is to exhibit a subset $A$ of $P$ that contains $P_\ell$ and is definitely compact. Then $P_\ell$, as a closed subset of the compact space $A$, will be compact.
Every polygon is congruent to one that has the origin as a vertex. No polygon whose perimeter is $\ell$ and has one vertex at the origin can have any of its other vertices outside of the disc of radius $\ell$ about the origin. Thus every polygon of radius $\ell$ is congruent to one that is represented by an $n$-tuple that looks like $(0,p_2,\dots,p_n)$ with $p_2,\dots,p_n$ all within the closed disc of radius $\ell$ about the origin. Let $A'$ be the subset of $(\mathbb{R}^2)^n$ consisting of $n$-tuples of this form: i.e. $A'$ is
$$\{0\}\times \overline{B}_\ell(0) \times\dots\times \overline{B}_\ell(0)$$
(where $\overline{B}_\ell(0)$ is the closed disc of radius $\ell$ about the origin). $A'\subset (\mathbb{R}^2)^n$ is clearly compact. But also, because every polygon of perimeter $\ell$ is represented by such an $n$-tuple, the image of $A'$ under the quotient map $(\mathbb{R}^2)^n\rightarrow (\mathbb{R}^2)^n/(D_n\times E(2))$ contains $P_\ell$. This image is the desired $A$. It is compact because $A'$ is compact, the quotient map is continuous (by construction of the quotient topology), and the continuous image of a compact set is compact.
This does it: now $P_\ell\subset A$ is a closed subset of a compact set, thus compact.
(c) Incidentally, I agree with your dimension count for $P_\ell$. In terms of the above discussion, $P$ is the quotient of $2n$-dimensional $(\mathbb{R}^2)^n$ by $3$-dimensional $D_n\times E(2)$; thus $P$ is $(2n-3)$-dimensional. ($E(2)$ is 3-dimensional because its quotient by its translation subgroup is the orthogonal group $O_2$, which is $1$-dimensional because it is homeomorphic to two copies of $S^1$, and the translations are $2$-dimensional because they are homeomorphic/isomorphic to $\mathbb{R}^2$. $D_n$ is finite, thus zero-dimensional, so doesn't affect the dimension.) Fixing the perimeter knocks off $1$ more dimension.
A: Dan Schved's step $2$ is not entirely correct.  Instead, I believe we must consider the convex hull of the non-convex polygon.  If we label the vertices $P_1$ through $P_n$, it could be that $P_i$ and $P_j$ are consecutive in transversing the convex hull, but not consecutive in transversing the polygon.  That is to say, there could be numerous vertices inside the convex hull, not just 1, as it appears to be the case in step $2$.  If the "shape" of the the polygonal line in the interior of the convex hull is something like a "W," then reflection around the line $P_iP_j$ will not yet make it convex, because this part of the new polygon will look like an "M," although the new polygon (still not convex) will indeed have a larger area. The key point that I think is missing is that an additional reflection about the new convex hull will make it convex, in this one case when the interior polygonal line looked like a "W."  In general, it will take multiple reflections to make the polygon convex (as many as there are "humps" in the interior polygonal line), but it is clear it must be finite, so the process will terminate eventually and with a convex polygon.    
