Home > Analysis, Geometry, Measure Theory, shape optimization > Existence of Maximal Area Polygon with given sides

## Existence of Maximal Area Polygon with given sides

Suppose ${a_1,..a_n,\ n \geq 3}$ are given positive real numbers such that there exists a polygon with ${n}$ vertices having ${a_1,..,a_n}$ as lengths of its size. Then among all polygons having these given side lengths there exists one which has maximal area.

Since I have used in the linked post the fact that the maximal area polygon with given sides exists, I will give a proof in the following.

Proof: First, let’s denote with ${\mathcal{P}}$ the set of all polygons having these given sidelengths. Denote ${K}$ the closed disk centered in the origin with radius ${a_1+..+a_n}$. It is easy to see that every polygon in ${\mathcal{P}}$ has a representative in ${K}$ (i.e. there is a translation which brings it inside ${K}$), and that is because we can translate any polygon ${P \in \mathcal{P}}$ such that the origin is contained in its interior, and since the diameter of the polygon is less than ${a_1+..+a_n}$ it follows that the translated of ${P}$ is included in ${K}$. Therefore we can restrict our attention to polygons ${P \in \mathcal{P}}$ which are in ${K}$. For simplicity denote with ${\mathcal{P}}$ the set of all these polygons.

The area of every polygon in ${K}$ is bounded by the measure of ${K}$ which is finite. Therefore the areas of the polygons ${P \in \mathcal{P}}$ are bounded above, and there is a least upper bound ${S>0}$ for these areas. This means that there exists a sequence of polygons ${(P_n) \subset \mathcal{P}}$ such that

$\displaystyle \text{Area}(P_n) \rightarrow S.$

Now all we should do is to find a polygon in ${\mathcal{P}}$ which has the area ${S}$. We call an orientation of the sides of the polygon an order on the appearance of the sides ${a_i,i=1..n}$ starting from ${a_1}$:

$\displaystyle a_1 \rightarrow a_{\sigma(2)} \rightarrow .. \rightarrow a_{\sigma(n)}$

where ${\sigma}$ is a permutation of ${\{2,3,..,n\}}$. Since there are only finitely many such permutations it follows that there exists one permutation which appears infinitely often in the sequence ${(P_n)}$. For simplicity, suppose that orientation given by the identity permutation appears infinitely often. Denote by ${P_1^iP_2^i...P_n^i}$ the ${i}$-th polygon from the sequence ${P_n}$ which has the desired orientation. Since all sequences of points ${(P_k^i)_{i \in \Bbb{N}},\ k=1..n}$ are contained in ${K}$, by a diagonal argument we can assume that each of these sequences has a limit ${P_i \in K}$. Then it is easy to see that ${P_1..P_n \in \mathcal{P}}$ because of the continuity of the distance function. Moreover, the characteristic functions of the polygons ${P_1^iP_2^i...P_n^i}$ converge to the characteristic function of ${P_1..P_n}$, which by integration gives us that the area of ${P_1..P_n}$ is ${S}$, i.e. the maximal area.