## Existence of Maximal Area Polygon with given sides

Suppose are given positive real numbers such that there exists a polygon with vertices having 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 the set of all polygons having these given sidelengths. Denote the closed disk centered in the origin with radius . It is easy to see that every polygon in has a representative in (i.e. there is a translation which brings it inside ), and that is because we can translate any polygon such that the origin is contained in its interior, and since the diameter of the polygon is less than it follows that the translated of is included in . Therefore we can restrict our attention to polygons which are in . For simplicity denote with the set of all these polygons.

The area of every polygon in is bounded by the measure of which is finite. Therefore the areas of the polygons are bounded above, and there is a least upper bound for these areas. This means that there exists a sequence of polygons such that

Now all we should do is to find a polygon in which has the area . We call an orientation of the sides of the polygon an order on the appearance of the sides starting from :

where is a permutation of . Since there are only finitely many such permutations it follows that there exists one permutation which appears infinitely often in the sequence . For simplicity, suppose that orientation given by the identity permutation appears infinitely often. Denote by the -th polygon from the sequence which has the desired orientation. Since all sequences of points are contained in , by a diagonal argument we can assume that each of these sequences has a limit . Then it is easy to see that because of the continuity of the distance function. Moreover, the characteristic functions of the polygons converge to the characteristic function of , which by integration gives us that the area of is , i.e. the maximal area.