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.

Advertisements
  1. No comments yet.
  1. No trackbacks yet.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: