# Maximizing area of a pentagon inscribed inside a circle

A pentagon is inscribed inside a circle. Prove that the area of the pentagon to be maximum, it must be a regular one.

I have read When is the area of a pentagon inscribed inside a fixed circle maximum?, but am not satisfied with the answer...

My approach:
We can divide the pentagon into a triangle and a cyclic quadrilateral by joining any two vertices. Now we can fix this chord and argue on the maximization of the areas of these two figures...but are we allowed to fix that chord?

• Any help from anyone? Dec 17, 2017 at 17:23
– Nate
Dec 17, 2017 at 17:32
• That is not a proof... Dec 17, 2017 at 17:37

The area of pentagon is given by:

$$A =\frac12R^2\theta_1+\frac12R^2\sin \theta_2+\frac12R^2\sin \theta_3+\frac12R^2\sin \theta_4+\frac12R^2\sin \theta_4=\frac12R^2(\sin \theta_1+\sin \theta_2+\sin \theta_3+\sin \theta_4+\sin \theta_5)$$

$$\theta_1+\theta_2+\theta_3+\theta_4+\theta_5=2\pi$$

Since $\sin x$ is concave (assuming $0<\theta_i<\pi$):

$$\frac{\sum \sin \theta_i}{5} \leq \sin \left(\sum \frac{\theta_i}{5}\right)=\sin \frac{2\pi}{5}$$

equality holds only if $$\theta_1=\theta_2=\theta_3=\theta_4=\theta_5=\frac{2\pi}{5}$$

Thus the area of the pentagon is maximum when it is regular $\square$.

NOTE

The same argument can be used to prove the statement for any polygon inscribed inside a circle.

• Got it....Thanks for help Dec 18, 2017 at 4:52

Take an angle $2\alpha$ and divide it into two parts $\alpha+\delta, \alpha-\delta$ as shown in this sketch.

The area of the triangles $A'OP$ and $POA$ will be $$\left\{ \matrix{ A_{\,POA} = \cos \left( {\left( {\alpha - \delta } \right)/2} \right)\sin \left( {\left( {\alpha - \delta } \right)/2} \right) = {1 \over 2}\sin \left( {\alpha - \delta } \right)\quad \left| {\;0 \le \alpha - \delta \le \pi } \right. \hfill \cr A_{\,A'OP} = \cos \left( {\left( {\alpha + \delta } \right)/2} \right)\sin \left( {\left( {\alpha + \delta } \right)/2} \right) = {1 \over 2}\sin \left( {\alpha + \delta } \right)\quad \left| {\;0 \le \alpha + \delta \le \pi } \right. \hfill \cr} \right.$$ and their sum $$A_{\,tot} = A_{\,POA} + A_{\,A'OP} = {1 \over 2}\left( {\sin \left( {\alpha - \delta } \right) + \sin \left( {\alpha + \delta } \right)} \right) = \sin \alpha \cos \delta \quad \left| {\;\left\{ \matrix{ 0 \le \alpha \le \pi \hfill \cr - \alpha \le \delta \le \alpha \quad \left| {\;\alpha \le \pi /2} \right. \hfill \cr - \left( {\pi - \alpha } \right) \le \delta \le \pi - \alpha \quad \left| {\;\pi /2 < \alpha } \right. \hfill \cr} \right.} \right.$$ which is clearly maximum when $\delta =0$, i.e. when the angle $2\alpha$ is split symmetrically, whatever be its value $0 \le \alpha \le \pi$.
The bounds on $\delta$ are just preserving that the two triagles do not overlap.

Therefore starting with a triangle, you can demonstrate that any n-agon ($3 \le n$) inscribed in a circle has maximal area when it is regular.