TITLE: Proving that out of all the possible n-gons that exist in the unit circle, the one with the maximum possible perimeter is the regular $n$-gon.
QUESTION [12 upvotes]: So in the context of my Convex Analysis studies, I have come across this problem:

First I have to prove that $ - \sin x $ is convex over $[0, \pi]$. That's easy enough using the second derivative theorem.
Now, using the answer from above, I have to prove that  out of all possible $n$-gons that can be specified on the unitary circle, the one with the maximum possible perimeter is the normal $n$-gon.

My geometric intuition is not good, and I cannot figure out how to do this using the tools from Convex Analysis. Any help would be greatly appreciated.

REPLY [2 votes]: Convexity Approach
Let $\theta_k$ be the angle subtended by side $k$ as viewed from the center of the circle. The perimeter of the polygon is
$$
2\sum_{k=1}^n\sin(\theta_k/2)
$$
$\sin(x)$ is concave on $[0,\pi]$. This means that
$$
\sin\left(\frac{\theta_k/2+\theta_{k+1}/2}2\right)\ge\frac{\sin(\theta_k/2)+\sin(\theta_{k+1}/2)}2
$$
If two adjacent sides are not equal, then replacing them with equal sides subtending the average of their angles will not decrease the perimeter. Thus, we see that a regular polygon will give the maximum perimeter.

Variational Approach
For simplicity of notation, let the indices be cyclic; i.e. $x_0=x_n$ and $x_1=x_{n+1}$.
For all $\delta x_k$ that maintain $|x_k|=1$, that is,
$$
x_k\cdot\delta x_k=0\tag1
$$
we want
$$
\begin{align}
0
&=\delta\sum_{k=1}^n|x_k-x_{k-1}|\tag{2a}\\
&=\sum_{k=1}^n\frac{x_k-x_{k-1}}{|x_k-x_{k-1}|}\cdot\delta(x_k-x_{k-1})\tag{2b}\\
&=\sum_{k=1}^n\frac{x_k-x_{k-1}}{|x_k-x_{k-1}|}\cdot\delta x_k-\sum_{k=0}^{n-1}\frac{x_{k+1}-x_k}{|x_{k+1}-x_k|}\cdot \delta x_k\tag{2c}\\
&=\sum_{k=1}^n\left(\frac{x_k-x_{k-1}}{|x_k-x_{k-1}|}+\frac{x_k-x_{k+1}}{|x_{k+1}-x_k|}\right)\cdot\delta x_k\tag{2d}
\end{align}
$$
Explanation:
$\text{(2a)}$: to maximize the perimeter, the variation of the perimeter must be $0$
$\text{(2b)}$: chain rule
$\text{(2c)}$: split the sum and reindex the second sum
$\text{(2d)}$: apply the cyclic indices and combine the sums
Orthogonality says that to get $(2)$ for all $\delta x_k$ that satisfy $(1)$, there must be a constant $\lambda$ so that
$$
\begin{align}
\lambda x_k
&=\frac{x_k-x_{k-1}}{|x_k-x_{k-1}|}+\frac{x_k-x_{k+1}}{|x_{k+1}-x_k|}\tag{3a}\\[3pt]
&=2\cos(\theta_k/2)\,x_k\tag{3b}
\end{align}
$$
$\text{(3a)}$: compare $(1)$ and $(2)$
$\text{(3b)}$: the sum of two unit vectors is a vector of length
$\phantom{\text{(3b):}}$ $2\cos(\theta/2)$ in the direction of their bisector,
$\phantom{\text{(3b):}}$ where $\theta$ is the angle between the unit vectors

Since $2\cos(\theta_k/2)=\lambda$, we know that $\theta_k$ is the same for all $k$. This makes each triangle $(0,x_k,x_{k+1})$ isosceles and congruent. Thus, the polygon is regular.