# 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.

So in the context of my Convex Analysis studies, I have come across this problem:

1. First I have to prove that $$- \sin x$$ is convex over $$[0, \pi]$$. That's easy enough using the second derivative theorem.
2. 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.

• Here is a potentially simpler argument, although I'm leaving this in the comments since I don't have time to flesh it out. We know by compactness that, as $n$ cyclically ordered points $z_1,\ldots,z_n \in S^1$ vary, there must be a maximum to the total polygon length $|z_1-z_2|+...+|z_{n-1}-z_n| + |z_n-z_1]$. So if suffices to show that if there exists $i = 1,\ldots,n$ such that $|z_{i-1}-z_i| \ne |z_i - z_{i+1}|$ then the total polygon length is not maximized. Apr 30 at 19:43
• So now you just have to work with two consecutive terms, fixing the points $z_{i-1}$ and $z_{i+1}$ and letting just the single point $z_i$ vary between them. This should be easier than an argument letting all the $z_i$ points vary. Apr 30 at 19:43

Unfortunately I do not know Convex Analysis too well, but here is one possible answer using Jensen's inequality. Each one of the sides of a polygon must be a chord of the circle, and each chord has an angle which corresponds to it.

Consider $$n$$ such angles, $$\{\theta_1,\theta_2, \dots ,\theta_n\}; \sum_{k=1}^n \theta_k = 2\pi; \theta_i < \pi$$ , which correspond with these chords, which form the perimeter of a polygon. It is not too hard to prove that the length of each chord is $$2\sin(\theta_i/2)$$.

The perimeter would be the sum of the chords: $$2\sum_{k=1}^{n}\sin(\theta_k/2)=P$$

Since sine is concave from 0 to $$\pi$$, we may use the concave version of Jensen's: $$f(\frac1n\sum_{k=1}^nx_i)\geq\frac1n\sum_{k=1}^nf(x_i)$$ $$f(x)=2\sin(x/2), x_i=\theta_i$$: $$2\sin(\frac1{2n}\sum_{k=1}^n\theta_k))\geq\frac1n2\sum_{k=1}^{n}\sin(\theta_k/2)$$ $$P=2\sum_{k=1}^{n}\sin(\theta_k/2)\leq2n\sin(\frac1{2n}\sum_{k=1}^n\theta_k))$$ $$\sum_{k=1}^n \theta_k = 2\pi$$: $$P=2\sum_{k=1}^{n}\sin(\theta_k/2)\leq2n\sin(\pi/n)$$ Now the current RHS is exactly what would happen if all of the $$\theta_i$$s were equal, and since every other case is lesser than that, we can see that the maximum possible perimeter is the regular polygon.

EDIT: To prove that the length of the cord is $$2 \sin(x/2)$$ consider the following figure:

Where, given a circle of radius 1, we want to find the cord $$CD$$.

$$AD=\cos(x)$$, and so $$DB=1-\cos(x)$$.

$$ABC$$ is isosceles, so $$\angle ACB=\angle ABC=\frac{\pi-x}2$$

$$\frac{DB}{\cos(\angle ABC)}=CB$$, so substituting everything in, and using the double angle identity: $$CB=\frac{1-\cos(x)}{\cos(\frac{\pi-x}2)}=\frac{1-(1-2\sin^2(x/2))}{\sin(x/2)}=\frac{2\sin^2(x/2)}{\sin(x/2)}=2\sin(x/2)$$

• That's an amazing answer, and I cannot really accept your claim that "you do not know Convex Analysis too well" when you successfully utilized every tool CA has to arrive at a conclusion. I will run the checks myself when I get home and I will green you asap! Many thanks! EDIT1: Can you give me a run down of how exactly can you prove that $\frac12\sin(\theta_i/2)$? Apr 30 at 16:53
• Perhaps I know it by another name then, I will add the proof when I have a bit more time. For now, here is the wikipedia article I refer too :en.wikipedia.org/wiki/Chord_(geometry) Apr 30 at 17:17
• I may have missed something, but a chord is analyzed as a single-variable function as $\mathrm{crd} \theta = 2 \sin (\frac{\theta}{2})$ as a result of The Pythagorean Theorem. Again, I might be wrong but looks like Wiki has the same information also. At any rate, looking forward to your input! Apr 30 at 17:20
• Apologies, I mistakenly wrote $\frac12\sin(x/2)$ instead of $2\sin(x/2)$, the proof remains valid though, as the difference of a constant simply cancels out. I have added the proof I remember to the answer. Apr 30 at 17:41
• It checks out now. Many thanks! May 1 at 18:14

I can't really improve on @person's answer other than some convex stuff.

The problem is $$\min \{ N(x) | 0 \le x_k \le \pi, \sum_k x_k = 2 \pi \}$$, where $$N(x) = -\sum_k 2 \sin { x_k \over 2}$$. Given the restriction on the arguments, we see that $$N$$ is convex. Since the feasible set is compact, there is a solution. Note that if $$x$$ is a solution, then so is $$\Pi x$$ for any permutation $$\Pi$$. Let $${\cal P}$$ be the set of permutations. Since $$N$$ is convex, we have $$N( {1 \over |{\cal P}|}\sum_\Pi \Pi x) \le {1 \over |{\cal P}|} \sum_\Pi N(\Pi x) = N(x)$$, and so a solution of the form $$(\theta,...,\theta)$$ exists ($$\theta = {2 \pi \over n}$$, of course).

A little more work shows that $$N''>0$$ and so $$N$$ is strictly convex, hence the solution is unique.

• If $\theta_j>\theta_{j+1}$ for some $j$ (with $\theta_{n+1}=\theta_1)$ then $\sin (\theta_j/2)+\sin (\theta_{j+1}/2)<2\sin ((\theta_j+\theta_{j+1})/2)$. Because if $\pi/2>u\ge v>0$ and $u+v$ is constant then $d (\sin u +\sin v)/du=\cos u -\cos v<0$ except when $u=v.$ So the perimeter is not maximized unless $\theta_j=\theta_{j+1}$ for every $j$, May 1 at 7:29

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.

• Nice. Your first approach is what I outlined in my comment. May 1 at 20:43
• The second approach can be extended slightly to show that of all polygons with $n$ sides and a given perimeter, the greatest area is given by a regular $n$-gon.
– robjohn
May 5 at 18:48