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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.

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  • $\begingroup$ 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. $\endgroup$
    – Lee Mosher
    Apr 30 at 19:43
  • $\begingroup$ 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. $\endgroup$
    – Lee Mosher
    Apr 30 at 19:43

3 Answers 3

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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.

enter image description here

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:

enter image description here

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: $$CB=\frac{1-\cos(x)}{\cos(\frac{\pi-x}2)}=\frac{1-(1-2\sin^2(x/2))}{\sin(x/2)}=\frac{2\sin^2(x/2)}{\sin(x/2)}=2\sin(x/2)$$

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  • $\begingroup$ 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 $\frac12\sin(\theta_i/2)$? $\endgroup$ Apr 30 at 16:53
  • $\begingroup$ 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) $\endgroup$
    – person
    Apr 30 at 17:17
  • $\begingroup$ 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! $\endgroup$ Apr 30 at 17:20
  • $\begingroup$ 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. $\endgroup$
    – person
    Apr 30 at 17:41
  • $\begingroup$ It checks out now. Many thanks! $\endgroup$ May 1 at 18:14
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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.

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    $\begingroup$ 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 $j$, $\endgroup$ May 1 at 7:29
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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

enter image description here

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.

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  • $\begingroup$ Nice. Your first approach is what I outlined in my comment. $\endgroup$
    – Lee Mosher
    May 1 at 20:43
  • $\begingroup$ 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. $\endgroup$
    – robjohn
    May 5 at 18:48

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