Isoperimetric problem.

Question

Introduction.

The isoperimetric problem is to determine a plane figure of the largest possible area whose boundary has a specified length (wiki).

---

I know several ways to solve this problem and I do not ask about any of them. I want to ask about my own solution which I describe below. What do you think is it valid?

Solution

It is quite obvious that figure must be convex in order to bound the largest area.

Suppose we have convex figure $\Phi$. Without loss of generality we can assume that length of $\Phi$ is equal to 1, so $l = 1$. Next step is to draw a straight line (call it $L_1$) that intersects $\Phi$ so points of intersections, say, $p_1$ and $p_2$ are split $\partial\Phi$ into two curves, say, $l_1$ and $l_2$ which have equal length. $L_1$ splits $\Phi$ into two figures, say, $\Phi'$ and $\Phi''$. Now the following algorithm should be executed:

Algorithm

1. If $S(\Phi')\geq S(\Phi'')$ then substitute ∂Φ″ with the reflection of ∂Φ′ around the line dividing Φ″ from Φ′.
   Else vice-versa.

2. Choose another line $L_i$ which is do the same job as the line $L_{i-1}$.

3. Return to step 1.

Where $S(\Phi)$ is area bounded by $\Phi$. It requires to note that $L_i$ can be chosen in the following way: every iteration point $p_1$ shifts along $\partial \Phi$ by $1/\sqrt{2}$. The second point $p_2$ then appear in order to fulfill the above condition.

It is obvious that $\Delta(S(\Phi))\geq 0$ after each iteration of the above algorithm. Also obvious that if initial figure was a circle then $$\lim_{n\to\infty}{S(\Phi_0) - S(\Phi_n)} = 0.$$

BUT. It is not obvious for me the opposite statement: $$\forall \Phi_0: \lim_{n\to\infty}{\Phi_n} = \mathrm{Circle}.$$

Answer 1

This is not an answer: I'm just showing some results I obtained with Mathematica from the algorithm described above.

Starting from a square, one readily ends with a circle (50 iterations are shown):

Starting from a rectangle (base : height = 100 : 1) one gets a circle again (50 iterations are shown). Notice that at a concave polygon might arise in the process, in which case one of its halves could be a self-intersecting polygon.

To prove that the iteration always tends to a circle, one could argue as follows:

1) polygon area is a not decreasing and bounded sequence, so it must tend to a limit;

2) the limiting figure must be reflection symmetric across any line dividing its perimeter into two equal pieces, but only a circle possesses such a symmetry.

This reasoning is flawed: area tends to a limit, but we don't know that there is a single shape corresponding to that limiting value. There could be, for instance, two or more limiting figures, such that applying the algorithm to one of them one gets another one. I found an example of that in the "minimum" algorithm, see below.

Just for fun, I also considered a "minimum" algorithm, where at each step one reflects the polygon having lower area, instead of the larger one. I was quite surprised when I saw that, starting with a square, a circle would apparently still be the limiting shape (200 iterations are shown):

On closer inspection, however, I realized it is not a "smooth" circle, because it has a jagged perimeter (so I presume the limiting shape could be some kind of fractal).

Starting with a rectangle I got another strange limiting behaviour: not a single shape, but two polygons of different area, with one of them transforming into a rotated copy of the other (200 iterations are shown):

There is a caveat: the area for a self-intersecting polygon can be defined in several ways. I used a formula which counts the different loops with different sign, according to their chirality. Using a different definition could lead to different results.