# Repeating an operation infinitely makes any convex $n$-gon a regular $n$-gon?

For any convex $n$-gon $P_{0,1}P_{0,2}\cdots P_{0,n}$, let us consider the following operation :

Operation : Let $k=0,1,\cdots$. Take $n$ points $P_{k+1,i}\ (i=1,2,\cdots,n)$ outside of $n$-gon $P_{k,1}P_{k,2}\cdots P_{k,n}$ such that every $P_{k,j}P_{k,j+1}P_{k+1,j}\ (j=1,2,\cdots, n)$ is an isosceles right triangle whose right angle is $\angle P_{k,j}P_{k+1,j}P_{k,j+1}$ where $P_{k,n+1}=P_{k,1}$.

Then, here is my question.

Question : Is the following true for any convex $n$-gon ?

$$\lim_{k\to\infty}\frac{P_{k,2}P_{k,3}}{P_{k,1}P_{k,2}}=\lim_{k\to\infty}\frac{P_{k,3}P_{k,4}}{P_{k,2}P_{k,3}}=\cdots=\lim_{k\to\infty}\frac{P_{k,n}P_{k,1}}{P_{k,n-1}P_{k,n}}=\lim_{k\to\infty}\frac{P_{k,1}P_{k,2}}{P_{k,n}P_{k,1}}=1.$$

In other words, "If we repeat the operation infinitely for any $n$-gon $P_{0,1}P_{0,2}\cdots P_{0,n}$, then do we get a regular $n$-gon ?"

Remark : This question has been asked previously on math.SE without receiving any answers.

Motivation : I've found that the answer for $n=3$ is YES. However, I have no good idea for $n$ in general. Can anyone help?

Proof for $n=3$ : Let $S_k$ be the area of a triangle $P_{k,1}P_{k,2}P_{k,3}$. By law of cosines in a triangle $P_{k,1}P_{k,2}P_{k,3}$, we get $${P_{k+1,2}P_{k+1,3}}^2={P_{k,1}P_{k+1,2}}^2+{P_{k,1}P_{k+1,3}}^2-2P_{k,1}P_{k+1,2}\cdot P_{k,1}P_{k+1,3}\cos\angle P_{k+1,2}P_{k,1}P_{k+1,3}.$$ Since $\cos\angle P_{k+1,2}P_{k,1}P_{k+1,3}=\cos(\angle P_{k,2}P_{k,1}P_{k,3}+90^{\circ})=-\sin\angle P_{k,2}P_{k,1}P_{k,3}$, we get \begin{align}{P_{k+1,2}P_{k+1,3}}^2=\frac 12{P_{k,3}P_{k,1}}^2+\frac 12{P_{k,1}P_{k,2}}^2+2S_k\qquad(1)\end{align}. By the same argument above, we get \begin{align}{P_{k+1,3}P_{k+1,1}}^2=\frac 12{P_{k,1}P_{k,2}}^2+\frac 12{P_{k,2}P_{k,3}}^2+2S_k\qquad(2)\end{align}. \begin{align}{P_{k+1,1}P_{k+1,2}}^2=\frac 12{P_{k,2}P_{k,3}}^2+\frac 12{P_{k,3}P_{k,1}}^2+2S_k\qquad(3)\end{align}. By $(1)(2)(3)$ and $S_k\gt 0$, we get $${P_{k+2,2}P_{k+2,3}}^2\gt \frac 12{P_{k+1,3}P_{k+1,1}}^2+\frac 12{P_{k+1,1}P_{k+1,2}}^2$$$$\gt\frac 12\left(\frac 12{P_{k,1}P_{k,2}}^2+\frac 12{P_{k,2}P_{k,3}}^2\right)+\frac 12\left(\frac 12{P_{k,2}P_{k,3}}^2+\frac 12{P_{k,3}P_{k,1}}^2\right)\gt\frac 12{P_{k,2}P_{k,3}}^2.$$ Hence, we know that $2^n{P_{k,2}P_{k,3}}^2\ge 2^{\frac n2}{P_{0,2}P_{0,3}}^2$ when $n$ is even and that $2^n{P_{k,2}P_{k,3}}^2\ge 2^{\frac{n-1}{2}}{P_{1,2}P_{1,3}}^2$ when $n$ is odd. In either case, we can see $\lim_{k\to\infty}2^n{P_{k,2}P_{k,3}}^2=\infty.$

By $(2)-(1)$, since we get ${P_{k+1,3}P_{k+1,1}}^2-{P_{k+1,2}P_{k+1,3}}^2=-\frac 12\left({P_{k,3}P_{k,1}}^2-{P_{k,2}P_{k,3}}^2\right)$, we get $${P_{k,3}P_{k,1}}^2-{P_{k,2}P_{k,3}}^2=\left(-\frac 12\right)^n\left({P_{0,3}P_{0,1}}^2-{P_{0,2}P_{0,3}}^2\right).$$ Hence we get $$\left(\frac{P_{k,3}P_{k,1}}{P_{k,2}P_{k,3}}\right)^2=1+\frac{(-1)^n\left({P_{0,3}P_{0,1}}^2-{P_{0,2}P_{0,3}}^2\right)}{2^n{P_{k,2}P_{k,3}}^2}\to 1$$ when $k\to\infty$, which leads $$\lim_{k\to\infty}\frac{P_{k,3}P_{k,1}}{P_{k,2}P_{k,3}}=1.$$ By the same argument above, we get $$\lim_{k\to\infty}\frac{P_{k,1}P_{k,2}}{P_{k,2}P_{k,3}}=1.$$ Since we can easily get $$\lim_{k\to\infty}\frac{P_{k,2}P_{k,3}}{P_{k,1}P_{k,2}}=\lim_{k\to\infty}\frac{P_{k,3}P_{k,1}}{P_{k,2}P_{k,3}}=\lim_{k\to\infty}\frac{P_{k,1}P_{k,2}}{P_{k,3}P_{k,1}}=1,$$ we now know that the proof for $n=3$ is completed.

Update : I added a word 'isosceles'.

It's true if $n$ is prime.

Represent the vertices of the polygon as a vector of complex numbers $(z_1,\dots,z_n)$. Then the operation replaces this by a scalar multiple of $(z_1-iz_2,z_2-iz_3,\dots,z_n-iz_1)$. This linear map has a basis of eigenvectors $(1,\omega,\omega^2,\dots,\omega^{n-1})$, for $\omega$ taking values in the set of $n$th roots of 1, and with corresponding eigenvalues $1-i\omega$.

Expressing the original polygon as a linear combination of eigenvectors, we can assume (up to a translation) that it doesn't involve the eigenvector $(1,1,\dots,1)$. If $n$ is odd, no two eigenvalues have equal absolute value, so in the limit, after appropriate length normalization, the shape tends to that of the eigenvector of the largest eigenvalue, which is a regular $n$-gon if $n$ is prime.

I haven't used the fact that the vertices are listed in consecutive order reading around the polygon, or that the polygon is convex, and if you do, then I think this argument may work for any odd $n$ (or maybe even any $n$ not congruent to 2 (mod 4)?).

I suspect j.c.'s example is essentially the eigenvector $(1,\omega,\dots,\omega^5)$, where $\omega=e^{\pi i/3}$, which has eigenvalue $1+e^{-\pi i/6}$, representing a regular convex hexagon, plus a small multiple of the eigenvector whose eigenvalue is $1+e^{\pi i/6}$.

Edit: Having realized that the operation doesn't necessarily preserve convexity or even simplicity, I should clarify that in interpreting the question I assume that at each stage we put the right triangle on the right hand side as we go from each vertex to the next. This certainly agrees with what the questioner intends if the polygon is simple with vertices numbered anti-clockwise.

Although my argument shows that, if $n$ is prime, the limiting behaviour is a regular $n$-gon, it may not be a simple regular $n$-gon (e.g., each vertex could be joined not to the adjacent vertices, but to those that are $k$ vertices away).

In fact, if $n\geq 7$ is odd (but not necessarily prime), this will be the usual behaviour, as then the largest eigenvalue does not occur for the eigenvector $(1,\omega,\dots,\omega^{n-1})$ (where $\omega=e^{2\pi i/n}$) that corresponds to a simple regular $n$-gon, but for $(1,\omega^k,\dots,\omega^{(n-1)k})$, where $k$ is the closest integer to $n/4$. So if we start with a configuration that involves this eigenvector, then the limiting behaviour will be the vertices of a regular $n$-gon with each vertex joined to those that are $k$ vertices away.

If $n$ is not prime, then it is possible to design convex starting polygons that approach polygons with a smaller number of edges (dividing $n$) in the limit. For example, if $n=9$, and $\omega=e^{2\pi i/9}$, and we take as our starting configuration $(1,\omega,\dots,\omega^8)$ plus a small multiple of $(1,\omega^3,\dots,\omega^6)$, which is the eigenvector with second smallest eigenvalue, then the limiting behaviour degenerates to a triangle.

• It's just occurred to me that the operation doesn't preserve convexity or even the property of being a simple polygon: consider what happens if you start with the convex quadrilateral with vertices at $(-1,0)$, $(1,0)$, $(x,x)$ and $(-x,x)$ for very small positive $x$. I think my answer still works if you interpret the question appropriately, but I'm no longer confident about what happens if $n$ is odd but not prime. – Jeremy Rickard Oct 29 '13 at 18:11
• Nice answer! I hope you don't mind that I've added some illustrations to your answer. – j.c. Oct 30 '13 at 15:09
• @j.c.: I don't mind at all! Very nice pictures! It might be fun to look at some examples where there are two maximal eigenvalues involved, with equal coefficients. – Jeremy Rickard Oct 30 '13 at 16:25

Here's a counterexample, a hexagon that falls into a nontrivial 2-cycle:

The coordinates of one of the hexagons is approximately:

{{-2.0951, 1.04954}, {-1.86954, -0.931927}, {-0.480952, -2.1886}, {1.92123, -1.76496}, {2.57605, 1.13906}, {-0.0516838, 2.69689}}

For those who want to experiment more, here's a link to the Mathematica notebook I used to generate it.

edit: Jeremy Rickard points out that (with points interpreted as complex numbers) this example is (up to scaling and rotation) a sum of eigenvectors $z_\omega=(1,\omega,\omega^2,\dots,\omega^5)$ with $\omega=e^{2\pi i/6}$ and $z_{\omega^2}=(1,\omega^2,\omega^4,\dots,\omega^4)$. Indeed, I found that (after rotation by -0.416 radians and a permutation of the coordinates) the above hexagon is $z_\omega+(0.142)z_{\omega^2}$.

I have also updated the above notebook with the code I used to make the animated images that I've edited into his answer.

• Strictly speaking I guess I mean 2-cycle after appropriate length normalization. – j.c. Oct 29 '13 at 17:03

This is just a request for clarification, an attempt to penetrate the notation. Is this what you have in mind? If so, the operation is not definite: any point $P_{k+1,j}$ on the circle arc between $P_{k,j}$ and $P_{k,j+1}$ forms a right angle.

After the OP's correction of the problem statement:

• Nice drawing. Also, related to the indefiniteness of each step of the process: one can choose the new points arbitrarily close to any one of the endpoints of the hypothenus of the triangle, and so end up with something that degenerates to a polygon with less vertices (by choosing $P_{j,k+1}$ and $p_{j+1,k+1}$ at distance $1/(k+1)$), or a polygon arbitrarily close to the original one (by choosing $P_{j,k+1}$ at distance $m^{-k-1}$ of $P_{j,k}$, say). – Jean Raimbault Oct 29 '13 at 12:10
• @Joseph O'Rourke: Thank you very much for pointing it out. I forgot to write a very important word 'isosceles'. – mathlove Oct 29 '13 at 12:31