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'.