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I've found on the internet this interesting claim

If you repeatedly take the midpoint polygon from a random polygon you will approximate a regular polygon

To support the claim, the following animation was provided

iterated midpoint polygon of pseudo-random hexagon

If true, this is quite an amusing result! But how to prove it?

First, let's formalize the statement. Let $E$ denote the euclidean plane and consider the function $M_n: E^n\to E^n$ given as

$$M_n \begin{pmatrix} X_1\\ X_2\\ \dots\\ X_n \end{pmatrix} = \begin{pmatrix} X_1+X_2\\ X_2+X_3\\ \dots\\ X_n+X_1 \end{pmatrix}$$

This function takes an $n$-gon to its midpoint polygon (scaled up by a factor of $2$ for simplicity). One wishes to prove successive applications of $M_n$ will get us closer to a regular polygon. Of course, we should normalize the polygon's area after each application for this convergence to make sense.

It is worth noting $M_n$ is the linear transformation given by $$M_n = \begin{pmatrix} 1 & 1 & 0 & \dots & 0\\ 0 & 1 & 1 & \dots & 0\\ 0 & 0 & 1 & \dots & 0\\ \vdots & \vdots & \vdots & \ddots & \vdots\\ 1 & 0 & 0 & \dots & 1\\ \end{pmatrix}_{n\times n}$$

Obs.: Notice $M_n$ is a circulant matrix, thus, it comutes with any other circulant matrix. Circulant matrices are not necessarily orthogonal, therefore, they do not necessarily preserve similarities. This could've been a problem, but it turns out regular polygons are preserved by circulant transformations, suggesting they are valid convergence points a priori.

One not-so-promising fact is that triangles in general are preserved by $M_3$ (in a similarity sense). In fact, this means the statement is false for triangles. Less obviously, the statement is also not true for general quadrilaterals: if you start with a rectangle, then every second application of $M_4$ will be a similar rectangle. The user Intelligenti pauca provided a similar counter-example for hexagons.

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  • $\begingroup$ Your claim is also wrong for some hexagons. Consider for instance the hexagon with vertices $(\pm a,\pm b)$ and $(0, \pm2b)$. $\endgroup$ Commented May 14 at 10:37
  • $\begingroup$ I also made some experiments with Geogebra and found that your claim seems to be false, in general: the shape of the polygons does indeed converge to some limit shape, but not necessarily to the shape of a regular polygon. $\endgroup$ Commented May 14 at 14:20
  • $\begingroup$ Related: Limit of iterated midpoint polygons $\endgroup$ Commented May 14 at 14:50
  • $\begingroup$ +1 for the fascinating animation! $\endgroup$ Commented May 14 at 16:13

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Not a full answer, just a comment with image.

Experimenting with GeoGebra I found that - at least in the case of a hexagon - odd-numbered polygons in the sequence tend to a shape, while even-numbered tend to another shape.

enter image description here

Here's a zoom on the inner region, showing the angles. The limit hexagons aren't regular, but opposite sides are parallel.

enter image description here

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  • $\begingroup$ This also seem to happen with quadrilaterals $\endgroup$ Commented May 14 at 17:23

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