# A puzzle by Andrei Zelevinsky

While I was reading Tanya Khovanova‘s blog I came across this problem which Andrei Zelevinsky reportedly put in his list of important problems that undegraduate students should think of. The problem reads:

Consider a procedure: Given a polygon in a plane, the next polygon is formed by the centers of its edges. Prove that if we start with a polygon and perform the procedure infinitely many times, the resulting polygon will converge to a point.

I’d like to present one solution in the rest of this post.

One way to solve this problem is to use complex coordinates. Denote by $z_1,...z_n \in \mathbb{C}$ the points. One iteration of the process yields the points $(\frac{z_1+z_2}{2},\frac{z_2+z_3}{2},...,\frac{z_{n-1}+z_n}{2},\frac{z_n+z_1}{2})$. More generally, the $m$-th iteration yields the points $\Big( \sum_{l=1}^n a_{i,l}^{(m)} z_l \Big)_{1\leq i\leq n}$, where
$a_{i,l}^{(m)}= \frac{1}{2^m} {\displaystyle \sum_{\begin{array}{c} 0\leq k\leq m\\ i+k\equiv l[n] \end{array} }}\binom{m}{k}$.

Now we can prove that the $a_{_i,l}^{(m)}$ tend to $\frac{1}{n}$ as $m\to \infty$, independently of $i$ and $l$. To see this let us take the example of $i,l=0$, WLOG. A straightforward calculation using the $m$-th roots of unity gives $a_{0,0}^{(m)}=\frac{1}{2^m}\sum _{0 \leq k \leq n-1}(1+\omega ^k)^m= \frac{1}{K}{\displaystyle \sum_{k=0}^{n-1}}\cos^{m}\frac{k\pi}{n}e^{\frac{ik\pi m}{n}}$;

This means that the procedure converges to a unique point which is the barycenter of the $n$ original points, which concludes the proof.

In the next variation, instead of using the centers of edges to construct the next polygon, use the centers of gravity of $k$ consecutive vertices.

## One thought on “A puzzle by Andrei Zelevinsky”

1. Youssouf Emin says:

Excellent !

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