The sum of the perimeter of regular polygons inscribed inside of regular polygons

Question

This is a question combining number theory and geometry. I am asking it purely from curiosity, but I think it might be a useful and interesting question.

Start with an equilateral triangle of perimeter 1. Inside of that triangle, inscribe the largest square possible, and add its perimeter to the perimeter of the triangle. Then, inscribe a regular pentagon inside of the square, and add its perimeter. Keep on doing this infinitely, and what would be the sum of the perimeters of all of the polygons?

After some thought, my guess is that it diverges to infinity, very slowly, because even as the polygons get smaller, the rate at which they get smaller decreases. But it might be a convergent series.

Answer 1

Let $P_n$ the $n$-agon built with this procedure and $p_n$ its perimeter. Let we consider the centre $O_{n+1}$ of $P_{n+1}$ and the rays departing from $O_{n+1}$ through the vertices of $P_{n+1}$: they meet the boundary of $P_n$ at $n+1$ points, delimiting $n+1$ polygonal arcs on $\partial P_n$. Let $L_n$ be the polygonal arc with the greatest length and $l_{n+1}$ be the side length of $P_{n+1}$.

Reasonable conjecture: $L_n \leq \frac{l_{n+1}}{\cos\frac{\pi}{n+1}}.$

If this conjecture (or something similar) holds, the perimeter of $P_{n+1}$ is shorter than the perimeter of $P_{n}$ (that is obvious since $P_{n+1}$ is a convex subset of $P_{n}$), but: $$ p_{n+1}\geq \cos\left(\frac{\pi}{n+1}\right) p_n, $$ and since $\cos\left(\frac{\pi}{n}\right)$ behaves like $e^{-\frac{\pi^2}{2n^2}}$ for large values of $n$, the product: $$ \prod_{n\geq 3}\cos\left(\frac{\pi}{n}\right) $$ is convergent to a positive constant, giving that $$ \lim_{n\to +\infty} p_n \geq C p_3 > 0 $$ and ensuring that $\sum_{n\geq 3}p_n$ is not converging.

Now it is enough to prove the above reasonable conjecture. $p_{n+1}$ is for sure greater than the perimeter of the regular $(n+1)$-agon inscribed in the incircle of $P_n$, hence:

$$ p_{n+1} \geq p_n \cdot \color{red}{\frac{(n+1)\sin\frac{\pi}{n+1}}{n\tan\frac{\pi}{n}}} $$

and the red term behaves like $e^{-\frac{\pi^2}{2n^2}}$ for large $n$, hence we may state that $\sum_{n\geq 3}p_n$ is divergent by the same argument as above.

Answer 2

In each polygon inscribe a circle, and then inscribe the next polygon inside that circle.

Clearly each polygon fits inside the one before it, and its perimeter is less than maximal.

The radius of each circle:

$r_{n+1} = r_n \cos \frac {\pi}{n}$

The perimiter of the polygon inside of it:

$P_{n+1} = 2r_{n+1}(n+1) \sin \frac {\pi}{n+1}$

$2(n+1) \sin \frac {\pi}{n+1} > 2n \sin \frac {\pi}{n}$

$\sum P_k > P_3\sum \frac{r_k}{r_3} $

$r_n = r_3\prod_\limits{k=3}^{n} \cos \frac {\pi}{k}$

$\cos \frac {\pi}{k} > (1-\frac{\pi^2}{2k^2})>\frac k{k+1}$ when k > 3

$\prod_\limits{k=3}^{n} \cos \frac {\pi}{k} > \prod_\limits{k=3}^{n} \frac{k}{k+1} = \frac {1}{n+1}$

$\prod_\limits{k=3}^{n} \cos \frac {\pi}{k} > \frac {1}{n}$ for all n.

$\sum P_k>P_3 \sum \frac 1n$

$\sum P_k$ diverges by the comparison test