What is the area of the white region if the total area is 1?
I wrote a code a while ago to try to answer this, and the answer I got was that the area of the white region approached π/10 of the total area. However, to confirm that, I would have needed to let the code run indefinitely to account for all infinity polygons. I was wondering if anyone could prove (or disprove) the area is π/10 rather than relying on code to approximate it.
By the way, I came up with this problem almost a year ago, and I showed a friend of mine and he posted it here, but it didn't get a concrete answer. Here's the link to his post: Area of the shaded region of a infinitely circumscribed set of polygons.
The way I would proceed is to first assume the smallest circle has radius $r_3 = 1$, calculate the limiting radius of the bounding circle, then rescale accordingly.
In general, let $r_n$ be the inradius of the corresponding $n$-gon. Then $r_{n+1}$ is the circumradius of the enclosed $n$-gon, and we have the relationship $$\cos \frac{\pi}{n} = \frac{r_n}{r_{n+1}}.$$ Thus $$r_n = r_3 \prod_{m=3}^{n-1} \sec \frac{\pi}{m},$$ and $$r_\infty = r_3 \prod_{n=3}^\infty \sec \frac{\pi}{n}.$$ We also note that the white area enclosed by a given $n$-gon excluding any nested polygons within its incircle is simply $$A_n = \left(n \tan \frac{\pi}{n} - \pi \right) r_n^2.$$ Thus the total white area is $$A = \sum_{n=3}^\infty A_n,$$ and after rescaling such that $r_\infty = 1$, we have $$A = \frac{1}{r_\infty^2} \sum_{n=3}^\infty A_n.$$ This expression has no closed-form value but it is approximately $0.97$. I will update this answer with a more precise value later.