Area of the shaded region of a infinitely circumscribed set of polygons.

Question

The problem itself.

Recently, a friend of mine has sent me this image, and asked me if I could figure out the area of the white region in this circle; initially, I thought it was simply a infinite summation of a n-gon area, but I found myself stuck, and as such, I wanted some help.

The puzzle is: We have a infinite set of circumscribed polygons, each one in a increasingly larger circle; a infinite amount of them form this larger circle. What is the area of all of the polygons?

My attemp boiled down to, first, describing the problem:

$$(\ \sum^{\infty}_{?}W_a +\sum^{\infty}_{?}C_a\ )=\pi r_1^2$$ This is just a statement that the infinite sum of the white shades and the circles give off the area of a circle, which is just $\pi r^2$; my next step was delineating this to a more specific sum. I decided that the white shades should be the series of a circumscribed n-gon:

$$ \sum^{\infty}_{3}\frac{1}{2}nr_n^2 sin(\frac{2 \pi}{n})$$

So, to find the actual area would just be: $$\sum^{\infty}_{3}\frac{1}{2}nr_n^2 sin(\frac{2 \pi}{n})=\pi r_1^2- ( \sum^{\infty}_{n} \pi r_n^2)$$ However, I realized that the actual radii would be changing in as well, in the form of a infinite series: $$\sum^{\infty}_0 r_n+y$$ After that, I couldn't figure it out anymore. Does anyone know how to deal with something like this?

Answer 1

Notice that two consecutive circles are the inscribed and circumscribed circles of a regular n-gon. You can show that their radii are related by $$r_n \cos \frac{\pi}{n} = r_{n-1}.$$

Hence $$r_n = r_2 \prod_{k=3}^n \frac{1}{\cos \frac{\pi}{k}},$$ where $r_2$ is the radius of the inscribed circle of the triangle. To see that $\lim_{n\to\infty} r_n$ exists, consider $$\ln r_n = \ln r_2 - \sum_{k=3}^n \ln \cos\frac{\pi}{k}.$$ Using this answer one can see that $\ln r_n$ indeed converges, and so does $r_n$.

Edit: Actually, this question is closely related to the problem at hand.