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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.

Vigo Hornblower
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    The diagram is amazing! Which software did you use to draw it? – Sarvesh Ravichandran Iyer Sep 23 '20 at 05:42
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    tbh, I don't remember where I found it. It's not mine. I just saw it and thought of this problem. But I do remember this diagram was not made with this problem in mind. – Vigo Hornblower Sep 23 '20 at 05:54
  • Neat question. Have you tried computing something easier, like the radius of the outer circle? – Thomas Andrews Sep 23 '20 at 05:55
  • Yes I think I did do that and got r = 8.7000323368 if we're saying the center circle has a radius of one, but I can't remember if I did that analytically, or got that from somewhere else. May have actually found that answer online. Keep in mind I came up with this like a year ago. – Vigo Hornblower Sep 23 '20 at 06:01
  • Assuming the starting radius is $1,$ in think the outer radius is something like $$\prod_{n=3}^{\infty}\sec(\pi/n)$$ Not sure if there is a closed form for that. The product converges since $\sec x=1+O(x^2).$ – Thomas Andrews Sep 23 '20 at 06:01

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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.

heropup
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  • When I solved it numerically, with code, I got it to be very close to pi/10 (up to 5 decimal places). 0.97 gives us ~0.3088 – Vigo Hornblower Sep 23 '20 at 06:25
  • @VigoHornblower I apologize, I misunderstood that you wanted the area of the enclosing circle to be $1$, not the radius. In that case, you would divide the last equation by $\pi$, which is what you have done. However, without getting more precision first, it may be premature to assert the value is exactly $\pi/10$. I will try to get more precision. – heropup Sep 23 '20 at 06:31
  • @VigoHornblower I got a more precise result of approximately $0.3141339$. Unfortunately I can't verify the precision since I had to use a polynomial fit for this very slow-converging series, and I haven't had time to figure out how to rewrite $A$ in terms of a more quickly converging series. But I am reasonably confident that the first five digits are correct. – heropup Sep 24 '20 at 01:36