Inscribed To Infinity

Geometry Level 4

Take a circle of radius 1. Inscribe a regular triangle in this circle. Inscribe a circle in this triangle. Inscribe a square in it. Inscribe a circle, regular pentagon, circle, regular hexagon and so forth.

What is the radius of the circle formed when the process is carried forever (the radius of the limiting circle)?

Give the answer correct to 3 decimal places.


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The answer is 0.114.

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3 solutions

Digvijay Singh
Apr 13, 2015

The equation relating the inradius and the circumradius of a regular polygon, r = R cos ( π n ) r=R\cos\bigg(\frac{\pi}{n}\bigg) gives the ratio of the radii of the final to initial circles as K = r final circle r initial circle = cos ( π 3 ) cos ( π 4 ) cos ( π 5 ) \large K^{'}=\frac{r_\text{final circle}}{r_\text{initial circle}}=\cos\bigg(\frac{\pi}{3}\bigg)\cos\bigg(\frac{\pi}{4}\bigg)\cos\bigg(\frac{\pi}{5}\bigg)\cdots Numerically, K = 1 K = 1 8.7000366252 = 0.1149420448 K^{'}=\frac{1}{K}=\frac{1}{8.7000366252}=0.1149420448

Moderator note:

Well done!

I calculated by hand until the radius of the 90th circle: 0.1211718

In the end I used wolfram Alpha...

It looks really cool by the way: Here

Julian Poon - 6 years, 1 month ago

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How did u calculate the 90th by hand. That's insane.

Trevor Arashiro - 6 years, 1 month ago

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I meant manually.. with a calculator.... opps...

Julian Poon - 6 years, 1 month ago

Yea I tried doing the last step with a calculator to but it wouldn't converge fast enough

Curtis Clement - 6 years, 1 month ago

Please Post Solution Here ... @Digvijay Singh

Karan Shekhawat - 6 years, 2 months ago

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done! :) hope you like it....

Digvijay Singh - 6 years, 2 months ago

Solving this for hours then I found out that this was only a constant XD

Rindell Mabunga - 6 years, 1 month ago
Paulo Carlos
May 8, 2015

Kepler-Bouwkamp constant.

Got on Wikipedia

Can anyone see a nice way to solve this analytically?

I reluctantly gave up and used Python to approximate the product series

n = 3 n = 10000000 sin ( ( n 2 ) π 2 n ) = 0.11494210157473356 \prod_{n=3}^{n=10000000}\sin(\frac{(n-2)\pi}{2n}) = 0.11494210157473356

Here is the code :

import numpy as np
def product_series(N):
product = 1.0
for i in range(3,n+1):
    product*= np.sin((n-2)*np.pi/(2*n))
return product

product_series(10000000)

I did the same with excel. got answer 0.115.

Nikhil Karkare - 6 years ago

It's possible to get this in terms of the Riemann zeta function, but I have no idea how.

Joe Mansley - 2 years, 9 months ago

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