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In this experiment, I am adding the inradius (let's call it $A$) and circumradius (let's call it $B$) of different polygons with equal sides each equal $1$ (starting with a square and adding one side each time). The result is $A+B=C$ when side of polygon = 1.

When comparing the $C$ of one polygon with the $C$ of a polygon with one side more, the difference seems to go smaller, as if approaching a version of $\pi$ number with $0.$ before (possibly such as 0.314159265359...).

Can anyone confirm it or elaborate on it?

I can not go over a polygon with 1000 sides in my computation power, and would like to know what to expect while going towards a polygon with infinity sides.

Here are some examples:

4 sided polygon: $0.5 + 0.707106781 = 1.207106781$

5 sided polygon: $0.68819096 + 0.850650808 = 1.5388417680000002$ (Difference of 0.33173498700000015 from previous result)

6 sided polygon: $0.866025404 +1 = 1.866025404$$ (Difference of 0.3271836359999998 from previous result)

7 sided polygon: $1.0382607 + 1.15238244 = 2.1906431399999997$ (Difference of 0.32461773599999977 from previous result)

8 sided polygon: $1.20710678 + 1.30656296 = 2.51366974$ (Difference of 0.3230266000000004 from previous result)

9 sided polygon: $1.37373871 + 1.4619022 = 2.83564091$ (Difference of 0.32197116999999986 from previous result)

10 sided polygon: $1.53884177 + 1.61803399 = 3.15687576$ (Difference of 0.3212348500000002 from previous result)

11 sided polygon: $1.70284362+ 1.77473277 = 3.47757639$ (Difference of 0.3207006299999997 from previous result)

12 sided polygon: $1.8660254+ 1.93185165 = 3.79787705$ (Difference of 0.32030066 from previous result)

13 sided polygon: $2.02857974+ 2.08929073 = 4.11787047$ (Difference of 0.31999341999999986 from previous result)

14 sided polygon: $2.19064313 + 2.2469796 = 4.43762273$ (Difference of 0.31975226000000045 from previous result)

15 sided polygon: $2.35231505+ 2.40486717 = 4.757182220000001$ (Difference of 0.3195594899999996 from previous result)

...

999 sided polygon: $158.995264 + 158.99605 = 317.991314$

1000 sided polygon: $159.154419 + 159.155205 = 318.309624$ (Difference of 0.31830999999999676 from previous result)

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Consider that the apothem $a$, circumcircle radius $c$, and edge of the polygon form a right triangle.

That is, we have two legs, one of length $1/2$, the other of length $a$, and a hypotenuse of length $c$. Then we have:

$$a = \frac{\tan\left(\frac{(n-2)\pi}{2n}\right)}{2}$$

$$c = \frac{1}{2}\csc\left(\frac{\pi}{n}\right)$$

Since the angle $\angle ac$ is always $\pi/n$.

We may simplify $a + c = \frac{1}{2}\cot\left(\frac{\pi}{2n}\right)$. Then what you seek to compute is:

$$\lim_{n \rightarrow \infty} \frac{1}{2}\left(\cot\left(\frac{\pi}{2n}\right) - \cot\left(\frac{\pi}{2n-2}\right)\right)$$

This in fact converges to $1/\pi$, which is $\approx 0.318$.

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  • $\begingroup$ Nice! But how do you evaluate that $\cot$ limit? $\endgroup$ – jlammy Jan 22 at 20:43
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Let $C_n$ be the value of $C$ for a regular $n$-gon. I claim that $C_n/n=1/\pi+O(n^{-2})$ as $n\to\infty$.

Consider the triangle formed by two adjacent vertices and the centre of the polygon. This is an isosceles triangle with apex angle $2\pi/n$.

The inradius is $\frac{1}{2\tan(\pi/n)}$ and the circumradius is $\frac{1}{2\sin(\pi/n)}$. As $h\to0$, we have the results $\tan h=h+O(h^3)$ and $\sin h=h+O(h^3)$. So as $n\to\infty$, $$n\tan(\pi/n)=\pi+O(n^{-2})\quad\text{and}\quad n\sin(\pi/n)=\pi+O(n^{-2}).$$ So we have $$\frac{C_n}{n}=\frac{1}{2n\tan(\pi/n)}+\frac{1}{2n\sin(\pi/n)}=\frac{1}{2\pi}+\frac{1}{2\pi}+O(n^{-2})=\frac{1}{\pi}+O(n^{-2})$$ as claimed. Now $$C_n-C_{n-1}=\frac{n}{\pi}-\frac{n-1}{\pi}+O(n^{-1})=\frac{1}{\pi}+O(n^{-1}).$$ So we get convergence to $\frac{1}{\pi}$.

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Let the sum of the inradius and circumradius of a regular polygon with $n$ sides be $C_n$. As $n\rightarrow \infty $, the regular polygon resembles a circle with circumference $\pi C_n$. The actual circumference is $n$ for an $n-$sided polygon with side length 1. Thus $n \rightarrow \pi C_n$ as n gets bigger. This implies $$C_n \rightarrow \dfrac{1}\pi(n)$$ as $n$ gets bigger. Consequently, $C_{n+1} -C_{n} \rightarrow 1/\pi = 0.318309\dots$ as $n$ tend towards infinity.

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