# Comparing between regular polygons

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$$ (Difference of 0.31830999999999676 from previous result) ## 3 Answers Consider that the apothem $$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$$. • Nice! But how do you evaluate that$\cot\$ limit? Jan 22, 2021 at 20:43

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 $$\frac{1}{\pi}$$.

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.