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 3


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 $\approx 0.318$.

  • $\begingroup$ Nice! But how do you evaluate that $\cot$ limit? $\endgroup$
    – jlammy
    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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.