# 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 π number with 0. before (possibly such as 0.314159265359...). Can anyone confirm it or elaborate on it?

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)
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relatatt9
Step 1
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:
$a=\frac{\mathrm{tan}\left(\frac{\left(n-2\right)\pi }{2n}\right)}{2}$
$c=\frac{1}{2}\mathrm{csc}\left(\frac{\pi }{n}\right)$
Step 2
Since the angle $\mathrm{\angle }ac$ is always $\pi /n$.
We may simplify $a+c=\frac{1}{2}\mathrm{cot}\left(\frac{\pi }{2n}\right)$. Then what you seek to compute is:
$\underset{n\to \mathrm{\infty }}{lim}\frac{1}{2}\left(\mathrm{cot}\left(\frac{\pi }{2n}\right)-\mathrm{cot}\left(\frac{\pi }{2n-2}\right)\right)$
This in fact converges to $1/\pi$, which is $\approx 0.318$
###### Did you like this example?
Emilio Calhoun
Step 1
Let ${C}_{n}$ be the value of C for a regular n-gon. I claim that ${C}_{n}/n=1/\pi +O\left({n}^{-2}\right)$ as $n\to \mathrm{\infty }$.
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 $\frac{1}{2\mathrm{tan}\left(\pi /n\right)}$ and the circumradius is $\frac{1}{2\mathrm{sin}\left(\pi /n\right)}$. As $h\to 0$, we have the results $\mathrm{tan}h=h+O\left({h}^{3}\right)$ and $\mathrm{sin}h=h+O\left({h}^{3}\right)$. So as $n\to \mathrm{\infty }$,
$n\mathrm{tan}\left(\pi /n\right)=\pi +O\left({n}^{-2}\right)\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}n\mathrm{sin}\left(\pi /n\right)=\pi +O\left({n}^{-2}\right).$
Step 2
So we have $\frac{{C}_{n}}{n}=\frac{1}{2n\mathrm{tan}\left(\pi /n\right)}+\frac{1}{2n\mathrm{sin}\left(\pi /n\right)}=\frac{1}{2\pi }+\frac{1}{2\pi }+O\left({n}^{-2}\right)=\frac{1}{\pi }+O\left({n}^{-2}\right)$ as claimed. Now
${C}_{n}-{C}_{n-1}=\frac{n}{\pi }-\frac{n-1}{\pi }+O\left({n}^{-1}\right)=\frac{1}{\pi }+O\left({n}^{-1}\right).$
So we get convergence to $\frac{1}{\pi }$