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?

Alexander Lewis 2022-10-20 Answered
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 π 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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Answers (2)

relatatt9
Answered 2022-10-21 Author has 12 answers
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 = tan ( ( n 2 ) π 2 n ) 2
c = 1 2 csc ( π n )
Step 2
Since the angle a c is always π / n.
We may simplify a + c = 1 2 cot ( π 2 n ) . Then what you seek to compute is:
lim n 1 2 ( cot ( π 2 n ) cot ( π 2 n 2 ) )
This in fact converges to 1 / π, which is 0.318
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Emilio Calhoun
Answered 2022-10-22 Author has 2 answers
Step 1
Let C n be the value of C for a regular n-gon. I claim that C n / n = 1 / π + O ( n 2 ) as n .
Consider the triangle formed by two adjacent vertices and the centre of the polygon. This is an isosceles triangle with apex angle 2 π / n.
The inradius is 1 2 tan ( π / n ) and the circumradius is 1 2 sin ( π / n ) . As h 0, we have the results tan h = h + O ( h 3 ) and sin h = h + O ( h 3 ). So as n ,
n tan ( π / n ) = π + O ( n 2 ) and n sin ( π / n ) = π + O ( n 2 ) .
Step 2
So we have C n n = 1 2 n tan ( π / n ) + 1 2 n sin ( π / n ) = 1 2 π + 1 2 π + O ( n 2 ) = 1 π + O ( n 2 ) as claimed. Now
C n C n 1 = n π n 1 π + O ( n 1 ) = 1 π + O ( n 1 ) .
So we get convergence to 1 π
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