The Formula, solution
Start with a circle with radius r and a n-gon with a known side \(s_0\).
In the picture to the right, \(OB=OC=r\).
Using Pythagoras' theorem on the triangle \(\Delta OBD\) we get
\[OD=\sqrt{r^2-\frac{s_0^2}{4}}\]
The length
\[CD=r-OD=r-\sqrt{r^2-\frac{s_0^2}{4}}\]
Using Pythagoras' theorem on the triangle \(\Delta BCD\) we get
\[BC=\sqrt{BD^2+CD^2}=\sqrt{\frac{s_0^2}{4}+\left(r-\sqrt{r^2-\frac{s_0^2}{4}}\right)^2} \]
This can be simplified as below
\[BC=\sqrt{\frac{s_0^2}{4}+r^2-2r\sqrt{r^2-\frac{s_0^2}{4}}+\left(r^2-\frac{s_0^2}{4} \right)} = \sqrt{2r^2-2r\sqrt{r^2-\frac{s_0^2}{4}}}\]
The last simplification is not needed in order to enter it into a spreadsheet.
A hexagon
A regular hexagon can be divided into six equilateral triangles with the interior angles 60°. If the hexagon is inscribed in a circle with radius 1, the side length of the hexagon is 1. We get the recursive formula for the side lengths of the successive polygons
\[ \left\{ \begin{eqnarray} s_0 &=& 1 \\ s_{n+1} &=& \sqrt{2-2\sqrt{1-\frac{s_n^2}{4}}} \end{eqnarray} \right. \]
where \(s_0\) is the side of the hexagon and \(s_n\) is the side of a 6⋅2n-gon when n>1.
by Malin Christersson under a Creative Commons Attribution-Noncommercial-Share Alike 2.5 Sweden License
