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I have been checking Gamow's question and noticed something! If we draw all diagonals there is a little polygon in the middle of the original polygon if it is odd-gon (odd-gon is defined as the number of vertices is an odd number.). For example;

enter image description here

So if you draw a regular pentagon on a piece of paper and draw all diagonals, there is a new pentagon in the middle of our original pentagon (shown as darker).

The area-ratio between the small and original pentagon is 6.854.

Question 1: What is the area-ratio between the small and original 99-gons of 99-gon where 99 is number of vertices of the polygon?

Question 2: Is there any formula that you can derive for the area-ratio between the small and original x-gons of x-gon where x is number of vertices of the polygon?

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up vote 10 down vote accepted

Inscribe the original $n$-gon in a circle of radius 1. The apothem of the large $n$-gon is $\cos(\frac{\pi}{n})$ and the apothem of the small $n$-gon is $\sin(\frac{\pi}{2n})$. Therefore the ratio of their areas is $\left(\frac{\cos(\frac{\pi}{n})}{\sin(\frac{\pi}{2n})}\right)^2$. For $n=99$ this is about $3968.53$.

Demonstration on a heptagon:

$OA=1$, $\angle OAB=\frac{\pi}{2n}$, $\angle AOC=\frac{\pi}{n}$ so $OB=\sin(\frac{\pi}{2n})$ and $OC=\cos(\frac{\pi}{n})$.

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good answer, thanks @f" – Oray Mar 28 at 18:11

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