Puzzling Stack Exchange is a question and answer site for those who create, solve, and study puzzles. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|improve this question
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 $OC=\cos(\frac{\pi}{n})$.

share|improve this answer
good answer, thanks @f" – Oray Mar 28 at 18:11

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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