# Internal polygon formed by drawing diagonals in a regular polygon

In an n-sided (n>4) regular polygon, label the vertices {0, 1, ..., n-1}. For each vertex i, draw a pair of diagonals:

• from i to (i+2) mod n and
• from i to (i-2) mod n

Question: What is the ratio of the area of the internal polygon to the external polygon?

For n=5 and n=6, this process will result in the following diagrams. For the pentagon, the ratio would be area(ONRQP)/area(CBFED). (source: uga.edu) Background: This page has a discussion about pentagons and notes that if CB=1 (unit length) then ON=1/$$\phi^2$$ where $$\phi=(1+\sqrt5)/2$$ is the golden ratio. Since the area of a pentagon is proportional to the square of its side length (see here), area(ONRQP)/area(CBFED) = 1/$$\phi^4$$ ~ 0.146.

But I don't know how to generalize it to a regular n-gon. It seems obvious that as n increases, the ratio approaches 1 but I wonder if there is a closed form expression as a function of n.

## 1 Answer

You mentioned that the area is proportional to the the square of the side length, but it could be difficult to calculate the side lengths that are involved.

Hint: The area is proportionate to the distance from the center.

Let $\omega$ be a $n$th root of unity, and $\omega^i$ be the vertices of the polygon.

What is the distance of a side of the regular polygon from the origin?

$| \frac{ \omega^i + \omega^{i+1} } { 2} |$

What is the distance of a diagonal of the regular polygon from the origin?

$| \frac{ \omega^{i} + \omega^{i+2} } {2}|$

Hence, the ratio of areas is

$\left| \frac{ \omega^1+\omega^{-1}}{\omega^{\frac{1}{2}}+\omega^\frac{1}{2}} \right|^2 = ...$

• Given $\omega^i$ as the vertices I can see that the length of a side of the polygon is |$\omega^i$ - $\omega^{i+1}$| and a similar expression for the diagonals. I don't know how to compute the distance of a side. Also, I just don't see how that will help me compute the areas. – user2602740 Sep 17 '13 at 17:43
• @user2602740 The distance from the side to the origin is the distance from the midpoint of the side to the origin, namely $\frac{\omega^i + \omega^{i+1} } { 2}$. What about the distance from the diagonal to the origin? – Calvin Lin Sep 17 '13 at 19:09
• As the number of sides increases and the ratio of inner area/outer area goes to 1, we are computing $\pi.$ – Fred Kline Dec 26 '14 at 11:19
• @FredKline Can you please show me how we are computing pi? – user2602740 May 9 '15 at 18:43
• ![en.wikipedia.org/wiki/… should have a little info. Scroll down to Polygon approximation to a circle – Fred Kline May 9 '15 at 18:45