# 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.

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. Sep 17, 2013 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? Sep 17, 2013 at 19:09
• As the number of sides increases and the ratio of inner area/outer area goes to 1, we are computing $\pi.$ Dec 26, 2014 at 11:19
• @FredKline Can you please show me how we are computing pi? May 9, 2015 at 18:43
• ![en.wikipedia.org/wiki/… should have a little info. Scroll down to Polygon approximation to a circle May 9, 2015 at 18:45