Internal polygon formed by drawing diagonals in a regular polygon

Question

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.

Answer 1

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

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