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
.