Good n-gon Part II

This was a beautiful problem. Although my answer is not exactly same as the one given.

We will use a small concept in physics to solve this question: Moment of inertia.

Take the center of circle as origin $O$ . Now, we want to find out sum of squares when one vertex of smaller polygon is connected to all the vertices of bigger one. Assume mass of each vertex as unity.

Moment of inertia $I$ of larger polygon about $O$ :

$I_{O}=2015x^2$ , where $x$ is distance of one vertex from $O$ .

Moment of inertia about one of the vertices of the smaller polygon is given by:

$I_{v}=I_{O}+2015r^2$ , where $r$ is radius of circle.

$\Rightarrow I_{v}=$ sum of squares of lengths $=\alpha=2015(x^2+r^2)$

Sum of areas of the polygons is simply: $\gamma=\pi(x^2+r^2)$ , since a $2015$ -gon can be approximated to a circle.

Hence: $\alpha/\gamma=\boxed{641.394}$

Brilliant accepted this approximate value.

@Vraj Mehta