Solve for intersecting chords inside a 9 sided polygon

Question

I am trying to solve for the number of intersecting points inside a sphere with 9 points. Each of the nine points has 8 chords running to adjacent points.

Answer 1

Let $G$ be a $9$-gon inscribed in a circle. Assume that three or more chords never pass through a point in the interior of the polygon. This is true in particular for the regular $9$-gon. We will count the number of chord intersection points in the interior of $G$.

Choose $4$ vertices from the $9$, say $A,B,C,D$. These form a convex $4$-gon. There are $6$ chords determined by these points, and $3$ ways to divide them into chord pairs. For exactly one of these divisions, the chord pair meet in the interior of $G$. (If the $4$-gon's vertices are $A,B,C.D$ in counterclockwise order, then chord pairs are $AC$ and $BD$.)

So the number of intersection points in the interior of $G$ is the same as the number of choices of $4$ vertices. This is the binomial coefficient $\binom{9}{4}$, which turns out to be $126$.

Remark: Let $G$ be an $n$-gon, not necessarily regular, inscribed in a circle. Suppose that there is no point $P$ in the interior of $G$ through which $3$ or more chords pass. Then the number of chord intersection points in the interior of $G$ is, by the same argument, $\binom{n}{4}$.