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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.

Captric
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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, $\binom{n}{4}$.

André Nicolas
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  • Perhaps I either don't understand or I did not explain the problem well. So I have a flat circle drawn on a piece of paper. Around the circle are 9 points equally spaced. Then I draw a line from EACH point to all of the 8 other points on the circle and I do this for EACH point. So the interior of the circle now looks like a matrix. Where the lines cross each other or may an intersection is what I would like to predict. In fact I counted the intersections by hand and there are 126 intersections. I cannot develop a mathematical model however that would PREDICT the number of intersections. – Captric Oct 13 '15 at 12:29
  • Well, I wrote a solution that says the number is $\binom{9}{4}$. This is $\frac{9!}{4!5!}$, which turns out to be $126$. And the points do not have to be equally spaced. So the procedure I used gives the answer you found by counting. In general, if you have $n$ points so that there are no multiple intersection points, the number of intersection points in the interior is $\binom{n}{4}$. Note this does not apply to things like the regular hexagon, since it has a multiple intersection point at the centre. – André Nicolas Oct 13 '15 at 13:41
  • You are welcome. There are other ways of doing the count, but this one is very quick, very slick. The argument is not "mine." I saw it somewhere many years ago. – André Nicolas Oct 13 '15 at 20:28