Circle With Points & Intersections: Solution
### Caution - spoiler!!!

This is the solution to this puzzle.

There is an elaborate way to calculate the number intersections like:

- given n points on a circle, all connected = n/2 (n-1) lines
- for every point, we look at each connection via index i running from 1 (nearest neighbor) to n-1 (nearest neighbor other side)
- for every i, the line will divide the circle into two pieces, having (i-1) points on one side, and (n-i-1) points on the other side, therefore there will be (i-1)(n-i-1) lines crossed by the line in current consideration
- so we have have for that one point the following number of intersections

Sum_{i=1}^{n-1} (i-1) (n-1-i)

- now if we do that for all points, we gain a factor of n, but we do some double countings - first in the number of considered lines, second in the number considered intersecting lines, therefore we have to correct by dividing by 4
- Result

C(n) = n/4 Sum_{i=1}^{n-1} (i-1) (n-1-i)

The sum can be summed up explicitly, and the result is

C(n) = n/4 ( n^3/6 - n^2 + 11/6 n -1)

But there is also a nice short solution: An intersection needs 4 points, so the problem reduces to the number of sets of 4 points, which is

C'(n) = 1/24 (n (n-1) (n-2) (n-3))

which happens to be the same as C(n).