I'm part of a tournament where teams can compete in either two or three way competitions. The goal is to have everyone face everyone, and maximize the number of 3 way competitions, without anyone facing anyone twice. Ideally we would then optimize for an equal or near equal number of 3 way and 2 way competitions for each team.
It occurred to me that this is the same as what I stated in the title. Given n points, how many triangles can be drawn between n points without repeating any lines? I know that you're looking at $ \dfrac{n(n-1)}{2} $ potential lines between the points, and each triangle would take up 3 of those lines. Because of that, you would have, at most, $ \dfrac{n(n-1)}{6} $ triangles without repeat.
However, is there a way to find out how many triangles can be drawn without brute force trial and error? If so, is there a way to show what works? If not has someone already looked at this?
So far, here's what a friend and I have found:
| n | ideal | found |
|:---- |:------:| -----:|
| 3 | 1 | 1 |
| 4 | 2 | 1 |
| 5 | 3.33 | 2 |
| 6 | 5 | 4 |
| 7 | 7 | 5 |
| 8 | 9.33 | 8 |
| 9 | 12 | 12 |
| 10 | 15 | 12 |
Thank you.