Let $S$ be a set of $n$ points in the plane, no $3$ collinear. Determine the maximum number of right-angled triangles with all three vertices as points in $S$.
This is a slightly more difficult and precise question than IMO 1970. For $n=3$, clearly the maximum is one. For $n=4$, we can have four triangles, etc. However, I don't know how to continue.
Best Answer
For 5 points the maximal number of right triangles is 7.
In the following figure, $\phi$ is the golden ratio. Drop two and one of the extremal points to obtain 12 and 16 right triangles on 6 and 7 points. Partially based on the Kepler triangle.
For 8 points, 24 right triangles are possible with a regular octagon. Dropping a point yields only 15 triangles, so the above solution is better.
The regular decagon gives 40 right triangles.
The regular dodecagon gives 60 right triangles.