There are $75$ points in the plane, no three collinear. Prove that the number of acute triangles is no more than $70\%$ from the total number of triangles with vertices in these points.
Source: Moldova 2017 TST, day 3, B12
Tags: combinatorics
There are $75$ points in the plane, no three collinear. Prove that the number of acute triangles is no more than $70\%$ from the total number of triangles with vertices in these points.