Problem 6

In a plane there are

100

points, no three collinear. Consider all triangles whose vertices are among these points. Prove that no more than

70\%

of these triangles are acute-angled.

Step 2 of 4: Bound five-point configurations

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5\cdot3=15\text{ acute-triangle incidences in 4-subsets},\quad 2A_5\le15\implies A_5\le7

Detailed analysis

A five-point set contains five four-point subsets, each with at most three acute triangles, so there are at most $15$ incidences. Every triangle belongs to exactly two of those four-point subsets. Hence if $A_5$ is the number of acute triangles, $2A_5\le15$ , and since $A_5$ is an integer, $A_5\le7$ .