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 3 of 4: Count five-point subsets of the 100-point set

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A_{100}\binom{97}{2}\le7\binom{100}{5}

Detailed analysis

Let $A_{100}$ be the number of acute triangles among the $100$ points. Each five-point subset contains at most $7$ acute triangles, giving at most $7\binom{100}{5}$ incidences. Each fixed acute triangle is contained in exactly $\binom{97}{2}$ five-point subsets. Therefore $A_{100}\binom{97}{2}\le7\binom{100}{5}$ .