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 4 of 4: Convert the count to the required percentage

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\frac{7\binom{100}{5}}{\binom{97}{2}}=\frac7{10}\binom{100}{3}

Detailed analysis

Using $\binom{100}{5}/\binom{97}{2}=\binom{100}{3}/10$ , the preceding inequality gives $A_{100}\le(7/10)\binom{100}{3}$ . Since $\binom{100}{3}$ is the total number of triangles, at most $70\%$ are acute.