Mock AIME 6 2006-2007 Problems/Problem 4

Problem

Let $R$ be a set of $13$ points in the plane, no three of which lie on the same line. At most how many ordered triples of points $(A,B,C)$ in $R$ exist such that $\angle ABC$ is obtuse?

Solution

For every triangle with all of its vertices included within the $13$ points, at most one angle can be obtuse. This means that at most $\frac{1}{3}$ of all of the angles can be obtuse. Since there are a total of $13\cdot12\cdot11$ angles, the maximum number of them that can be obtuse is $\frac{13\cdot12\cdot11}{3}=\boxed{572}$ . This is obtainable if the $13$ points are $13$ consecutive vertices of a regular $1000-gon$ (because every triangle out of these $13$ points has an obtuse angle).

~alexanderruan

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Mock AIME 6 2006-2007 Problems/Problem 4

Problem

Solution