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Difference between revisions of "Mock AIME 6 2006-2007 Problems/Problem 4"

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For every triangle with all of its vertices included within the <math>13</math> points, at most one angle can be obtuse. This means that at most <math>\frac{1}{3}</math> of the angles can be obtuse. Since there are a total of <math>13\dot12\dot11</math> angles, the maximum number of them that can be obtuse is <math>\frac{13\dot 12\dot 11}{3}=\boxed{572}</math>. This is obtainable if the <math>13</math> points are <math>13</math> consecutive vertices of a regular <math>1000-gon</math>.
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Given any triangle with all of its vertices included within the <math>13</math> points, at most one of its three angles can be obtuse. This means that at most <math>\frac{1}{3}</math> of all of the angles can be obtuse. Since there are a total of <math>13\cdot12\cdot11</math> angles, the maximum number of them that can be obtuse is <math>\frac{13\cdot12\cdot11}{3}=\boxed{572}</math>. This is obtainable if the <math>13</math> points are <math>13</math> consecutive vertices of a regular <math>1000-gon</math> (because every triangle out of these <math>13</math> points has an obtuse angle).
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~alexanderruan

Latest revision as of 01:42, 3 January 2024

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

Given any triangle with all of its vertices included within the $13$ points, at most one of its three angles 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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