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 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 has an obtuse angle). | + | 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 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). |
~alexanderruan | ~alexanderruan | ||
Revision as of 02:41, 3 January 2024
Problem
Let 
be a set of 
points in the plane, no three of which lie on the same line. At most how many ordered triples of points 
in exist such that 
is obtuse?
Solution
For every triangle with all of its vertices included within the points, at most one angle can be obtuse. This means that at most 
of all of the angles can be obtuse. Since there are a total of 
angles, the maximum number of them that can be obtuse is 
. This is obtainable if the points are consecutive vertices of a regular 
(because every triangle out of these points has an obtuse angle).
~alexanderruan
