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\ | + | 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>. |
Revision as of 01:38, 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
This problem needs a solution. If you have a solution for it, please help us out by adding it.
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 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 
.
