View source for 1970 IMO Problems/Problem 6

==Problem==

In a plane there are <math>100</math> points, no three of which are collinear.  Consider all possible triangles having these point as vertices.  Prove that no more than <math>70 \%</math> of these triangles are acute-angled.

==Solution==
At most <math>3</math> of the triangles formed by <math>4</math> points can be acute. It follows that at most <math>7</math> out of the <math>10</math> triangles formed by any <math>5</math> points can be acute. For given <math>10</math> points, the maximum number of acute triangles is: the number of subsets of <math>4</math> points times <math>\frac{3}{\text{the number of subsets of 4 points containing 3 given points}}</math>. The total number of triangles is the same expression with the first <math>3</math> replaced by <math>4</math>. Hence at most <math>\frac{3}{4}</math> of the <math>10</math>, or <math>7.5</math>, can be acute, and hence at most <math>7</math> can be acute.
The same argument now extends the result to <math>100</math> points. The maximum number of acute triangles formed by <math>100</math> points is: the number of subsets of <math>5</math> points times <math>\frac{7}{\text{the number of subsets of 5 points containing 3 given points}}</math>. The total number of triangles is the same expression with <math>7</math> replaced by <math>10</math>. Hence at most <math>\frac{7}{10}</math> of the triangles are acute.

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[[Category:Olympiad Geometry Problems]]