Difference between revisions of "1970 IMO Problems/Problem 6"
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| − | 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></math> | + | 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. | 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. | ||
Latest revision as of 18:26, 14 July 2017
Problem
In a plane there are 
points, no three of which are collinear. Consider all possible triangles having these point as vertices. Prove that no more than 
of these triangles are acute-angled.
Solution
At most 
of the triangles formed by 
points can be acute. It follows that at most 
out of the 
triangles formed by any 
points can be acute. For given points, the maximum number of acute triangles is: the number of subsets of points times 
. The total number of triangles is the same expression with the first replaced by . Hence at most 
of the , or 
, can be acute, and hence at most can be acute.
The same argument now extends the result to points. The maximum number of acute triangles formed by points is: the number of subsets of points times 
. The total number of triangles is the same expression with replaced by . Hence at most 
of the triangles are acute.
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