Difference between revisions of "1970 IMO Problems/Problem 6"

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==Solution==
 
==Solution==
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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>7</math> can be acute.
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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.
  
 
{{IMO box|year=1970|num-b=5|after=Last question}}
 
{{IMO box|year=1970|num-b=5|after=Last question}}
  
 
[[Category:Olympiad Geometry Problems]]
 
[[Category:Olympiad Geometry Problems]]

Latest revision as of 18:26, 14 July 2017

Problem

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

Solution

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

1970 IMO (Problems) • Resources
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