Difference between revisions of "1970 IMO Problems/Problem 6"
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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. | 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. | ||
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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. | 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. | ||
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| + | ==Remarks (added by pf02, December 2024)== | ||
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| + | 1. The solution above contains an error (a typo?) and skips too many steps, | ||
| + | which make it very hard to understand. For the benefit of future readers, | ||
| + | and as a public service, I re-write the proof below. | ||
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| + | 2. Note the commonality with [[1969 IMO Problems/Problem 5]]. In fact, | ||
| + | the solution to this problem has a strong commonality with Solution 2 | ||
| + | of [[1969 IMO Problems/Problem 5]]. | ||
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| + | ==Solution re-written== | ||
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| + | Define a <math>n</math>-gon to be a configuration of <math>n</math> points, no three | ||
| + | of which are collinear. Given an <math>n</math>-gon draw all the triangles | ||
| + | whose vertices are among the <math>n</math> points. We say that a triangle | ||
| + | <math>\triangle ABC</math> is in the <math>n</math>-gon, or that the <math>n</math>-gon has | ||
| + | <math>\triangle ABC</math> in it, if <math>A, B, C</math> are points in the <math>n</math>-gon. | ||
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| + | We start by proving that a <math>4</math>-gon has <math>4</math> triangles, out of which | ||
| + | at most <math>3</math> are acute. | ||
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{{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]] | ||
Revision as of 15:34, 11 December 2024
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.
Remarks (added by pf02, December 2024)
1. The solution above contains an error (a typo?) and skips too many steps, which make it very hard to understand. For the benefit of future readers, and as a public service, I re-write the proof below.
2. Note the commonality with 1969 IMO Problems/Problem 5. In fact, the solution to this problem has a strong commonality with Solution 2 of 1969 IMO Problems/Problem 5.
Solution re-written
Define a 
-gon to be a configuration of points, no three
of which are collinear. Given an -gon draw all the triangles
whose vertices are among the points. We say that a triangle
is in the -gon, or that the -gon has
in it, if 
are points in the -gon.
We start by proving that a -gon has triangles, out of which
at most are acute.
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