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I encounter the following problem with solution. But I do not quite understand the argument for 5, 10 points and eventually 100 points. Can someone elucidate the details?


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.


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 no. of acute triangles is: the no. of subsets of 4 points x 3/the no. of subsets of 4 points containing 3 given points. The total no. of triangles is the same expression with the first 3 replaced by 4. Hence at most 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 no. of subsets of 5 points x 7/the no. of subsets of 5 points containing 3 given points. The total no. of triangles is the same expression with 7 replaced by 10. Hence at most 7/10 of the triangles are acute.

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Linked – Alex Ravsky Dec 13 '14 at 9:32
up vote 3 down vote accepted

First, show that out of 4 points, at most 3 triangles are acute.

Out of 5 points, there are ${ 5 \choose 4 } =5$ sets of 4 points. In each of these sets, there are at most 3 triangles that are acute. This gives us $5 \times 3 = 15 $ acute triangles, but we double counted.
Notice that each triangle is counted exactly twice. Hence, there are at most $ \lfloor \frac{15}{2} \rfloor = 7$ acute triangles.

Now do the same, bootstrapping your way up to 10 points.

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Thank you. My follow-up question is, how does one pick the bootstrapping sequence, e.g., in this case 4,5,100? Trial and error? – Hans Apr 11 '14 at 20:02

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