1970 IMO Problems/Problem 6

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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