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.
Clearly there are exactly triangles. We have two cases. If
the convex hull of the -gon is a quadrilateral, then at least
one of 
is 
(if they were all 
then
, which is false).
Let us say that 
. Then 
is obtuse, so at most the other triangles are acute. (Note
that it is possible to have acute triangles.)
If the convex hull of the -gon is a triangle, let us say that 
is in the interior of 
. Then at least two of
are
(if they were all
then their sum would be 
, which is false).
So, at least two of 
are obtuse.
Note: We would be tempted to use this result, and the argument which
follows below to prove the problem. But that would only yield the
result that at most 75% of the triangles in an -gon (in particular
a -gon) are acute. We need to do better.
Now, we will prove that a -gon has triangles, out of which
at most are acute. It is easy to see that there are
triangles in a -gon, either by counting, or by remembering that
the number of triangles 
. Now, we could
proceed like in the case of a -gon, by looking at many possible
cases and chasing angle sizes. Instead, we will give a different
argument, using an idea we will use again later.
A -gon has -gons in it. Each of these -gons has
triangles in it, of which at most are acute. Make a list
of all the triangles, and a sub-list 
of all the acute triangles. The size of 
,
and the size of 
. These lists contain
duplicates. Each triangle is counted exactly twice (corresponding to
the two points which are not in the triangle; we can see this as
). Dividing by 
to
eliminate duplicates, we get that out of the 
triangles, at most 
are acute. We conclude that
at most out of the are acute.
Now we are ready to prove that given a -gon with 
, at most
of its triangles are acute. Consider all the
-gons in the given -gon, and consider all the triangles in each
-gon. Make a list of all these triangles, and a
sub-list of all the acute triangles. We have that
(according to what we showed above). These lists contain
duplicates. Each triangle is counted the same number of times, say 
times (in fact, 
, but we
don't need this). Dividing both the numerator and the denominator by
to eliminate duplicates, we have that at most of
the triangles are acute.
[Re-writing by pf02, December 2024]
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