Junior Balkan Mathematical Olympiad 2007 - Problem 3

This was the third problem of the JBMO 2007 held in Bulgaria. I find it very beautiful and I think it has a very elegant solution. Again, like all JBMO problems in my opinion, it only takes a few observations to reach a solution.

Using the Pigeon Hole Principle, we can express 50 in terms of 4. Like this:

From this statement, we can reach the conclusion that there are at least 13 points of the same colour in the plane.

Now, the key to solving this problem is to find out how many triangles we can create out of 13 points of the same colour and then reach more conclusions.

The amount of triangles we can create is equal to:

 Note that we would not be able to use this formula if it was not stated in the hypothesis that no three points are collinear.

What is left to prove, is that there exist at most 156 isosceles triangles, so that the remainder which is equal 130 will be scalene triangles.

Pay attention now, because this part is tricky.

Consider a segment created by two points. Then we can say that we can create at most 2 isosceles triangles per segment. The reason for that is that there can be at most 2 points belonging to the semi-perpendicular of that segment*. The semi-perpendicular of a segment is a straight line that bisects the segment and is perpendicular to it. Geometry tells us that the distances between any point on the semi-perpendicular of a segment and the points creating that segment are equal. For a better understanding, I have prepared a shape:

*That is due to the fact that no three points are collinear.

So now that we have proven that there can be up to 2 isosceles triangles created by a segment, we have to find the amount of segments we can create and multiply it by 2.

So the total amount of segments is 78. Should we multiply that by 2, we receive 156 isosceles triangles, which gives us 130 scalene triangles, as we wanted.