2007 JBMO Problems/Problem 3

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Problem 3

Given are $50$ points in the plane, no three of them belonging to a same line. Each of these points is colored using one of four given colors. Prove that there is a color and at least $130$ scalene triangles with vertices of that color.

Solution

By the Pigeonhole Principle, there exists at least one set of 13 points all colored the same color. Because no three points are collinear, any combination of three points is associated with a unique nondegenerate triangle, so there exist ${13\choose3}=286$ triangles with vertices of a common color. Now we will bound the number of isosceles (and equilateral) triangles. Given any two points, a third point will create an isosceles triangle if and only if it lies on the perpendicular bisector of the segment connecting the first two points. Because no three points are collinear there is a maximum of two points that can satisfy this property. There exist ${13\choose2}=78$ segments in this same color set for a maximum of $156$ isosceles triangles. Therefore there exist at least $286-156=130$ scalene triangles with all vertices the same color, and we are done.

See Also

2007 JBMO (ProblemsResources)
Preceded by
Problem 2
Followed by
Problem 4
1 2 3 4
All JBMO Problems and Solutions