Tons of Triangles
All points on a Cartesian plane are colored either or .
For all positive numbers there exists a non-degenerate isosceles triangle with legs of length whose vertices are all colored .
Is the statement above true for at least one of the colors?
Note: A degenerate triangle is one where all three vertices lie on a straight line.
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1 solution
Assume that none of the colors are required to have this property. Then there will exist two lengths r ≥ b such that no three red points form an isosceles triangle whose legs have length r , and no three blue points form an isosceles triangle whose legs have length b . (note that we may have to swap the colors to have this inequality hold)
Consider a circle of radius r centered at a red point. Then at most two points on this circle are colored red (diametrically opposite points won't form a triangle) and the rest of the circle is blue.
However, since b ≤ r , we can pick three blue points on the circle P , Q , R such that P Q = Q R = b , making a blue isosceles triangle whose legs have length b , a contradiction. The conclusion follows that at least one of the colors has the desired property.