Suppose you have an infinite plane, and each point on the plane has been arbitrarily painted one of two colors.

Prove that there exists an equilateral triangle whose vertices are all the same color.

What is the fewest number of points needed to prove this?

for this proof, it is important to note that the plane has already been (randomly) colored.  the points however do not HAVE to be chosen at random.  someone choosing points on the plane can choose based on either a)position or b) color.  (because the coloring is random, there is no way to control both at the same time.)  in the five point proof, the first point is chosen at random.  the second point is chosen based on COLOR.  the other three points are picked based on POSITION.  bryan has proven that using certain conditions for picking the points, a group of five points can be picked where an equilateral triangle of one color exists.  (we don't know which color when we start the proof- it depends on the color of the last three points, which were picked based solely on their position) 

if you could rearrange the color any way you want (or for that matter, the position) you would only need three points!  but this is not the case.  kelsey- bryan's proof uses more than just position; it is correct that five points could be arranged thus that there is no equilateral triangle.  but those five points would not have fit the criteria.

Posted by keiko on 2004-06-27 18:41:15