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?
(In reply to
re(3): 5-point proof by Kelsey)
The very first part of the proof is that you start with two points that are different in color. In your diagram, those two points have the same color.
The proof lies not only on the number of points or even solely by virtue of their arrangement, but it hinges on how you select the points.
You have to start with two differently-colored points, and similarly, the last two points chosen depend on whatever the color of the midpoint happens to be.
The problem doesn't say that all points have to be randomly selected, and indeed, they cannot be.
|
Posted by DJ
on 2003-08-26 21:14:49 |