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?
I have a proof by contradiction that uses 7 points:
Suppose there are no such monochrome equilaterial triangles.
Call the colors Red and Blue. Pick two Red points a and b in the plane
(if you can't do this, Blue triangles are pretty easy to
come by...)
Rotate/translate/whatever your coordinate system (or your brain) so
that one of these points a is at (0,0) and the other b is at (1,0). Now
consider the points below (Sorry about the strange formatting, please ignore the _, I couldn't get this system to line things up nicely):
__1
_2_3
a_b_4
_5
which are spread out at the corners of equilaterial triangles. Since
(a,b,2) is not a Red triangle, 2 is Blue. Similarly for 5.
Now, (2,4,5) form an equilaterial triangle, so that 4 must be Red.
So far then we have:
__1
_B_3
R_R_R
_B
Where R=Red, B=Blue. 3 must be Blue, since (3,b,4) form a triangle
and b and 4 are Red. But then if 1 is Blue, we have a Blue triangle
[(1,2,3) in the original numbering]. If 1 is Red, we have a
Red triangle [(1,a,4) in the original numbering]. So either 1 is
some third color, or monochrome triangles do in fact exist.
Another proof
