suppose that every point in the plane is colored either red or blue. prove that there exists an?

Question

equilateral trangle whose three vertices all have the same color

Answer 1

you have an equalateral triangle ABC .
if you throw the triangle in any place in the room, you will have either 2 points on a red and one on a blue, or 2 points on a blue and one on a red.
*this is proven by using pigeon hole theory

Answer 2

Proof by contradiction: assume no equilateral triangle has three same-color vertices.

Start with any regular hexagon ABCDEF on the plane and its center O. The hexagon must have two adjacent vertices of the same color, otherwise the colors would alternate and ACE (equilateral) would be same-colored.

So without loss of generality (WLOG) let's assume both A and B are red. Then ...
→ O is blue (otherwise ABO would be same-colored)
→ at least one of D or E is red (ODE), assume WLOG that D is red
→ F is blue (BDF)
→ E is red (OFE)
→ C is blue (ACE)

Now let X be the reflection of O across AB. We have
X can't be red (ABX)
X can't be blue (CFX). Contradiction.

Answer 3

Given your initial statement, it is possible for EVERY point to be the same color.

Thus such a triangle easily exists. Proof is by simple inspection.

Right??? Anybody care to dispute this?

(case closed)

Answer 4

That's a good example of a poorly posed problem. You could color the origin red and the rest of the plane blue and there'd be no triangles.