Pigeonhole principle (I think): colored points in the plane Suppose that each points in $\Bbb R^2$ is colored red, green or blue. 
Prove that either there are two points of the same color a distance $1$ unit apart, or there is an equilateral triangle of side length $\sqrt3$ all of whose vertices are the same color.
I thought I would separate each point into cases. Like, the combinations you can have are $(R,R),(R,B),(R,G)$, etc. and something about there being an infinite number of points, so the ones that are $(R,R),(G,G)$ or $(B,B)$ have to not be empty? Not sure if this helps.
I also tried putting all the points on the unit circle and looked at the $\sqrt3$ triangles created by tangent circles all with centers of the same color.
 A: You can actually prove more. You can prove that there are two points of distance $1$ part with the same color in all cases. Suppose not, and that every two points of distance one apart have different colors. Suppose the origin is colored blue. I claim that every point on a circle of radius $\sqrt{3}$ centered at the origin is also colored blue. Draw the radius out to any point and put two equilateral triangles on the radius as in BaronVT's picture where you have a dotted line (representing the radius) bisecting two back-to-back triangles. The two middle vertices are colored red and green (in some order) because the origin is colored blue. But this forces the vertex lying on the circle to be blue. Now that we have a circle of radius $\sqrt{3}$, just pick a chord of length $1$. This will give two points colored blue of distance one apart, which is a contradiction.
A: Assume that the first condition fails; every time you construct an equilateral triangle with side length $1$, the three vertices are of different colors. 
The short version: draw a regular hexagon of side length $1$, start labelling the vertices and center of the hexagon according to the above assumption, you will necessarily construct two $\sqrt3$ equilateral triangles satisfying the second property.
Longer: Start "gluing" triangles together, i.e. draw one such triangle, then draw another congruent triangle that shares a side with the first triangle. There will be two same-colored vertices $\sqrt 3$ distance apart from each other. 
Continue drawing adjacent triangles; you will know the color of the new vertex. After two such triangles, you will find the third point with the same color.

A: Assume no points 1 unit apart are the same color. 
Construct a regular hexagon ABCDEF with center O so that AB = BC = CD = EF = FA = OA = OB = OC = OD = OE = OF = 1.  
Note: the triangle ACE is equilateral with side of length $\sqrt 3$.
Wolog, assume O is BLUE.  As ABCDEFG are all one unit from O, none of ABCDEF are BLUE so all are RED or GREEN.  Wolog, assume A is RED.  Then B (being one unit from O and A) is GREEN.  C being one from B is RED.  D is GREEN.  E is RED.
So ACE are all red.  And ACE are the vertices of an equilateral triangle with side $\sqrt 3$.
A: The right proof,


*

*First, if there are two points A, B distance $2$ are different color, then the midpoint O, must be either same as A or B; O,A same color, then draw $2$ triangles using side OA, we have two points C, D with same color with B which is $\sqrt{3}$ side equilateral triangle

*If there are no two points are different color, then easy deduct all the points must be same color. QED
