Prove that there there exists a triangle similar to T. 
Suppose all the points of euclidean plane are colored either red or blue. Given any triangle T, show that there exists a triangle similar to T in this plane for which all the vertices have the same color.

I know a bit about Euclidean plane but wasn't able to reach anywhere near the solution. 
 A: Start by doing it for equilateral triangles (the general case is essentially the same) 
First Remark:  You can find two points $P,Q$ such that they, and their midpoint $M$, are all the same color.
Pf:  Take a line.  We can find two points $P_1,Q_1$ on it of the same color (let's say blue).  Then consider their midpoint $M_1$.  If it is also blue we are done, so assume it it is red.  Now consider $P_2$ to the left of $P_1$, and $Q_2$ to the right of $Q_1$ such that $\overline {P_2P_1}=\overline {P_1,Q_1}=\overline {Q_1,Q_2}$.  If either $P_2, Q_2$ are blue then we are done (if, say, $P_2$ is blue then $P_2,P_1,Q_1$ works), so assume that $P_2,Q_2$ are both red.  But then $P_2,M_1,Q_2$ works.
Now consider the standard decomposition of an equilateral triangle into four equilateral triangles and use the first remark to require that all three vertices on, say, the bottom side are all blue. It is easy to see that we are done if any of the other three vertices are blue, but if they are all red then the top triangle is all red, so either way we are done.
To do the general case, start with three collinear blue points $P,M,Q$ as in the first remark.  Now construct a point $N$ such that $\Delta PNQ$ is similar to your given triangle.  Taking midpoints of the sides $PN, QN$ we get four triangles, each similar to the original one, and we are done as before.
A: 
wlog, we must have two vertices the same colour (the red ones). This forces the two deep blue ones. Pink vertex number 1 must now be red, otherwise the light blue triangle is similar (the blue lines bisect each edge they cross, and therefore the lines from vertex to midpoint of the opposite edge are all similar).
Vertex 2 must be blue, and vertex 3 can be neither, otherwise the yellow triangle is similar.
A: Fill in the details yourself.
Hint: Prove that we can always find a monochromatic equilateral triangle with 1 base parallel to the x-axis, and the other vertex above this base.
(We cannot fix the side length. There are 2-colorings of the plane without a unit monochromatic equilateral triangle. However, we can guarantee that it either has length 1 or 2.)
Then, apply an affine transformation that maps our given triangle to the equilateral triangle with 1 base parallel to the x-axis and the other vertex above this base.
Find the monochromatic equilateral triangle with 1 base parallel to the x-axis and the other vertex above this base, and pull it back to get a monochromatic triangle similar to $T$.
Hence we are done. 

More generally, the hint applies for $n-$colored $\mathbb{R}^2$, and so the question extends to $n-$colors too. Again, the important part here is doing it for the equilateral triangle. 
