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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.

Aditya Jain
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3 Answers3

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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.

lulu
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triangles

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.

JMP
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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.

Calvin Lin
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  • Thank you foe the answer but I don't think I know enough about affine transformation to use it. Can you provide the solution or an alternate method? – Aditya Jain Apr 12 '20 at 15:35
  • It's just saying that there is a linear transformation that maps an arbitrary triangle onto an equilateral triangle. IE Can you find a map of $T(x, y ) -> (Ax+By+C, Dx+Ey+F)$ that sends $(x_1, y_1 ) \rightarrow (0,0), (x_2, y_2, ) \rightarrow (1, 0), (x_3, y_3 ) \rightarrow (1/2, \sqrt{3}/2)$? We have 6 variables and 6 equations, so a solution should exist. – Calvin Lin Apr 12 '20 at 15:40
  • Why should there be a color preserving affine map of the right form? – lulu Apr 12 '20 at 16:17
  • @lulu It's not a "color preserving affine map". The details are 1) Prove the statement for equilateral triangles, 2) Prove that an affine map that sends T to the equilateral triangle will send all triangles similar to T to an equilateral triangle, 3) Map the current 2-coloring under this affine map, to get a new 2-coloring with a monochromatic triangle, pull it back to the current 2-coloring. (It's very similar to yours. The main crux is proving it for the equilateral triangle.) – Calvin Lin Apr 12 '20 at 17:06
  • But why should any affine map preserve the coloring? Suppose you have constructed an all blue equilateral triangle, and that these three are the only blue points in the plane. Then no non-trivial affine map will preserve the coloring. (of course the problem is trivial in this case, but I don't see how the affine construction has a chance of working). – lulu Apr 12 '20 at 17:09
  • (I do not care about color preserving affine maps, and never made that claim ever.) Take the map $M: \mathbb{R}^2 \rightarrow \mathbb{R}^2 $ such that $M(T) = $equilateral triangle. Apply the claim to $M(\mathbb{R}^2)$, which is a 2-coloring that has a monochromatic triangle $S$. Then $M^{-1} (S)$ is our monochromatic triangle that is similar to $T$ in the original 2-coloring. – Calvin Lin Apr 12 '20 at 17:12
  • Is it true that $\textit {every}$ equilateral triangle pulls back to one similar to the given one? – lulu Apr 12 '20 at 17:28
  • Ah, you're right. It only works for those that are parallel (in the required order). So, I'd need to add a "directional" element to it. (Let me edit.) – Calvin Lin Apr 12 '20 at 17:31
  • To me, this problem is mostly combinatorial, with very little geometric structure. – lulu Apr 12 '20 at 17:31
  • I agree that it's mostly combinatorial (even the extension to n-colors). – Calvin Lin Apr 12 '20 at 17:32