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We start with this

Example: No matter in which way you color the points of $\mathbb R^4$ with two colors, you can always find an equilateral triangle with vertices of the same color. In fact in $\mathbb R^4$ the regular hyper-tetrahedron (the regular 4-simplex) has $5$ vertices, and every $3$ of them form an equilateral triangle as a face. If you are using only two colors, in every coloring of $\mathbb R^4$ there must be $3$ points of the regular hyper-tetrahedron of the same color, and thus there is a monochromatic equilateral triangle in $\mathbb R^4$. Of course this holds for every $\mathbb R^n$ with $n \geq 4$.

Curiosity: Using $k$ colors, every $2k$-dimensional coloring contains a monochromatic equilateral triangle because the regular hyper-tetrahedron has $2k+1$ vertices.

Problem: Find the best dimension $n \geq 2$ such that every coloring with $2$ colors of the points of $\mathbb R^n$ contains a monochromatic equilateral triangle with side length 1.

Since we already solved the case $n\geq 4$, which is the best dimension? 2, 3 or 4?

enter image description here

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  • $\begingroup$ Your terminology is lacking. A tetrahedron with 5 vertices? And your example for $\mathbb{R}^4$ also holds for all dimensions $>4$, so I guess the word "triangle" in the statement of the problem also needs a replacement to avoid triviality. $\endgroup$
    – Leen Droogendijk
    Commented Mar 18, 2016 at 17:49
  • $\begingroup$ @LeenDroogendijk I edited the question. As every Ramsey number of course if it works for $n=4$ it works for every $n \geq 4$. I meant hyper-tetrahedron, beacuse they are a good finite graph to find a counterexample, but we are looking for monochromatic triangles in $\mathbb R^n$. I hope it is more clear now!! $\endgroup$
    – Maffred
    Commented Mar 18, 2016 at 18:04
  • $\begingroup$ Thanks, the problem is clear now. I am still wondering about the relevance of the picture. The problem is about 2-colorings and the picture uses 4 colors. It does not seem to relate to your "curiosity" either. $\endgroup$
    – Leen Droogendijk
    Commented Mar 18, 2016 at 18:36
  • $\begingroup$ You may want to have a look at arxiv.org/pdf/math/0701940v1.pdf. This shows at least that $n>2$, so the answer will be that $n=3$ is the "best" value. $\endgroup$
    – Leen Droogendijk
    Commented Mar 18, 2016 at 18:45

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I can show it's true in dimension $3$.

Suppose we had a $2$-colouring of $\mathbb R^3$ with no monochromatic triangle of side $1$.
Given any two points of opposite colour, there is a path joining them consisting of an even number of segments of length $1$, and therefore there are two points $a$, $b$ of the same colour (let's say white) with $\|a-b\|=1$. Now the circle $C(a,b)$ consisting of points $c$ with $\|a-c\| = \|b-c\| = 1$ is all of the opposite colour (black). That circle has radius $\sqrt{3}/2 > 1/2$, so for any point $c$ on $C(a,b)$ there is are others $c'$ with $\|c - c'\| = 1$. The circles $C(c,c')$ for $c \in C(a,b)$ form a monochromatic surface looking like this:

enter image description here

This surface does contain points forming an equilateral triangle of side $1$ (shown in blue in the picture). If $a = [0,0,1/2]$ and $b = [0,0,-1/2]$, the vertices of the triangle can be taken to be $[\sqrt{2}/2, 0, \sqrt{3}/2]$, $[(\sqrt{33}+3\sqrt{2})/10, 1/2, (-\sqrt{22} + 3 \sqrt{3})/10]$, $[(\sqrt{33}+3\sqrt{2})/10, -1/2, (-\sqrt{22} + 3 \sqrt{3})/10]$.

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  • $\begingroup$ Even if it's hard to check if this is correct since we don't have the equation of the surface I understand the proof. I don't understand the even segments part, by the way there are of course two points of the same color of distance 1, just consider a random equilater triangle and use pigeonhole. $\endgroup$
    – Maffred
    Commented Mar 18, 2016 at 19:46

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