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Let $\Gamma$ be the unit distance graph of $\Bbb R^3$: points $(x,y)$ form an edge if $|x,y|=1$.

Let $(A,B,C,D)$ be a unit side rhombus in the plane, with a transcendental diagonal, e.g. $A = (\alpha,0)$, $B = (0,\beta)$, $C=(-\alpha,0)$, $D = (0,-\beta)$, where $\alpha^2+\beta^2=1$ and $\alpha \notin \overline{\Bbb Q}$.

QUESTION: Is it possible to color $\Gamma$ with three colors so that no triangle (in $\Gamma$) is rainbow (has all three colors) and the colors of $A,B,C,D$ are $1,1,2,3$ respectively?

This question came up in my work in Discrete Geometry and is crucial to resolving an open problem. It is similar to many questions in Euclidean Ramsey Theory, but seems to be sufficiently different and perhaps has never been studied. Having spent many months on this, we are hoping someone can help.

NOTE: We do know some examples of such colorings of the plane, but none of them extend to the whole $\Bbb R^3$. Similarly, we don't know any such coloring where the rhombus $(A,B,C,D)$ as above is replaced by any $4$-cycle in $\Gamma$.

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  • $\begingroup$ how does the triangle relate to ABCD? $\endgroup$ – JonMark Perry Apr 14 '15 at 11:30
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    $\begingroup$ @IgorPak : What do you know about the case where $\alpha$ is algebraic? $\endgroup$ – Timothy Chow Apr 14 '15 at 17:14
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    $\begingroup$ @TimChow. As I say in the note, this is not an important condition - we will be happy with any $\alpha$. However, we can rule out some values such as $\alpha = 1/\sqrt{2}$ when (ABCD) is a square - it's easy to see that such coloring does not extend from (ABCD) to $\Bbb R^3$. We cannot rule out any transcendental values. $\endgroup$ – Igor Pak Apr 14 '15 at 17:23
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    $\begingroup$ In spite of your efforts, it is not clear what you ask of the coloring. My wording: Given R^3, A,B,C,D on a rhombus as stated, and start a three coloring with the assigned colors to the four points, can the coloring be extended to all of R^3 so that no unit equilateral triangle is rainbow? In particular, we need at least three circles around three edges to receive only two colors. I think this formulation avoiding Gamma is clear, since you are not asking for a proper graph coloring (neighboring vertices get different colors). Gerhard "Gamma Free For More Clarity" Paseman, 2015.04.14 $\endgroup$ – Gerhard Paseman Apr 14 '15 at 18:05
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    $\begingroup$ @GerhardPaseman : Igor Pak's formulation seemed perfectly clear to me. Since he defined $\Gamma$, I thought it was obvious that "triangles" meant 3-cliques in $\Gamma$. In any case, your formulation is equivalent. $\endgroup$ – Timothy Chow Apr 14 '15 at 22:58

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