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 $\Gamma$.

UPDATE (8/8/2020): The problem we wanted to solve using these $3$-colorings was later posted in this MO question, and was largely resolved in this paper. The connection between these two problems is explained in $\S5.4$. It's only in one direction, so the $3$-coloring problem remains open.

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