There are many sort of equivalent theorems about monochromatic configurations in finite colorings, such as Van der Waerden, Hales-Jewett or Gallai's theorem, the latter of which states that in a finite coloring of $\mathbb Z^d$ or $\mathbb R^d$, there is a homothetic (i.e., scaled and translated) copy of any finite configuration $S$. Motivated by this problem, I wonder if similar statements hold if instead we require that in a configuration $(S,s_0)$ all points in $S$ are monochromatic, while $s_0$ has a different color from the rest.

Obviously, we need to impose some conditions on the coloring and the configuration. About the coloring, I only want to demand that it is non-monochromatic, i.e., not all points of the space are colored with the same color. About $(S,s_0)$, I want to require that $s_0\notin conv(S)$, i.e., $s$ is not in the convex hull of some points from $S$, as then we might not have a solution if the "first" half of the space is red, while the "second" half is blue.

Is there always an almost monochromatic copy (homothetic or isometric) of any finite $(S,s_0)$ with $s_0\notin conv(S)$ in a non-monochromatic finite coloring of $\mathbb R^d$?

Note that the answer is no for $\mathbb Z$ if $s_0=0$ and $S=\{1,2\}$ as shown by coloring odd numbers red and even numbers blue. This particular configuration, however, is easy to find in $\mathbb Q$.

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    $\begingroup$ What prevents you from taking the "convex" configuration $(0,s_0,1)$ with some transcendental $s_0\in(0,1)$ and the "half-plane" coloring and apply a field automorphism of $\mathbb C$ over $Q$ that moves $s_0$ to a negative number to create a non-convex configuration and some crazy coloring that just encode the old ones? $\endgroup$ – fedja May 20 '18 at 0:53
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    $\begingroup$ You might be interested in mathoverflow.net/q/3322/806, which was a problem in a similar spirit in $[n]$. $\endgroup$ – Boris Bukh May 20 '18 at 12:58
  • $\begingroup$ @fedja Sorry, but my field automorphisms would need some polishing, though your idea certainly seems good. Could you give me an example of how you want to derive a coloring? $\endgroup$ – domotorp May 20 '18 at 20:39
  • $\begingroup$ If I understand things right, you can design a field automorphism $\tau$ of $\mathbb C$ over $\mathbb Q$ so that $\tau(s_0)<0$. Your configuration is then $0,1,\tau(s_0)$ and in your coloring $\tau(z)$ is blue if $\Re z\ge 0$ and red if $\Re z<0$. The point is that the "similar" configurations are obtained by linear transformations and $a\tau(s_0)+b=\tau(\tau^{-1}(a)s_0+\tau^{-1}(b))$ $\endgroup$ – fedja May 20 '18 at 21:55
  • $\begingroup$ @fedja I'm not sure, but it feels like your reasoning goes the other way. Let me try. We know that there is a bad coloring (of even $\mathbb C$) for $(0,1,s_0)$ whenever $0<s_0<1$ and from this we want to show that for some $\tau$ there is also a bad coloring (of $\mathbb R$) for $(0,1,\tau(s_0))$ where $1<\tau(s)$. And then what do you require of $\tau$? $\endgroup$ – domotorp May 21 '18 at 5:49

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