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
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    $\begingroup$ 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
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    $\begingroup$ Didn't I spell it out? $\tau$ is a field automorphism of $\mathbb C$ preserving $\mathbb Q$. $\endgroup$
    – fedja
    May 21 '18 at 11:27
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    $\begingroup$ @fedja I've just discovered that essentially the same problem was posed by some dude called Erdős and his pals in '75, and they made the same conjecture, so you might want to spell out your argument for their sake; see Conjecture 4 in old.renyi.hu/~p_erdos/1975-12.pdf. $\endgroup$
    – domotorp
    Jun 1 '19 at 20:40

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