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Source. Theorem 8 and Figure 2, printed pp. 345–346, physical pp. 5–6 of the published paper.

For every set $T$ of three distinct points,

$$R(T,3,2)\ \text{is true}. \tag{1}$$

Exact six-triangle gadget

Let the three mutual distances in $T$ be $a,b,c$, with $a>0$. Identify a plane with $\mathbb C$ and put

$$\omega=e^{i\pi/3},\qquad B=0,\qquad C=a,\qquad A=a\omega. \tag{2}$$

The triangle inequalities give a point $z\in\mathbb C$ such that

$$|z|=b,\qquad |z-A|=c. \tag{3}$$

Indeed, these are two circles whose center distance is $a$, and they intersect exactly when $|b-c|\le a\le b+c$. Define

$$E=z,\qquad D=\overline\omega z,\qquad F=z+a,\qquad G=z+a\overline\omega,\qquad H=a+\omega z. \tag{4}$$

Using

$$1-\overline\omega=\omega,\qquad \omega-1=-\overline\omega,\qquad \overline\omega^{\,2}=-\omega, \tag{5}$$

one obtains the following exact side-length table:

$$\begin{array}{c|ccc} \text{triangle}&\text{side of length }a&\text{side of length }b& \text{side of length }c\\ \hline ABE&AB&BE&EA\\ DBC&BC&BD&DC\\ GFC&GF&FC&CG\\ EFH&EF&FH&HE\\ ACH&AC&CH&HA\\ DEG&EG&DE&DG. \end{array} \tag{6}$$

More explicitly, with $\delta=z-a\omega$, the six rows are verified by

$$\begin{aligned} ABE:\quad& |AB|=a,\quad |BE|=|z|=b,\quad |EA|=|\delta|=c,\\ DBC:\quad& |BC|=a,\quad |BD|=|\overline\omega z|=b,\quad |DC|=|\overline\omega\delta|=c,\\ GFC:\quad& |GF|=|a(1-\overline\omega)|=a,\quad |FC|=|z|=b,\quad |CG|=|\delta|=c,\\ EFH:\quad& |EF|=a,\quad |FH|=|(\omega-1)z|=b,\quad |HE|=|a-\overline\omega z|=c,\\ ACH:\quad& |AC|=a,\quad |CH|=|\omega z|=b,\quad |HA|=|\overline\omega(H-A)|=|\delta|=c,\\ DEG:\quad& |EG|=a,\quad |DE|=|(1-\overline\omega)z|=b,\quad |DG|=|\overline\omega(G-D)|=|\delta|=c. \end{aligned} \tag{7}$$

Here, for example, $D-C=\overline\omega\delta$, $\overline\omega(H-A)=\delta$, and $\overline\omega(G-D)=\delta$; the remaining displayed identities follow directly from (2)–(5). Thus all six listed triangles are congruent to $T$.

Color forcing

Two-color $\mathbb R^3$. By Theorem 6, there is a monochromatic equilateral triangle of side $a$. Place it as $A,B,C$ in (2), and call its color red. Suppose no congruent copy of $T$ is monochromatic.

The copies $ABE,DBC,ACH$ force $E,D,H$ blue. Then $DEG$ forces $G$ red. The copy $GFC$ forces $F$ blue, whereas $EFH$ forces $F$ red, a contradiction. This proves (1).

If $T$ is collinear, one triangle inequality in (3) is an equality and the two circles are tangent. The construction still exists. Since $T$ consists of three distinct points, $a,b,c$ are positive, so every triangle in (6) has three distinct vertices. Coincidences between vertices belonging to different triangles only force an earlier color contradiction and do not affect the argument. Thus no nondegeneracy assumption was used.

Used by. Theorem 9 and Theorem 11.