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Statement and coordinates

Suppose the plane has no red unit-distance pair and no blue $\ell_5$. Every red copy of $T_3$ is contained in a red copy of $T_6$.

Use the [[discrete_geometry/tsaturian_2017_euclidean_ramsey_result_plane/configurations|triangular coordinates and definitions of $T_i$]]. The proof successively adjoins red points, keeping the original three. Each progression listed below has one of the six unit steps, and the tables identify a red unit neighbor for every immediately forced blue point.

From three to four points: Figure 4

Normalize the initial red triangle to $A=(0,0)$, $B=(1,1)$, $C=(2,-1)$. Each of $X=(3,0)$, $Y=(1,-2)$ and $Z=(-1,2)$ completes it to a copy of $T_4$. Suppose none of these points is red. The following points are also blue:

Set $K=(-1,-1)$, $L=(-1,-2)$, $M=(0,-2)$, $N=(-1,3)$, $P=(-1,4)$ and $Q=(0,3)$. If $K$ were red, its unit neighbors $L,M$ would be blue, producing the blue progression $L,M,Y,G,H$ with step $(1,0)$. Thus $K$ is blue. The progression $K,J,I,Z,N$, with step $(0,1)$, forces $N$ red. Its unit neighbors $P,Q$ are blue, so $P,Q,F,E,X$ is a blue progression with step $(1,-1)$. This contradiction proves that at least one of $X,Y,Z$ is red, giving a red $T_4$.

From four to five points: Figure 5

Relabel and move the red $T_4$ so that $A,B,C$ have the preceding coordinates and its fourth point is $D=(3,0)$. Consider the possible extensions $X=(2,2)$, $F=(1,-2)$ and $G=(4,-2)$. Adjoining any one gives a copy of $T_5$. Suppose all three are blue. The following unit neighbors are blue:

The progression $F,H,I,G,P$ with $P=(5,-2)$ has step $(1,0)$, so $P$ is red. Hence $Q=(5,-1)$ and $R=(6,-2)$ are blue unit neighbors of $P$. Now $X,N,M,Q,R$ has unit step $(1,-1)$ and is entirely blue, a contradiction.

Thus one of these extensions is red. All three yield the stated $T_5$: in the notation $w=(1,1)$, $t=(2,-1)$, the red $T_4$ is $\{0,w,t,w+t\}$, and the candidates are $2w,t-w,2t$. Reflection interchanging $w,t$ exchanges $2w,2t$, and the half-turn $z\mapsto w+t-z$ exchanges $2w,t-w$. These are symmetries of the four-point set, justifying normalization of the new fifth point to $E=2w=(2,2)$ in the next step.

From five to six points: Figure 6

We now have red points

$$A=(0,0),\ B=(1,1),\ C=(2,-1),\ D=(3,0),\ E=(2,2).$$

The required sixth point is $F=(4,-2)$. Suppose it is blue. By Lemma 3, $X=(-1,2)$ and $Y=(0,3)$ are blue. Indeed, $X,E,C$ and $Y,A,D$ are equilateral triangles of side $3$ with center $B$; their other vertices and centers are red. Further blue points are

Put $P=(1,-2)$ and $Q=(0,-2)$. If $P$ were blue, the progression $Q,P,K,L,F$ with step $(1,0)$ would force $Q$ red. Then $T=(-1,-2)$ and $U=(-1,-1)$ would be blue unit neighbors of $Q$. The five points $T,U,G,H,X$ would form a blue progression with step $(0,1)$. Therefore $P$ is red.

We also claim $R=(4,1)$ is red. If it were blue, the progression $F,M,N,R,S$, where $S=(4,2)$, would force $S$ red. Its unit neighbors $V=(4,3)$ and $W=(3,3)$ would be blue, so $V,W,J,I,Y$ would be a blue progression with step $(-1,0)$. This proves the claim without an unexpanded symmetry step.

The seven red points $A,B,C,D,E,P,R$ form a copy of $T_7$: $P,C,D,R$ are four successive points with step $w=(1,1)$, while $A,B,E$ form the adjacent three-point row with the same step. The displacement from $P$ to $A$ is $w-t=(-1,2)$, of length $\sqrt3$ and at $60^\circ$ to $w$, giving the reflected orientation of the defining strip. This contradicts Lemma 4. Thus $F$ is red, and the six points are $T_6$.

All adjoined points belong to the same unit triangular lattice as the initial triangle. The relabelings above preserve the already selected set, so the final red configuration contains the original triangle.

Source and dependencies

Lemma 5, Figures 4–6, published pp. 4–5; Lemma 2.4 in arXiv v2. The coordinates are read from those figures and their TeX diagram definitions. The proof uses Lemmas 3–4, unit neighbor forcing, and the prohibition of a blue unit-step $\ell_5$. All figure and symmetry steps needed for these extensions are expanded above; no external theorem is used.

Bears on. #188.