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Coordinates and forcing rules

Write $(a,b)$ for $au+bv$, where $u=(1,0)$ and $v=(1/2,\sqrt3/2)$ in Cartesian coordinates. Thus

$$\|(a,b)\|^2=a^2+ab+b^2.$$

The unit triangular lattice is $\mathbb Zu+\mathbb Zv$. Its six unit directions, in these coordinates, are $\pm(1,0)$, $\pm(0,1)$ and $\pm(1,-1)$. Throughout the contradiction argument, there is no red unit-distance pair and no blue $\ell_5$, where $\ell_5$ consists of five consecutive points of a straight arithmetic progression with unit step. Therefore every unit neighbor of a red point is blue; and if four points of an $\ell_5$ are blue, the remaining point must be red.

Coordinates can be transported by any Euclidean isometry. Applying a forcing argument after reflecting or rotating its configuration does not assume that the original coloring has that symmetry.

The configurations in Figure 2

Put $w=(1,1)$ and $t=(2,-1)$. Both have length $\sqrt3$, as does $w-t$, and they make an angle of $60^\circ$. Representatives of the paper's configurations are

$$\begin{aligned} T_3&=\{0,w,t\},\\ T_4&=\{0,w,t,w+t\},\\ T_5&=\{0,w,t,w+t,2w\},\\ T_6&=\{0,w,t,w+t,2w,2t\},\\ T_7&=\{0,w,2w,3w,t,t+w,t+2w\}. \end{aligned}$$

Here $T_6$ is an equilateral triangle of side $2\sqrt3$ together with its three side midpoints. The seven-point set is a strip with four points on one row and three on the adjacent row, at spacing $\sqrt3$. These descriptions specify congruence classes, so the different orientations and labelings in later figures give the same configurations.

Four completions of a fixed small triangle

Let $S=T_6$. The complete list of copies of $T_6$ containing $\{0,w,t\}$ is

$$S,\qquad S-w,\qquad S-t,\qquad -S+w+t. \tag{1}$$

For completeness, the only triples of vertices of $S$ all of whose pairwise distances are $\sqrt3$ are

$$\{0,w,t\},\quad\{w,2w,w+t\},\quad \{t,w+t,2t\},\quad\{w,t,w+t\}.$$

This follows either by checking the six listed points with the norm formula or by separating the three corner triangles from the central triangle. The symmetries of the large triangle permute its three corner triangles and all vertex labelings of the central triangle. Mapping one of these four small triangles onto the fixed triangle therefore gives precisely the three corner completions and one central completion in (1). Explicitly their extra points are

In particular, if $t-w$ and $w-t$ are blue, only the first completion can be entirely red. This is the finite geometric check used in Lemma 6.

Source

Figure 2 on published p. 3 and the completion step in Figure 8 on pp. 6–7. These coordinates rewrite the paper's figures; the enumeration expands its geometric step rather than adding an external result.

Bears on. #188.