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Statement

Suppose the plane is colored red and blue, with no red pair at distance $1$ and no blue $\ell_5$. There is no blue equilateral triangle of side $3$ whose center is red.

Proof

Use the triangular coordinates and forcing rules. After an isometry, a hypothetical triangle and its center have coordinates

$$A=(0,0),\quad B=(-3,3),\quad C=(0,3),\quad O=(-1,2).$$

The points $D=(-1,1)$, $E=(-2,2)$, $F=(0,1)$ and $G=(0,2)$ are all unit neighbors of $O$, so all are blue. Set $X=(1,-1)$ and $Y=(0,-1)$. The five points $X,A,D,E,B$ form an arithmetic progression with step $(-1,1)$ of length $1$. Since its last four points are blue, $X$ is red. Likewise $Y,A,F,G,C$ has unit step $(0,1)$, forcing $Y$ red. But $X-Y=(1,0)$ is a unit vector, contradicting the absence of a red unit pair.

Source and correction

Lemma 2, Figure 1(a), published p. 2; Lemma 2.1 in arXiv v2. Both versions mistakenly call the forbidden progressions $XADEB$ and $YAFGC$ red. They must be blue in those conditional statements, as the listed colors and the hypothesis show. The proof above makes this correction explicit. No external theorem is used.

Bears on. #188.