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Statement

In a red-blue coloring of the plane with no red unit-distance pair and no blue $\ell_5$, there is no red copy of the [[discrete_geometry/tsaturian_2017_euclidean_ramsey_result_plane/configurations|configuration $T_7$]].

Proof

Here use Cartesian coordinates. A hypothetical red $T_7$ can be labeled as in Figure 3 by

$$\begin{aligned} A&=(-\sqrt3,0),&B&=(0,0),&C&=(\sqrt3,0),&D&=(2\sqrt3,0),\\ E&=(-\sqrt3/2,-3/2),&F&=(\sqrt3/2,-3/2),& G&=(3\sqrt3/2,-3/2). \end{aligned}$$

Let $X=(\sqrt3/2,3/2)$ be the reflection of $F$ in $BC$. The triangles $XAF$ and $XDF$ have side $3$ and centers $B$ and $C$, respectively; each vertex has distance $\sqrt3$ from its center.

Let $R$ be clockwise rotation through $\theta=2\arcsin(1/(2\sqrt3))$. Rotate $X,A,F$ about $B$ by $R$, obtaining $X',A',F'$. Every point moves distance $1$, so $A',F'$ are blue. If $X'$ were blue, this side-$3$ triangle with red center $B$ would contradict Lemma 2. Hence $X'$ is red. Applying the same rotation about $C$ to $X,D,F$ gives $X'',D'',F''$, with $D'',F''$ blue and consequently $X''$ red by the same lemma.

It remains to verify their separation exactly. The affine rotation formulas give

$$X'=B+R(X-B),\qquad X''=C+R(X-C),\qquad X'-X''=(I-R)(B-C).$$

For any vector $z$, $\|(I-R)z\|=2\|z\|\sin(\theta/2)$. Since $|BC|=\sqrt3$, it follows that $|X'X''|=1$. Thus these two red points contradict the hypothesis.

Source and dependencies

Lemma 4, Figure 3, published pp. 3–4; Lemma 2.3 in arXiv v2. The affine calculation replaces the proof's $60^\circ$ rotation argument and uses an exact angle instead of the rounded plotting angle in the TeX diagram. The points $E,G$ identify the full source configuration, although this contradiction already uses its other five red points. No external Ramsey theorem is used.

Bears on. #188.