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Source. Theorem 11, printed p. 347, physical p. 7 of the published paper.

Let

$$L=\{(-1,0),(0,0),(1,0),(1,1)\}\subset\mathbb R^2.$$

Then

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

Proof

Color $\mathbb R^3$ red and blue. By Theorem 8, there are three collinear, unit-spaced points of one color. After an isometry and an exchange of colors, write them as

$$A=-e,\qquad B=0,\qquad C=e, \tag{2}$$

where $e$ is a unit vector, and suppose they are red. Let $P=e^\perp$, and write

$$\mathcal C_A=A+\{w\in P:|w|=1\},\quad \mathcal C_B=\{w\in P:|w|=1\},\quad \mathcal C_C=C+\{w\in P:|w|=1\}. \tag{3}$$

Assume for contradiction that there is no monochromatic copy of $L$. Every point of $\mathcal C_A\cup\mathcal C_C$ is blue: a red point on either circle, together with $A,B,C$, would supply the perpendicular unit edge at an endpoint of the red collinear triple.

The circle $\mathcal C_B$ is red. Indeed, suppose $w\in\mathcal C_B$ were blue. Choose $w'\in P$ with $|w'|=1$ and $|w'-w|=1$. The four blue points

$$A+w',\qquad A+w,\qquad B+w=w,\qquad C+w \tag{4}$$

are congruent to $L$: the last three are collinear and unit-spaced, while $w'-w$ is a unit vector in $P$ and hence is perpendicular to their direction $e$. This contradiction proves the claim.

Let $S$ be the sphere of radius $\sqrt2$ about $B$, and let $S'$ be the set of points of $S$ whose distance from $\mathcal C_B$ is at most $1$. Write a point $s\in S$ as $s=te+w$, with $w\in P$. Since $t^2+|w|^2=2$,

$$\operatorname{dist}(s,\mathcal C_B)^2 =t^2+(|w|-1)^2=3-2|w|. \tag{5}$$

Consequently

$$S'=\{te+w:t^2+|w|^2=2,\ |w|\ge1\} =\{te+w:t^2+|w|^2=2,\ |t|\le1\}. \tag{6}$$

Every point of $S'$ is blue. For $s=te+w\in S'$, choose a unit vector $x\in P$ with $w\mathbin\cdot x=1$; this is possible because $|w|\ge1$. Then $x\in\mathcal C_B$ and

$$|s-x|^2=|s|^2+|x|^2-2s\mathbin\cdot x=1, \qquad (s-x)\mathbin\cdot x=0. \tag{7}$$

The points $-x,B,x$ form a red collinear unit-spaced triple. If $s$ were red, (7) would attach a perpendicular unit edge at its endpoint $x$, giving a red copy of $L$. Hence $s$ is blue.

Choose orthonormal vectors $f,g\in P$ and put

$$p=2f,\qquad q=\frac54f+\frac{\sqrt7}{4}g,\qquad r=2q-p=\frac12f+\frac{\sqrt7}{2}g. \tag{8}$$

The point $p$ is blue. Otherwise $B,f,p$ would be a red collinear unit-spaced triple. Choose a unit vector $g_0\in P$ perpendicular to $f$. Since $g_0\in\mathcal C_B$ is red, the point $g_0$, attached to the endpoint $B$, would complete a red copy of $L$. Direct calculation gives

$$|q|=|r|=\sqrt2, \qquad |p-q|=|q-r|=1. \tag{9}$$

Both $q$ and $r$ lie in $P\cap S'$ and are blue. Thus $p,q,r$ are a blue collinear unit-spaced triple. The circle

$$\Gamma=\{r+u:|u|=1,\ u\mathbin\cdot(q-p)=0\} \tag{10}$$

must therefore be red: a blue point of $\Gamma$ would attach the required perpendicular unit edge at $r$.

It remains to see explicitly that $\Gamma$ meets the blue band $S'$. Set $d=q-p$. Then $|d|=1$, $r\mathbin\cdot d=1/2$, and, for

$$r_\perp=r-\frac12d,$$

one has $r_\perp\perp d$ and $|r_\perp|^2=7/4$. Define

$$u=-\frac27r_\perp+\sqrt{\frac67}\,e. \tag{11}$$

Because $e\perp P$, equations (8)–(11) give

$$|u|=1,\qquad u\mathbin\cdot d=0, \qquad r\mathbin\cdot u=-\frac12. \tag{12}$$

Thus $s=r+u$ belongs to $\Gamma$ and satisfies

$$|s|^2=|r|^2+|u|^2+2r\mathbin\cdot u=2, \qquad |s\mathbin\cdot e|=\sqrt{6/7}<1. \tag{13}$$

By (6), $s\in S'$. It is blue by the band argument and red because it lies on $\Gamma$, the final contradiction. This proves (1).