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Source. Theorem 9, printed p. 346, physical p. 6 of the published paper. The displayed statement is planar: its ambient space is $\mathbb R^2$.

Let $d>0$. Let $T_1,T_2,T_3$ be triangles such that $T_1$ has a side of length $d$, $T_2$ has a side of length $\sqrt3d$, and $T_3$ has a side of length $2d$. Every two-coloring of $\mathbb R^2$ contains a monochromatic triangle congruent to at least one of $T_1,T_2,T_3$.

Equilateral forcing

Put

$$u=d(1,0),\qquad v=d(1/2,\sqrt3/2). \tag{1}$$

By the positive part of Theorem 5, there are same-colored points at distance $d$. Apply an isometry and call $0,u$ red, exchanging the color names if necessary. Suppose there is no monochromatic equilateral triangle with side length $d$, $\sqrt3d$, or $2d$.

The points $v$ and $u-v$ complete the two unit-scale equilateral triangles on $0,u$, so both are blue. The three points

$$v,\qquad u-v,\qquad 2u \tag{2}$$

form an equilateral triangle of side $\sqrt3d$; hence $2u$ is red. The triangle $0,2u,2v$ has side $2d$, so $2v$ is blue.

Now $u,2u,u+v$ form an equilateral triangle of side $d$, so $u+v$ cannot be red. But $v,2v,u+v$ is another such triangle, so $u+v$ cannot be blue. This contradiction proves that a monochromatic equilateral triangle exists at one of the three scales.

Passing to the prescribed triangles

The coordinate construction in Theorem 8 has the following planar consequence: from a monochromatic equilateral triangle whose side equals one side of a prescribed triangle $T$, its six congruent planar copies force a monochromatic copy of $T$. Apply that gadget to $T_1$, $T_2$, or $T_3$ according to which scale was forced above. This proves the theorem.

Used by. Corollary 10.

Bears on. #173: the proof's first step shows that no two-coloring of the plane misses the equilateral triangles of all three sides $d$, $\sqrt3d$ and $2d$. The problem's page, through Theorem 1 of the 1975 sequel, reduces the question to whether a coloring can miss equilateral triangles of two different sides; this theorem excludes only missing all three sides in these ratios at once. It forces a single prescribed triangle only when one triangle has all three sides, the case of Corollary 10.