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Source. P. Erdős, R. L. Graham, P. Montgomery, B. L. Rothschild, J. Spencer and E. G. Straus, Euclidean Ramsey Theorems. I, J. Combin. Theory Ser. A 14 (1973), 341–363; the conjecture is unnumbered, at the top of printed p. 347, after Corollary 10 on p. 346. The edition read is named on the source card.

Statement

Here $R(T,2,2)$ means that every coloring of the plane in two colors contains a monochromatic triangle congruent to $T$ (pp. 343–344). The paper first records what was known in 1973: the $30^\circ$–$60^\circ$ right triangle is the only triangle known to satisfy $R(T,2,2)$, and the equilateral triangle the only one known to fail it. It then states (p. 347):

We conjecture that $R(T, 2, 2)$ holds unless $T$ is equilateral, and, moreover, that any 2-coloring of $E^2$ with no monochromatic equilateral triangle of side $d$ in fact has monochromatic equilateral triangles of side $d'$ for all $d' \neq d$.

The conjecture has two clauses. The first concerns each non-equilateral triangle across all two-colorings; the second bounds, within one coloring, the equilateral triangles that can be missed to a single side length. The paper does not say whether degenerate (collinear) triples count as triangles here; its Theorem 8 on p. 345 does allow them.

Scope

This is a conjecture, not a result proved in the paper. It follows the planar Theorem 9 and Corollary 10. The strip coloring of the introduction (p. 342) has no monochromatic unit equilateral triangle, so the equilateral exception in the first clause is needed.

The two clauses together imply the statement of Problem 173 (a deduction drawn here, not printed in the paper). In a given two-coloring, the first clause supplies every non-equilateral triangle; if some equilateral side $d$ is missed, the second clause supplies all other equilateral sides; so at most one triangle, up to congruence, is missed.

Read depth. Claims checked: the conjecture and the sentence before it were read clause by clause on the page image of p. 347.

Bears on. #173: the conjecture, with both clauses, implies the problem's statement, as shown above. The paper proves neither clause.