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Statement

Conjecture 11.1.2 (p. 282), labeled "(stronger)", is posed as:

For any partition $\mathbb{E}^2 = C_1 \cup C_2$, every triangle occurs (up to congruence) in $C_1$, or else the same holds for $C_2$, with the possible exception of a single equilateral triangle.

Read literally, the conjecture says that in every two-coloring of the plane one of the two classes, by itself, contains a congruent copy of every triangle, except possibly one equilateral triangle. The label "stronger" sets it against Conjecture 11.1.1, which it implies.

The equilateral exception (p. 282). The chapter shows that the exception is needed: the partition with $C_1=\{(x,y): 2m\le y<2m+1,\ m\in\mathbb{Z}\}$ and $C_2=\mathbb{E}^2\setminus C_1$, alternating half-open strips of width 1, has no class containing an equilateral triangle of side $\sqrt3$. It adds that other two-colorings avoid a monochromatic unit equilateral triangle, the "zebra-like" colorings of Jelínek, Kynčl, Stolař and Valla, and reports from the same authors that when the plane is split into an open set and a closed set, every equilateral triangle occurs in at least one of the two.

Scope

This is a conjecture, not a result proved in the chapter. As for Conjecture 11.1.1, the chapter does not say whether degenerate triples count as triangles.

Source. R. L. Graham, Euclidean Ramsey theory, Chapter 11 of J. E. Goodman, J. O'Rourke and C. D. Tóth (eds.), Handbook of Discrete and Computational Geometry, 3rd edition, CRC Press, Boca Raton, FL, 2017; the conjecture, the strip coloring and the remarks after it on p. 282. Pages are those printed on the edition named on the source card.

Read depth. Claims checked: the quotation was compared word for word with the printed page, and the remarks after it were read clause by clause.

Bears on

• Problem 173: the conjecture implies the problem's statement (a deduction drawn here, not printed in the chapter). If one class of a two-coloring contains every triangle except possibly one equilateral triangle, then every triangle but at most one has a monochromatic congruent copy. The converse need not hold, since the problem lets different triangles lie in different classes and does not require the missed triangle to be equilateral. The chapter proves neither statement.