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Source. The paragraph on p. 46 of P. Erdős, Combinatorial problems in geometry, Math. Chronicle 12 (1983), 35--54, the transcript of an invited address at the 17th New Zealand Mathematics Colloquium (Dunedin, 17--19 May 1982), as named on the source card. The lecture numbers none of its statements; the pages are the journal's own.

Statement

Conjecture (p. 46). If the plane is split into two sets $S_1$ and $S_2$ with $S_1\cup S_2$ the whole plane, then for every triangle that is not equilateral, $S_1$ or $S_2$ contains a triangle congruent to it. In the lecture's more general wording: every 2-colouring of the plane has a class containing a triangle congruent to any given triangle, "with the sole exception of a single equilateral triangle".

The exception (p. 46). Colour the plane in parallel strips of equal width alternately $1$ and $2$; an equilateral triangle whose height equals the strip width can be placed in no way with all three vertices of one colour, according to the lecture. The conjecture says this is the only exception.

Reported state (p. 46). The authors proved the conjecture in their first publication for all triangles with angles $120^\circ$, $30^\circ$, $30^\circ$; many special cases were settled, the general case not.

Read depth. Claims checked: the passage was read clause by clause on the page images of the print. A second reader checked the statement, hypotheses, label and page against the print.

Proof pointer

None in the paper.

Dependencies

None.

Bears on

• Problem 173: the conjecture is the problem's statement with the one allowed exception named as an equilateral triangle, which the strip colouring shows is needed. The paper reports the $120^\circ$, $30^\circ$, $30^\circ$ case as proved and proves nothing itself.