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Claim. L. E. Shader, All right triangles are Ramsey in $\mathbb{E}^2$!, J. Combin. Theory Ser. A 20 (1976), no. 3, 385--389. A triangle is Ramsey in the plane when every two-coloring of $\mathbb R^2$ contains a monochromatic congruent copy of it. Theorem 2 states that every right triangle is Ramsey in this sense; Corollary 4, that every triangle with sides $a$, $b$ and $\sqrt{b^2+2a^2}$ with $2b>a$ is Ramsey; and Corollary 5, that every triangle with sides $a$, $b$ and $\sqrt{4b^2-a^2}$ with $\sqrt{3/2}\,b<a<\sqrt{5/2}\,b$ is Ramsey. The engine is Lemma 1: for every real $a>0$ and every two-coloring of the plane there is a monochromatic equilateral triangle of side $ka$ for some $k\in\{1,3,5,7\}$, proved by a case analysis on the coloring of a lattice. Theorem 3 states that every parallelogram has a congruent copy with three vertices of one color, which decides no triangle. The source is carded at shader_1976_all_right_triangles_are_ramsey_e2.

Covers. The statement of Problem 173 restricted to the right triangles, to the triangles with sides $a$, $b$, $\sqrt{b^2+2a^2}$, $2b>a$, and to the triangles with sides $a$, $b$, $\sqrt{4b^2-a^2}$, $\sqrt{3/2}\,b<a<\sqrt{5/2}\,b$: in every two-coloring of the plane each of them has a monochromatic congruent copy, so none of them is the exceptional triangle of any coloring. The paper says nothing about whether one coloring can miss two other triangles, which is the question.

Depends on. No page of this wiki.

Acceptance. Refereed: the paper is a journal publication in the Journal of Combinatorial Theory, Series A, volume 20, issue 3 (May 1976), the refereed evidence; the issue carries no day, so this page is dated to the first day of that month. The site's curator credits the right-triangle case to [Sh76] in the problem's commentary, but the site labels the problem OPEN, so that credit is not reviewed evidence. The proofs are not checked by this corpus, and nothing is independently reviewed by this project.