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Source: Imre Leader, Paul A. Russell and Mark Walters, Transitive sets and cyclic quadrilaterals, Journal of Combinatorics 2 (2011), no. 3, 457--462: Conjecture 3 on p. 458. The edition read is identified on the source card.

Statement

Conjecture 3, printed p. 458, states:

Let $-1<a<1$ be transcendental. Then the cyclic quadrilateral with vertices $(-1,0),(1,0),(a,\sqrt{1-a^2}),(a,-\sqrt{1-a^2})$ is not Ramsey.

Scope

This is a conjecture, not a consequence proved in the paper. The paper's Corollary 2 proves that the same set does not embed in a finite transitive set. The conjecture would follow from Corollary 2 together with the authors' rival conjecture that the Ramsey sets are exactly the subsets of finite transitive sets (p. 457), which is itself only a conjecture in the paper.

The two five-page prepublication versions accidentally quantify $\alpha$ while the displayed vertices use $a$. The published version changes the quantified variable to $a$, as recorded here.

Bears on. Problem 174: the conjecture asserts that the kites of Corollary 2, which are spherical, are not Ramsey. If true, it would refute Graham's conjecture that every spherical set is Ramsey. The paper proves nothing toward it.