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Statement

Remark (§1, p. 3, unnumbered). Every triangle embeds into a transitive set. A right-angled triangle lies in a rectangle and an acute-angled one in a cuboid in three dimensions. For a general triangle $ABC$, the paper takes a point $D$ on the perpendicular from $C$ to $AB$ such that the angle $AOB$, $O$ the circumcentre of $ABD$, is a rational multiple of $\pi$; then $A$ and $B$ lie on a regular polygon $\Pi$ centred at $O$, and the "twisted prism" $\Pi\cup\Sigma$, with $\Sigma$ a copy of $\Pi$ translated perpendicular to its plane and rotated about its centre, is transitive and, for suitable translation and rotation, contains a copy of $ABC$.

Derivation

The paper gives the construction only. Its details, in the corpus's words: the triangle is taken nondegenerate, since three collinear points are not spherical. With $D$ at height $t$ below the height $h$ of $C$, the circumradius of $ABD$ varies with $t$, so the angle $AOB$ takes a rational multiple of $\pi$ for some $t\in(0,h)$. Rotate $\Pi$ so that a vertex goes to $D$ and translate it by $\sqrt{h^2-t^2}$; that vertex and $A$, $B$ form a copy of $ABC$. The rotation through $2\pi/q$ of both layers and an isometry exchanging them generate a finite dihedral group transitive on the $2q$ points.

Note

The paper remarks that the transitive sets into which Frankl and Rödl embed triangles are very different. With Kříž's soluble-group theorem, the dihedral group above also makes each triangle Ramsey, which Frankl and Rödl proved first.

Source. Imre Leader, Paul A. Russell and Mark Walters, Transitive sets in Euclidean Ramsey theory, J. Combin. Theory Ser. A 119 (2012), no. 2, 382--396, doi:10.1016/j.jcta.2011.09.005; pages from the arXiv version 1012.1350v1 identified in the source digest.

Read depth. Claims checked: the remark was read against the print; the derivation above is the corpus's own.

Bears on

• Problem 174: an example of the pattern the paper builds Conjecture A on, that known Ramsey sets are proved so by embedding them in a transitive set.