 Spherical point set

A finite set $X=\{x_0,\dots ,x_m\}\subseteq {\mathbb{R}}^d$ is spherical if all its points lie on one sphere: there are a center $c$ and radius $r$ with $\Vert x_i-c\Vert =r$ for every $i$. Subtracting the squared-distance equation for $x_0$ gives the equivalent linear equationsThus consistency of a finite linear system tests whether a common center exists. A nondegenerate triangle is spherical, with center its circumcenter. Three distinct collinear points are not spherical: the equal-distance equations along their line are inconsistent. This is the geometric obstruction behind three-term unit arithmetic progression is not Euclidean Ramsey.

Every Euclidean Ramsey set is spherical. The converse was historically conjectured. A September2026 preprint by Dömötör Pálvölgyi reports the seven-point configurationas a counterexample. These points do lie on the circle $(x-r{)}^2+y^2=r^2$; the non-Ramsey assertion is Theorem1 of that dated preprint.^[ref]