= Spherical point set
A finite <set> $X=\{x_0,\ldots,x_m\}\subseteq\mathbb R^d$ is spherical if all its <points> lie on one <sphere>: there are a center $c$ and <radius> $r$ with $\|x_i-c\|=r$ for every $i$. Subtracting the squared-distance equation for $x_0$ gives the equivalent <linear equations>
$$
2\langle c,x_i-x_0\rangle=\|x_i\|^2-\|x_0\|^2\qquad(1\leq i\leq m).
$$
Thus 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 configuration
$$
\{(0,0)\}\cup\left\{\left(j,\pm\sqrt{j(2r-j)}\right):j\in\{1,3,4\}\right\},\qquad r>2\text{ transcendental},
$$
as 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.<A cyclic non-Ramsey heptagon by Dömötör Pálvölgyi>[Theorem1, arXiv:2609.23327v1, 20 September 2026]{ref}
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