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Source. Published p. 349, Lemma 14 (published scan).

Statement. For $K=\{v_0,\ldots,v_k\}$, nonsphericity is equivalent to existence of real $c_1,\ldots,c_k$, not all zero, with

$$\sum_{i=1}^k c_i(v_i-v_0)=0,\qquad b:=\sum_{i=1}^k c_i(\|v_i\|^2-\|v_0\|^2)\ne0.$$

Complete proof. If $w$ is a sphere center, then

$$\|v_i\|^2-\|v_0\|^2=2\langle w,v_i-v_0\rangle.$$

Every relation in the first display therefore makes $b=0$.

Conversely, take a minimal nonspherical subset of $K$ and relabel one of its points as the base point. Its difference vectors are linearly dependent: otherwise the independent equations $2\langle w,v_i-v_0\rangle=\|v_i\|^2-\|v_0\|^2$ have a solution, a sphere center. Choose a nonzero relation and an index $j$ with $c_j\ne0$. The proper subset omitting $v_j$ has a sphere center $w$ and radius $R$. Translating the squared-norm relation by $w$ changes it by $-2\langle w,\sum_i c_i(v_i-v_0)\rangle=0$. Hence

$$b=c_j\bigl(\|v_j-w\|^2-R^2\bigr)\ne0:$$

equality would put the omitted point on the same sphere and contradict nonsphericity. Extend the coefficients by zero to the other points of $K$. To return to any originally specified base point, write the relation with coefficients $\lambda_v$ over all points and $\sum_v\lambda_v=0$; then changing which point is the base changes neither vector relation nor $b$. $\square$

The same identities are invariant under orthogonal transformations and translations. By the Gram extension in definitions, they also hold for any congruent copy in a different ambient dimension.

Bears on. #174.