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Source. Published p. 358, example preceding Theorem 23 (published scan).

Statement. There are five distinct points in $\mathbb R^3$ for which every angle determined by three points is nonobtuse, but which cannot be embedded among the vertices of a brick in any dimension.

Complete proof. Take three points $a_1,a_2,a_3$ on a circle of radius $\sqrt{2/3}$ in the plane $z=0$, equally spaced by $120$ degrees about the origin. They form an equilateral triangle of edge length $\sqrt2$. Add $b_+=(0,0,1/\sqrt3)$ and $b_-=(0,0,-1/\sqrt3)$. Then

$$\|a_i-a_j\|^2=2,\quad \|a_i-b_\pm\|^2=1,\quad \|b_+-b_-\|^2=4/3.$$

Each possible triangle type has squared side lengths $(2,2,2)$, $(2,1,1)$ or $(1,1,4/3)$. In each case the largest squared length is no larger than the sum of the other two, so all angles are nonobtuse.

A point equidistant from $a_1,a_2,a_3$ lies on the $z$-axis. Equality of its distances to $b_+$ and $b_-$ then forces it to be the origin. But $\|a_i\|^2=2/3$ and $\|b_\pm\|^2=1/3$, so the five points have no common sphere in their affine hull. Projection of a higher-dimensional center onto this hull would still give a common center, so no sphere in any ambient dimension contains a congruent copy.

Every brick is spherical about its coordinatewise midpoint. Any of its subsets lies on that sphere, so our nonspherical configuration cannot be a brick subset. $\square$

The source calls the example a “5-point simplex,” but these five points have affine dimension three and are affinely dependent. It is not a counterexample involving a nondegenerate four-dimensional simplex, all of which are spherical. The obstruction concerns the stated five-point configuration and the insufficiency of squared triangle inequalities.

Bears on. #174.