An empty-sphere tetrahedron can be arbitrarily flat

For every $n\ge 1$, ${\epsilon }_n\in \mathbb{D}$ and the four sites therefore satisfy the dyadic-coordinate input condition of [F1]. The first three points are noncollinear in $z=0$ and $p_4$ has positive height, so the four points are affinely independent and form a nondegenerate tetrahedron. Its volume is one third of base area $1\mathrm{/}2$ times height ${\epsilon }_n$, namely $V_n={\epsilon }_n\mathrm{/}6$. With only these four sites, its circumsphere contains no other site in its interior, so [F1] makes it an empty-sphere Delaunay tetrahedron.

The total face area $S_n$ is at least the base area $1\mathrm{/}2$ and is bounded above uniformly in $n$, since every edge length is at most $2$ and each of the four face areas is at most half the product of two such edge lengths; in particular $S_n\le 8$. The inradius is $r_n=3V_n\mathrm{/}{S}_n={\epsilon }_n\mathrm{/}(2S_n)$, so ${\epsilon }_n\mathrm{/}16\le r_n\le {\epsilon }_n$ and $r_n\to 0$. The diameter is at least $\mathrm{\mid }{p}_1-p_2\mathrm{\mid }=1$. Thus empty-sphere Delaunay membership imposes no positive uniform inradius-to-diameter bound in three dimensions.