I posted those "simplex extra-sphere" puzzles under the impression that I actually knew the answers, a delusion quickly exposed. Now that I finally do know (most of) them, I shall try to make amends. ________________________________________________________________________________ Understanding what's going on here requires adopting notions from contact geometry. A general "cycle" in inversive n-space comprises a sphere equipped with an extrinsic orientation, naturally represented by the corresponding open ball (positive radius), or the closed ball's complement (negative radius). Special cases of cycles include points (unoriented), primes (hyperplanes), and inversive infinity. Two cycles are "tangent" when their spheres touch and their half-spaces intersect. "Eversion" of a cycle changes just its orientation. The generalised Apollonian problem (GAP) demands construction of a cycle bearing given relations (in terms of tangent length or intersect angle) to n+1 given cycles. It is a theorem that this problem has in general just two solutions, which --- in traditional fashion --- may be real, double or complex; they may also be congruent (equal except for orientation), or one or both at infinity. [ The only construction known to me involves the Clifford algebra representation of the Lie-sphere group of transformations preserving tangency. ] A particular extra-sphere (extended exsphere) of a simplex occupies some convex region bounded by facet hyperplanes, exterior to a given subset, and interior to the remainder. In terms of contact-geometry, it can (almost) be identified with some cycle tangent to every facet prime, interior facets oriented positively, exterior negatively. Of the two GAP solutions, one lies at infinity (to which all primes are tangent) and may be discarded. However, notice that the Euclidean configuration might equally well be represented by everting every cycle. Since the unoriented solutions of both problems are congruent, regions fall into complementary pairs, such that just one of each pair contains an extra-sphere! ____________________________________________________________ Previously I discussed the construction of a simplex similar to the original, inscribed to all but one of its facet primes and exscribed to the other. By induction on m an analogous construction applies in the more general situation of m negative facets, the original instead scaled by a factor (-1)^(n+m) / (n-2m) . Using the circumradius of this "extra-medial" simplex as an upper bound immediately yields constraint 0 < |r(m)| <= R/(n - 2m) on extra-radius r(m) in terms of circumradius R , at any rate for 2m < n . I conjecture that when 2m = n , the upper bound is not attained. Which now implies that for 2m <> n+1 , the nearer one ( 2m < n+1 ) of a complementary pair of regions is always occupied by an extra-sphere, while the further one ( 2m > n+1 ) is always empty! For 2m = n+1 , things are more complicated. There are binomial(n+1, m)/2 Euclidean spheres, to be allocated to twice as many regions: so in general, just half those regions may be occupied, while their complements lie empty! Worse still, symmetry may cause a pair of complementary regions to be congruent, in which case the only possible resolution involves both spheres retreating to infinity: this occurs for example when the simplex is regular. In fact, expressing simplex content two ways in terms of facet contents and scribed radii leads immediately to relation between extra-radius and inradius (n+1-2m) |r(m)| = (n+1) r(0) for a regular simplex: eg. r(2) = oo for a tetrahedron, as painfully deduced earlier. _____________________________________________________________ Fred Lunnon [14/04/16]