I've long been confused about the exact metric definition of "dual polyhedron". Sure, define one "new vertex" for each old face, and one "new edge" for each pair of old faces that share an edge — that gives the dual combinatorially. But as for the exact shape & size, I'm deeply confused. (We're in R^3.) But nevertheless, Definition of "round polyhedron": A convex polyhedron P in R^3 is "round" if ----- a) all vertices of P lie on its circumsphere, and b) all faces of P touch its insphere. ----- Which convex polyhedra are round? (E.g., Platonic solids are round.) Vert(P) determines P uniquely, and Vert(P) is a finite subset Vert(P) = {z_j in C u {oo} | 1 <= j <= n} of the Riemann sphere S^2 C u {oo} on which much complex analysis takes place. Question: Suppose Vert(P) is a subset of the unit sphere S^2. ----- Does it follow that the insphere touches all faces? Or equivalently, Does it follow that P is round? Or vice versa. —Dan Let R = inradius, S = circumradius, so R < S as of course it should be (since alphanumeric inequality inconsistencies are rarely mnemonic). Now suppose again we sandwich a polyhedron P between its insphere and its circumsphere. Bill Gosper wrote: ----- ... If a Platonic solid is scaled to have the same inradius as its dual, then they will also have the same circumradius. ... -----