The case of our fifth platonic solid, the tetrahedron, is unique in that its respective truncation is another platonic solid, the octahedron:
Unique also in that its dual is the same as itself:
The arcs of the dual tetrahedra and their truncation, the octahedron, will tile the sphere thus:
So what do we have? Looking at the five regular, platonic solids (the reader will find joy in discovering for themselfs that these are the only possible regular solids), applying the least arbitrary (i.e. most reasonable) truncation (bisecting each edge), generates three other solids: the cuboctahedron, the icosidodecahedron, and the octahedron. These truncations of the platonic solids are the same three solids we came across above, when investigating the most regular way of dividing the sphere with 3, 4, and 6 great circles; the most regular division of the sphere with 5 great circles generates the golden pentagramma mirificum, where the division of each great circle is not equal.
Reinverting the investigation, if we divide the sphere in the most regular way with 3 great circles an octahedron is defined, which itself defines two tetrahedra, which are duals of eachother, and are both regular, platonic solids. The most regular division of the sphere using 4 great circles generates a cuboctahedron, which defines both the cube and the octahedron, two more platonic solid duals. Using 5 great circles, the most regular division of the sphere generates the golden pentagramma mirificum solid. And with 6 great circles, dividing the sphere in this most regular way, we have the icosidodecahedron, which defines the two other platonic solid duals, the icosahedron and the dodecahedron.
The pentagramma mirificum asserts itself again, when investigating the least arbitrary (i.e. inherent) divisions of the sphere; more evidence, in addition to what was seen in part 1 of this pedagogy, that the pentagramma mirificum is an artifact of the fundamental nature of the sphere itself!