wendy wrote:When you set the icosahedral symmetry over the octahedral symmetry, then these symmetries match: o3a5e and a3o4e. That is, the ID becomes an octahedron, and the D becomes a tetrahedron or cube. The linkage is actually more facinating than it looks.
Here "a" and "e" just represent equivalent, but Independent instances of either "x" or "o" nodes.
E.g. into o3x5o an fq = (1+sqrt5)/sqrt2 scaled x3o4o can be vertex inscribed.
Or into o3o5x an f = (1+sqrt5)/2 scaled o3o4x can be vertex inscribed.
This in fact is due to the common pyritohedral subsymmetry. (Which sadly has no representation as mere reflectional Group. And therefore cannot be displayed by a Dynkin symbol either.)
Note that o3x5o has 30 vertices, x3o4o only has 6. Fixing a pair of opposite ones, there are no further orientations possible (except of the symmetries of oct itself). Thus we get 30/6 = 5 possible ways to inscribe it. - Or, put it an other way, there is a compound of 5 octs, which has its vertices right in the directions of those of an id. (That one is described as [5{3,4}]2{3,5} and known as
small icosiicosahedron.)
Similarily, o3o5x has 20 vertices, o3o4x has only 8. Fixing a pair of opposite ones of the latter within the former, then there will be exactly 2 ways to inscribe it (connecting to either of the non-neighbouring pentagon vertices). Thus we get again 2*20/8 = 5 possible ways to inscribe it. - Or, within that other view, there is a compound of 5 cubes, which has its Vertices right in the directions of those of a doe. And then 2 cubes each will capture the same vertex position. (That one is described as 2{5,3}[5{4,3}] and known as
rhombihedron.)
For sure you could consider o3x5x (tid) and x3o4x (sirco) too. And there will be a corresponding compound of 5 sircoes with a global icosahedral symmetry (
rhombisnub rhombicosicoahedron). But because of the above mentioned different scaling factors, here the hull will NOT be the uniform tid, in fact not even some variant o3y5z therefrom (y and z to be determined accordingly). Rather you'll get some more general variant x3y5z of grid here for its hull!
So, concluding, "a" and "e" finally are NOT Independent choices of "x" and "o". Rather they seem to be "XOR" ones only. - Or, alternatively, you would have to choose the lengths A to D for o3A5B and C3o4D quite properly, so that these could match.
--- rk