Re: Polyhedra with congruent faces

In news:<1177999879.295913.212570@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>
schrieb Bill Taylor <w.taylor@xxxxxxxxxxxxxxxxxxxxx>:

This question of congruent-faced polyhedra is a very interesting one.

I thought I'd add my 10c worth, as the combinatorial side of the issue
is an important one, IMHO perhaps the most important, and has not
really been addressed yet.

And I at least am *very* glad that you started the combinatorial analyis. -
I personally thought that the result would be very much more complicated.

Well, it *is* _somewhat_ more complicated, because
you forgot to take the diagonals into account
for the cases of the cube and the dodecahedron:

Even if, e.g., _all side lengths of the pentagonal faces_
of the dodecahedron _are equal_, the diagonal lengths of
these pentagons might all be pairwise distinct, and then
there will be 2*5 = 10 ways to fit a face into its position,
which makes a total of 12^10 combinations.
Of course the symmetries of the dodecahedron make that only
a fraction of these variations are distinct;
but there are not enough symmetries to reduce the combinatorics
to 10^3 (maybe even 10^6) different combinatorial types
of dodecahedra with all edges of equal lenghts.

My guess, however, is that most of these combinatorial types
will *not* be realizable as polyhedra.
-
Just look at the cube:
If you want to build a (convex!) cube-type hexagon
with congruent non-square rhombi as faces, then
the way how you arrange three of these faces in one vertex of the
hexagon already determines the positions of the other three faces -
and the resulting polyhedron will always become a parallelepiped.

This makes that, given the rhombic faces,
only two non-congruent parallelepipeds can be build from them,
and these only differ by changing all interiour angles of
the faces from phi to (Pi-phi). - What a difference
to the "6^2 mod symmetries" combinatorially imaginable solutions!

Actually, I lied a bit: If the given rhombus is to small -
if the smaller interior angle of the given rhombus
is <= Pi/3 (60 degrees), then
no cube-like hexahedron can be build from it.


As has been observed already, any such polyhedron must be graphically
isomorphic to a regular polhedron, only five types. To make the sides
geometrically congruent is then no great difficulty, because there is
a huge amount of flexibility in the frameworks once we are beyond
the tetrahedon, and provided the lengths aren't TOO disparate,
(as the "impossibly-overflat" tetrahedron was noted), we are
essentially down to making them combinatorially congruent.

This means, in effect, we merely take a regular polyhedron, and color
its edges in some number of colors, (nothing beyond five colors),
making sure all the faces have the same cyclic color order,
which OC includes allowing the exact opposite cyclic order.
Efectively, red edges are then 1cm long, yellow edges 1.1 cm,
green edges 1.2 cm and so on.
[ ... ]

The distance-geometric part of the problem
is not half as simple as you seem to think it is.

But I have not got time to comment on this now.

Hope to be able to write some more later in the day...
.