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anton sherwood wrote:
> All a fair die needs to be is a dual of an (extended) Archimedean
> polyhedron - all vertices alike. That way all the faces will be
> alike, but not necessarily regular.
I'll put that another way. Dice can be made like this:
intersection {
plane { V0, 1 }
[...]
plane { V9, 1 }
}
For a fair die, there must be a rotation group S relating all the
vectors. For each (i,j) there must be a rotation in S such that the
image of Vi is Vj; and every image of Vi must be a member of the set V.
The vertices of a semiregular (Archimedean) solid satisfy this
condition, but so do some nonregular intermediate forms. For example:
take two pentagons and place them parallel, so that their vertices are
those of a prism. Now rotate one of the pentagons on the axis of the
prism: the symmetry group remains the same, and thus the vertices of the
two pentagons remain a valid set V, as described above, for a fair 10-die.
> The 10-die is the dual of the pentagonal prism or antiprism.
Or something between. The one I have is anti.
--
"How'd ya like to climb this high without no mountain?" --Porky Pine
Anton Sherwood *\\* +1 415 267 0685 *\\* http://www.jps.net/antons/
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