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From: moroney@jon.DEC
Newsgroups: net.math
Subject: Re: Polyhedral dice (duals of Archemedians)
Message-ID: <3710@decwrl.UUCP>
Date: Sat, 22-Sep-84 13:11:31 EDT
Article-I.D.: decwrl.3710
Posted: Sat Sep 22 13:11:31 1984
Date-Received: Wed, 26-Sep-84 02:19:30 EDT
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>In fact, there is a fairly large (infinite!) set of these fair polyhedral dice.
>For example, the duals of the Archemedian (semi-regular) polyhedra are all
>fair dice.  The dual d(P) of a polyhedron P has a vertex at the center of
>each face of P, a face centered on to each vertex of P, and edges `perpendicular'
>to the edges of P (i.e. there is an edge joining two vertices of d(P) if the
>corresponding two faces of P share an edge.)
>It is not hard to prove that if P is Archemedian, then there is a rotation
>mapping d(P) onto itself which maps any face onto any other face.  I.e., you
>can't tell one face from another by looking at it.  Write numbers on the faces
>and you have a fair die.

Only one problem with that, if you take a dual of an Archemedian, what you get
is another Archemedian.  The dual of a cube is an octahedron, the dual of an
octahedron is a cube, dodecahedron <--> icosahedron, and the dual of a 
tetrahedron is another tetrahedron.  So you don't gain any more fair dice by
taking duals of Archemedians.

	There is a permutation of polyhedra which is similar to taking duals
I will call it the edge-dual.  It is somewhat hard to visualize and explain,
but here goes.  The dual can be visualized by taking a polyhedron and filing
down all the vertices evenly until you just reach the point where none of the
original faces remain.  Similarly, the edge dual can be formed by taking a
polyhedron and filing down all the edges evenly until you just reach the
point where none of the original faces remain.  You will then have a figure
which has one face for each of the original edges and one vertex for each 
of the original vertices and faces.  When this is done on an Archemedian,
you get only 2 new polyhedra.  When you do this to a cube or an octahedron,
you get the rhombic dodecahedron.  When you do this to a (pentagonal)
dodecahedron or an icosahedron you get the "30" I mentioned in a previous
article.  (probably called a rhombic {whatevertheprefixfor30is}ahedron)
The edge dual of a tetrahedron is a cube standing on a corner. (Try visualizing
that process!)


>I believe that the rhombic dodecahedron mentioned in elecvax.328 is one of these
>Archemedian duals, although offhand I can't think which.

The edge-dual of a cube or octahedron.

					Mike Moroney
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