Ian Burgmyer wrote:
> NOT a regular polyhedron?  Odd. . .
> 
> Larry Fontaine <lfo### [at] isdnet> wrote
> > The 10-sided isn't a regular polyhedron, either. ...

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.  The 10-die is the dual of the pentagonal
prism or antiprism.

-- 
"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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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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