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- Newsgroups: sci.math
- Path: sparky!uunet!secapl!Cookie!frank
- From: frank@Cookie.secapl.com (Frank Adams)
- Subject: Re: Three-sided coin
- Message-ID: <1992Nov11.192804.140054@Cookie.secapl.com>
- Date: Wed, 11 Nov 1992 19:28:04 GMT
- References: <5413@daily-planet.concordia.ca> <1992Nov10.204733.25206@massey.ac.nz>
- Organization: Security APL, Inc.
- Lines: 95
-
- In article <1992Nov10.204733.25206@massey.ac.nz> news@massey.ac.nz (USENET News System) writes:
- > If you want to throw a randomising device, you'd better
- >stick to Platonic solids.
-
- Actually, Platonic solids is too constrained a condition. A Platonic solid
- must have all faces isomorphic (strictly, in a single orbit under the
- symmetries of the object), and likewise for edges and vertices.
-
- All that's required for a randomizing device is that the faces are
- isomorphic; the constraints on edges and vertices are not relevant. And, as
- it turns out, there other solids which work.
-
- There are two infinite families of these, each with 2n sides for each n>2.
- The easiest to visualize consists of two n-sided pyramids, with their faces
- stuck together. The regular octahedron is a special case of this, for n=4.
- (We count polyhedra as equivalent if they have the same structure.
- Polyhedra in this family can be deformed by changing the steepness of the
- pyramids without changing the equivalence of the faces.)
-
- The second class is a bit harder to visualize. It can be described as two
- n-sided pyramids, stuck together skew, with the faces extended to form
- quadrilaterals. The cube is a special case of this, for n=3. For actual
- randomizing devices, this form is better for n odd, since it then has a top
- face; the first class has a top face for n even. 10 sided dice of this form
- are commercially available.
-
- The other polyhedra of this type can all be "built" from Platonic solids.
-
- One way to build them is to build a short pyramid on each face of the solid
- (short enough that the previous edges remain convex). This produces five
- distinct polyhedra, with respecively 12, 24, 24, 60, and 60 faces for the
- tetrahedron, cube, octahedron, dodecahedron, and icosohedron. These are all
- distinct from each other and from the Platonic solids.
-
- If the pyramids are made a bit higher, until the old edges become flat,
- polyhedra with half as many sides are created: 6, 12, 12, 30, and 30.
- However, there are only 2 new solids here: the one created from the
- tetrahedron is a cube, and the other pairs with the same number of faces are
- in fact the same. These are called rhombic polyhedra. The 30-sided rhombic
- polyhedron is commercially available, and the 12-sided one may still be.
-
- It is also possible to create polyhedra with 12, 24, 24, 60, and 60 faces
- with quadilateral faces. Create a shallow "bump" in the center of each
- edge, add a new vertex in the middle of each face (raised), and connect
- these new vertices to the "bumps". This gives us 2 new polyhedra: the pairs
- derived from dual regular polyhedra are equivalent, and the one derived
- from the tetrahedron is a rhombic dodecahedron.
-
- Yet another way to get size 12, 24, 24, 60, and 60 faces creates pentagonal
- faces. Turn each edge into a zig-zag:
-
- /\
- \/
-
- and connect the centers of the faces to the "outdents" on the adjacent
- edges. This only produces 2 new polyhedra; again the duals produce
- equivalent solids, and the tetrahedron produces the regular dodecahedron.
- These two are the only polyhedra with equivalent faces which have a
- "handedness"; they cannot be transformed into their mirror images in
- 3-space.
-
- Finally, one can put a "bump" on each edge, and connect each old face to
- both the bumps and the old vertices, giving polyhedra with 24, 48, 48, 120,
- and 120 faces. Again, there only 2 new polyhedra here. The duals are again
- equivalent, and the polyhedron derived from the tetrahedron is the same as
- the one derived before by adding bumps to the cube or octahedron.
-
- That these are the only possibilities can be shown by considering the
- sequence of valences of the vertices around a face. (This must, of course,
- be the same for each face.) The basic condition is that
-
- sum(1/2 - 1/v_i) < 1
-
- which is required for the network to close off into a polyhedron. The
- actual proof involves checking each solution to this, showing that each
- produces at most one polyhedron.
-
- Note that tilings of the plane with all regions equivalent satisfy the above
- condition, with "<" replaced by "=". For example, 3,6,3,6 gives a tiling
- into rhombuses:
-
- *---* *---* *---* *---*
- / \ \ / / \ / \ \ / / \
- / \ \ / / \ / \ \ / / \
- * *---*---* * *---*---* *
- \ / / \ \ / \ / / \ \ /
- \ / / \ \ / \ / / \ \ /
- *---* *---* *---* *---*
- / \ \ / / \ / \ \ / / \
- / \ \ / / \ / \ \ / / \
- * *---*---* * *---*---* *
- \ / / \ \ / \ / / \ \ /
- \ / / \ \ / \ / / \ \ /
- *---* *---* *---* *---*
-
-