For each of the five Platonic solids, decide whether or not there exists such a solid in R^3 such that each...

Cube:  Yes.  (Obvious)

Tetrahedron:  Yes.  As mentioned in a previous post, we may
appropriately choose four of the vertices of a cube.

Octahedron:  Yes.  Take a cube with edge length 2, centered at the
origin, and with faces parallel to the coordinate planes; then the
centers of the faces are the vertices of an octahedron.

Dodecahedron and Icosahedron:  No.

Lemma:  In R^n, for n >= 2, it is not possible to construct a pentagon
such that all vertices have integral coordinates.

Proof:  Let a, b, c be three consecutive vertices of a pentagon whose
vertices have integral coordinates.  Let [ab], [bc], and [ac] be the
lengths of the sides of the triangle formed by these vertices.  Note
that [ab] = [bc].  Then by the law of cosines, we must have

    [ac]^2 = [ab]^2 + [bc]^2 - 2 [ab] [bc] cos 108 = 2 [ab]^2 (1 - cos 108).

Since the vertices have integral coordinates, [ac]^2 and [ab]^2 are
integers, and hence cos 108 is rational.  As this is false, there is no
pentagon whose vertices have integral coordinates.  Note that this
proof is completely independent of the dimension of the space in
consideration.

Back to the original problem:

Since the dodecahedron has pentagonal faces, the conclusion is evident.

Now choose a vertex of the icosahedron.  Then the 5 vertices adjacent
to this vertex form a pentagon.  Q.E.D.

-- Vince Matsko

About a month ago, David Wilson asked:
>       For each of the five Platonic solids, decide whether or not
>       there exists such a solid in R^3 such that each of its vertices
>       lies at a point with all integer coordinates.

which was correctly answered (with proof) to be just the tetrahedron,
cube, and octahedron.

The new question is: Which of the 13 semi-regular (Archimedian) solids
can be put in a regular (3-D) lattice?

(For an answer, with proofs and some references, see "Equiangular
Lattice Polygons and Semiregular Lattice Polyhedra", by P.R.Scott, in
the College Mathematics Journal, Vol. 18, #4, Sept, 1987.)

-- Stan Isaacs

[ Editor: the answer to the Archimedean question is not in the collected
postings beyond this reference. ]