Why are there only five regular polyhedra, and why is the solid made of two tetrahedra glued face to face...

Because the angles between the edges of adjacent faces aren't all the
same.

-- Prem

In article <768@dover.uucp> hume@dover.uucp (Chris Hume) writes:
>There are five classical Regular Polyhedra.  My understanding is
>that a Regular Polyhedron is a convex polyhedron each of whose faces
>is comprised of instances of the same Regular Polygon.

This understanding is wrong.  A regular polyhedron in the sense you
seem to mean--- more commonly called a `platonic solid'--- is one in
which all the vertices are congruent and all the faces are congruent.
Certainly there are, up to similarity, only five such animals.  Your
`five-pointed hexahedron' fails the first test.  Another polyhedron,
somewhat more `regular' in general appearance, which also fails this
test is the rhombic dodecahedron, which may be obtained by cutting a
cube into six square pyramids (each with its top in the middle of the
cube) and gluing the bases of the resulting pyramids onto the faces of
another cube congruent to the original one.

It is possible to prove that a polyhedron meeting the two given
conditions has regular faces, but it's not part of the definition.
Thus there is

>an additional "symmetry requirement"...

and moreover not all the ones you give are required.  I'm pretty sure
that there are, in your sense of the words, infinitely many regular
polyhedra.

-- fred (...!cornell!oravax!fred)

To be precise, to be a regular polyhedron, you must have ALL angles the
same; that is, you must have the angles of the corners of each face
equal to the angle between any two faces.  If the requirement you gave
were all there were, there could be "regular" polyhedra of 30 or 100
sides, or 10, or just about any number; it's getting all those angles
equal that's the tricky part.

-- John C. George

Charles Mills (aka fred@oravax.odyssey.UUCP) has raised the question of
how many convex polyhedra with congruent regular faces there are.  He
and another poster express suspicion that there might be very many of
these.  Perhaps surprisingly, in addition to the platonic solids, there
are only five more:

   Triangular dipyramid             2 Tetrahedra
   Pentagonal dipyramid             2 Pentagonal Pyramids
   Gyroelongated square dipyramid   Square pyramid + 4-Antiprism
                                      + Square pyramid
   Triaugmented triangular prism    3 Tetrahedra + 3-prism
   Snub Disphenoid                  ?? Hard to describe ??
    (aka Siamese Dodecahedron)      (12 triangular faces)

For information about these see:

   H. Freudenthal and B. L. van der Waerden, Over een bewering van
     Euclides, Simon Stevin, 25 (1947), 115-121.

       To my knowledge, the first enumeration of the convex polyhedra
       with congruent regular faces.

   Viktor A. Zalgaller, Convex Polyhedra With Regular Faces, Seminars
     in Mathematics, V.A. Steklov Mathematical Institute, Leningrad,
     Volume 2.  (Translated from Russian by Consultants Bureau, New
     York, 1969)

      An astonishing enumeration of all of the convex, regular faced
      polyhedra and a "proof" of the completeness of the list.  In
      addition to the regular polyhedra, the semiregular polyhedra, and
      the infinite family of prisms and antiprisms, there are but 92
      more.  (I say "proof" because he resorts to numerical
      approximation at several points.)

   N. W. Johnson, Convex Polyhedra With Regular Faces, Canad. J. Math.,
      18 (1966), 169-200.

      The same list as above, without proof, together with some
      interesting observations.

If you are interested in building models of regular faced polyhedra
(convex or not), you can obtain interlocking plastic triangles,
squares, and pentagons from Polydron USA INC., 700 Fee Fee Road, St.
Louis, MO 63043, (314) 991-3589.  They are somewhat pricey, but they
make visualizing polyhedra quite easy.  (I have no connection with this
company other than as a satisfied customer.)

-- James R. Driscoll <j.r.driscoll@dartmouth.edu>, Dartmouth College

> I am unsure of the proper Algebraic Terminology, but an Octahedron
> does not exhibit the same level of symmetry as the rest of its
> family.

What do you mean?  The octahedron is just as symmetric as the cube
(sometimes called a hexahedron.)  In fact they are duals of each other.

> Why isn't this "Five Pointed Hexahedron" among the Regular Polyhedra?

Because three vertices have 4 triangles each, and two have 3 triangles
each.  The basic definition of the platonic solids is that each face
and vertex is congruent, and the faces are regular polygons.

> I would be curious to learn how the such a proof is most efficiently
> approached.

That there are no more than 5 regular polyhedra is easily proved by
looking at the possibilities for a solid angle with regular polygons -
6 triangles lie flat, as do 4 squares or 3 hexagons.  Fewer than 3
polygons don't make a solid angle.  So the only possibilities are 3, 4,
or 5 triangles, 3 squares, and 3 pentagons.  Then you have to show
existance, which is usually done by construction.

> Secondly: is there an approach of sufficient power that it extends
> to 2-dimensional manifolds with one or more Handles?

I don't remember the status of regular polyhedra of higher genus.  You
might try a book by Bonnie Stewart called something like "Adventures
among the Toroids" (???) about toruses with regular polygons for faces.
For lots of other generalizations (star polyhedra, semi-regular
polyhedra, etc) try books by Coxeter, or Mathematical Models by Cundy
and Rollett??, or Ball's "Mathematical Recreations and Essays".  There
is also an interesting paper by Branko Grunbaum in Aequationes
Mathematicae 16 (1977) p.1-20 "Regular polyhedra-old and new".

-- Stan

[ Editor: a related problem and further questions from the same
collection. ]

In article <IYs6qry00XojMAMEt3@andrew.cmu.edu> vm0h+@andrew.cmu.edu
(Vincent J. Matsko) writes:
>Show that if four regular tetrahedra are ``glued'' onto the faces of a
>regular octahedron such that no two tetrahedra are glued to adjacent
>faces, the result is a regular tetrahedron with edge-length twice that
>of the original octahedron.

Proof:  Take the larger tetrahedron to have vertices
(1,1,1),(1,-1,-1),(-1,-1,1),and (-1,1,-1).  The tetrahedron is regular
because the vertices have all requisite symmetries.  The set of
midpoints of the edges, (+-1,0,0),(0,+-1,0),(0,0,+-1) is evidently a
regular octahedron because it also has all requisite symmetries.  It is
also clear that the remainder is four tetrahedra congruent to the
original one.

Now for a harder problem:  Show that there are exactly six Platonic
solids in four dimensions.  Find a unified proof.

Also show that three of these six Platonic solids tile Euclidean space.

If that's too easy, try this one:  Each Platonic solid in any dimension
has an analogue in hyperbolic space which is uniquely determined up to
volume.  Show that in three dimensions, exactly four of these analogues
tile hyperbolic space, and in four dimensions, exactly five of these
analogues tile.

Add to these six existence problems and twelve tiling problems the fact
that there exist five Platonic solids in three dimensions, and the fact
that exactly one of these solids tiles Euclidean space, and you have a
total of 24 cases.  Find (something close to) a unified proof for all
cases.

Definition:  A Platonic solid is a convex polyhedron whose sides are
congruent Platonic solids.  A 2-dimensional Platonic solid has all
equal angles.

-- Greg

[ Editor: no solutions to Greg's four-dimensional and hyperbolic
problems were found in the collected postings.  For the question
whether the Platonic solids can have vertices at integer lattice
points, see geometry/lattice/platonic.solids. ]