**tetrahedral symmetry**- {3,3} tetrahedron, rhombic dodecahedron [2]
- (3,6,6) truncated tetrahedron, 12 zones
**octahedral symmetry**- {3,4} octahedron, cube [1]
- {4,3} cube, rhombic dodecahedron [2]
- (3,4,3,4) cuboctahedron, 6 zone truncated octahedron [3]
- (3,8,8) truncated cube, 12 zones [9]
- (4,6,6) truncated octahedron, 12 zones [10]
- (3,4,4,4) rhombicuboctahedron, 12 zones, [11]
- (4,6,8) truncated cuboctahedron, 24 zones (illustrated below) [15]
- (3,3,3,4) snub cube, 24-zones, (illustrated above at left)
**icosahedral symmetry**- {3,5} icosahedron, 6 zone rhombic triacontahedron [4]
- {5,3} dodecahedron, 10 zone rhombic enneacontahedron [8]
- (3,5,3,5) icosidodecahedron, 15 zone truncated icosidodecahedron [13]
- (5,6,6) truncated icosahedron, 30 zones
- (3,10,10) truncated dodecahedron, 30 zones
- (3,4,5,4) rhombicosidodecahedron, 30 zones
- (4,6,10) truncated icosidodecahedron, 60 zones
- (3,3,3,5) snub dodecahedron, 60 zones

One interesting thing aspect of zonohedra is that we can create new ones by a systematic process starting with any given polyhedron. There aren't too many interesting general processes which input an arbitrary polyhedron and output another polyhedron. Truncation, stellation, dualization, compounding, and zonohedrification are the main ones that I can think of.

Shown here is a 31-zone 242-sided zonohedron. Its star is the 31 axes of symmetry of the icosahedron. (It has been proposed as the basis of an architectural system --- akin to the geodesic dome, but with a wider variety of forms and a smaller inventory of parts --- by Steve Baer.) Notice it contains thirty irregular twelve-sided faces, but they are centrally symmetric; each edge is opposite an equal parallel edge. The sixty hexagons which border decagons are also slightly irregular, but the twenty hexagons on the 3-fold axes are regular.

The term for a polygon which is not necessarily regular, but opposite
sides are equal and parallel, is a *zonogon*. The general definition
of a zonohedron is *a polyhedron bounded by zonogons*. In the special
case where the zonogons happen to be 4-sided, they are parallelegrams.
(A *zonotope* is an *n*-dimensional object for which the zonogon
and the zonohedron are the 2- and 3-dimensional instances.)

(Note the original 1893 definition of a zonohedron by Fedorov only required faces in which opposite edges are parallel --- not necessarily equal --- but modern usage adds the equality condition. See the paper by Jean Taylor in the references.)

**Exercise: **In three of the Archimedean
solids, all the faces have an even number of sides. As these faces
are zonogons, these three polyhedra are zonohedra by this definition. Which
three ?

**Answer: **This, this,
and this. Note that
they are the truncated quasiregular
polyhedra.

The vertices of a zonohedron are determined from a star of vectors in
space by considering all possible ways of summing a subset of those vectors.
For example, given any three edge-direction vectors, *p*, *q*,
and *r*, the eight vertices of a parallelepiped are defined as sums
of the form *a p + b q + c r*, where {*a*, *b*, *c*}
is one of the eight boolean triples: {0,0,0}, {0,0,1}, {0,1,0}, {0,1,1},
{1,0,0}, {1,0,1}, {1,1,0}, {1,1,1}. (Here {0,0,0} gives a vertex at the
origin and {1,1,1} gives its opposite vertex.. Use -1 in place of the 0's
for coordinates of a larger zonohedron centered on the origin.) With more
than three vectors, the same idea gives 2 to the *n* vertex coordinates,
of which some are in the inside and so are not of interest. One can take
the "convex hull" of the set to find just the visible vertices.
For example, with four vectors, chosen as the vertices of a tetrahedron,
there are sixteen defined points using a boolean {*a,b,c,d*}, but
only fourteen are vertices visible on the outside of the rhombic
dodecahedron. Points {0,0,0,0} and {1,1,1,1} lie at the center and
are eliminated. Using an algorithm which does essentially this, but in
a much faster manner, I have created the zonohedrifications listed
below.

To see how larger zonogon faces arise, observe that if three vectors
of the star lie in a plane, the three parallelegram faces flatten out into
a hexagon: For example, if the star which produces the rhombic dodecahedron
is adjusted so that three of the vectors are almost coplanar, the resulting
zonohedron is a distorted rhombic dodecahedron, consisting of twelve
rhombi which approximate a hexagonal prism. If the star is adjusted
further so three vectors are exactly coplanar, the result is a hexagonal
prism, because sets of parallelegrams merge into hexagons. In general,
a star with *k* vectors in a plane produces a 2*k*-sided polygon
in which opposite sides are equal and parallel.

And
now for *zonohedrification*. The vertices of any polyhedron can be
thought of as vectors which define a star relative to the polyhedron's
center. We use that star to create the zonohedron which we call the zonohedrification
of the original polyhedron. As a result, for any two vertices of the original
polyhedron, there are two opposite planes of the zonohedrification which
each have two edges parallel to the vertex vectors. At left is the 24-zone
solid defined by the vertices of the snub
cube. Note that this zonohedrification
of the snub cube has octahedral symmetry, but is not chiral like the
snub cube. All zonohedra have central symmetry by construction, and so
are not chiral.

**Exercise: **Because there are** **twenty-four 8-sided zonogons
in the zonohedrification shown, there must be twenty-four different planes
through the center of the snub cube which
each contain four vertices. Study the snub cube and find four of its vertices
which lie in a common plane through its center.

**Answer: **These four,
for example. (No other planes through the center contain more than two
vertices, because all the other faces of the zonohedrification are 4-sided.)

Here is a list of zonohedrifications of the platonic and archimedean
solids, categorized according to the symmetry of the starting solid. (In
the tetrahedral case, the resulting zonohedra have octahedral symmetry.)
Each line lists a starting polyhedron and its zonohedrification. In a few
cases, the result is one of the familiar Archimedean solids pointed out
in the exercise above. Some of these zonohedra correspond with photos of
paper models (made by Paul Donchian) in Plate II of Coxeter's *Regular
Polytopes*. For reference, his figure number is given here [in brackets].
All the other ones here have never been constructed before, as far as I
know.

**octahedral symmetry**- three 4-fold axes (same as octahedron), cube [1]
- four 3-fold axes (same as cube), rhombic dodecahedron [2]
- six 2-fold axes (same as cuboctahedron), 6 zone truncated octahedron [3]
- 3-fold and 4-fold axes, 7 zone truncated rhombic dodecahedron [5]
- 2-fold and 4-fold axes, 9 zone truncated cuboctahedron [6]
- 2-fold and 3-fold axes, 10 zones [7]
- 2-fold, 3-fold, and 4-fold axes, 13 zones [12]
**icosahedral symmetry**- six 5-fold axes (same as icosahedron), 6 zone rhombic triacontahedron [4]
- ten 3-fold axes (same as dodecahedron), 10 zone rhombic enneacontahedron [8]
- fifteen 2-fold axes (same as icosidodecahedron), 15 zone truncated icosidodecahedron [13]
- 3-fold and 5-fold axes, 16 zones
- 2-fold and 5-fold axes, 21 zones [14]
- 2-fold and 3-fold axes, 25 zones
- 2-fold, 3-fold, and 5-fold axes, 31 zones (illustrated at top of page)

(At present, the last two crash Mathematica when I try to run them.)

Another method of creating symmetric zonohedra is to choose as the star selected groups of axes of symmetry from a spatial symmetry group (and take unit vectors in those directions). In some cases this gives one of the examples above (if a polyhedron has its vertices sitting on the axes).

One can always construct polyhedra with vertices on these axes at a unit distance from the origin, so the above are each zonohedrifications of some polyhedron, but not usually a standard polyhedron. For example, the rhombic dodecahedron or tetrakishexahedron have vertices on the 3-fold and 4-fold axes of the octahedral symmetry group (but not a unit distance from the origin). A similar polyhedron can be constructed which zonohedrifies into the 7-zoner listed above. Similarly, because the rhombic triacontahedron has its vertices on the 3-fold and 5-fold axes of the icosahedral symmetry group (but not at unit distance), the 16-zoner above is very close to the zonohedrification of the rhombic triacontahedron. It is interesting that the zonohedrification of either the rhombic dodecahedron or triacontahedron is also a truncation of the same solid.

Here is the 24-zone solid based on the truncated
cuboctahedron. Its 552 faces are all parallelegrams. This appears to
be the same as the biggest paper model [15] illustrated in *Regular Polytopes*
but it is hard to be certain as the book does not describe the models precisely.
(In some cases two similar zonohedra may have only slightly differing stars,
e.g., the above-listed 12-zone zonohedrifications of the truncated tetrahedron
and the rhombicuboctahedron. To see the difference in their underlying
stars, look at the squares of vertices in the compound
of the truncated tetrahedron with its reflection.)

By the way, I made up the word *zonohedrification,* but its
not such a bad word.

**Hint: **icosahedron = pentagonal
antiprism + 2 pentagonal
pyramid

**Answer: **rhombic icosahedron

The algorithm which I used for constructing these large zonohedra is described in a paper which I can provide to the interested reader. Send me e-mail if you would like a copy.