How many sides might a deltahedron have ? Clearly four is the minimum number of faces, giving the tetrahedron. Because no vertex can have more than five equilateral triangles meeting, the largest convex equilateral deltahedron is the icosahedron, giving twenty as the maximum number of triangles. It is also not hard to show that any deltahedron has an even number of sides.

**Exercise:** *Show that any deltahedron has an even number of
faces.*

**Answer: ***Answer.*

The above arguments eliminate all numbers except the even numbers from 4 to 20. It turns out, when one examines the possibilities, that for all of these but one there is a convex deltahedron with that many faces. Various triangular construction toys are ideal for working on this problem, or you can search among the list of Johnson solids.

**Exercise:** *For which of these numbers, n, is it impossible
to find a convex equilateral deltahedron with n faces: 4, 6, 8, 10, 12,
14, 16, 18, 20 ?*

**Answer:** *Answer.*

The one with twelve sides is the least obvious. It was called the *Siamese
dodecahedron* in the original reference
by Freudenthal and van der Waerden. It can be thought of as a tetrahedron
split into two wedges of two adjacent faces, which are then joined together
by a band of eight triangles. Accordingly, Norman Johnson named it the
snub disphenoid.

Here is a list of all eight.

Note that if we do not restrict ourselves to the convex case, there are an infinite number of deltahedra. For example, glue two icosahedra together face to face, or replace each pentagon of a dodecahedron with a dimple of five equilateral triangles. A nice example, based on an octahedron with its faces divided into 16 sub-triangles, is this 128-hedron sent to me by John Futhey.