2. How many convex Deltahedra possibly exist?

An n-face Deltahedron is a polyhedron with n congruent equilateral triangular faces, conventionally denoted Dn. (The name comes from the Greek capital letter Delta, which has a triangular shape.)

An interesting problem is to find how many such Convex Polyhedra there are.

In trying to exhaust the set of convex Deltahedra, we note, first of all, that this set includes three of the five Regular/Platonic solids:

 D4  - the tetrahedron; D8  - the octahedron; D20 - the icosahedron

Other convex Deltahedra might be obtainable by attaching several triangles side to side in different spatial configurations.

Our question is: How many convex Deltahedra can thus be obtained?

2.1 How many faces can convex Deltahedra possess?

To answer this question we first observe that in each vertex 3, 4, or 5 triangular faces only, can meet. These numbers define the valence of a vertex. Clearly, the valences of all vertices in any given Deltahedron are not necessarily the same. Denote V3, V4, V5 the number of vertices of valence 3, 4, 5 respectively.

Next, note that

(i)   The minimum number of faces, is four, as in the Tetrahedron.

(ii)  Since no vertex can have more than five equilateral triangles meeting in it (as six give a flat 3600 angle sum around the vertex), the largest possible convex Deltahedron is a solid having 5 equilateral triangles meet in each of its vertices, i.e., the Icosahedron.

Thus twenty is the maximum number of faces for any Deltahedron.

Therefore, The total number of faces of any existing convex Deltahedron is in the range between 4 and 20.

2.2 Candidates for existing Deltahedra

First observe that the number of faces in any Deltahedron must be even.

To establish it note that each face has three edges and each edge is shared by two faces, hence the relationship between the number of faces F and the number of edges E, satisfies: .

Substituting this in Euler’s formula for simple Polyhedra:

F(aces)+V(ertices)=E(dges)+2 ,

we get

F=2(V-2)

which implies the evenness of the total number of faces.

Thus,we conclude that there are only 9 candidates for Deltahedra – those having 4, 6, 8, 10, 12, 14, 16, 18 and 20 (triangular) faces.

To determine the various possible Deltahedra, we employ Descarte's formula for the total angular deficit [2] of a convex polyhedron. First observe that:

(i)  The angular deficit of a vertex of valence 3 is 1800

(ii)  The angular deficit of a vertex of valence 4 is 1200

(iii)  The angular deficit of a vertex of valence 5 is 600

Descarte's formula for the total angular deficit of a convex polyhedron yields

1800V+ 1200V + 600V5  = 7200

or

3V3 + 2V + V5  = 12

Every Deltahedron must satisfy this (Diophantine) equation. This is a necessary but yet insufficient condition for the existence of a Deltahedron.

2.3 The solution

Every solution of the above mentioned (Diophantine) equation, which takes into account the evenness of the total number of faces, is a candidate for an existing Deltahedron.  Altogether, there are nineteen possible solutions to the equation. They are listed in Table 1 (Lichtenberg, 1988).

However, the only solutions that correspond to actually existing Deltahedra are 1, 3-7,16, and 19, total of eight convex Deltahedra.

Visual and verbal description of the eight Deltahedra

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[1]   Interestingly, if we do not restrict ourselves to convex deltahedra, there are infinitely many deltahedra. E.g., the stella octangula.

[2]   The Angular Deficit or spherical deviation d - at a vertex of a polyhedron, defined as the difference between 3600 and the sum of the angles of the polygons surrounding the vertex. The total angular deficit taken over each vertex of a convex polyhedron equals 7200