MATHEMATICS AND ORIGAMI

Jesús de la Peña Hernández

The faces are planes concurrent in the vertices, i.e. convex pentagons. These pentagons

have species e = 1 (they are not stellate).

5 of those planes pass through each vertex: 5 × 12 = 60. But as each one of these 60

planes is common to 5 vertices, we end up with 12 faces (60 : 5). C = 12.

Number of sides: 5 per vertex; making a total of 5 × 12 = 60 sides. The sides of the sunk

trihedral must be ignored. But as each side is common to a pair of vertices, the result will be 30

sides (60 : 2). A = 30.

Summary. The polyhedron we have obtained has these features:

C = 12 ; e = 1 ; V = 12 ; E = 2 ; A = 30

The species ε for this polyhedron nº 2, according to Euler ́s general theorem is:

3

2

12 1 2 12 30

2

=

× + × −

=

+ −

=

Ce EV A

ε

Therefore what we have got is a dodecahedron (C = 12); stellate of 3rd species (ε = 3);

with 1st species (e = 1) convex pentagonal faces; having V = 12 stellate pentahedral angles, i.e.

of 2nd species (E = 2).

Fig. 5 is another view of the dome in H to make it clearer.

18.12.3 STELLATE REGULAR POLIHEDRON nº 3

It is similar to nº 1 (Point 18.12.1). The difference consists in the dissimilitude of their

polyhedral angles; though in both cases they are pentahedral, in nº 1 they are stellate whereas in

nº 3 they are convex (E = 1). We start with an auxiliary convex dodecahedron for nº3 as we did

for nº 1.

∆BCH in Fig. 1 is the same as the one equally named in Fig.1, Point 18.12.1. Present

Fig. 1 is the folding diagram of the pentagonal pyramid (only one lap joint is shown) to be set

on each face of the auxiliary dodecahedron; therefore we need 12 of these pyramids to get

polyhedron nº 3.

Fig. 2 is the complete polyhedron nº 3. One can see that it has 12 vertices (V = 12) be-

ing E = 1 as said before.

5 stellate pentagons start from each vertex, i.e. of e = 2.

There are 5 × 12 = 60 faces, each of them common to 5 vertices; therefore they produce

60 / 5 = 12 faces: C = 12.

In turn, from each vertex start 5 sides, each of them common to two vertices, which

gives 5 × 12 / 2 = 30; i.e., A = 30.