MATHEMATICS AND ORIGAMI

Mathematics and Origami

The species ε according to Euler is:

3

2

12 2 1 12 30

2

=

× + × −

=

+ −

=

Ce EV A

ε

To resume: polyhedron nº 3 has:

ε = 3 ; C = 12 ; e = 2 ; V = 12 ; E = 1 ; A = 30

What means that we have got a dodecahedron (C = 12) with stellate faces (e = 2) and

convex pentahedral angles (E = 1); its species is ε = 3.

We shall raise now the following exercise: Let ́s consider polyhedron nº 3 not as a stel-

late one, but as a mere irregular polyhedron in order to apply to it Euler ́s basic theorem. In that

case we shall have:

Faces: C = 5 × 12 = 60

Vertices: we shall add to the 12 extreme points, the vertices of the 12 pyramid ́s bases,

i.e. the 20 vertices of the auxiliary dodecahedron. V = 12 + 20 = 32.

Sides: The stellate polyhedron had 30 sides, but in the polyhedron we are considering

now, each one of them produces 3: one on the base of the pyramid and two lateral sides of an-

other two pyramids, hence: A = 30 × 3 = 90

Applying Euler ́s basic theorem we have:

C + V = A + 2 ; 60 + 32 = 90 + 2

18.12.4 STELLATE REGULAR POLYHEDRON nº 4

The starting auxiliary polyhedron is the icosahedron shown in Fig. 1 as a wirework ver-

sion. In it we can see the outlined triangular face ABC that is homothetic to triangle A ́B ́C ́

(dashed line). The sides of this triangle are the diagonals of the pentagons that in turn have as

side the icosahedron ́s side L.

That homothety with center at H, is segregated in Fig. 2 where one can see how seg-

ments AA ́; BB ́; CC ́ are also sides of the icosahedron.

Consequently, to construct the wanted polyhedron nº 4, all we need is 20 pyramids like

HABC (Fig. 2), to set on the faces of the auxiliary icosahedron.

Let ́s figure out the lateral side HB in the pyramid HABC (Fig. 2):

1 , 618034

́ ́ ́

= =

BC

BC

HB

HB

(Point 18.6.1)

1 , 618034 1 , 618034

HB ́ HB L

HB

+

= = hence: HB = 1,618034 L

H

B

C

1