MATHEMATICS AND ORIGAMI

Jesús de la Peña Hernández

HD= OH^2 +OD^2 = 1 , 4142135 L (curiously HD = 2)L

BD=DC= 0 , 618034 L (equal value than AC ́, Point 18.6.3)

With the information gathered till now (the three sides HB, BD, DH of the triangles

forming the lateral faces) we are in the position to construct stellate pyramids with H as vertex

and a stellate pentagonal base. These pyramids can be placed on the faces of a convex dodeca-

hedron that in turn will be used as the auxiliary structure needed to build the wanted stellate

icosahedron.

Fig. 2 shows, besides the triangles with bases DB and DC, and H as upper vertex, the

triangles BD ́H ́ and D ́CH ́ associated to the other facial pentagon having also BC as a side.

We can note in it that those mentioned triangles intersect in the interior of the dodecahedron

taken as auxiliary structure. So we ought to find out the length of segments type DF and FC.

Fig. 3 will help us for the former and Fig 4 for the latter.

To make up Fig. 3 we should draw ∆HOD (right triangle whose legs we know) and

segment OE. Producing HE we get Ang. OEO ́ = 116,56505º (ε in Point 18.6.1). The bisector

of Ang. OEO ́ passes through the center of the dodecahedron, therefore the pyramid in H` will

be the symmetric of pyramid H with respect to the plane formed by the side through E, i.e. BC,

and said center.

Coming back to Fig. 3, ∆EDF is determined because

DE = OE – OD = 0,3632713 L ; Ang. E = 116,56505 / 2 = 58,282525

Ang. D =

OD

OH

arctg = arctg 4,2360692 = 76,717478º

from what we get that Ang. F = 45º and hence Ang. HFH ́ = 90º. Consequently:

sen 58 , 282525 sen 45

DF DE

= ; DF = 0,3632713 L 0 , 437016 L

0 , 7071067

0 , 8506508

=

HF = HD + DF = 1,4142135 L + 0,437016 L = 1,8512296 L

Now we can make up Fig. 4 drawing in first place ∆HDC whose three sides we know:

HD = 2 L ; HC = BH ; DC = BD

Then we shall produce HD to get vertex F (DF being also known). This way we get

∆HFC: with 10 of these triangles we construct the polyhedral angle in H over one face

of the auxiliary dodecahedron. Extending the operation to its 12 faces we get the com-

plete stellate polyhedron we are looking for.