MATHEMATICS AND ORIGAMI

Mathematics and Origami

In the argentic rectangle of fig. 2, and because of its own definition, angles α and γ will ap-

pear in the four corresponding right triangles. That is, in the straight angle in O containing the

media parallel of the rectangle, are occupied 2γ degrees.

Let ́s figure out the difference d = 180 – 2 γ

Looking at Fig. 1 we have:

360 = 10 γ ; 180 = 5 γ

so it is:

d= 5 γ− 2 γ= 3 γ

What means that around the center of the rectangle we can trace 10 γ angles by pleat

folding of the four we began with.

Let ́s see now what is in Fig. 3. Broken line AOBC determines in O and B the β angles

of a pentagon whose diagonal is OC (Ang. OBC = Ang. BOF = 3γ).

Vertex D is the symmetric of C with respect to the bisector of ε, and E is the symmetric

of D with respect to the bisector of Ang. OBC.

Fig. 4 shows what the resultant stellate pentagon looks like. We may note that the center

O of the rectangle becomes one of the points of the stellate pentagon.

11.2 STELLATE HEPTAGON

Begin with the convex heptagon of Fig. 1 making the partial folds of its diagonals. To

fold-stellate that polygon we would have to be able to gather the paper of triangles like KMN.

Since that is not possible in a flat single paper figure, we have to yield some gatherable paper

by reducing the size of the given heptagon while producing a twist on it.

To produce that torsion we need first to build the small heptagon of side AB (Fig. 2), in

this way:

• Fold over each side:
  
  
  • Its perpendicular bisector.
  
  
  • The perpendicular bisector of half a side (in the figure, anticlockwise sequence).
  
  
  
  


• The successive intersections of those perpendicular bisectors produce points like A,
  B and, in consequence, the complete heptagon.

234

F

O

E

B

C

B

C

D

A A

O

F