MATHEMATICS AND ORIGAMI

Jesús de la Peña Hernández

Vertex ρω

i – 1 









−

6

cos^2

i π

a ()

6

1 4

π

i− +

i 









−

6

cos^1

i π

a ()

6

4

π

i+

Substituting in (1) the last four values of ρ,ω, we ́ll have:











− 

6

cos^2

i π

a

()+ 3 π 6

=

mi

ke (2)











−

6

cos^1

i π

a

()+ 4 π 6

=

mi

ke (3)

Dividing (2) by (3):

6

6

ln cos

π

π













m= = - 0.2747161 (4)

Taking (4) to (3) and assigning any value to i (e.g. i = 1), we get the other constant k:

(^62). 052801
5 lncos
= = ×
−  
k e a
π
(5)
Once we know the values of m, k in (1) we may verify that equation in its particular ap-
plication to Figs. 1,2,3. First one shows the series of tangents to the spiral; the second is the spi-
ral as envelope, and the third one is the folding process that will be as follows.

1. To produce the three diametrical folds of the hexagon (Fig. 1).


2. To fold each radii to lie on their contiguous one to get the six apothems (Fig. 1).


3. OB → OB ; A → A
   Result: AC.


4. OD → OD ; C → C
   Result: CE.

1

2

3

4

1

a a

3

2

4

1

2

a

B

A

3

C

E

D H

G I

O

F