Mathematics and Origami

In it,

n

360

2

1

α= ; 2 αis the central angle of that polygon ant the segments x, symmetri-

cal about the perpendicular bisector of L, determine the parallel lines to that perpendicular bi-

sector which in turn will condition the twist to flattening.

Fig. 2 is obtained by folding flat Fig. 1. To accomplish this, x has to be such that PM =

P ́M (Fig. 1). Fig. 3 is the reverse side of Fig. 2.

Then it will be:

cos 2 α

́

x

PM=

Being

2

L

PM= we shall have:

cos 2 2

x L

=

α

n

L

x

360

cos

2

= (1)

The relation between x and l (side of the small central polygon of n sides around which the torsion is
performed), can be found out in Fig. 2 (Point 11.2):

()

n n

n

l

x

L

180

cos

2

180 2

(^2) sen =
−

−
M
P
P ́
L
x x
1
M
P ́
(^23)
M
P
x