MATHEMATICS AND ORIGAMI

Jesús de la Peña Hernández

JARITA”, the bulletín of the AEP entitled Introduction to creation). On the contrary, we are

going to recall some of its geometric grounds.

Each side of the triangle in fig. 1 has been divided in two parts by projecting incenter I

over those sides.

Thus:

AB = a + b

BC = b + c

CA = c + a By subtracting the two last expressions:

BC – CA = b – a

so:

b + a = AB

(1)

b – a = BC - CA

To fold a certain figure, say a quadruped, a previous task is to get an adequate folding

base. There exist many of them as “pre-designed”, but sometimes we will need to build one of

our own according to specific requirements.

A folding base is, at the end, a triangulation of the starting paper (usually a square), by

means of mountain and valley folds.

The requirements mentioned above may refer to figure tips, keeping due proportions:

tail, legs, snout, ears, horns, etc.

Thus, what we really know is the pair of segments a, b in side AB (fig. 1) rather than the

two other sides CA, CB; a / b must keep the tips proportion already mentioned.

Therefore, in ∆ABC we only know a, b and the expressions of system (1), i.e., side AB

and the fact that vertex C is situated on the hyperbola with focuses A and B, because the dis-

tance difference from C to B and A is a constant with the known value of b – a (fig. 2).

Of course, the hyperbolas ́s branch will pass through V (its vertex) for VB - VA = b – a.

Its center O, is the mid-point of AB.

To draw the hyperbola in the Cartesian plane OXY we ́ll start up with its equation

2 1

2

2

2

− =

b

y

a

x

(2)

such as (a ≠ a; b ≠ b): a = VO ; b = VD ; c = AO = OD

so it ́ll be c^2 =a^2 +b^2

The only thing left is to assign values to x in (2) to get the corresponding values of y

(^1) b (^2) x (^2) a (^2) b 2
a
y= −
A
B
1
C
I
B
2
A
I
C
V
O
D
X
Y
C 1
C 3
C 2
a
b
a
c
c
b