MATHEMATICS AND ORIGAMI

Mathematics and Origami

This way we have got points on the hyperbola such as C 1 C 2 C3, besides C and V.

The design of a folding base consists of finding wanted point C as the intersection of a

branch of one hyperbola (fig. 2), and another branch obtained similar way.

We must add that VD, perpendicular to AB at V is the locus of the incenters of triangles

with vertex Ci on the hyperbola. That is evident in fig. 1, but is also true for any other triangle

because AB is fixed.

8.2.8.7 FLATTENING OF A QUADRILATERAL

Point 8.2.8.4 showed how to flatten a triangle through its incenter. Now, the association

of incenter and hyperbola (point 8.2.8.6) will allow us to fold flat a quadrilateral under these

conditions:

two of its opposite vertices must be on the same branch of a hyperbola whose

focuses are the other two vertices of said quadrilateral.

Let focuses be the vertices B and C (Fig. 1), O the center of the hyperbola and OV = 2a

(a being the parameter of that hyperbola).

To obtain the other two vertices A, A ́ of the quadrilateral, let ́s draw the circumferences

with centers C, B, and radii CV and BV, respectively. Then, draw arbitrary cicumferences si-

multaneously tangent to the other two circles: their centers A and A ́ will seat on the hyperbola

and therefore are the other pair of wanted vertices of the quadrilateral.

It is so because for any of them (e.g., A ), it is:

AB – AC = VB – VC = 2VO for

()

2 2

BV VC BV VC

OV BV OB BV

−

=

+

= − = −

The quadrilateral ABA ́C so constructed, shows in Fig. 2 the four bisectors of its angles:

they are concurrent in I, the quadrilateral incenter.

Points A y A ́ are not only points of the hyperbola: they also are the tangency points on

it, of lines AI and A ́ I, for they are the bisectors of radius vectors AB, AC; A ́C, A ́ B. On the

other hand, in point 13.3.4 (Poncelet ́s theorem) will be demonstrated that IC is the bisector of

angle A ́CA.

A

B OV C

A ́

1

2

A

C

A ́

B OV

I