MATHEMATICS AND ORIGAMI

Jesús de la Peña Hernández

Let ́s recall what is the division of a segment in media and extreme ratio.

When that division takes place, it happens that the ratio of the total segment to the great sub-

segment is the same as the ratio of the latter to the small subsegment. It ́s shown in Fig. 1

Fig. 2 shows how to get an auric rectangle through folding. Its small side is taken for the great

subsegment. We ́ll begin with a paper strip of adequate dimensions. Folding process and results

are as follows:

1- a ́ → a. We get E.

2- D → b; E → E. We get O.

3- Fold over AO.

Till now we have constructed ∆AOD which is the same in Fig. 1: To get total segment AB

(larger side of the auric rectangle) OB should be added to hypotenuse AO, OB been equal to

small leg OD.

4- A → a; O → O. We get B ́.

5- a → a; B ́ → B ́. We get B: AB is the greater side of the auric rectangle.

6- B ́ → a ́; B → B. We get C.

The sides of the auric rectangle are AB (great) and AD (small); subsegments are AC y CB.

Since Fig. 2 gets by folding the segments of Fig. 1, we ́ll have:

Total segment AB is divided into the great CB

and the small AC, in such a way that:

AC

CB

CB

AB

=

Or:

CB^2 =AB×AC

As CB = AD, we can write:

AD^2 =AB×AC

which is true as a consequence of the power of

point A with respect to the circumference of

center O and radius OD.

a

a ́

b

A C B

O B ́

D

E

1

2

3

4

5

2 6

A D

O

B

C

1