MATHEMATICS AND ORIGAMI

Jesús de la Peña Hernández

9.17 DIVISION OF A PAPER STRIP (Fujimoto ́s method)

It deals with a system of successive approximations, which allows dividing the starting

object in 3, 5, 7, etc equal parts through a quick, precise and practical convergence.

For the sake of simplicity we shall show only two examples: the division in 3 and 5

parts respectively, by folding a strip of paper.

Something similar could be done to divide an angle if we start up with a wider paper

surface. In the case of a paper strip the little side of the strip is taken to lie each time on the

previous fold. If we deal with angles, it is a ray of the angle what is revolved over its vertex to

lie on the previous fold.

If a paper strip is folded from end to end, we get its two halves. Repeating the operation

over one of these halves, the

4

1

strip segment will appear. Continuing the same way, we ́ll see

how the segments

8

1

;

16

1

, etc. show up. That is, fractions such as:

2

1

;

4

1

;

8

1

;

16

1

;

32

1

; ... ; n

2

1

n being the number of folds produced.

Fujimoto ́s method is an application of the n

2

1

procedure, to the division in an odd

number of parts. To divide in three equal parts we ́ll make n = 1, i.e. the folds will take place

each time in simple halves. On the contrary, to divide in 5 equal parts we ́ll make n = 2: each

fold will be made double each time.

9.17.1 DIVISION IN THREE EQUAL PARTS

Before commencing, I should like to remind the reader about the division of a segment

in three equal parts such as was treated in Point 8.2.8.3

Let AB = 1; first fold is produced at C distant x from A; x may be of any measure

though in practice, and in order to get a quicker convergence it should be as near as possible to

1 / 3. Nevertheless this is not an indispensable condition for the method by itself is highly con-

vergent. In fact, fig 1, displays x strikingly smaller than 1 / 3 just to show how well the method

works. Therefore it is:

AC=x ; AB= 1 ; CB= 1 −x

Fold B over C to get D in such a way that:

2

1

2

CB x

BD CD

−

= = =

C B

1

A

x

2 A C D B