MATHEMATICS AND ORIGAMI

Jesús de la Peña Hernández

9.15 A SQUARE IN 9 EQUAL PARTS (two solutions)

An approximate solution

It is based in the assumption that CB is divided in segments CE, EG and GB keeping propor-

tions 4,2,3 respectively, that is, adding up to 9:

4 2 3 9

CE EG GB CB

= = =

Of this, the only certain is that GB CG

2

1

= :

∆ACG and GBD are similar; the sides in the small triangle are half the size of their homolo-

gous in the big one (BD AC

2

1

= ). Therefore it will be also GB GC

2

1

=.

This means that folding CG in halves, CB will be divided into three parts equal to GB.

Hence, of the four ratios made equal above, the only exact is the last proportion, which, for one

unit side, ends up to be:

0. 124226

3 9

= =

GB CB

We ́ll see later that

0. 1252249

4

=

CE

and 0. 122228

2

=

EG

which makes evident the inexactitude.

Justification:

α = arc tang 2 = 63.434949º ; α + 30 = 93.434949º ; α + 45 = 108.43495º

CH = AC tang 30 = tang 30 = 0.5773502

In ∆CEH:

sen 60 sen()+ 30

=

α

CE CH

;

sen() 30

sen 60

+

=

α

CE CH = 0.5008998

In ∆GBD:

sen 45 sen() 180 − − 45

=

α

GB BD

;

()

0. 372678

sen 45

sen 45

2

1

=

+

= ×

α

GB

On the other hand,

1. 118034

2

5

4

1

CB= 1 + = =

EG = CB – CE – GB = 0.2444561

The supposed equalities

4 2 3 9

CE EG GB CB

= = = have these real values:

A C

D

H

B

= = A = = C

H

B D

A = = C

H

B D

E F

G I

45

F

H

E

C

(^3060)
30