MATHEMATICS AND ORIGAMI

Jesús de la Peña Hernández

7. In 4 appears
   12

2

6

1

a= =. In said square 4 fold its right lower vertex to lie on the upper side,

while its lower side lies on

12

2

. This way we get
11

1

x=.

8. Let ́s recapitulate the x values obtained till now:
   Square nº x

6

7

1

5

6

1

4

11

1

9. With a similar procedure we would get:
   Square nº x

3

10

1

2

19

1

1

37

1

10. The whole process is contained in the following table that exhibits the successive a / x val-
    ues:

a x

2

1

↓

8

2

4

1

=

7

1

7

1

↓

7

2

6

1

12

2

6

1

=

11

1

11

1

↓

11

2

10

1

REMARK.- One could ask: If the end of the reverse way is al-

ways

2

1

, which is the difference, if any, from one process to

another to arrive in the direct way from that

2

1

to the possible

different fractions given as solution?

The answer is in the different possible combinations to

get a fraction double or half of one already obtained.

You may compare the direct way shown to reach

37

1

with the other we need to get

27

1

: the latter is shown in the

following table at right.