MATHEMATICS AND ORIGAMI

Mathematics and Origami

6 So we get EH = EG + 1 = 2EF + 1 = 5. With successive folds similar to those of step 5, I

and J are obtained. Result is EI = 5 + 2 y EJ = 5 + 3

7 To get O: K → LP; M → M

8 LMNO is the wanted rectangle because LM = EI = 2 + 5 ; MO = MK = EJ = 3 + 5.

That makes: LM = 4,236068 ; MO = 5,236068 ; MN = 3,0776835

9 Discard lower and lateral correspondent strips.

10 NOTE: the rectangle so obtained is a must to get a perfect convex regular pentagon.

6.6.8 3 : 1

In ∆DBC, it is: BC = 1 −( 1 / 2 )^2 =

2

3

In rectangle ABCD:

2

3

BC= ;

2

1

DC= , therefore = 3

DC

BC

: 1

6.6.9 AURIC RECTANGLE

It is that in which the greater side is divided by the smaller one in media and extreme ratio, i.e.,

the small side is the geometric mean of the length of the greater side and the difference in

length between both sides.

A MZU LKB

G H I J Y

X E F

N

D

O

P C

D C

B

D C

A B