Mathematics and Origami

HI=EF ;

16

1

32

2 2

GH= 2 HI= =

In ∆HAJ: ang H = 45º ; ang A = 22. 5 º

2

45

= ; ang J = 180 – 45 – 22.5 = 112.5º

sen 112. 5 sen 22. 5

16

1

(^1) HJ

−
; HJ = 0.3883252
cos 45 0. 3370873
16
1
GK=GH+HK= +HJ =
inexact result: 0. 33333 .....
3
1

9.6 A SQUARE IN THREE EQUAL PARTS (Haga ́s theorem)
Beginning with a square with side one unit, the result is:
3
1
EG= ;
3
2
EF=
Justification:
FH= 1 ; angABH=Arctg 2 ;
2
1
angDBA=angBAD=Arctg
2
1
angHBD=Arctg 2 −Arctg ; angFEB=angHBD (perpendicular sides)
in ∆FEB:
FE
FEB
×

2
1
tg ;
3
2
2
1
2 tg tg 2 tg
1







−

Arc Arc
FE ;
3
1
3
2
EG= 1 − =
G H
I
K A
C B
D
F
E
J
AB= 1 ; AC= 2 ;
16
2
CD=
= − = − = −()+ = −
2
2
2
2
2
2
EF AF AE ED EF FD
() 






− + − = − + −
16
2
2
2
2
2
EF FC CD EF
EF= −EF
16
2
;
32
2
EF=
A
B
C
C
D D
A
E E
B F H
G
= =