MATHEMATICS AND ORIGAMI

Mathematics and Origami

Solution 2

Solution 3

CD = 1 ; ∆ACD, equilateral ; ang. ACD = 60º ; ang. ECD = 30º

DE=tg 30 ;

tg 30

1

=

EF

BF

;

2

1

tg 30

2

1

EF=ED− = −

2 tg 30

1

1

tg 30

= = −

EF

BF ; 15 º

tg 30

1

tg tg (^2) =





= =Arc −
FG
BF
angBGF Arc
Solution 4
This solution is not perfect, though very ingenious and close to perfection.
It is a matter of folding M and N, and then, over mid-point O. A perfect solution calls for a
point different from O: an undetermined point between O and M.
AC= 1 ;
2
2
CB=AB= ; angNCB= 22. 5 º ; tg 22. 5
2
2
NB=
D
F B
A
C
b a
c
2
3
C D
b
A
D
a
A
F B
G
E
= H =
1
C
c
Folds:
1- D → GH; B → B. We get a and E.
2- B → GD; a → a. We get C and A.
3- Fold AC. We get simultaneously b
and c.
DB = EB = FB : ∆BEF, equilateral.
Then ang. EBF = 60º and CBD = 15º.
In the last figure, obtained after folding the former, and be-
cause of the symmetry, we have:
Ang. ABF = CBD =15º; ang. CBA = 90 – 30 = 60º. There-
fore ∆ABC is isosceles with an angle of 60º: it is equilateral.
A
C D
F
B E
A
F
B
G