Mathematics and Origami

Solution 2

Beginning as usual, with a square of side 1, let ́s see how much precision we get: it is supposed

that

2

HC

be equal to

7

1

.

Comparing the last figure of present Solution 2 with Fig 2 in Point 4 (demonstration of Haga ́s

theorem), we have:

DE = x ; EB = z ; FG = GI = f

Now it is DE = x = 0.5; in Point 4 we had for this value of x:

z = 0.375 = EB ; f = 0.125 = FG = GI

From these data we can study ∆ABC.

ang ABC = arc tg

EC EB GI z f

IC

− −

=

− − 1

1

= arc tg

1 0. 375 0. 125

1

− −

ang ABC = arc tg 2 = 63.434949º

ang BAC = 180 – ang ABC - 











+

2

45

45 = 49.065051º

sen sen 63. 434949

1 AC

BAC

z

=

−

; AC = () 0. 7399749

sen 49. 065051

sen 63. 434949

1 − 0. 375 =

0. 2831761

2

3

cos (^45) =





CH=AC × ; 0. 141588
2

CH
The reader can see the difference between the last value and 0. 1428571
7
1

= =

=
C C
B
C
B
A A

=

=
A

C
B
DE=
H
G
I
F 45º

=
(^1234)
(^5678)