MATHEMATICS AND ORIGAMI

Mathematics and Origami

7- OV. Getting F; OF= 4

8- A → OF; O → O. Getting X

9- AO → AO; X → X. ,, : U

10- OU. Getting H; OH= 5

Justification of procedure 2 (Fig. 3)

• Let ́s go back from end to beginning.

In ∆AOH,

senγ

OA

OH= ; senγ = 1 −cos^2 γ

In ∆ION, IN IJ

OI

IN

cosγ= = = 1 −

To reach the end we should have to get the successive values BZ, ZW, WY, YV, VX, XU,

UI, IJ, etc. as functions of the unit (width of the strip). Let ́s see how to do it.

• As OI = OX = OY = OZ = OA = 1, we can draw a circumference (whose first quadrant is
  shown) with center O and radius OA.


• Power of B with respect to the circumference at O:

12 =BZ(BZ+ 2 ) ; BZ^2 + 2 BZ− 1 = 0 ; BZ 2 1

2

2 4 4

= −

− ± +

= (discarding

the negative value of BZ).

In the isosceles rectangled triangle BZW we have:

WZ

2

1

1

2

2 1

2

= −

−

= =

BZ

• Power of W with respect to the same circumference:
  WZ(WK+KZ)=WY(WO+OY)

()() 1 () 1 1 1

2

1

(^1)  + − = + +





− WZ WY WY
2 2
2
1
1
2
1
1 = WY+WY





 +





−
Y
3
O
X V
a
Z
W
ABDFH
U
K
L
M
N
I J