This is in continuation of
Origamic.
A sheet of paper has the exact shape of a rectangle (denoted by ABCD) where AB is the longer side and AD is the shorter side. The vertex A is folded onto the vertex C, resulting in the crease EF (E on AB and F on CD).
The paper is thereafter unfolded and, the vertex A is folded onto F, resulting in the crease KJ.
Determine separately the ratio of the longer side (AB) of the rectangle to the shorter side (AD), whenever:
(i) K coincides with D and, J is on AE.
(ii) J coincides with B and, K is on AD.
(iii) J coincides with E and, K is on AD.
Let a coordinate system be applied such that
the points have the following coordinates:
A(0,0), B(b,0), C(b,d), D(0d),
E(e,0), F(fjd), J(j,0), and K(0,k)
where b, d, e, f, j, and k are real numbers
greater than zero and b > d > 0.
The creases EF anf JK coincide with the
perpendicular bisectors of line segments
AC an AF respectively.
EF: y = (b/d)x + q
d/2 = (b/d)(b/2) + q
y = (b/d)x + (b^2 + d^2)/(2d)
E: 0 = (b/d)e + (b^2 + d^2)/(2d)
e = (b^2 + d^2)/(2b)
F: d = (b/d)f + (b^2 + d^2)/(2d)
f = (b^2  d^2)/(2b)
JK: y = (f/d)x + p
d/2 = (f/d)(f/2) + p
y = (f/d)x + (d^2 + f^2)/(2d)
J: 0 = (f/d)j + (d^2 + f^2)/(2d)
j = (d^2 + f^2)/(2f)
= (b^2 + d^2)^2/[4b(b^2  d^2)]
K: k = (f/d)(0) + (d^2 + f^2)/(2d)
= (d^2 + f^2)/(2d)

i) K = D > k = d
> (d^2 + f^2)/(2d) = d
> d = f
> d = (b^2  d^2)/(2b)
> 0 = b^2  2db  d^2
> b/d = 1 + sqrt(2)
~= 2.4142
ii) J = B > j = b
> (b^2 + d^2)^2/[4b(b^2  d^2)] = b
> 0 = 3b^4  6d^2b^2 = d^4
> b/d = sqrt(1 + 2*sqrt(3)/3)
~= 1.4679
iii) J = E > j = e
> (b^2 + d^2)^2/[4b(b^2  d^2)]
= (b^2 + d^2)/(2b)
> b/d = sqrt(3)
~= 1.7321

QED

Posted by Bractals
on 20120318 13:24:13 