This is in continuation of
Origamic II.
A sheet of paper has the exact shape of a rectangle (denoted by ABCD) where the length of AB is greater than or equal to the length of AD. 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 so that, the length of the resulting crease is equal to AB.
Is ABCD always a square? If so, prove it - otherwise, give a counter example.
Clearly if ABCD is a square, then the length
of the second fold will equal the length of AB.
But, there exists another rectangle ABCD such
that the length of the second fold equals the
length of AB.
The fold EF is perpendicular to AC and passes
through its midpoint X. The second fold GH
( G on AB and H on AD ) is perpendicular to AF
and passes through its midpoint Y.
Let |AB| = x and |AD| = 1.
Right triangles ABC and AXE are similar,
therefore,
|EX| |EX| |CB| 1
--------------- = ------ = ------ = ---
sqrt(x^2+1)/2 |AX| |AB| x
or
|EX| = sqrt(x^2+1)/(2x).
|AE| |AE| |AC|
--------------- = ------ = ------
sqrt(x^2+1)/2 |AX| |AB|
sqrt(x^2+1)
= -------------
x
or
|AE| = (x^2+1)/(2x).
Right triangles AXE, AXF, and CXF are concruent,
therefore,
|CF| = |AF| = |AE| = (x^2+1)/(2x) and
|FD| = |CD|-|CF| = x-(x^2+1)/(2x)
= (x^2-1)/(2x).
Right tringles ADF, AYH, and GAH are similar,
therefore,
|YH| |YH| |DF|
-------------- = ------ = ------
(x^2+1)/(4x) |AY| |AD|
(x^2-1)/(2x)
= --------------
1
or
|YH| = (x^4-1)/(8x^2).
|AH| |AH| |AF|
-------------- = ------ = ------
(x^2+1)/(4x) |AY| |AD|
(x^2+1)/(2x)
= --------------
1
or
|AH| = (x^2+1)^2/(8x^2).
(x^2+1)^2/(8x^2) |AH| |YH|
------------------ = ------ = ------
x |GH| |AH|
(x^4-1)/(8x^2)
= ------------------
(x^2+1)^2/(8x^2)
or
(x^2+1)^3 = 8x^3(x^2-1). (*)
Equation (*) has four complex roots, one
real root which does not have H on AD,
and
x ~= 1.510577.
This agrees with Geometer's Sketchpad.
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Posted by Bractals
on 2012-10-08 20:09:52 |
Solution
