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Topic: Bisect Rectangle by Folding (Read 1029 times) |
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Barukh
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Bisect Rectangle by Folding
« on: Oct 11th, 2007, 10:33am » |
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You are given a rectangular sheet of paper with arbitrary dimensions. You need to fold a border of constant width such that the inner rectangle has an area exactly half that of the sheet.
You have no ruler or compass, not even a pencil. You must determine the inner rectangle purely by paper folding.
Source: Colin Singleton.
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towr
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Re: Bisect Rectangle by Folding
« Reply #1 on: Oct 11th, 2007, 11:18am » |
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Can you use folds/sides of the rectangle to 'measure'? Or are you only allowed to make fold by putting one defined point onto the other? (i.e. where folds/sides cross)
The width we want is one fourth of the sum of the sides minus the diagonal
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Barukh
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Re: Bisect Rectangle by Folding
« Reply #2 on: Oct 11th, 2007, 12:15pm » |
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on Oct 11th, 2007, 11:18am, towr wrote:| Can you use folds/sides of the rectangle to 'measure'? Or are you only allowed to make fold by putting one defined point onto the other? (i.e. where folds/sides cross) |
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The latter.
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FiBsTeR
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Re: Bisect Rectangle by Folding  BRF.bmp
« Reply #3 on: Oct 11th, 2007, 4:44pm » |
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Is this kind of a fold allowed?
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Barukh
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Re: Bisect Rectangle by Folding
« Reply #4 on: Oct 12th, 2007, 12:16am » |
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on Oct 11th, 2007, 4:44pm, FiBsTeR wrote:| Is this kind of a fold allowed? |
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Yes.
You may want to refer to the following page which under "Huzita's Origami Axioms" describes what operations are allowed. Your operation appears as O3.
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FiBsTeR
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Re: Bisect Rectangle by Folding
« Reply #5 on: Oct 12th, 2007, 4:30am » |
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The problem isn't as restricted as I first thought, then.
on Oct 12th, 2007, 12:16am, Barukh wrote:
Quote:| This should seem amazing - origami actually allows us to do simple calculus! |
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Quote:| Moral of the exercise: Axiom (O6) can solve cubic equations. This is something that SE&C cannot do. Further, the problems of trisecting angles and doubling cubes all reduce to solving cubic equations. Thus axiom (O6) should allow us to solve these problems via origami. Indeed, it can! |
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| « Last Edit: Oct 12th, 2007, 4:34am by FiBsTeR » |
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FiBsTeR
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I'm in class now so I'll post the figures later; most of this is coming from my head, so I could've easily transposed a few points:
For some reason the areas in Sketchpad don't agree with the calculations, but close enough.
Given a rectangle with vertices ABCD labeled from the top-left clockwise:
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Fold A to B, and label the intersection of the crease and AB point E. Fold E to B and label the crease line m and label the intersection of m and AB point F. FB = (1/4)AB and m is perpendicular to AB.
Fold C to B, and label the intersection of the crease and BC point G. Fold G to B and label the crease line n and label the intersection of n and BC point H. BH = (1/4)BC and n is perpendicular to BC.
Label the intersection of m and n point I. Fold I onto AF such that the crease line intersects F (O5). Label the image of I on AB point J. BH = IF = JF, and JB = JF + FB = (AB+BC)/4
Fold FH onto n, and label the image of F on n as K. Fold KH onto AB, and label the image of K on AB as L. Then:
JL = JB - LB
= (AB+BC)/4 - KH
= (AB+BC)/4 - FH
= (AB+BC)/4 - (sqrt(AB2+BC2))/4
= (AB+BC-(sqrt(AB2+BC2)))/4
Fold a crease line perpendicular to AB through L (O4), and label the image of J on AB point M. JM = 2JL. Fold a crease line perpendicular to AB through J (O4), and label this line q. Fold M ontoq such that the crease intersects J, and label the image of M on q point N. JN = JM = 2JL.
Fold J to N, and the resulting crease line will create a border of width JN/2 = JL.
Creating this border similarly on all four sides yields an internal rectangle of area:
[ Let D = sqrt(x2+y2) ]
(AB - 2JL)(BC - 2JL) = (AB - (AB + BC - D))/2)(BC - (AB + BC - D)/2)
= (1/4)(ABxBC - AB2 + ABxD - BC2 + ABxBC - BCxD + BCxD - ABxD + AB2 + BC2)
= (1/4)(2ABxBC)
(AB - 2JL)(BC - 2JL) = ABxBC/2,
and thus the area of the internal rectangle is half the area of ABCD. |
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| « Last Edit: Oct 12th, 2007, 9:31am by FiBsTeR » |
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Barukh
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Nice solution, FiBsTeR! I hope it did not interrupt you from learning your class material? 
I followed your construction step-by-step. Seems like it may be somewhat simplified. For instance, after folding points I, J, L, make a crease IJ, and mark the image of L as P. Then, just apply O4 on line BC and point P.
Hope you liked the problem.
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FiBsTeR
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Re: Bisect Rectangle by Folding
« Reply #8 on: Oct 13th, 2007, 6:58am » |
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Ah, yes, that simplifies the last folds a bit.
And yes, it was a very interesting problem! 
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