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Topic: Folded sheet of paper (Read 8447 times) |
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NickH
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Folded sheet of paper
« on: Oct 5th, 2002, 2:10pm » |
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A rectangular sheet of paper is folded so that two diagonally opposite corners come together. The crease thus formed is as long as the longer side of the rectangle. What is the ratio of the longer side of the rectangle to the shorter?
Nick
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| « Last Edit: Nov 3rd, 2002, 1:47am by william wu » |
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Carl_Cox
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Re: NEW PUZZLE: Folded sheet of paper
« Reply #1 on: Oct 5th, 2002, 7:02pm » |
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Here's a picture
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Now, you're asking, if the diagnal is equal to the verticle, what is the ratio of the vertical to the horizontal, yes?
(heh, ascii art rules)
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Jeremy
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Re: NEW PUZZLE: Folded sheet of paper
« Reply #2 on: Oct 5th, 2002, 7:55pm » |
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i think it's more like this :
A _____
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C
well, it's a pretty crappy drawing, but angle CAB is 45 degrees... so what is the ratio of AC/AB...
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ok you know what? that didn't work at all
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| « Last Edit: Oct 5th, 2002, 7:56pm by Jeremy » |
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Carl_Cox
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Re: NEW PUZZLE: Folded sheet of paper
« Reply #3 on: Oct 5th, 2002, 8:16pm » |
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Ah, that makes more sense. Picking up a piece of paper and trying it does to. I am a fool! Fear my foolishness!
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Carl_Cox
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Re: NEW PUZZLE: Folded sheet of paper
« Reply #4 on: Oct 5th, 2002, 8:36pm » |
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and regardless, if you draw a line on the crease, then unfold the paper and draw a line parallell to the crease starting at one corner of the paper, the two lines on the paper will be the same length (since this is rectangular paper) which is to say, will be the same length as the longer side of the paper. Which imples to me that the ratio of longer side to shorter is
<answer>infinite, since this line cannot exist on a finite piece of paper (rules of sum of sides of triangles applies here, I believe)</answer>
Why must I seek confirmation? Is it that I don't trust myself?
Am I right? Or am I wrong?
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NickH
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Re: NEW PUZZLE: Folded sheet of paper
« Reply #5 on: Oct 6th, 2002, 3:50am » |
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I'm not sure I'm reading the diagrams above correctly, but only when the sheet of paper is a square does the crease pass through either corner. Otherwise, the crease will run from one longer side to the other. (Apologies if this is what you meant!)
It's possible to investigate the question empirically. The table below shows: ratio of longer side to shorter side, typical dimensions in inches, approximate length of crease. The answer lies between 1.25 and 1.3!
1.0 10x10 14
1.25 8x10 10 1/4
1.3 8.5x11 10 3/4
2.0 5x10 5 1/2
Nick
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Garzahd
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Re: NEW PUZZLE: Folded sheet of paper
« Reply #6 on: Oct 7th, 2002, 1:13pm » |
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Here's an answer.... don't read if you don't want to be spoiled.
High school geometry and trig collectively save the day.
Draw a rectangle
AB
DC
such that AB<BC. Say that AB=1 and BC=k, so k is the ratio we want to find. (k>1)
Draw segment BD, mark its midpoint E.
Now draw a segment FEG, such that F is on AD, G is on BC, and FEG is perpendicular to BD. This segment is the fold line. Our goal is to find k such that k=BC=FG.
Convince yourself that angles ABD, BGF, BDC, and GFD are all congruent. Call that angle X. (shorter to type than theta)
Add point H on DF such that GC=DH. Draw GH (parallel to CD).
Now angle FGH is (90-X), and by applying SOHCAHTOA to triangle FGH we get
cos(90-X) = 1/FG which reduces to sin(X)=1/k.
And we knew from our original diagram that
tan(BDC=X) = k. And since tan=sin/cos, our equations are:
sin(X)=1/k
sin(X)/cos(X)=k
Substituting, 1/cos(X) = k^2, so cos(X) = 1/(k^2).
All the trig conveniently vanishes when we use sin^2+cos^2=1:
(1/k)^2 + (1/(k^2))^2 = 1
k^2 + 1 = k^4
k^4 - k^2 - 1 = 0
Now the old quadratic formula gives k^2 = (1+sqrt(5))/2 or (1-sqrt(5))/2. The second one is negative, so k must be:
sqrt((1+sqrt(5))/2), which is about 1.272.
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NickH
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Re: NEW PUZZLE: Folded sheet of paper
« Reply #7 on: Oct 8th, 2002, 11:33am » |
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That's the answer I got. Here's an alternative solution.
Let the sheet of paper have sides a, b, where a <= b. Let x be the length of the fold and d the length of the diagonal.
Draw a line between the two corners used to make the fold. It's clear by symmetry that this line intersects the fold at right angles. Further, also by symmetry, both lines meet at the center of the rectangle, and bisect each other.
By similar triangles, we have a/b = (x/2)/(d/2) = x/d. Therefore x = (a/b) sqrt(a^2 + b^2).
If x = b, then we soon reach b/a = sqrt[(1 + sqrt(5)) / 2].
Nick
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| « Last Edit: Oct 8th, 2002, 11:35am by NickH » |
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Jonathan_the_Red
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Re: NEW PUZZLE: Folded sheet of paper
« Reply #8 on: Oct 9th, 2002, 11:23am » |
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Ah, our old friend phi (aka the Golden Ratio) makes another appearance.
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