I've been thinking about the following problem:
We have a $1\times 5$ rectangle: how to cut it and reassemble it such that it forms a square?
Thanks a lot!
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asked Jan 5, 2012 at 22:40
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1$\begingroup$ Are you allowed to overlap the pieces? Do you require that the resulting square have the same area as the rectangle? $\endgroup$– user2468Commented Jan 5, 2012 at 22:51
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$\begingroup$ yes... total area of the square should be 5 $\endgroup$– Amihai ZivanCommented Jan 5, 2012 at 23:04
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1$\begingroup$ The inverse of this question can be found at mathoverflow.net/questions/15181/…. $\endgroup$– TonyKCommented Jan 10, 2012 at 22:44
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6 Answers
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You can do it with four pieces, and translations only (no rotations).
answered Jan 5, 2012 at 23:59
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$\begingroup$ While this is pretty impressive, how would you figure this out if you didn't already know the answer? Is there a general technique being employed here, or is this problem-specific? $\endgroup$ Commented Jan 6, 2012 at 2:08
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9$\begingroup$ The area of your original rectangle is 5. A square with the same area must have sides of length $\sqrt{5}$. So thinking of the pythagorean theorem, we need $a^2+b^2=5$. The only positive integer solution is $a,b=1,2$. This led me to chop off a rectangle of size $1\times 2$ and slice it diagonally. It then made sense to do it again and then sliding things around we get the above answer. That's how I thought about it anyway. The hard part was drawing it in Microsoft Paint :) $\endgroup$ Commented Jan 6, 2012 at 2:50
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A standard approach to finding solutions to problems like this is to overlay two tilings. In the following image, the bright yellow rectangles give one 5-piece and two 4-piece solutions requiring only translations (one of which is the same as Robert Israel’s above):
$\hspace{1.15in}$ 
answered Jan 10, 2012 at 21:52
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Here's a method that is surely far from optimal.
First, cut it into 5 $1\times1$ squares, and arrange them into a $2\times2$ square sitting next to a $1\times1$. Now there's a standard way to cut two squares into a total of four pieces that rearrange to form a single square. I wish I knew how to draw pictures. Anyway, let the small square be $ABCD$ With $C$ a vertex of the big square and $CD$ along the side $CEFG$ of the big square. Find $H$ on $CG$ such that $GH=AB$. Cut along $FH$ and along $AH$. Then the bits $FHG$, $ABH$, and $ADEFHA$ can be moved to form a square.
answered Jan 5, 2012 at 22:56
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$\begingroup$ Googling for Pythagoras' theorem gives a couple of usefull pictures, e.g this one: goo.gl/PuAz7 $\endgroup$– MyselfCommented Jan 5, 2012 at 22:59
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$\begingroup$ If you want to draw pictures on MSE, draw them on anything you want as long as you can get them in a file format that allows you to upload them here. There's a button to upload images directly in your answers when you're answering... find it =) personally I am a fan of picturing my blackboard with my Android phone! $\endgroup$ Commented Jan 5, 2012 at 23:26
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$\begingroup$ @Patrick, thanks, but I'm on the road, using whatever computers I can find, and the only thing I can draw on is a napkin. But I'll keep your advice in mind. $\endgroup$ Commented Jan 6, 2012 at 12:50
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$\begingroup$ @Gerry Myerson : You have no idea what can come out of a napkin... =P $\endgroup$ Commented Jan 6, 2012 at 16:24
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Cut a small square out of it and throw the rest away. If you are not allowed to throw any part away, then crumple the rest into a ball and stack it on the square. If you are not allowed to throw any part away and you are not allowed to overlap pieces, then I don't know the answer.
Edit:
Now there's a standard way to cut two squares into a totasl of four peices that rearrange to form a single square. I wish I knew how to draw pictures.
(source: cut-the-knot.org)
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5$\begingroup$ When you copy something from another post, common practice is to acknowledge the source. But thanks for supplying the picture. $\endgroup$ Commented Jan 5, 2012 at 23:17
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We can consider this problem in terms of a tiling of the plane with squares. Or equivalently, we treat the large square as a torus, i.e., its top edge is glued to its bottom edge, and its left edge is glued to its right edge. Thus all the current solutions on this page are essentially equivalent.
Note that all of those right triangles are similar to the triangle with sides $1, 2, \sqrt5$.