# dissection of a $1\times 5$ rectangle to a square

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!

• Are you allowed to overlap the pieces? Do you require that the resulting square have the same area as the rectangle? – user2468 Jan 5 '12 at 22:51
• yes... total area of the square should be 5 – Amihai Zivan Jan 5 '12 at 23:04
• The inverse of this question can be found at mathoverflow.net/questions/15181/…. – TonyK Jan 10 '12 at 22:44

You can do it with four pieces, and translations only (no rotations).  • Very optimal. +1! – Patrick Da Silva Jan 6 '12 at 16:24 • 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? – templatetypedef Jan 6 '12 at 2:08
• 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 :) – Bill Cook Jan 6 '12 at 2:50

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):

$\hspace{1.15in}$ 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.

• Googling for Pythagoras' theorem gives a couple of usefull pictures, e.g this one: goo.gl/PuAz7 – Myself Jan 5 '12 at 22:59
• ... or this one: goo.gl/DcLWS – Myself Jan 5 '12 at 23:04
• 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! – Patrick Da Silva Jan 5 '12 at 23:26
• @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. – Gerry Myerson Jan 6 '12 at 12:50
• @Gerry Myerson : You have no idea what can come out of a napkin... =P – Patrick Da Silva Jan 6 '12 at 16:24

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.

• When you copy something from another post, common practice is to acknowledge the source. But thanks for supplying the picture. – Gerry Myerson Jan 5 '12 at 23:17