Can you divide one square paper into five equal squares? You have a scissor and glue. You can measure and cut and then attach as well. Only condition is You can't waste any paper.

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    $\begingroup$ This question is hardly appropriate for MO, despite that AnonymousQQJ's answer is actually informative. I'm voting to close. $\endgroup$ – José Figueroa-O'Farrill Feb 13 '10 at 14:36
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    $\begingroup$ Jose: it's not immediately obvious that this is possible with finitely many cuts is it? One has to think a little, which, to me, means the question is one of those "here's the question here's the answer now everyone moves on" questions rather than one of those "here's the question but any mathematician who looks at it instantly recognises that it's trivial so it gets closed" questions. $\endgroup$ – Kevin Buzzard Feb 13 '10 at 17:31
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    $\begingroup$ @Kevin: I agree with your assessment, perhaps I should have written not that the question was inappropriate, just inappropriately posed. I think we ought to discourage questions which are unmotivated or not properly contextualised. This one has attracted some interesting answers, but in general such questions do not. $\endgroup$ – José Figueroa-O'Farrill Feb 13 '10 at 18:31
  • $\begingroup$ Someone on math.stackexchange asked the converse. My answer there (read backwards) gives a concrete answer to this question: math.stackexchange.com/questions/96776/… $\endgroup$ – Bill Cook Jan 11 '12 at 0:52

The Wallace-Bolyai-Gerwien Theorem theorem says:

Any two simple polygons of equal area are equidecomposable

(where simple means no self intersections and equidecomposable means finitely cut and glued).

For your problem you can take the first polygon to be a unit square and the second to be a sqrt(5) by 1/sqrt(5) rectangle and apply this theorem. Then perform the remaining four cuts.

Also, the generalisation of your question is the 2d analogue of Hilbert's 3rd Problem which asks whether given any two polyhedra with equal volume can one be finitely cut and glued into the other. The answer here, unlike in the 2d case, is "no" which was proved by Dehn using Dehn invariants in 1900.

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    $\begingroup$ More information about geometric dissections and equidecomposability can be found in: Boltyanskii, V., Hilbert's Third Problem, Wiley, New York, 1978. Boltyanskii, V., Equivalent and Equidecomposable Figures, D. C. Heath, Boston, 1963. Frederickson, G., Dissections: Plane and Fancy, Cambridge U. Press, New York, 1997. $\endgroup$ – Joseph Malkevitch Feb 13 '10 at 14:56

Cut from (0,0) to (1,1/2), and from (0,1/2) to (1,1). We can glue these three pieces together to get a ring with circumference $\sqrt 5$ and height $\sqrt 5 / 5$. Now it's easy!

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    $\begingroup$ That's really neat! $\endgroup$ – Alon Amit Feb 13 '10 at 17:03
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    $\begingroup$ That is a very simple and elegant solution. $\endgroup$ – Darsh Ranjan Feb 13 '10 at 17:47

Since $1+2i$ has length $\sqrt5$, you can lift a square fundamental domain of $\mathbb C/\mathbb Z[i]$ to $\mathbb C/(1+2i)\mathbb Z[i]$. Overlay a square fundamental domain for the larger torus to get a way to divide a square into 5 smaller squares.

It's pretty easy to decompose any rectangle into a square geometrically, but the general decomposition is not as nice.

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    $\begingroup$ I think this is the best answer yet. It is basically computation-free! $\endgroup$ – Darsh Ranjan Feb 14 '10 at 6:10
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    $\begingroup$ With one choice of fundamental domain, it can be the same as TonyK's construction. This shows why that type of construction works for divisions into 2, 4, 5, 8, 9, 10, or 61 squares, since these are norms of Gaussian integers, but doesn't adapt easily to divisions into 3, 6, 7, 11, or 91 squares, although other constructions work then. $\endgroup$ – Douglas Zare Feb 14 '10 at 17:15
  • $\begingroup$ Very nice answer! Will have to remember this next time I teach number theory. $\endgroup$ – Cam McLeman Aug 24 '10 at 12:49

protected by Lucia Apr 14 '15 at 20:30

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