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7

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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.

edited Feb 13 2010 at 17:39

Darsh Ranjan
3,547●21●35

asked Feb 13 2010 at 10:43

sanz
145●1●5

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This question is hardly appropriate for MO, despite that AnonymousQQJ's answer is actually informative. I'm voting to close. – José Figueroa-O'Farrill Feb 13 2010 at 14:36

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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. – Kevin Buzzard Feb 13 2010 at 17:31

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@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. – José Figueroa-O'Farrill Feb 13 2010 at 18:31

Someone on math.stackexchange asked the converse. My answer there (read backwards) gives a concrete answer to this question: math.stackexchange.com/questions/96776/… – Bill Cook Jan 11 2012 at 0:52

Q.Q.J. · Accepted Answer · 2010-02-13 14:07:32Z UTC

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.

TonyK · Answer 2 · 2010-02-13 12:33:50Z UTC

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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!

answered Feb 13 2010 at 12:33

TonyK
1,453●6●11

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That's really neat! – Alon Amit Feb 13 2010 at 17:03

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That is a very simple and elegant solution. – Darsh Ranjan Feb 13 2010 at 17:47

Douglas Zare · Answer 3 · 2010-02-13 18:00:33Z UTC

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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.

answered Feb 13 2010 at 18:00

Douglas Zare
13.2k●3●36●58

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I think this is the best answer yet. It is basically computation-free! – Darsh Ranjan Feb 14 2010 at 6:10

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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. – Douglas Zare Feb 14 2010 at 17:15

Very nice answer! Will have to remember this next time I teach number theory. – Cam McLeman Aug 24 2010 at 12:49

Divide a square into 5 equal squares

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