Divide a square into 5 equal squares - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T14:11:17Z http://mathoverflow.net/feeds/question/15181 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/15181/divide-a-square-into-5-equal-squares Divide a square into 5 equal squares sanz 2010-02-13T10:43:53Z 2010-02-13T18:00:33Z <p>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.</p> http://mathoverflow.net/questions/15181/divide-a-square-into-5-equal-squares/15183#15183 Answer by TonyK for Divide a square into 5 equal squares TonyK 2010-02-13T12:33:50Z 2010-02-13T12:33:50Z <p>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!</p> http://mathoverflow.net/questions/15181/divide-a-square-into-5-equal-squares/15184#15184 Answer by Q.Q.J. for Divide a square into 5 equal squares Q.Q.J. 2010-02-13T13:15:04Z 2010-02-13T14:07:32Z <p>The Wallace-Bolyai-Gerwien Theorem theorem says: </p> <p>Any two simple polygons of equal area are equidecomposable </p> <p>(where simple means no self intersections and equidecomposable means finitely cut and glued).</p> <p>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.</p> <p>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.</p> http://mathoverflow.net/questions/15181/divide-a-square-into-5-equal-squares/15198#15198 Answer by Douglas Zare for Divide a square into 5 equal squares Douglas Zare 2010-02-13T18:00:33Z 2010-02-13T18:00:33Z <p>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. </p> <p>It's pretty easy to decompose any rectangle into a square geometrically, but the general decomposition is not as nice.</p>