You are watching: Divide a square into 5 equal parts

The Wallace-Bolyai-Gerwien theorem theorem says:

Any two an easy polygons of equal area room equidecomposable

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

For your trouble you deserve to take the an initial polygon to be a unit square and the 2nd to it is in a sqrt(5) by 1/sqrt(5) rectangle and also apply this theorem. Then carry out the remaining four cuts.

Also, the generalisation that your concern is the 2d analogue the Hilbert"s third Problem which asks even if it is given any type of two polyhedra with equal volume can one it is in finitely cut and glued into the other. The answer here, unlike in the 2d case, is "no" i m sorry was confirmed by Dehn utilizing Dehn invariants in 1900.

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edited Feb 13 "10 in ~ 14:07

answered Feb 13 "10 at 13:15

Q.Q.J.Q.Q.J.

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Cut native (0,0) come (1,1/2), and also from (0,1/2) come (1,1). We have the right to glue these 3 pieces with each other to get a ring with circumference $\sqrt 5$ and also height $\sqrt 5 / 5$. Currently it"s easy!

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reply Feb 13 "10 at 12:33

TonyKTonyK

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Since $1+2i$ has length $\sqrt5$, you deserve to lift a square fundamental domain of $\mathbb C/\mathbb Z

*$ come $\mathbb C/(1+2i)\mathbb Z*

*$. Overlay a square basic domain because that the larger torus to acquire a way to division a square into 5 smaller sized squares.*It"s pretty easy to decompose any rectangle into a square geometrically, however the general decomposition is no as nice.

See more: Why Is Wu Hou Considered A Significant Individual? Wu Zhao: Ruler Of Tang Dynasty China

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answer Feb 13 "10 at 18:00

Douglas ZareDouglas Zare

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## Not the price you're spring for? Browse various other questions tagged mg.metric-geometry discrete-geometry tiling polygons or questioning your very own question.

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