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Partitioning a Rectangle...


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#1 Anza Power

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Posted 11 June 2012 - 12:50 PM

I've been pondering this for a few months now and haven't got a clue yet what to do...

Prove that if you have a rectangle and you partition it into smaller rectangles such that every rectangle has at least 1 edge of integer length, then the large rectangle has 1 edge of integer length.

The proof is supposed to be simple by using the fact that in a graph the number of nodes with odd degrees is even, and it's generalized for Rn, but I'm still stuck even on R2...

Edited by Anza Power, 11 June 2012 - 12:51 PM.

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#2 Rox53

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Posted 11 June 2012 - 01:33 PM

If you start with a rectangle with no integer edges, then at least one of the smaller rectangles will also lack integer edges. This is heuristically true - but I can't prove it yet.
If my first statement can be proved, then it follows that the original proposition (If you have a rectangle and you partition it into smaller rectangles such that every rectangle has at least 1 edge of integer length, then the large rectangle has 1 edge of integer length) must be true.
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#3 curr3nt2

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Posted 12 June 2012 - 11:07 PM

Spoiler for maybe...?

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#4 witzar

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Posted 19 June 2012 - 07:17 PM

Look here for proofs (there are many):
http://mathdl.maa.or...agon601-617.pdf
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