Given a Pythagorean triple $(a,b,c)$ satisfying $a^2+b^2=c^2$, how to calculate the least number of polyominoes of total squares $c^2$, needed, such that both the square $c^2$ can be build by piecing them together, as well as the two separate squares of side length $a$ and $b$?
Tiling pythagorean triples with minimal polyominoes
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combinatorics
geometry
puzzle
computational-geometry
tiling
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0Is it just me, or does this problem sound a little too close to Project Euler problem 139? – 2012-04-24
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0I suppose it means that Mike has looked at Project Euler problem 139 and is wondering whether your problem is very similar. If it is, that's a bad thing, since the PE people don't want anyone getting help on this (or any other) website. It would perhaps have been helpful had Mike linked to PE 139. – 2012-04-24
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0Here's a link: http://projecteuler.net/problem=139. Personally, I don't see much resemblance between the two problems. – 2012-04-24
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0There is a well-known way to cut any two squares into a total of 5 pieces which can then be used to form a single square. However, the 5 pieces are not polyominoes, so this is of no help. – 2012-04-24
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0http://en.wikipedia.org/wiki/Polyomino – 2012-04-26