Packing Rectangles
Here is a puzzle to which I dunno the answer yet.
Claim: Given N (N>1) rectangles all of the same area but all with different perimeters.
We cannot pack them without gaps to form a large rectangle.
IOW, if a rectangle is broken into N smaller rectangles all of same area, at least two of the pieces are identical.
Remark: it is not too hard to show rectangles which can be divided into N equal area rectangles with just 2 pieces identical, and all others having different perimeter.
Perhaps the claim may be easier to prove under the constraint: the sides of the rectangle and the pieces ought to be integers/rationals.
The following pages may be of interest.
http://mathdl.maa.org/images/upload_library/22/Ford/Wagon601-617.pdf
http://www.math.uwaterloo.ca/navigation/ideas/articles/honsberger2/index.shtml
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Update (March 27th 2012) - (The portion above was not touched in this edit)
The claim above *does not* hold if the tiles can have incommensurate sides - if the ratio between the lengths can be irrational. http://www.brand.site.co.il/riddles/201203q.html has details. As of now, if all tiles ought to have rational side lengths, the claim remains open.
We posted the problem as we knew it a few days back at http://maven.smith.edu/~orourke/TOPP/P78.html
And then, came a MAJOR SURPRISE: I just gathered from http://mathworld.wolfram.com/BlanchesDissection.html that the problem was known at least in its basic form long back (in the late 1930s!) to the 4 mathematicians - Brooks, Smith, Stone and Tutte - who did plenty of work on 'squaring the square'. This 'Blanche dissection' was not known to even some of the experts I approached with above question - a possible reason might be its odd name. I am not sure if the quartet attempted the rational edge lengths case or left it for us to ask.
Claim: Given N (N>1) rectangles all of the same area but all with different perimeters.
We cannot pack them without gaps to form a large rectangle.
IOW, if a rectangle is broken into N smaller rectangles all of same area, at least two of the pieces are identical.
Remark: it is not too hard to show rectangles which can be divided into N equal area rectangles with just 2 pieces identical, and all others having different perimeter.
Perhaps the claim may be easier to prove under the constraint: the sides of the rectangle and the pieces ought to be integers/rationals.
The following pages may be of interest.
http://mathdl.maa.org/images/upload_library/22/Ford/Wagon601-617.pdf
http://www.math.uwaterloo.ca/navigation/ideas/articles/honsberger2/index.shtml
---------------------
Update (March 27th 2012) - (The portion above was not touched in this edit)
The claim above *does not* hold if the tiles can have incommensurate sides - if the ratio between the lengths can be irrational. http://www.brand.site.co.il/riddles/201203q.html has details. As of now, if all tiles ought to have rational side lengths, the claim remains open.
We posted the problem as we knew it a few days back at http://maven.smith.edu/~orourke/TOPP/P78.html
And then, came a MAJOR SURPRISE: I just gathered from http://mathworld.wolfram.com/BlanchesDissection.html that the problem was known at least in its basic form long back (in the late 1930s!) to the 4 mathematicians - Brooks, Smith, Stone and Tutte - who did plenty of work on 'squaring the square'. This 'Blanche dissection' was not known to even some of the experts I approached with above question - a possible reason might be its odd name. I am not sure if the quartet attempted the rational edge lengths case or left it for us to ask.
posted by R.Nandakumar @ 3:27 AM
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