# Dissecting rectangle into two equal area parts to form noncongruent rectangle

A rectangle can be cut into two connected pieces of equal area so that they can be rearranged to get a new rectangle with different side lengths than the original one. Show that this can be done in countable many ways. Can this be done in uncountable many ways?

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A rectangle is a product of finitely many closed real intervals? – shuhalo Nov 18 '10 at 12:31
Of course you can split a rectangle into finitely many parts but you can split it to infinitely many part too. – Jaska Nov 18 '10 at 14:30
For the last question, you need the Millikan Theorem of Math Problems. Any time a problem is stated as "Show how xxx can be done. Can yyy be done?" where yyy is a harder case of xxx, the answer is no. – Ross Millikan Nov 18 '10 at 15:29

Let n ≥ 1 be an integer. Divide the width of an H-by-W rectangle into n+1 equal parts, and the height into n equal parts. Separate the rectangle into two connected and congruent pieces along a path stepped as necessary from a side divided at the 1/(n+1) part of its length, down a distance of 1/n of its height, and so on alternating until the opposite side of the rectangle is reached. The two parts may then be "slid" across one another to form a rectangle of height (n+1)/n H and width n/(n+1) W.

That gives a countable number of ways to rearrange the rectangle, at most one of which could be congruent to the original (i.e. when W = (n+1)/n H for some n, or conversely).

I believe these are the only such dissections of a rectangle into two pieces, and if so there are not uncountably many ways to do it (though of course the roles of height and width can be exchanged). The locations of the four original corners (right angles) are constrained so that they can only appear on the boundary of the rearranged rectangle, and the rearrangement works both ways. The dissection must thus operate so as to separate the boundary of both rectangles into two parts (because an intact boundary of a rectangle cannot be encompassed by a different rectangle of equal area). A bit of case analysis should then show that the two rectangles share a pair of opposite corners and that the path dividing the pieces must enter and exit at the "new" pair of corners.

regards, hm

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There are however uncountably many ways to dissect a rectangle into two noncongruent pieces of equal area that can be rearranged to give a rectangle with the same side lengths as the original. The idea here is to remove an L-shaped "gnomon" that leaves a subrectangle of the original, then rotate the L-shape around to encompass of the opposite corner of the subrectangle. Care is needed to make such pieces have equal area. – hardmath Nov 18 '10 at 15:26