A rectangle with sides x and y is cut with straight cuts into pieces that are reassembled, by rotation and...
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Problem
A rectangle with sides x and y is cut with straight cuts into pieces that are reassembled, by rotation and translation only (no flipping), into a square of the same area. Is this always possible, even when x/y is irrational? What is the minimum number of pieces?
Show solution
If P and Q are polygons of equal area, then P can always be cut up into
a finite number of pieces with straight cuts and rotated, translated,
and reassembled into Q (Bolyai's theorem.)
WLOG we can let x>=y. As x/y goes to infinity, so will the minimum
number of pieces (since we need to shorten the longer side by a factor
of at least sqrt(x/2y) to fit it into the square, meaning at least
sqrt(x/2y) pieces.)
If x/y=1, we need 1 piece.
If 1<x/y<4, we need 3 pieces:
---------------
------------------------- |.. |
| .. | | ... B |
| ... B | | A ... |
| C | ... | ===> |________... |
| | ... | | ..|
| | A ..| | |
------------------------- | C |
| |
---------------
(1) (2)
If x/y=4 then we need 2 pieces, and the dissection is obvious.
For bigger x/y, we can stack rectangles D, E, ... to the left of C in
(1) and below C in (2).
9/4<=x/y<=9 1 extra rectangle, 4 pieces
25/4<=x/y<=16 2 extra rectangles, 5 pieces
...
(n+1/2)^2<=x/y<=(n+2)^2 n extra rectangles, n+3 pieces
I don't know if these are best possible.
-- David Moews <moews@math.berkeley.edu>
Your problem belongs to the field of "geometric dissections" and the
standard reference on the subject is the book of that name by H.
Lindgren. Your particular problem may be simple enough for provably
minimal solutions to be easily obtainable, but if not, check out
Lindgren's book before sitting down to work on the problem in earnest.
-- Tim Chow <tycchow@math.mit.edu>
[ Editor: a proof of the theorem quoted above, from the thread
"Partitioning equalarea rectangles ?" (geometry.puzzles, February
1995). ]
It is indeed a venerable enough chestnut - when Hilbert was
investigating the foundations of geometry, he went to the trouble of
proving that given any two polygons of the same area one could dissect
one into a finite number of polygonal pieces that could be reassembled
to form the other.
Here's a quick proof:
1) any triangle can be chopped into two pieces that will form a
parallelogram:
Proof _____
/| /| /
/ | / | /
/ | / | /
/ | / | /
/ | / |/
/ / / /
/ /| / /|
/ / | / /
/ / | / /
/ / | / /
/_____/____| /_____/
2) any two parallelograms of the same area can be "inter-chopped".
Proof. Let the longer side of one be AB, and the longer side of the
other AC, in the figure below, where the angle CAB is adjusted so that
the parallelogram ABDC of this figure has the same area as each of the
two given ones:
--------A--------B--------X--------Y--------Z--------
/ / / / / /
/ / / / / /
C---a----D---b----P---x----Q---y----R---z----
/ / / / /
/ / / / /
c--------d--------p--------q--------r
/ / / /
/ / / /
---------------------------------
Then we can find points a and b on the line CD so that the first
parallelogram has shape ABba.
This shows we can chop it into ABDC. But we can similarly chop this
into the second one of the two givens.
In fact you can draw both parallelograms as period-parallelograms for
the lattice defined by A,B,C,D, and directly chop one into the other,
so saving quite a lot of cuts.
3) The theorem.
Proof: chop each figure into triangles, then chop these into
rectangles with one side = 1, and stack these into a 1-by-@ rectangle,
where @ is the common area.
Since I feel I haven't described 2) too well, let me just show you how
to chop any parallelogram into a square.
----A-------B-------C-------D-------E-------F
| | | | | |
| | | | | |
| | | | | |
-------a-------b-------c-------d-------e-------f
| | | | | |
| | | | | |
| | | | | |
P-------Q-------R-------S-------T-------U-------
| | | | | | |
| | | | | | |
| | | | | | |
-------p-------q-------r-------s-------t-------u-------
| | | | | |
Since one side of the given parallelogram exceeds the side of the
corresponding square, we can find a point a on the lower edge of that
square (possibly produced), so that Aa is that side, in length. Now
chop the above figure by the straight lines like AaQq... BbRr...
CcSs... DdTt...
Each square gets cut into finitely many pieces, copies of which can be
assembled to form a strip like that bounded by the lines AaQq... and
BbRr... in the figure. But now we can find some points X and x (so
that Xx = Aa = Bb = ...) on the line AaQq... so that BbxX is the same
shape as the given parallelogram.
Cutting the diagonal strip just mentioned by the lines BX and bx then
produces some pieces that can be assembled either to form the square
or to form the parallelogram.
I remark that if we demand that the pieces be translated rather than
also rotated to form the second parallelogram, then the problem
becomes affinely invariant. It should therefore be easy to find the
minimal number of pieces for any pair of parallelograms.
-- John Conway <conway@math.Princeton.EDU>