How does one dissect a $10\times2$ rectangle into four pieces that can be reassembled to form a square?
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1$\begingroup$ can the rectangle only be cut into unit sizes? does the square need to be completely filled in? an area of 20 square units can't be evenly divided into a square with whole-number sides. need more info $\endgroup$– dfperryCommented Dec 9, 2015 at 15:54
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1$\begingroup$ @dperry: This is a standard dissection problem, without any strange requirements. This site contains many other puzzles of this type, as for instance puzzling.stackexchange.com/questions/22615/the-challenge-square $\endgroup$– GamowCommented Dec 9, 2015 at 16:13
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2$\begingroup$ @dperry: I'm assuming the square needs to be completely filled in, otherwise you could just do four 2 x 2.5 blocks and make a 4.5 x 4.5 square with a hole in the middle. $\endgroup$– orpCommented Dec 9, 2015 at 16:13
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1$\begingroup$ @Hugh >!yourText $\endgroup$– user14478Commented Dec 9, 2015 at 16:18
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1$\begingroup$ @Hugh: take a look at meta.stackexchange.com/questions/72877/… for a nice little guide $\endgroup$– IrishpandaCommented Dec 9, 2015 at 16:26
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4 Answers
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Index your rectangle from (0,0) to (10,2). Then cut from
(3,0) to (4,2); (4,2) to (8,0); (8,0) to (9,2).
These four pieces can be used to make the square.
Note that this dissection works without any flipping or even any rotation of the pieces! To show that it's a square is also relatively simple. Easy geometry shows that the angles are right angles, and that the sides are $2\sqrt{5}$.
answered Dec 9, 2015 at 16:42
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6$\begingroup$ Any reason why you do not shift all the cut to the left by one unit so that the outer parts are symmetric? =) $\endgroup$– justhalfCommented Dec 10, 2015 at 5:21
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2$\begingroup$ That's a good observation, which probably reveals something about my brain. Or at least the way I came to the solution! $\endgroup$ Commented Dec 10, 2015 at 5:43
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1$\begingroup$ If you make the long cut first and slide the pieces appropriately, you can make this in two cuts instead of three! :D $\endgroup$– corsiKaCommented Dec 10, 2015 at 17:40
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$\begingroup$ That's true. Also, the initial long cut can be anywhere between a vertical line at 1 and a vertical line at 9. All it does is shift where the line in the central parallelogram of the square goes. $\endgroup$ Commented Dec 10, 2015 at 17:43
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$\begingroup$ @corsiKa: But it's still four pieces, and the piece count is what the puzzle specifies. $\endgroup$ Commented Dec 10, 2015 at 18:24
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Well, I can do it with 4 cuts, 5 pieces
Cut two 4x2 sections (2 cuts). Cut them diagonally (4 total cuts). The diagonals have length $\sqrt{20}$. Assemble them so the diagonals are the sides of the new square, length 4 against length 2. That leaves a 2x2 hole in the middle, for your remaining 2x2 piece.
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2$\begingroup$ So inspired by this answer I tried it myself $\endgroup$ Commented Dec 9, 2015 at 16:40
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Big square: Actually 2 - Inside (10x10) and outside (10.5x10.5).
2 cuts (or 3 without stacking) Cut piece in half parallel with the long side, giving 2 10x1 pieces. Stack these pieces together, and cut in the same manner, giving 4 10x0.5 pieces. Arrange these pieces so that the short ends are against then end of the next piece, forming a large square
Picture:
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2$\begingroup$ I like this answer! I guess it's a square perimeter, but you can't argue that this doesn't answer this question. $\endgroup$ Commented Dec 9, 2015 at 21:46
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$\begingroup$ You can make it two cuts without stacking by simply cutting along the horizontal middle then the vertical middle to make four 5x1s. $\endgroup$ Commented Dec 9, 2015 at 22:09
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Here's a four-piece solution, taken from here (the link only has pictures, but I supplemented with cut locations):
Cut from (0, 2) to (4, 0), from (4, 0) to (4 + $\sqrt{3}$, 2), and from (10 - $\sqrt{3}$, 0) to (10, 2).
Image of the cut-up rectangle:
Then just slide the four pieces into a (rotated) square:
Despite the irrational locations of the cuts in the original rectangle, the vertices of all pieces in the joined square are at integer locations.
