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I have cut a $10\times10$ chessboard into $20$ pieces. Each of these pieces consists of exactly five chessboard squares. All my cuts were made along grid lines (bounding lines of the chessboad squares), and the total length of my cuts was equal to twice the perimeter of the chessboard.

True or false? Then the $20$ chessboard pieces are all congruent to each other.

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Does the colouring of the squares matter? – Question Asker Mar 28 at 13:47
@QuestionAsker: I suspect not, because otherwise the puzzle would be too easy. If the pieces are all congruent in shape and coloring, then each pentomino will have three squares of one color and two of the other, for a total of 60 squares of one color and 40 of the other. But a chessboard with standard coloration has a 50-50 division of squares. – Michael Seifert Mar 28 at 14:07
If it's 10 x 10 then it's not really a chessboard is it? ;) – JoeMalpass Mar 28 at 20:06
@JoeMalpass Especially if it doesn't even have colors ;) More like a giant chocolate bar. – AlexR Mar 28 at 20:07
mmmmmm... chocolate... – JoeMalpass Mar 28 at 20:09
up vote 26 down vote accepted


It is true.


The perimeter of the board is: $$P_{board}=4\cdot10=40$$ This means the length of all cuts is: $$C=2\cdot P_{board}=80$$ The sum of all perimeters of all pieces is the sum of the perimeter of the board and twice the length of all cuts, because each cut is the perimeter of 2 pieces: $$P_{allPieces}=P_{board}+2\cdot C=200$$ This means the average perimeter of a piece is: $$P_{pieceAvg}=\frac{P_{allPieces}}{20}=10$$ There is no pentomino with a perimeter less than 10 and only one with a perimeter equal 10, namely the P pentomino. Since the average perimeter of the pieces is 10, there consequently cannot be a piece with perimeter larger than 10.


Therefore all pieces must be (rotated and/or mirrored) P pentominoes and therefore all are congruent.

For the sake of completeness, here is one of the possible tilings:

enter image description here

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Also, it is trivial to produce such a cutting (Split the board in slices of 2x10, halve these along the short axis and it should be clear) – AlexR Mar 28 at 17:34
@AlexR A trivial example of such a cutting does not solve the puzzle. – ricksmt Mar 28 at 19:46
@ricksmt Together with Sleafar's answer it does. The answer without a proof of existence is not complete, however. This trivial cutting thus finishes the proof. – AlexR Mar 28 at 19:47
@KevinWells I recounted and ended up with a cut length of 80 as required. So my tiling is in fact valid and completes the proof. EDIT: You probably misunderstood my description. To avoid spoilers I left out how you have to cut the ten 2x5 tiles for the tiling. – AlexR Mar 28 at 19:57
@KevinWells This confirms my point. I left out which cuts you need to make on each of the 2x5 tiles to get the P-tiling ^^ – AlexR Mar 28 at 20:06

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