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Can you find 3 similar geometrical figures (common shape but can be different sizes) A, B and C with the following property: If you remove any square from a 8x8 chess board, then the remaining area can be exactly covered with A, B and C.

How do A, B and C look like?

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  • $\begingroup$ Maybe add the constraint that the covers are not allowed to stick out over the sides of the board? $\endgroup$
    – jochen
    Jan 28 '20 at 9:08
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    $\begingroup$ I'd be interested to see a proof, or counterproof, that this set of 3 figures is the only set of 3 figures (similar or otherwise) that has this interesting property with respect to an 8x8 chessboard. Anyone feel like tackling this bonus, or should I post a new puzzle? $\endgroup$ Jan 28 '20 at 21:15
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These should do it:

enter image description here
For example, like so:
enter image description here
Just orient the large block so that the missing piece is in the missing corner of the block, then do the same for each successively smaller block.

Just to show another example:

enter image description here

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    $\begingroup$ Perfect explanation! $\endgroup$
    – z100
    Jan 27 '20 at 16:25
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    $\begingroup$ So it's basically a bpgerr (rot13)? $\endgroup$
    – QBrute
    Jan 27 '20 at 17:23
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    $\begingroup$ Yes, although the 2-d version is called a dhnqgerr I think. $\endgroup$
    – hdsdv
    Jan 27 '20 at 17:58
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    $\begingroup$ @Fogmeister yeah, I think that was the requirement for this puzzle :) $\endgroup$
    – hdsdv
    Jan 29 '20 at 0:58
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    $\begingroup$ @hdsdv oh. 😂 I misread the question to begin with. $\endgroup$
    – Fogmeister
    Jan 29 '20 at 8:56
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I think this is the answer.

Three L's. Each L is made up of 3 squares. The first L is 3 $4\times 4$ squares, the second is 3 $2 \times 2$, the third is 3 $1\times 1$ squares. Any one square of the chess board is in one of four quadrants, so you align the L so that the empty section matches with that quadrant. Repeat until only the taken square isn't covered.

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This is a generalisation:

For any $N \times N$ board with one square missing and $N=2^k$ (some power of two), we can solve the puzzle with $k$ L-shaped pieces as described in the accepted answer. The first piece has width 1, the next has width 2, each piece doubling in width until we reach the largest piece with width $2^{k-1}$.

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