Consider a standard $8\times 8$ chessboard where a pawn is placed on each of the squares $d1,d2,d3,d4$ . Dissect the board into $4$ congruent pieces (reflections are allowed) such that each piece contains exactly one pawn.
[Math] Dissection of a chess board into 4 congruent pieces
combinatoricsgeometrypuzzlerecreational-mathematics
Related Solutions
Draw a hamiltonian cycle on the chessboard. Any cycle will do, and the following one suffices:
Pick (any) one square and call it $s_0$. Number the following squares in the hamiltonian cycle $s_1, s_2,\ldots s_{63}$:
Note that the even $s_i$ are all one color (say pink) and the odd $s_i$ are the other color (say brown). So far this is all very easy.
Now suppose two squares of the opposite colors have been deleted; for example:
The cycle has been cut into two paths, both of which have even length. Such a path is easily covered with dominoes: say one path is $s_k, s_{k+1}, \ldots, s_{k+2j+1}$, where the subscripts are understood mod 64. Then the dominoes easily go onto $(s_k, s_{k+1}), (s_{k+2}, s_{k+3}), \ldots (s_{k+2j}, s_{k+2j+1})$. Similarly the other path is tiled by $(s_{k+2j+3}, s_{k+2j+4}), \ldots, (s_{k-3}, s_{k-2})$.
This theorem and the proof I have given appear on pages 111–112 of Solomon W. Golomb Polyominoes (revised ed.), Princeton University Press, 1994. The proof is attributed to Ralph Gomory.
If $n$ is even, put $n^2/2-1$ knights on the other squares of the same colour of that on which the 1st knight stands.
If $n=2k+1$ is odd, there are $2(k^2+k)+1$ of one colour (say, white) and $2(k^2+k)$ black squares. If the original knight stands on a white square, just put $2(k^2+k)$ knights on the unoccupied white squares. If the original knight stands on a black square it controls at least 3 white squares (because all corner squares are white) and so occupying the remaining $2(k^2+k)-1$ squares gets the maximum.
Best Answer
I think growing spirals from each corner simultaneously does the trick.