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Suppose we have a board of 121 squares ($11\times11$) but with a black square in the tile in the middle (6,6). What is the maximum number of $3\times4$ tiles we can tile the checkerboard with?? (without putting anything on the black one). We can rotate the $3\times4$ pieces.

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Have you tried the standard coloring? What results did you get? – Calvin Lin Sep 28 '13 at 23:34
what is the standard coloring? – dREaM Sep 29 '13 at 0:08
What is the source of this problem, please? – Gerry Myerson Sep 29 '13 at 0:26
It was on a mexican math olympiad from two years ago – dREaM Sep 29 '13 at 0:43
The standard coloring alternates black and white squares. If you number the rows and columns $1,\ldots,11$ from top to bottom and left to right, respectively, and label a square $\langle r,c\rangle$ if it’s in row $r$ and column $c$, then all of the squarees $\langle r,c\rangle$ with $r+c$ even will be of one color, and all those with $r+c$ odd will be of the other. – Brian M. Scott Sep 29 '13 at 1:56
up vote 2 down vote accepted

Consider the rows 3, 6 and 9.
Show that at most 3 tiles can intersect rows 3 and 9.
Show that at most 2 tiles can intersect rows 6.
Show that any tile that's placed must intersect rows 3, 6 or 9.
Conclude that there are at most 8 tiles that can be placed.

Find such a construction with 8 tiles, which you should already have done.

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