Checkers Board Problem

Question

Here we consider a checkerboard expanded to size 12 × 12 instead of the ordinary 8 × 8 checkerboard. a) How many squares on this board contain more than a third of the total number of dark small squares? b) How many rectangles on this board contain at least two thirds of the total number of small squares?

Answer 1

A third of the number of small dark squares is 24, so we require at least 25 dark squares for the first problem. Since the number of dark squares is never more than 1 away from the number of white squares, this means each large square must be $7\times 7$ or larger; and if it is only $7\times 7$ its corners must be dark (half the $7\times 7$ squares have this property).

Then assuming the lower left corner square of the checkerboard is dark, the lower left corner of the large square can be:

• Only at (0,0) if the large square has side 12

• Anywhere in a $2\times 2$ square $(0,0),(0,1), (1,0), (1,1)$ if the large square has side 11.

• Anywhere in a $3\times 3$ square if the large square has side 10.

• Anywhere in a $4\times 4$ square if the large square has side 9.

• Anywhere in a $5\times 5$ square if the large square has side 8.
• At half the spots (the dark squares) in a $6\times 6$ square if the large square has side 8.

$$ 1 + 2^2 + 3^2 + 4^2 + 5^2 + \frac12\cdot 6^2 = 73 $$

For the second part, the same reasoning applies, but this time the large square will have to be $10 \times 10$ or bigger.

$$ 1 + 4 + 9 = 14$$

EDIT

It is pointed out in the comments that the second part includes rectangles as well as squares.

There are 2 rectangles of length and height $11 \times 12$1, 2 $12 \times 11$, 3 each of $10 \times 12$ and $12 \times 10$, and 6 each of $10 \times 11$ and $11 \times 10$, giving a total of $$ 4+6+12 = 22$$ extra rectangles. Adding this to the squares wee get a total answer of 36.