On a checkered square the cells of the upper left square are black and all the other cells are white. What is the maximal such that the original square can be dissected (along the borders of the cells) into polygons such that in each of them the number of black cells is three times less than the number of white cells? (The polygons need not be congruent or even equal in area.)
Solution
The answer is . We can see that there are only cells on the border of the black square that connect to the white area. Each of them belongs to at most polygon, so there are at most polygons.
An example as follows (each of cells belongs to one part that has the ratio of black:white is )
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