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Problem 1332

AIME late

Combinatorics Difficulty 5.6 Prove it Saudi Arabian Mathematical Competitions · Saudi Arabia

On a checkered square $10 \times 10$ the cells of the upper left $5 \times 5$ square are black and all the other cells are white. What is the maximal $n$ such that the original square can be dissected (along the borders of the cells) into $n$ 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.)

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

The answer is $9$ . We can see that there are only $9$ cells on the border of the black square that connect to the white area. Each of them belongs to at most $1$ polygon, so there are at most $9$ polygons.

An example as follows (each of cells belongs to one part that has the ratio of black:white is $1:3$ )

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