# Problem 8. (5 points)

A figure called a "half-bishop" moves one square diagonally. What is the maximum number of non-attacking half-bishops that can be placed on a $9 \times 9$ chessboard?
Answer: 45

Solution:

Consider all diagonals of one direction. On each diagonal of even length, no more than half of the cells can be occupied by pseudo-bishops, as it can be divided into pairs of cells that cannot be simultaneously occupied.

On each diagonal of odd length $2n+1$, no more than $n+1$ pseudo-bishops can be placed for the same reasons.

There are a total of 9 odd diagonals, so the maximum possible number of pseudo-bishops is $\frac{81-9}{2}+9=45$.

An example can be constructed by placing pseudo-bishops in all cells of the rows with odd numbers.

## 7th grade

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