We draw a diagonal on some squares of the chessboard in such a way that no two have a common point. What is the maximum number of diagonals we can draw this way?
Solution. The vertices of the chessboard squares are the lattice points of a $9 \times 9$ grid. In the rows with odd numbers (first, third, fifth, seventh, ninth), there are a total of $5 \cdot 9 = 45$ lattice points, and in the rows with even numbers, there are $4 \cdot 9 = 36$ lattice points. Any diagonal drawn in any square of the chessboard connects two adjacent, and thus one even and one odd numbered row lattice points. Therefore, the number of diagonals without common points cannot be more than the number of lattice points in the even-numbered rows, which is 36. However, 36 such diagonals can indeed be drawn, as the diagram shows an appropriate arrangement.

![](https://cdn.mathpix.com/cropped/2024_05_02_84cfeaf5ec2c47b99ce0g-1.jpg?height=431&width=425&top_left_y=392&top_left_x=839)

Note. The upper bound of 36 only uses the condition from the problem that no two diagonals can have a common endpoint. Therefore, the number of diagonals that can be drawn is not increased if we allow the existence of common interior points, that is, if two diagonals can intersect at the center of a square.