9.5. Three families of parallel lines have been drawn, with 10 lines in each. What is the maximum number of triangles they can cut out of the plane?
Solution. Consider 100 nodes - the intersection points of lines from the first and second directions. Divide them into 10 sectors: the first sector - nodes lying on the first lines of the first and second directions. The second sector - nodes lying on the second lines (excluding points lying in the first sector) and so on. Triangles with sides parallel to three fixed directions can have two orientations, and each of our 100 nodes can be the vertex of no more than one triangle of each orientation. Therefore, 10 lines of the third direction form no more than 2.25 triangles with the last five sectors, since these five sectors contain only 25 nodes in total.

Furthermore, note that each of the lines of the third direction forms no more than one triangle of each orientation with nodes belonging to one sector. Therefore, the number of triangles with vertices in the nodes of the remaining five sectors will not exceed \(10 \cdot 2 \cdot 5\).

In total, there are no more than \(100 + 50 = 150\) triangles.

An example with 150 triangles is shown in the figure.

Answer: 150 triangles

![](https://cdn.mathpix.com/cropped/2024_05_06_9e2567a4466362c958fbg-3.jpg?height=539&width=556&top_left_y=450&top_left_x=887)