Problem 2. Three families of parallel lines have been drawn, with 10 lines in each. What is the maximum number of triangles they can cut out from the plane?

Answer: 150 triangles

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# Solution.

Consider 100 nodes - the intersection points of lines in the first and second directions. Divide them into 10 sectors: the first sector - nodes lying on the first lines of the first and second

![](https://cdn.mathpix.com/cropped/2024_05_06_e225e7bd0bfdae79743fg-1.jpg?height=548&width=557&top_left_y=2374&top_left_x=1407)
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 in the third direction form no more than $2 \cdot 25$ triangles with the last five sectors, since these five sectors contain a total of 25 nodes.

Furthermore, note that each line in 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.