19. Given an arbitrary triangle. On each side of the triangle, 10 points are marked. Each vertex of the triangle is connected by segments to all the marked points on the opposite side. Into what maximum number of parts could the segments divide the triangle?
Answer: 331. Solution. We will draw the segments one by one. The number of added parts is equal to the number of parts into which the segment is divided by the points of intersection. This means, the more points of intersection, the more parts. The maximum number of parts will be when each segment intersects each of the segments drawn from other vertices, and at its own point. For this, it is sufficient to draw segments not through the existing points of intersection. For 10
![](https://cdn.mathpix.com/cropped/2024_06_04_8879c536e89d29e9be5ag-07.jpg?height=954&width=1790&top_left_y=1756&top_left_x=152)

Fig. 5: To problem 16
segments from the first vertex, each will add one part. For 10 segments from the second vertex, each will have 10 points of intersection, so each will add 11 parts. For 10 segments from the third vertex, each will have 20 points of intersection, so each will add 21 parts. In total, $1+10+10 \cdot 11+10 \cdot 21=331$.