3. Can 5 lines be drawn on a plane such that no three of them pass through the same point, and there are a) exactly $11;$ b) exactly 9 intersection points?
Answer: a) no; b) yes.

Solution. The maximum number of intersection points will be obtained if there are no two parallel lines among the lines and no three lines pass through the same point. In this case, there will be 4 intersection points on each of the 5 lines, and the total number of intersection points is $\frac{5 \cdot 4}{2}=10$, since each intersection point belongs to exactly two lines. Therefore, 11 intersection points are not possible. But 9 can be, if some two lines are parallel (an example is shown in the figure.)

![](https://cdn.mathpix.com/cropped/2024_05_06_65636c5c9319cd549e45g-3.jpg?height=391&width=277&top_left_y=1061&top_left_x=889)

Grading. 12 points for a correct solution. If only one of the parts is done, 6 points. In the example, it must be specified that there are parallel lines! (If any three lines pass through the same point, there will be no more than 8 intersection points).