30. What is the maximum number of intersection points that a closed broken line consisting of 7 segments can have? (The common ends of the segments of the broken line are not counted.)

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30. Take a broken line consisting of two segments. By adding a third segment, it can only intersect one of the two original segments, resulting in one point (1) (Fig. 30). Taking a fourth segment, it can intersect a maximum of two of the three existing segments, resulting in two more points $(2,3)$. A fifth segment can intersect a maximum of 3 segments, resulting in three more points $(4,5,6)$. A sixth segment can intersect a maximum of 4 segments, resulting in 4 more points $(7,8,9,10)$. A seventh segment can intersect a maximum of 4 segments, resulting in four more points $(11,12,13,14)$. Thus, the maximum possible number of self-intersection points is 14.

![](https://cdn.mathpix.com/cropped/2024_05_21_00604dc020e3721fc1f4g-182.jpg?height=320&width=414&top_left_y=1596&top_left_x=358)

Fig. 29

![](https://cdn.mathpix.com/cropped/2024_05_21_00604dc020e3721fc1f4g-182.jpg?height=477&width=782&top_left_y=1549&top_left_x=888)

Fig. 30