## 17. Hexagon

Consider a hexagon. How many diagonals, i.e., straight lines connecting two non-adjacent vertices, do you think it has?

If this question seems too easy to you, then extend the six sides of the original hexagon and determine how many points, different from the vertices of the hexagon, all these lines (extended sides and diagonals) will intersect?
17. 18) Through six different points, one can draw as many lines as there are ways to choose two out of six elements, i.e.,

$$
C_{6}^{2}=\frac{6!}{2!4!}=15
$$

Among these 15 lines, 6 are the sides of the hexagon; the remaining 9 lines are diagonals.

2) Generally speaking, the number of intersection points of 15 lines is equal to the number of different ways to choose 2 out of 15 elements, i.e.,

$$
C_{15}^{2}=\frac{15!}{2!13!}=105
$$

![](https://cdn.mathpix.com/cropped/2024_05_21_fe999c0fe2ad81fc5164g-159.jpg?height=856&width=851&top_left_y=1334&top_left_x=588)

But each vertex of the hexagon corresponds to the intersection of five different lines; thus, at each vertex, there are $C_{\overline{5}}^{2}=10$ intersection points, because

$$
C_{\underline{b}}^{2}=\frac{5!}{3!2!}=10
$$

Therefore, the six vertices correspond to 60 intersection points. Consequently, there are $105-60=45$ intersection points distinct from the vertices of the original hexagon ${ }^{1}$.