41. a) Each vertex at the base of the triangle is connected by straight lines to \( n \) points located on the opposite side. Into how many parts do these lines divide the triangle?

b) Each of the three vertices of the triangle is connected by straight lines to \( n \) points located on the opposite side of the triangle. Into how many parts do these lines divide the triangle if no three of them intersect at the same point?
41. a) First, let's draw $n$ lines connecting vertex $B$ with $n$ points located on side $AC$ - these lines will divide the triangle into $n+1$ parts. Now, we will draw lines connecting vertex $C$ with points located on side $AB$ (Fig. 57). Each of these lines (the number of which is $n$) will intersect the previously drawn $n$ lines.

![](https://cdn.mathpix.com/cropped/2024_05_21_68334347b64402497ba5g-177.jpg?height=313&width=361&top_left_y=399&top_left_x=179)

Fig. 57. and, consequently, will be divided into $n+1$ parts by the points of intersection. But each of these parts of the drawn line divides one of the already existing parts of the triangle in half, i.e., increases the number of parts of the triangle by one. From this, it follows that all the lines divide the triangle into $(n+1) + (n+1)n = (n+1)^2$ parts $\left.{ }^{1}\right)$.

b) The lines coming from vertices $B$ and $C$ divide the triangle into $(n+1)^2$ parts (problem a)). Each of the lines coming from vertex $A$ intersects all $2n$ lines coming from vertices $B$ and $C$ at $2n$ points (all intersection points are distinct, as no three of our lines intersect at the same point). Thus, each of the lines coming from vertex $A$ is divided by the other lines into $2n+1$ parts and, consequently, increases the total number of parts by $2n+1$. From this, it follows that the total number of parts is

$$
(n+1)^2 + n(2n+1) = 3n^2 + 3n + 1
$$