The polygons $\mathcal{K}$ and $\mathcal{L}$ are placed in the plane such that none of their vertices lie on the side lines of the other polygon. What is the maximum number of points where the sides intersect if $\mathcal{K}$ is a pentagon and
a) $\mathcal{L}$ is a triangle;
b) $\mathcal{L}$ is a quadrilateral.
If a line does not pass through any vertex of a polygon, then it intersects the sides of the polygon at an even number of points, because the segments of the line separated by the intersection points alternate between being outside and inside the polygon, and the first and last segments are two rays, which are infinite, so they cannot be inside the polygon (Figure 1).

![](https://cdn.mathpix.com/cropped/2024_05_02_d0f770eb5b9a0204e803g-1.jpg?height=347&width=464&top_left_y=306&top_left_x=836)

Figure 1

In case a), each side of the pentagon can intersect at most 2 sides of the triangles, so the number of intersection points can be at most $5 \cdot 2 = 10$. This is possible, as shown in Figure 2.

![](https://cdn.mathpix.com/cropped/2024_05_02_d0f770eb5b9a0204e803g-1.jpg?height=349&width=447&top_left_y=888&top_left_x=405)

Figure 2

![](https://cdn.mathpix.com/cropped/2024_05_02_d0f770eb5b9a0204e803g-1.jpg?height=396&width=726&top_left_y=1358&top_left_x=1144)

In case b), each side of the quadrilateral can intersect at most 4 sides of the pentagon, so the maximum number of intersection points is $4 \cdot 4 = 16$. This can be achieved as shown in Figure 3.

Based on the work of Dávid Papp (Budapest, Szent István Gymnasium, 12th grade)