\section*{Problem 6 - 021136}

In a plane, a quadrilateral and a pentagon are positioned such that none of their vertices lie on any side of the other figure.

What is the maximum possible number of intersection points of the sides of both polygons?

The polygons do not need to be convex.
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A general planar pentagon is defined as a geometric figure of five pairwise distinct points \(A_{1}, A_{2}, A_{3}, A_{4}, A_{5}\) in the same plane, such that no three consecutive points lie on the same line. These points, together with the segments \(A_{1} A_{2}, A_{2} A_{3}, A_{3} A_{4}, A_{4} A_{5}, A_{5} A_{1}\), form the pentagon.

First, we determine the maximum number of intersection points of a line \(g\) with the sides of the pentagon, considering the conditions given in the problem (the pentagon is not necessarily convex, and the vertices of the pentagon do not lie on any side of the quadrilateral).

The line \(g\) divides the plane \(\epsilon\) into two half-planes \(\epsilon_{1}\) and \(\epsilon_{2}\). Without loss of generality, we assume: \(A_{1} \in \epsilon_{1}, A_{1} \notin g\). From the definition and the conditions of the problem, it follows that:

- \(A_{1} A_{2}\) has a common intersection point with \(g\) only if: \(A_{2} \in \epsilon_{2}, A_{2} \notin g\)
- \(A_{2} A_{3}\) has a common intersection point with \(g\) only if: \(A_{3} \in \epsilon_{1}, A_{3} \notin g\)
- \(A_{3} A_{4}\) has a common intersection point with \(g\) only if: \(A_{4} \in \epsilon_{2}, A_{4} \notin g\)
- \(A_{4} A_{5}\) has a common intersection point with \(g\) only if: \(A_{5} \in \epsilon_{1}, A_{5} \notin g\)

The segment \(A_{5} A_{1}\) cannot have an intersection point with the line \(g\) because \(A_{1}\) and \(A_{5}\) lie in the same half-plane \(\epsilon_{1}\).

Thus, it is proven that a line and, consequently, a side of a quadrilateral can have a maximum of four intersection points with the sides of a pentagon. From this, it follows that the maximum number of intersection points between the sides of both polygons is \(4 \cdot 4 = 16\).

It suffices to show by example that 16 intersection points exist under the conditions of the problem, as indicated in the image.

![](https://cdn.mathpix.com/cropped/2024_06_06_b85a894aa8dbf722a7b3g-1695.jpg?height=451&width=485&top_left_y=754&top_left_x=794)

Let \(A B C D E\) be a concave pentagon and \(K L M N\) be a concave quadrilateral.

Remark: The sides of a planar \(n\)-gon have at most \(n-1\) common intersection points with a line \(g\) for odd \(n\) and \(n\) common intersection points for even \(n\).