Two pentagons are placed on the plane such that their vertices do not lie on the side lines of the other pentagon. What is the maximum number of points where the sides can intersect?
We will prove that the sides of the polygons can intersect in at most 18 points. This many intersection points are possible, for example, as shown in Figure 1.

To prove that more than 18 intersection points cannot occur, we use three lemmas.

1. **First Lemma.** A convex polygon can be intersected by a line in at most 2 points.

Proof. The segment connecting the two farthest intersection points, along with its interior, is part of the polygon, so this segment cannot intersect the boundary of the polygon.

2. **Second Lemma.** A pentagon can be intersected by a line, which does not pass through any of its vertices, in at most 4 points.

Proof. The line intersects each side of the pentagon at most once, so more than 5 intersection points are certainly not possible. The intersection points divide the line into two rays and several segments. These segments alternate between being inside and outside the pentagon; the two rays are certainly outside. This implies that the number of segments is odd, and the number of intersection points is even. Therefore, the number of intersection points cannot be 5.

3. **Third Lemma.** Two polygons, whose vertices do not lie on the sides of the other polygon, can intersect each other in an even number of points.

Proof. Traversing the perimeter of one polygon, at each intersection point, we transition from the interior to the exterior, or from the exterior to the interior of the other polygon. After completing the full traversal, we return to the same region of the plane, so the total number of transitions from the interior to the exterior is the same as from the exterior to the interior.

With these lemmas, we will prove that two pentagons can intersect in at most 18 points. Let the two pentagons be \( A \) and \( B \). Each side of \( A \) can intersect \( B \) in at most 4 points according to the second lemma; this totals at most 20 intersection points. According to the third lemma, 19 intersection points are not possible. We only need to prove that the number of intersection points cannot be 20.

If the number of intersection points is 20, then each side of \( A \) intersects 4 sides of \( B \), and vice versa, each side of \( B \) intersects 4 sides of \( A \). From the first lemma, it follows that \( A \) and \( B \) are concave.

Since \( A \) is concave, it has both convex and concave angles, and there are three consecutive sides: \( a, b, \) and \( c \) such that \( a \) and \( b \) meet at a convex angle, and \( b \) and \( c \) meet at a concave angle (Figure 2).

Let \( d \) be the side of \( B \) that is not intersected by \( b \). This side intersects 4 sides of \( A \), so it intersects \( a \) and \( c \) as well; similarly, \( b \) intersects the other 4 sides of \( B \). Since \( a \) and \( c \) are on opposite sides of the line \( b \), the segment \( d \) intersects the line \( b \). However, this means that the line \( b \) intersects all sides of the pentagon \( B \), which contradicts the second lemma. Therefore, the number of intersection points cannot be 20.

Thus, the two pentagons can intersect in at most 18 points.

Péter Várjú (Szeged, Radnóti M. Gimn., 11th grade)
![](https://cdn.mathpix.com/cropped/2024_05_02_f0b435785faebe9b17c2g-1.jpg?height=928&width=662&top_left_y=1495&top_left_x=688)