9.4. What is the maximum number of triangles with vertices at the vertices of a regular 18-gon that can be marked so that no two different sides of these triangles are parallel? The triangles can intersect and have common vertices, coinciding segments are considered parallel.
Answer: 5.

Solution. Estimating the number of triangles. Let's number the vertices of the 18-gon from 1 to 18 clockwise. The sides of the triangles are the sides and diagonals of the regular 18-gon. We will call a diagonal even if an even number of sides lies between its ends, and odd otherwise. The parity of a diagonal coincides with the parity of the difference in the numbers of its ends. Due to the evenness of the total number of sides of the polygon, it does not matter from which side of the diagonal to count the number of sides. We also consider the sides as odd diagonals. Two diagonals $\mathrm{AB}$ and $\mathrm{CD}$, where $\mathrm{AC}$ and $\mathrm{BD}$ do not intersect, are parallel if and only if the number of sides between A and C, and B and D, is the same, that is, the positive difference in the numbers of A and C is equal to the positive difference in the numbers of B and D. It is not difficult to notice that any odd diagonal is parallel to one of the nine sides of the 18-gon, and any even diagonal is parallel to one of the nine diagonals that cut off a triangle from the 18-gon (two sides of which are sides of the polygon). In total, there are 18 families of diagonals, any two diagonals of the same family are parallel, and any two diagonals of different families are not parallel. Nine of these families contain even diagonals and nine contain odd diagonals. As representatives of the odd families, we can take the sides with ends $1223, \ldots, 89,910$ and the diagonals $13,24, \ldots, 810,911$. Therefore, triangles with pairwise non-parallel sides, constructed on the vertices of a regular 18-gon, cannot use more than one from each of these families of diagonals, and the total number of such triangles does not exceed $18: 3=6$. Moreover, any triangle constructed on three vertices of the 18-gon can contain either three even diagonals or one even and two odd, since the sum of the parities of its three sides equals the parity of the number 18. Consequently, the total number of odd sides in any set of triangles with pairwise non-parallel sides must be even, and we will not be able to use all 18 families of diagonals for their sides. Thus, the number of triangles with pairwise non-parallel sides, constructed on the vertices of a regular 18-gon, does not exceed five.

Example. Five triangles with vertices $\{1,2,3\},\{3,4,5\},\{5,6,7\},\{7,8,9\},\{1,5,9\}$.

Grading criteria. It is shown that there are 18 pairwise non-parallel families of diagonals: 2 points. It is shown that all 9 non-parallel odd diagonals cannot be used as sides of triangles: 2 points. Example: 2 points. Justification of the example: 1 point.