In the plane, there are 5 points. Among the lines that connect these 5 points in pairs, there are no parallel, perpendicular, or coincident lines. From each of the given points, we draw perpendiculars to all the lines formed by connecting the remaining four points in pairs. What is the maximum number of intersection points of these perpendiculars, if we ignore the given 5 points?
The number of lines connecting the given 5 points pairwise is $5 \cdot 4 / 2 = 10$, because from each point, 4 lines extend to the other points, and in the product $5 \cdot 4$, each such line is counted at 2 points; these lines are all assumed to be in different directions. For each connecting line, we draw a perpendicular from 3 other points, which, by assumption, are all different, so the total number of perpendiculars is $10 \cdot 3 = 30$, and these are parallel in threes.

Selecting two of the connecting lines, the 3 perpendiculars to each of them intersect at 9 points. The 10 connecting lines can be chosen in $10 \cdot 9 / 2 = 45$ ways, similar to the above reasoning, so among the 30 perpendiculars, $45 \cdot 9 = 405$ pairs of intersections are created.

However, in many cases, the pair of perpendiculars intersects at one of the given points. Through each of the given points, we draw 6 perpendiculars - i.e., to the lines connecting the other 4 points pairwise, $4 \cdot 3 / 2$. Each pair of these 6 perpendiculars intersects at the selected given point, so we get 1 such intersection for each of the $6 \cdot 5 / 2 = 15$ pairs of lines. The 5 given points result in $5 \cdot 15 = 75$ such pairs of intersections that should be ignored.

Coincidences can also occur at points different from the given ones. Any three of the original points define a triangle, and among the perpendiculars being examined, each altitude of such a triangle appears, and every drawn perpendicular is an altitude of such a triangle. Thus, at the orthocenter of each triangle, three of the pairwise intersections coincide, and the number of expected intersection points should be reduced by 2 for this reason. - The number of triangles defined by the given points is the same as the number of pairwise connecting lines, i.e., 10 - i.e., the connected point pair uniquely determines the triangle of the remaining points, and vice versa. The orthocenter never falls on a vertex of the triangle, as this only happens in a right triangle, and the given points do not define a right triangle. The number of expected intersection points is thus reduced by $10 \cdot 2$, and the maximum number of distinct intersection points to be considered is $405 - 75 - 20 = 310$.

Klára Staub (Budapest, Apáczai Csere J. Secondary School III. grade)

Note. Many solutions started from the $30 \cdot 29 / 2 = 435$ pairs of intersections that can be imagined through the 30 perpendiculars, and by accounting for the 30 missing intersections due to parallelism, they arrived at the 405 pairs, which is the starting point of the above solution.

However, the following reasoning avoids counting the intersections falling into the given points. From the two selected given points, 6 perpendiculars each extend. By selecting one from each, we get 3 pairs of parallel lines, i.e., when we select the perpendiculars to the same side of the triangle formed by the other 3 points, and in the remaining $6 \cdot 6 - 3 = 33$ cases, an intersection point different from the original points is created. The 10 point pairs give 330 points, from which the further reduction due to the orthocenters must be subtracted.