Given 10 points on a plane. We draw the perpendicular bisectors of the segments determined by them. In how many points can these lines intersect at most?
In the intersection points that arise, the perpendicular bisectors of segments whose endpoints form either a triangle or a quadrilateral intersect. Since the perpendicular bisectors of the sides of a triangle intersect at a single point, we get one intersection point per triangle. For each quadrilateral, up to 3 intersection points can arise, as we can choose pairs of non-adjacent sides from the 6 segments connecting the vertices of a quadrilateral in 3 different ways. The number of intersection points thus arising is at most $\binom{n}{3} + 3\binom{n}{4}$ for $n$ points, which is 750 for $n=10$.

We will show that this upper bound is achievable for every $n$. For this, it is necessary that the possible $\binom{n}{3}$ triangles and $\binom{n}{4}$ quadrilaterals are formed, and that no intersection point is counted more than once in the above calculation. One might think that by placing the points in a "general position," the circumcenters of the triangles do not coincide with other intersection points; however, this must be proven, which we do by induction on the number of points.

For three points placed at the vertices of a triangle, we clearly get a suitable set of points. Now let $n > 3$ and suppose we have already successfully placed the points $P_{1}, P_{2}, \ldots, P_{n-1}$. Denote the perpendicular bisector of the segment $P_{i} P_{j}$ by $f_{i j}$. The new point $P_{n}$ must be chosen such that:

a) the new $f_{n i}$ lines intersect every new $f_{n j}$ and every old $f_{i j}$ line;

b) the new perpendicular bisectors do not pass through any old intersection points (Figure 1);

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Figure 1

c) the intersection point of the new perpendicular bisectors does not lie on any old perpendicular bisector (Figure 2), except for the circumcenter of the new $P_{n} P_{i} P_{j}$ triangles (so in the figure, $f$ is not the same as $f_{i j}$);

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Figure 2

d) no three new perpendicular bisectors pass through the same point (Figure 3).

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Figure 3

The perpendicular bisectors of given segments intersect if and only if the segments are not parallel. Therefore, $P_{n}$ cannot lie on the lines connecting any two previous points (so that the new perpendicular bisectors intersect each other) or on lines parallel to the lines connecting two previous points and passing through a previous point (so that the new perpendicular bisectors intersect the old ones). If $P_{n}$ is chosen outside these finitely many lines, then all necessary intersection points will be formed.

If $Q$ is an old intersection point (Figure 1), then $f_{n i}$ passes through $Q$ if and only if $P_{n}$ lies on the circle centered at $Q$ with radius $Q P_{i}$. There are only finitely many such circles, so $P_{n}$ can be chosen so that it does not lie on any of these circles or on the previously excluded lines.

Finally, we claim that with this choice of $P_{n}$, neither the situation in Figure 2 nor the situation in Figure 3 can occur. In Figure 2, the new $f_{n i}$ and $f_{n j}$ lines intersect at $Q$, and the old $f_{i j}$ also passes through $Q$, so $Q$ is the intersection of two old perpendicular bisectors, $f_{i j}$ and $f$. By the choice of $P_{n}$, no new perpendicular bisector passes through such a point. The point $Q$ in Figure 3, being the circumcenter of the old $P_{i} P_{j} P_{k}$ triangle, is also the intersection of old perpendicular bisectors and thus cannot lie on a new perpendicular bisector.

We have thus shown that the above choice of $P_{n}$ is suitable, and this completes the proof.

Therefore, it is possible to place 10 points in the plane such that all 750 intersection points arise.

Remark. Most solutions only contained the initial calculation, and generally lacked the proof that the given bound is indeed achievable. Many were satisfied with ensuring that no three of the given points are collinear and no four lie on a circle. This is insufficient, as it is still possible to choose 3 segments such that their perpendicular bisectors intersect at a single point.