$A_{1}, A_{2}, \ldots, A_{n}$ are distinct points in the plane. Color red the midpoints of all segments determined by them. What is the minimum number of red points created?
I. solution. The given points determine a finite number of lines. Thus, there exists a line that is not parallel to any of these. Draw a parallel line to one of these through each of the given points. These parallel lines are all different from each other. Intersect these parallel lines with another line. The set of intersection points consists of as many points as the given point set, and the elements of the two sets are mapped to each other by the parallel lines in a one-to-one correspondence. By coloring the midpoints of the segments determined by these intersection points blue, we can conclude that there are at most as many blue points as red ones, since if the midpoints of two pairs of the given points coincide, then the midpoints of the images of these pairs (their parallel projections) also coincide. Let the number of given points be $n$. Let the two extreme points among the intersection points be $A$ and $B$, and the midpoint of $AB$ be $F$.

The number of blue points is at least $2(n-2)+1=2n-3$, for by reducing the middle $n-2$ intersection points to half from the midpoint of $A$, and from the midpoint of $B$, we get all blue points, of which the former are all interior points of the segment $AF$, and the latter are all interior points of the segment $BF$, and the point $F$ is also blue. Consequently, the number of red points is also at least $2n-3$.

The number of red points can be $2n-3$, for example, if the points $A_1, A_2, \ldots, A_n$ are taken at the points $1, 2, \ldots, n$ of a number line. In this case, the midpoints are precisely the fractions with a denominator of 2 between 1 and $n$, so their number is exactly $2n-3$.

II. solution. If $n=1$, then no midpoint is created. We will show that if $2 \leq n$, then the minimum number of red points is $2n-3$.

First, if $A_i$ is the point on the number line that marks the number $i-1$, then exactly those points will be red that mark a positive number greater than 0 but less than $2n-2$; this is exactly $2n-3$ points.

Now we only need to show that for any arrangement of the points $A_i$, there will be this many red points. We prove this by complete induction on $n$.

If $n=2$, then there is always exactly $1=2 \cdot 2-3$ red point. If $n$ points always determine at least $2n-3$ red points, then starting from the points $A_1, \ldots, A_{n+1}$, we draw the convex hull $C$ of the set $P=\{A_1, \ldots, A_n\}$. Let one of its vertices be $B$, and let $e$ be a line that is tangent to $C$ at $B$. Let $B_1$ and $B_2$ be two points of the set $P \backslash \{B\}$ that are as close as possible to the line $e$. The set $P \backslash \{B\}$, by our inductive hypothesis, determines at least $2n-3$ red points; let this set be $F$. The midpoints of the segments $B_1B$ and $B_2B$ do not coincide, and they cannot be elements of $F$ either, since both points are closer to the line $e$ than the midpoint of the segment $B_1B_2$, while the points of $F$ are at least as far from $e$ as the midpoint of the segment $B_1B_2$. This is thus two more red points, so the set $P$ determines at least $(2n-3)+2=2(n+1)-3$ red points, as we needed to prove in the inductive step.