Pechkovsky A.N.

On a plane, there are $N$ points. Let's mark the midpoints of all possible segments with endpoints at these points. What is the smallest number of marked points that can result?
Consider all possible distances between the given $N$ points and select the greatest among these distances - let this be the distance between some two points $A$ and $B$. Connect point $A$ to the other $N-1$ points; we obtain $N-1$ segments. The midpoints of these segments are distinct and all lie inside or on the boundary of a circle centered at point $A$ with radius $\frac{1}{2}|A B|$. Similarly, connecting point $B$ to the remaining $N-1$ points, we obtain $N-1$ marked points (midpoints) located inside or on the boundary of a circle of the same radius centered at point $B$.

The two constructed circles have only one common point - the midpoint of segment $A B$. Therefore, there are at least $2(N-1)-1$ (since the midpoint of the segment is counted twice) marked midpoints, that is, no fewer than $(2N-3)$ marked points.

We will show that $2N-3$ is the smallest number of marked points. Suppose the given $N$ points lie on a straight line, and the distances between adjacent points are equal. It is easy to see that in this case, the number of marked points - midpoints is $(N-2)+N-1=2N-3$. Thus, everything is proven.

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