4. In the plane, there is a set $S$ of $n$ points ( $n>2$ ) with the property: if $A, B \in S$, then there exists $C \in S$ such that triangle $A B C$ is equilateral. How many can the number $n$ be?
Solution. Let us choose points $A, B \in S$ such that the distance $AB$ is as large as possible. Let point $C \in S$ be such that triangle $ABC$ is equilateral. We will prove that the set $S$ cannot contain more than three points, i.e., that $n=3$.

We construct circles with centers at $A$,

![](https://cdn.mathpix.com/cropped/2024_06_05_72b79cc38cc70f646757g-06.jpg?height=367&width=539&top_left_y=600&top_left_x=920)

$B$, and $C$, and radii equal to the segment $AB$, as shown in the diagram on the right. Due to the way points $A$ and $B$ are chosen, any other points in the set $S$, different from $A, B, C$, cannot lie outside any of these circles. We will prove that they cannot be inside their common interior region, nor on the arcs $AB, BC, CA$ of these circles. It is sufficient to prove that there are no points in the region bounded by segments $OB$ and $OC$ (where $O$ is the center of triangle $ABC$), and arc $BC$—let us denote this region by $F$. Suppose there is a point $D \in S$ in region $F$, different from $B$ and $C$. Then for some $D' \in S$, triangle $ADD'$ is equilateral, i.e., point $D'$ is obtained by rotating point $D$ around point $A$ by an angle of $60^\circ$ (in some direction). During this rotation, one of the points $B$ and $C$ is mapped to the other, for example, $B$ is mapped to $C = B'$. It is easy to prove that in this process, the segment $BO$ is mapped to a segment $B'O'$ which lies on the tangent to the circle $(B, AB)$ at point $C$. From this, it follows that the entire region $F$ except for the point $B$ is mapped to a region where we have already proven that there are no points of the set $S$. This means that $D' = B'$ and $D = B$, which is impossible. The contradiction obtained proves that the set $S$ does not contain any other points besides $A, B, C$.

## Fourth Year