12.10. (SFRY, 76). Find all values of $n \in \mathbf{N}$, greater than 2, for which it is possible to select $n$ points on a plane such that any two of them are vertices of an equilateral triangle, the third vertex of which is also one of the selected points.
12.10. We will prove that the condition of the problem is satisfied only by the value $n=3$ (in which case the points can be placed at the vertices of an equilateral triangle). Indeed, suppose that it is possible to arrange $n \geqslant 4$ points in the manner specified in the problem. We select two points $A$ and $B$, the distance between which is maximal, and a third point $C$, such that the triangle $A B C$ is equilateral. Then all the other points lie within the figure $M$, which is the intersection of three circles of radius $A B$ centered at points $A, B$, and $C$ (Fig. 68). If point $O$ is the center of the triangle $A B C$, then the segments $A O, B O$, and $C O$ divide the set $M$ into three equal parts, none of which can contain any of the $n$ points, except those already mentioned. Indeed, suppose, for example, that part $M_{B C}$, consisting of the triangle $B O C$ and the segment of the circle centered at $A$ with the arc $B C$, contains another point $D$. Then there exists a point $D^{\prime}$ such that the triangle $A D D^{\prime}$ is equilateral. Consequently, when rotating (in a certain direction) around point $A$ by an angle of $60^{\circ}$, point $D$ transitions to point $D^{\prime}$, which thus lies simultaneously in the image $M_{B C}^{\prime}$ of part $M_{B C}$ under this rotation and in the set $M$. But since $\angle B A C=60^{\circ}$, either $C^{\prime}=B$ or $B^{\prime}=C$. For definiteness, assume that $B^{\prime}=C$. Then the sets $M_{B C}^{\prime}$ and $M$ lie in different half-planes relative to the line $B^{\prime} O^{\prime}$, because

$$
\angle B B^{\prime} O^{\prime}=\angle B C A+\angle A B^{\prime} O^{\prime}=60^{\circ}+30^{\circ}=90^{\circ},
$$

i.e., the line $B^{\prime} O^{\prime}$ is tangent to the arc $A C$. Therefore, $M_{B C}^{\prime}$ and $M$ have only one common point $B^{\prime}$, from which it follows that $D^{\prime}=B^{\prime}$ and $D=B$, which contradicts the choice of point $D$. The statement is proved.