15 A finite set of points $M$ on the plane satisfies: for any two points $A, B$ in $M$, there must exist a third point $C$ such that $\triangle A B C$ is an equilateral triangle. Find the maximum number of elements in $M$.
15 It is evident that the set of three vertices of an equilateral triangle satisfies the requirement, so
$$
\max |M| \geqslant 3 .
$$

Since $M$ is a finite set of points, among the equilateral triangles formed by three points in $M$, there must be a largest one. Let it be $\triangle A B C$, then all points in $M$ are within the common part of the three circles centered at $A, B, C$ respectively, with $A B$ as the radius (as shown in Figure 37).

If there is a 4th point $P \in M$, without loss of generality, assume $P$ is within $\angle Q A C$, where $Q$ is the midpoint of the arc $\overparen{B C}$.
(1) $P$ is within the segment $B Q C$.

In this case, there is a point $P^{\prime} \in M$ that forms an equilateral triangle with $A$ and $P$. Since the tangent line $A T$ at point $A$ to the arc $\overparen{A C}$ forms a $30^{\circ}$ angle with $A C$, and $\angle P A T = \angle A C B = 60^{\circ} > 30^{\circ}$,
$$

thus $P^{\prime}$ is outside the segment formed by the arc $\overparen{A B}$. This is impossible.
(2) $P$ is within $\triangle A B C$.

Since $\angle P A B < 60^{\circ}$, $P^{\prime}$ must be within the segment formed by the arc $\overparen{A B}$ and $A B$. Due to (1) (replacing $P$ with $P^{\prime}$), this is also impossible. Therefore, the maximum number of elements in $M$ is 3.