[Example 1.4.4] For a set of $n$ points $M$ in a plane, if every three points in $M$ form the vertices of an equilateral triangle, find the maximum value of $n$.


Obviously, the set of points $M$ formed by the three vertices of an equilateral triangle has the property mentioned in the problem. This is the simplest case.
To find the maximum value of $n$, we can start from the simple case $n=3$ and gradually increase the number of points, so we can consider the case of $n=4$.
Starting from the simple case, we first consider a three-point set $\{A, B, C\}$, where $\triangle A B C$ is an equilateral triangle.
Since $n=4$, if every three points form the vertices of an equilateral triangle, there will be $C_{4}^{3} = 4$ equilateral triangles.
How do we construct the first equilateral triangle?
A natural idea is to find three points in $M$ that form the largest equilateral triangle. This way, the 4th point will be controlled within the range of this largest equilateral triangle. This strategy focuses on extreme cases, i.e., considering the two points in the point set $M$ that are farthest apart.
Assume $M$ is a point set with at least 4 points, and every three points can form the vertices of an equilateral triangle.
Since the finite point set $M$ has at least 4 points, there must be two points with the maximum distance. Let the distance between point $A$ and point $B$ be the maximum.
By the problem's condition, there exists a point $C$ such that $\triangle A B C$ is an equilateral triangle.
Draw arcs with $A, B, C$ as centers and $A B$ as the radius, forming a curved triangle $A B C$.
Since $A B = B C = C A$ is the maximum distance between any two points in $M$, the distance between any other point in $M$ and $A, B, C$ will not exceed $A B$, so all points in $M$ are within or on the boundary of the curved triangle $A B C$.

Assume the set $M$ has another point $P$ different from $A, B, C$. We consider the position of point $P$.
(1) If point $P$ is inside or on the boundary of $\triangle A B C$.
At this time, by the problem's condition, every three points in $M$ form the vertices of an equilateral triangle, so for points $A, P$, there is a third point $P_{1}$ such that $\triangle A P P_{1}$ is an equilateral triangle.

Since $\angle B A P \leqslant 60^{\circ}$, point $P_{1}$ cannot be inside $\triangle A B C$ but can only be inside or on the boundary of a certain segment of the arc.

Similarly, for points $C$ and $P_{1}$, there must be a third point $P_{2}$ such that $\triangle C P_{1} P_{2}$ is an equilateral triangle.

If point $P_{2}$ is on the same side of line $C P_{1}$ as point $A$, consider $\triangle C P_{2} A$ and $\triangle C P_{1} B$. Since
$$
\begin{array}{c}
C P_{2} = C P_{1}, \\
C A = C B, \\
\angle P_{2} C A = \angle P_{1} C B = 60^{\circ} - \angle A C P_{1},
\end{array}
$$

then
$$
\triangle C P_{2} A \cong \triangle C P_{1} B.
$$

Thus,
$$
\angle P_{2} A C = \angle P_{1} B C = \angle P_{1} B A + 60^{\circ} > 60^{\circ}.
$$

At this time, $P_{2}$ is outside the curved triangle $A B C$, so
$$
P_{2} B > A B.
$$

This means that the distance between point $P_{2}$ and $B$ in the point set $M$ is greater than the distance between $A$ and $B$, which contradicts the fact that $A$ and $B$ are the points with the maximum distance in $M$.
If point $P_{2}$ is on the same side of line $C P_{1}$ as point $B$, the same contradiction can be derived.
This shows that in the point set $M$, a fourth point $P$ different from $A, B, C$ cannot appear inside or on the boundary of $\triangle A B C$.
(2) If point $P$ is inside or on the boundary of a certain segment of the arc, at this time, point $P$ is equivalent to point $P_{1}$ in (1). In this case, using point $P$ and the point corresponding to the arc segment where point $P$ is located as two vertices to form an equilateral triangle, by the proof in (1), point $P_{1}$ will be outside the curved triangle, leading to a contradiction.
Therefore, point $P$ will not appear inside or on the boundary of a certain segment of the arc.
From the above argument, we can conclude that the maximum number of points in the point set $M$ we are looking for is 3.