Find the number of elements of set $M$ having: $\forall A, B \in M$ there is $C \in M$ such that the triangle $ABC$ is equilateral 
Let $M$ a finite set of points in a plane such that $\forall A, B \in
 M$ there is $C \in M$ such that the triangle $ABC$ is equilateral.
   Find the possible number of $M$ elements.


My guess is the number is 3. It is easy to show there isn't such a set with, let's say, 4 elements, but I cannot cope with the general case.
 A: Take the two points $A$ and $B$ farthest away in the set $M$, i.e. $|AB|=d$, the diameter of $M$. This is legit, since $M$ is a finite set, and thus there are two points achieving the diameter.
There must be some point $C$ such that $ABC$ is equilateral. Now, observe that since $AB=BC=CA=d$, the diameter of $M$, all points in $M$ must lie in the following figure $\Phi$ (I can't draw a diagram, but I can describe it explicitly).
$\Phi$ consists of the triangle $ABC$ together with three circular segments $s_A,s_B,s_C$, where $s_A$ is the segment of the circle centered at $A$ with radius $d$ bounded by the chord $BC$. $s_B,s_C$ are defined similarly.
Now suppose there's some other point $X$ in the set. First suppose $X$ is inside or on the sides of the triangle $ABC$. Split $ABC$ into three triangles by connecting the vertices to the center $O$, and suppose $X$ falls inside or on the sides of WLOG $ABO$. Then, there must be a point $Y\in\Phi$ such that $CXY$ is equilateral. Thus, $Y$ is inside the image of $ABO$ under rotation centered at $C$ by $60^\circ$. The two possible images of $ABO$ under such a rotation are two triangles that can be seen to lie outside of $\Phi$ except for the images of $A$ and $B$ under them, but we know $X\neq A,B$ by assumption. The key observation needed to see that is that the tangents to the circle centered at $C$ with radius $d$ at $A$ and $B$ respectively make $30^\circ$ angles with $AB$, and the image of $ABO$ is on one side of these tangents, while the segment $s_C$ is on the other. So, $X$ can't be inside $ABC$.
Similarly, suppose $X$ is WLOG in or on the boundary of $s_A$. Then there is some $Y\in\Phi$ such that $AXY$ is equilateral, i.e. $Y$ is in the image of $s_A$ under a rotation centered at $A$ with radius $d$. But again, it's easy to see that the two possible images intersect $\Phi$ only at $B$ and $C$, which are not allowed for $Y$.
So we conclude that there can be no other points in $M$, thus $|M|=3$.
A: Do the construction.  You choose $A$ and $B$.  Draw circles of radius $AB$ about the points $A$ and $B$.  The intersections of these circles have to be a distance $AB$ away from both.  We see two points of intersection.  Clearly, $|M| = 3$ will work by choosing either of the points of intersection.  However, if $|M| > 3$ then $M$ has to include the other point of intersection (because both points have to make an equilateral triangle with $A$ and $B$ as required by the definition of $M$).  But then do a similar analysis by drawing circles of radius $CD$ about $C$ and $D$ and see that neither intersect at $A$ and $B$ as required.  Thus, $|M|$ = 3.
A: 
We'll prove tham $|M|=3$
Suppose that $|M|\geq 4$ then say that points $A,B,C,E$ are different points in plane then:
$ABC$ is equilateral triangle.
Let D be such that CD is perpendicular to AB, then from equilateral triangle property we know that $CD$ is perpendicular bisector of $AB$, so also height of triangle $ABE$ should be also bisector of $AB$,since D is midpoint of AB then height of $ABE$ is ED, so ED will be the prependicular bisector of $AB$, which means that E should be on line CD. 
Since $\angle{DAE}=60$ then either E=C or is symmetric point of C with respect to AB, but since E and C different then the second holds.
Referring to this figure then $ABE$ is equilateral but then $$\angle{CBE}=\angle{CBA}+\angle{ABE}=60+60=120$$
which means that triangle $CBE$ is not equilateral triangle.

So $|M|=3$
 
The idea of solution is that for any 2 points $X,Y$ and point $Z$, $XYZ$ make equilateral triangle if $Z$ lies in perpendicular bisector line of $XY$  
