67th NMO Selection Tests for BMO and IMO

The required configurations consist of the three vertices of an equilateral triangle. Clearly, the three vertices of an equilateral triangle form a configuration satisfying the condition in the statement.

To prove the converse, let $a$ and $b$ be the end points of a diameter of $C$ and notice that exactly one of the two equilateral triangles erected on the segment $ab$ has the third vertex $c$ in $C$.

Suppose, if possible, $x$ is a fourth point in $C$. Since the segments $ab$, $bc$ and $ca$ are all three diameters of $C$, the point $x$ is interior to (at least) one of the angles $abc$, $bca$, $cab$, say the latter. Consider the equilateral triangles $axy$ and $axz$, where $b$ and $z$, respectively $c$ and $y$, both lie on the same side of the line $ax$. Since $C$ contains at least one of the points $y$ and $z$, both of which lie outside the triangle $abc$ (this is because the angles $bax$ and $cax$ are both less than $\pi /3$), we may and will further assume that so does $x$, so the quadrangle $abxc$ is convex.

Since $ax\le ac$, it follows that $\angle axc\ge \angle acx>\pi /3$, so $y$ is interior to the angle $axc$, and $\angle bxy<\angle bxc<\pi$. And since $\angle cay=\angle bax<\pi /3$, it follows that $\angle bay=\angle bac+\angle cay=\pi /3+\angle cay<2\pi /3<\pi$, so the quadrangle $abxy$ is convex. Hence $ax+by>ab+xy=ab+ax$, i.e., $by>ab$. Similarly, $cz>ac$, and since $ab$ and $ac$ are both diameters of $C$, and the latter contains at least one of the points $y$ and $z$, we reach a contradiction.