Exercise 6. Find the largest integer $n \geqslant 3$ for which there exists a set $\mathcal{S}$ of $n$ points in the plane with the following property: every triangle (even degenerate) whose vertices belong to $\mathcal{S}$ is isosceles but not equilateral.
Solution to Exercise 6 We say that a set $\mathcal{S}$ having the required property is cute. We will prove that any cute set contains at most 6 elements. Furthermore, we will denote by $\omega_{A, B}$ the circle centered at $A$ passing through $B$, and by $\ell_{A, B}$ the perpendicular bisector of the segment $[A B]$.

First, if $\mathcal{S}$ consists of the five vertices $P_{1}, \ldots, P_{5}$ of a regular pentagon and its center $O$, it is easily verified that $\mathcal{S}$ is cute.

Conversely, let $\mathcal{S}$ be a cute set with $n \geqslant 6$ points: we will prove that $n=6$. For any points $X, Y$, and $Z$ in $\mathcal{S}$, we know that $Z$ belongs to one of the three sets $\omega_{X, Y}, \omega_{Y, X}$, or $\ell_{X, Y}$. This observation already allows us to prove the following result.

Lemma 1. Any three points in $\mathcal{S}$ are never collinear.