TITLE: Self-Intersection of closed curves
QUESTION [9 upvotes]: Supoose I have a closed curve $\gamma$ in the plane such that for any isometry $g$ of $\mathbb{E}^2,$ such that $g(\gamma)\neq \gamma,$ $\gamma$ intersects $g(\gamma)$  in at most two points. It should be an easy corollary of the four-vertex theorem that $\gamma$ is a circle, but I am not quite seeing it.

REPLY [6 votes]: Igor Pak's argument takes care of the general convex case. So to complete the proof (without assuming any smoothness) it remains to consider the nonconvex case. To this end it is enough to note that if $\gamma$ is not convex, then it must intersect some line $L$ at least $4$ times. Let $g(\gamma)$ be the reflection of $\gamma$ with respect to $L$. Then $\gamma$ and $g(\gamma)$ intersect at least $4$ times.