This question spawned from this recent thread. The notorious square peg problem states that any continuous, simple and closed curve $\gamma$ in the plane contains the vertices of some square. It has been proved under many different regularity assumptions, but it still remains an open problem in its original formulation. I realized that the solution I proposed in the linked thread contains the main ideas of a solution of the $C^2$ case of the square peg problem: for any $B\in\gamma$, by considering a sequence of circles centered at $\gamma$ with slowly increasing radii we get a solution of $A,C\in\gamma$, $AB=BC$ and $AB\perp BC$. Given an isosceles and right triangle $ABC$, we may define $B'$ as the symmetric of $B$ with respect to the midpoint of $AC$. If for some $B\in\gamma$ we have $B'\in\gamma$ the square peg problem is solved, and solving the square peg problem with suitable smoothness assumptions boils down to showing that two curves intersect (like in Gauss' original proof of the fundamental Theorem of Algebra).

$\hspace{1in}$enter image description here

The green curve and the ellipse intersect, hence the square peg problem can be solved in an ellipse.

Context's over. Now I propose the following variation: let $\gamma$ be a (simple,) closed curve of class $C^2$ without any inflection point. Given some $P\in\gamma$, we define $P'$ as the symmetric of $P$ with respect to the center of curvature of $P$. We denote as $\gamma'$ the set $\{P':P\in\gamma\}$.

Q1. Is it true that $\gamma$ and $\gamma'$ always have an intersection?

Q2. Has this problem already been studied in the literature?


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