For every line in the 2d plane, we can construct a shape with an "inside and outside" (often a circle) such that the shape is cut by the line into two symmetrical parts.
Does this property of lines extend to all curves?
My idea is if you "zoom in" on a curve, if you get small enough, it might become a line segment, then it would obviously extend. But does anybody have any ideas on a real proof?
REVISION: Let me rephrase my question as the following:
Prove or disprove the following statement:
For any $C^1$ curve, we can construct a shape (an object with a boundary and thus a clear inside and outside) such that the smooth curve cuts the shape into two congruent pieces.
Best Answer
While it is false to say that a curve becomes a line segment if you zoom in small enough, still though the idea of that thought gave me an idea for a simple proof.
The symmetry that I get is not a reflective symmetry (as it is for a straight line cutting a circle in half), it is instead a translational symmetry. But you didn't specify the type of symmetry, so this is presumably acceptable.
Let $\Gamma$ be the given $C^1$ curve. Choose $P \in \Gamma$ with tangent line $L$. Choose a closed arc $\alpha \subset \Gamma$ that contains $P$ in its interior, and choose $\alpha$ so short that the orthogonal projection from $\alpha$ to $L$ is a $C^1$-diffeomorphism. Let $A \subset L$ be the image of this projection.
To simplify matters, we can apply a rigid Euclidean motion to $\Gamma$ so that $A = [-T,T] \times \{0\}$ for some $T > 0$. The arc $\alpha$ then becomes the graph of some $C^1$ function $f : [-T,T] \to \mathbb R$, $$\alpha = \{(t,f(t)) \mid t \in [-T,T]\} $$ Now pick $\rho > 0$ and consider the region $$R = \{(t,f(t)+r) \mid t \in [-T,T], r \in [-\rho,\rho]\} $$ If $\rho$ is sufficiently small, its intersection with the curve $\Gamma$ is equal to $\alpha$, and $\alpha$ cuts $R$ into two congruent pieces $R = R_- \cup R_+$ where $$R_- = \{(t,f(t)+r) \mid t \in [-T,T], r \in [-\rho,0]\} $$ $$R_+ = \{(t,f(t)+r) \mid t \in [-T,T], r \in [0,\rho]\} $$ The rigid motion $(x,y) \mapsto (x,y+\rho)$ demonstrates the translational symmetry between $R_-$ and $R_+$.