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Let $b >0$ and $\alpha : (-a, a) \rightarrow \mathbb{R^2}$ be a plane curve parametrized by arc length. Suppose that: $$k(s) = k(-s) \quad \forall s \in (-a,a)$$ where $k(s)$ denotes the curvature of $\alpha$ at the point $s$. Show that the trace of $\alpha$ is symmetric with respect to the normal line of $\alpha$ at $0$.

My try

I denote by $\phi:\mathbb{R}^2\rightarrow\mathbb{R}^2$ the function that takes a point $b$ to his symmetric relative to $\alpha(0) + L(N_{\alpha}(0))$ (normal line at zero). Then I define the curve $\beta(s)=\phi(\alpha(-s))$ and my attempt is to prove that $\beta=\alpha$.

I know that $\beta(0)=\alpha(0)$, so I tried to show that the curves have the same curvatures $K_{\alpha}=K_{\beta}$ and the same tangents at zero $T_{\alpha}(0)=T_{\beta}(0)$, but I don't know how to show this.

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Your approach seems great. So, if you differentiate carefully and correctly with the chain rule, you should find exactly that. (If we take $\alpha(0)=0$, then $\phi$ will be linear, and so $(\phi\circ\gamma)'(s) = \phi(\gamma'(s))$.) You should get $$\beta'(s) = -\phi(T_\alpha(-s)) \quad\text{and}\quad \beta''(s) = \kappa(s)\phi(N_\alpha(-s)).$$ So we have $T_\beta(0)=-\phi(T_\alpha(0)) = T_\alpha(0)$ and $\kappa_\beta=\kappa_\alpha$ (why?).

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  • $\begingroup$ $\phi$ is linear bacause its a rigid motion? $\endgroup$ – Eduardo Silva Jan 13 '18 at 1:39
  • $\begingroup$ Oh, I was assuming $\alpha(0)=0$, so it's reflection about a line through the origin. $\endgroup$ – Ted Shifrin Jan 13 '18 at 1:42
  • $\begingroup$ can i supose this without lost of generality? that's the only way that $\phi$ can be linear? $\endgroup$ – Eduardo Silva Jan 13 '18 at 2:12
  • $\begingroup$ It's not that big a deal, but yes, you can compose with an isometry that translates the point to the origin. Otherwise you need to write $\phi(x)=Ax+b$ and then, differentiating, $A$ will show up everywhere rather than the entire $\phi$. $\endgroup$ – Ted Shifrin Jan 13 '18 at 2:19
  • $\begingroup$ last question (i guess), $-\phi(T_{\alpha}(0)=T_{\alpha}(0)$ by the same reason as $-\phi(\alpha(0))=\alpha(0)$ ? $\endgroup$ – Eduardo Silva Jan 13 '18 at 2:28

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