# Proving that two curves that are symmetric about the origin have same curvature and same torsion (up to a sign)

This is an exercise in a differential geometry book I'm studying, and currently I can't fathom why it's there (and it doesn't feel at all intuitive why it would be true). A simple counter example would be the curves $\alpha(t) = (t, t^3,t^5)$ and $\beta(s) = (s,s,s)$. A quote of the exercise is as follows: "Prove that, if two curves are symmetric about the origin, then they have the same curvature and the torsion differs only up to a sign". What's happening here? Did I completely misinterpret something?

If $\alpha\colon I \to \Bbb R^3$ is a regular curve and $\beta :I \to \Bbb R^3$ is defined by $\beta(t) \doteq -\alpha(t)$, then $\kappa_\beta = \kappa_\alpha$ and $|\tau_\beta| = |\tau_\alpha|$.
You can assume they both have unit speed. Then $T_\beta = -T_\alpha$, hence $\kappa_\beta N_\beta = -\kappa_\alpha T_\alpha$, whence $\kappa_\beta = \kappa_\alpha$ and so $N_\beta = -N_\alpha$. Then $B_\beta = -B_\alpha$, and follows from this (differentiating) that $|\tau_\beta| = |\tau_\alpha|$.