I recently proved as excercise the following fact
Let C be a curve in $\mathbb{R}^3,$ with torsion $\neq 0$ everywhere; for $R$ an oriented line, the following are equivalent:
- The angle between the tangent line to $C$ in $u$ and $R$ is some constant $C\in (0,\pi)$ which does not depend on $u.$
- The normal $N(u)$ at $C$ in $u$ is orthogonal to $R$ for every $u$
- The angle between the binormal $B(u)$ at $C$ in $u$ and $R$ is some constant $C\in (0,\pi)$ which does not depend on $u.$
Moreover, the existence of line $R$ satisfying $(i)\iff (ii) \iff (iii)$ above is in turn equivalent to:
- The ratio $k(u)/\tau(u)$ between curvature and torsion is constant, i.e. it does not depend on $u.$
So far so good, and I managed to prove this without assuming anything about the curve $C$ more than what is stated above. An example of such a curve $C$ satisfying $4$ is given by an helix.
What I also find written, but I am not able to prove myself, is that any curve $C$ such that $4.$ is true is necessarily an helix. How to prove this ?
I add that by (cylindrical) helix I mean the curve in 3D space parametrised as $(r \cos t, r \sin t, ht)$.
I want to add that I know cylindrical helixes are uniquely characterized by the fact that they have both curvature and torsion constant, but I'm not seeing how to show from this the equivalence between point 4. and being a cylindrical helix.