# Show that two intersecting curves on a regular surface with the same osculating plane that is not the tangent plane have the same curvature

Let $\bf{p}$ be a point on a regular surface $S$. Let $\boldsymbol{\alpha}(s)$ and $\boldsymbol{\beta}(s)$ be two curves parametrized by arc length on the surface $S$ such that $\boldsymbol{\alpha}(0) = \boldsymbol{\beta}(0) = \bf{p}$.

Denote the Frenet trihedron of $\boldsymbol{\alpha}(s)$ at $\bf{p}$ by $\bf{t}_{\alpha}$, $\bf{n}_{\alpha}$, $\bf{b}_{\alpha}$; the Frenet trihedron of $\boldsymbol{\beta}(s)$ at $\bf{p}$ by $\bf{t}_{\beta}$, $\bf{n}_{\beta}$, $\bf{b}_{\beta}$.

Suppose both curves have the same osculating plane at $\bf{p}$ and the osculating plane is not equal to the tangent plane at $\bf{p}$.

i) Show that $\bf{t}_{\alpha}$ and $\bf{t}_{\beta}$ are parallel, i.e. $\bf{t}_{\alpha} = \pm\bf{t}_{\beta}$.

ii) Show that $\bf{n}_{\alpha}$ and $\bf{n}_{\beta}$ are parallel, i.e. $\bf{n}_{\alpha} = \pm\bf{n}_{\beta}$.

iii) Show that both curves have the same curvature at $\bf{p}$.

I have only succeeded in showing i), where the strategy was to first assume they are not parallel. Taking the cross product of the two tangent vectors would then produce the normal of the osculating plane, a non-zero vector parallel to the Gauss map at $\bf{p}$. Since the tangent plane at $\bf{p}$ is defined by the normal vector at $\bf{p}$, i.e. the Gauss map, this implies that the osculating plane is the same as the tangent plane, which contradicts the setup.

This question was from an exam I took recently. Thank you.

You've done the hard part. Since the osculating planes of $\alpha$ and $\beta$ at $\mathbf p$ are the same and the tangent vectors are ($\pm$) the same, ii) follows. Here's a major hint for iii): Apply Meusnier's formula, which relates the normal curvature of $S$ in the direction of $\mathbf t_\alpha = \pm\mathbf t_\beta$ and the curvature of $\alpha$ (or $\beta$). What's the missing ingredient?
EDIT: Meusnier tells us that $\kappa_n = \kappa \mathbf n\cdot \nu = \kappa\cos\phi$, where $\nu$ is the surface normal and $\phi$ is the angle between the curve's principal normal and the surface normal. So if $\kappa_n$ is the same for both curves (as it depends only on the tangent direction of the curve), now what do you have?