Suppose $\alpha(s)$ and $\beta(s)$ are two plane curves with non-zero curvatures. I want to show that if $\alpha$ and $\beta$ are Bertrand mates (two curves are called Bertrand mates if they have the same normal lines at each parameter), then $\alpha(s)=\beta(s)$ for each $s$, in other words $\alpha$ and $\beta$ are the same curve iff $s$ is the arclength parameter.
My progress:
First I’m gonna assume $s$ is the arclength parameter. So we get $\alpha’’(s)\perp \alpha’(s)$ and similarly $\beta’’(s)\perp\beta’(s)$. So the parametric equation of the normal lines to $\alpha(s)$ and $\beta(s)$ at each $s$ (denoted by $L_{\alpha}(t)$ and $L_{\beta}(t)$ respectively) are:
$L_{\alpha}(t)=t\alpha’’(s)+\alpha(s)$
$L_{\beta}(t)=t\beta’’(s)+\beta(s)$
Since the normal lines are the same at each $s$ we get:
$\forall s\,\forall t: t\alpha’’(s)+\alpha(s)= t\beta’’(s)+\beta(s)$
Or equivalently:
$\forall s\,\forall t: t(\alpha’’(s)-\beta’’(s))= \beta(s)-\alpha(s)$
Now if $\alpha\neq\beta$, then there exists an $s_0$ such that $\alpha(s_0)\neq\beta(s_0)$ implying $\|\beta(s_0)-\alpha(s_0)\|\neq 0$.
So we have:
$\forall t: t(\alpha’’(s_0)-\beta’’(s_0))= \beta(s_0)-\alpha(s_0)$
Implying:
$\forall t: |t|\|\alpha’’(s_0)-\beta’’(s_0)\|= \|\beta(s_0)-\alpha(s_0)\|$
Dividing both sides by the non-zero $\|\beta(s_0)-\alpha(s_0)\|$ and then by $|t|$ (assuming $t\neq0$) yields:
$\underbrace{\dfrac{\|\alpha’’(s_0)-\beta’’(s_0)\|}{\|\beta(s_0)-\alpha(s_0)\|}}_{\text{constant}}=\dfrac1{|t|}\quad\forall t\neq 0$
which is contradiction.
What’s wrong with my proof?
I kinda didn’t use the fact that both curves are parametrized by the arc length.
$\mathbf{Update}:$
I was wrong since for each $s$, the two normal lines need not have the same point for each parameter $t$ as Prof. Shifrin Stated. So I updated my progress.
Since the two curves are Bertrand mates we conclude that:
$\beta(s)=\alpha(s)+\lambda\mathbf{N}_{\alpha}(s)$
So the distance between the points $\alpha(s)$ and $\beta(s)$ is constant:
$\|\alpha(s)-\beta(s)\|=\lambda$
Now we have to prove $\lambda=0$
Let’s assume to the contrary that $\lambda\neq0$. Differentiating both sides yields:
$\dfrac{\langle\alpha’(s)-\beta’(s),\alpha(s)-\beta(s)\rangle}{\|\alpha(s)-\beta(s)\|}=0$
So we get:
$\|\alpha’(s)-\beta’(s)\|=0 \Longrightarrow \alpha’(s)=\beta’(s)\quad\forall s$
How can I go further?