I'm struggling with the injectivity of the parallel curve.
Let $\alpha:[0,\ell]\to\mathbb R^2$ with $0<\ell\in\mathbb R$ be a planar closed simple regular curve parametrized by arc length, and its normal always points inside the region it bounds. Let $0\neq r\in\mathbb R$ and define the parallel curve $\alpha_r:[0,\ell]\to\mathbb R^2$ as $\alpha_r(s):=\alpha(s)+rN(s)$, being $N(s)$ the unit normal vector of $\alpha$ in $\alpha(s)$.
I have to show that there is $r$ such that $\alpha_r$ is a simple curve.
Attempt: In the traditional way, let be $s_1, s_2 \in [0,\ell)$ where $\alpha_r(s_1)=\alpha_r(s_2)$ and let's try to prove that $s_1=s_2$. From the previous equation and by definition of $\alpha_r$ we can obtain $\alpha(s_1) - \alpha(s_2)=-r[N(s_1)-N(s_2)]$ and because we have planar curves $N(s) = JT(s)$ where $J$ is the rotation of 90° degress (in the desired direction) and $T(s)=\alpha'(s)$. Therefore $\alpha(s_1) - \alpha(s_2) = J(-r[\alpha'(s_1) - \alpha'(s_2)])$ which implies that $\alpha(s_1) - \alpha(s_2)$ and $\alpha'(s_1) - \alpha'(s_2)$ are orthogonal, this is $\langle\alpha(s_1) - \alpha(s_2),\alpha'(s_1) - \alpha'(s_2)\rangle=0$
Here is the stuck! I don't know how to proceed from there.