Cusps of parallel/offset curves Found this remark in Wikipedia: https://en.wikipedia.org/wiki/Parallel_curve
When parallel curves are constructed they will have cusps when the distance from the curve matches the radius of curvature. These are the points where the curve touches the evolute.
I would like to prove it, or at least have a better explanation of why this happens.
Thanks in advance!!
 A: Since I don't recall seeing this question on this site before and because it may be of interest to other people, I'm going to sketch the computation to which I alluded in the comments.
Let's consider the parallel curve $\beta(s)=\alpha(s)+rN(s)$ with $r=1/\kappa_0$ (here the subscript denotes the value at $s=0$). The local canonical form (see, e.g., p. 17 of my text) for $\alpha$ at $s=0$ is
$$\alpha(s)=\big(s-\tfrac16\kappa_0^2s^3 + o(s^3)\big)T_0 + \big(\tfrac12\kappa_0s^2 + \tfrac16\kappa'_0s^3 + o(s^3)\big)N_0.$$
Differentiating, we see that
$$\alpha''(s) = \kappa(s)N(s) = \big({-}\kappa_0^2s + o(s)\big)T_0+\big(\kappa_0 + \kappa'_0s + o(s)\big)N_0,$$
and so
$$\frac1{\kappa_0}N(s) = \frac{\big({-}s+o(s)\big)T_0 + \big(\frac1{\kappa_0}+\frac{\kappa'_0}{\kappa_0^2}s+o(s)\big)N_0}{\sqrt{\left(1+\frac{\kappa'_0}{\kappa_0}s\right)^2+\kappa_0^2s^2+o(s^2)}}.$$
Using the usual first-order approximation  $\frac1{\sqrt{1+x}}\approx 1-\frac12x$, we work out that
$$\beta(s) = \big({-}\tfrac16\kappa_0s^3+O(s^3)\big)T_0 + \big(\tfrac1{\kappa_0} + \tfrac12\kappa_0 s^2 + O(s^2)\big)N_0.$$
This makes it clear that $\beta$ has a cusp at the point $\frac1{\kappa_0}N_0$.
