A closed regular planar curve $C$ is said to have constant width $\mu$ if the distance between any pair of parallel tangent lines to $C$ is always $\mu$. If two points on $C$ have parallel tangent lines, call them opposite.
I've already shown the opposite point to $\alpha(s)$ is of form $\beta(s)=\alpha(s)+\mu N(s)$.
I'm trying to prove that the sums of the reciprocals of the unsigned curvatures at opposite points is $\mu$.
Here's my attempt:
$$ \beta(s)=\alpha(s)+\mu N(s)\implies \beta'(s)=\upsilon(s)T_\beta(s)=\alpha'(s)-\mu\kappa(s) T_\alpha(s)\implies \beta''(s)=\upsilon'(s)Τ_\beta(s)+\upsilon(s)\kappa_\beta(s)Ν_\beta=\kappa_\alpha(s)Ν_\alpha(s)-\mu(\kappa_\alpha'T_\alpha(s)-\kappa_\alpha^2Ν_\alpha(s))\implies \upsilon'(s)=\mu\kappa_\alpha'(s);\upsilon(s)\kappa_\beta=\kappa_\alpha^2\implies \upsilon(s)=\mu\kappa_\alpha(s);\upsilon(s)\kappa_\beta=\kappa_\alpha^2\implies \frac{1}{\kappa_\alpha}+\frac{1}{\kappa_\beta}=\frac{\mu}{\upsilon}+\mu $$
This is unfortunately not the correct answer. Can someone help me locate the errors I made? Thanks in advance.