Prove that the sum of the reciprocals of the curvature at opposite points in a planar curve of constant width $μ$ is equal to $μ.$
My attempt:- I can write $\beta(s)=\alpha(s)+\mu N(s).$ I know that For any regular parametrized curve $α$, we have $κ=\frac{\| \alpha' \times \alpha''\|}{\|\alpha'\|}.$
I found $\beta(s)=\alpha(s)+\mu N(s)\implies \beta'(s)=\alpha'(s)+\mu N'(s)=\alpha'(s)+\mu(-\kappa(s)T(s)+\tau(s)B(s))\implies \beta''(s)=\alpha''(s)+\mu(-\kappa'(s)T(s)-\kappa(s)T'(s)+\tau'(s)B(s)+\tau(s)B'(s))=\alpha''(s)+\mu(-\kappa'(s)T(s)-\kappa^2(s)N(s)+\tau'(s)B(s)-\tau^2(s)N(s)) $
I am not able to simplify. Could you help me?
A Geometric Approach
Since the width of the curve $\gamma$ at any point is $\mu$, for any $t$, there is a $\rho(t)$ so that $|\gamma(\rho(t))-\gamma(t)|=\mu$. Since the curvature of $\gamma$ is assumed to exist at both $t$ and $\rho(t)$, $\gamma'$ exists at both $t$ and $\rho(t)$. Since $\mu$ is the diameter, $$ \begin{align} 0 &=\mu\frac{\mathrm{d}}{\mathrm{d}s}|\gamma(s)-\gamma(t)|_{s=\rho(t)}\tag{1a}\\ &=(\gamma(\rho(t))-\gamma(t))\cdot\gamma'(\rho(t))\tag{1b} \end{align} $$ That is, $$ \gamma(\rho(t))-\gamma(t)\perp\gamma'(\rho(t))\tag2 $$ Similarly, $$ \begin{align} 0 &=\mu\frac{\mathrm{d}}{\mathrm{d}s}|\gamma(\rho(t))-\gamma(s)|_{s=t}\tag{3a}\\ &=(\gamma(t)-\gamma(\rho(t)))\cdot\gamma'(t)\tag{3b} \end{align} $$ Which says, $$ \gamma(\rho(t))-\gamma(t)\perp\gamma'(t)\tag4 $$ Thus, $$ \gamma'(t)\parallel\gamma'(\rho(t))\tag5 $$ Since the tangents at $\gamma(t)$ and $\gamma(\rho(t))$ are parallel, looking at $\gamma$ near $\gamma(t)$ and $\gamma(\rho(t))$, it looks like
where $\kappa$ is the curvature of $\gamma$.
Since the diameter is $\mu$, we have $$ \frac1{\kappa(\gamma(t))}+\frac1{\kappa(\gamma(\rho(t)))}=\mu\tag6 $$
Using the Formula in the Question
Using that $\gamma$ and $\gamma+\mu N$ are the opposing points on $\gamma$, and that the formula for curvature is $$ \kappa=\frac{\gamma'\times\gamma''}{|\gamma'|^3}\tag7 $$ Where the cross product in $\mathbb{R}^2$ is the dot product of the first vector and the second vector rotated $\frac\pi2$ clockwise. This means that when a curve is turning left, $\kappa\gt0$. Furthermore, for the unit left-pointing normal, $N$, $$ N'=-\kappa\gamma'\tag8 $$ which implies $(\gamma+\mu N)'=\gamma'(1-\mu\kappa)$ and $(\gamma+\mu N)''=\gamma''(1-\mu\kappa)-\gamma'\mu\kappa'$.
Thus, when we plug $\gamma+\mu N$ into $(7)$, we get $$ \begin{align} \frac{(\gamma+\mu N)'\times(\gamma+\mu N)''}{|(\gamma+\mu N)'|^3} &=\frac{\gamma'\times\gamma''|1-\mu\kappa|^2}{|\gamma'|^3|1-\mu\kappa|^3}\tag{9a}\\ &=\frac{\kappa}{|1-\mu\kappa|}\tag{9b} \end{align} $$ If $\kappa\ge\frac1\mu$, then $$ \frac1\kappa+\overbrace{\ \frac{\mu\kappa-1}{\kappa}\ }^{\substack{\text{reciprocal of}\\\text{the curvature at}\\\text{the opposite}\\\text{point}}}=\mu\tag{10} $$ We cannot have $\mu\lt\frac1\kappa$ since then $\gamma+\mu N$ is not a point of maximal distance from $\gamma$:
Note that $\gamma+\mu N$ is actually a point of minimal distance from $\gamma$.