Suppose we know that for every point $c(t)$ on a closed, strictly convex curve $c:I\rightarrow \mathbb{R}^2$, there is a unique point $c(t')$ such that $e_1(t)=-e_1(t')$. $c$ is said to have a constant width if $d(c(t), c(t') )=d$, a constant.
Prove: The circumference of a closed, strictly convex curve of contant width $=d$ is equal to $\pi d$.
Current Progress:
By reparametrization, we can let $\| c'(t) \| = 1$ for all $t\in I=[a,b]$. Suppose $c'(t)=(\cos \theta(t), \sin \theta(t))$, then the curvature $\kappa(t)=\theta'(t)$.
Since $c$ is strictly convex, w.l.o.g. $\kappa(t)>0$, and it is not hard to show that $\theta:I\rightarrow [0,2\pi]$ is diffeomorphism and increasing. Thus its inverse function $t(\theta)$ exists.
Let $e_1(\theta)=c'(t(\theta))$, write $\kappa(\theta)$ for $\kappa(t(\theta))$ when there is no confusion.
Let $f(\theta)=c(t(\theta + \pi)) - c(t(\theta))=\int_\theta^{\theta+\pi} e_1(\theta)/\kappa (\theta)d\theta\equiv d$.
$\frac d{d\theta}\|f(\theta)\|=\frac {[e_1(\theta+\pi)/\kappa(\theta+\pi) - e_1(\theta)/\kappa(\theta)]\cdot f(\theta)}{\|f(\theta)\|} = 0$.
But by our construction of $\theta$, $e_1(\theta + \pi)=-e_1(\theta)$,
Therefore, $\frac {[1/\kappa(\theta+\pi) /+ 1/\kappa(\theta)] e_1(\theta)\cdot f(\theta)}{\|f(\theta)\|}=0$ ,
thus $e_1(\theta)\cdot f(\theta)=e_1(\theta)\cdot (c(t(\theta + \pi)) - c(t(\theta)))=0$.
Up to now I manage to show that: if you draw a line through the points whose tangent vectors are in opposite directions, this line is perpendicular to the tangent vectors.
Now keep differentiating,
$\frac d{d\theta} e_1(\theta)\cdot (c(t(\theta + \pi)) - c(t(\theta)))$
$=Je_1(\theta) \cdot (c(t(\theta + \pi)) - c(t(\theta))) - e_1(\theta) \cdot e_1(\theta) [1/\kappa(\theta+\pi) + 1/\kappa(\theta)]=0$,
where $J=[\begin{array}{} 0 & -1 \\ 1 & 0\end{array}]$ is the rotation matrix of 90 degrees counter-clockwisely.
But the first term is just the measure of the diameter, which is $d$!
Therefore, we get $1/\kappa(\theta+\pi) + 1/\kappa(\theta)=d$ for arbitary $\theta$.
By the inverse function theorem, the circumference of $c$ is $b-a=\int_0^{2\pi}1/\kappa(\theta)d\theta=\int_0^{\pi}(1/\kappa(\theta)+1/\kappa(\theta + \pi))d\theta=\pi d$.
