If $S\subseteq \mathbb{R} ^2$ is closed and convex, we say $S$ is strictly convex if for any $x,y\in Bd(S)$ we have that the segment $\overline{xy} \not\subseteq Bd(S)$.
Show that if $S$ is compact, convex and constant width then $S$ is strictly convex.
Any hint? Than you.
The idea of celtschk works just fine. Suppose that the line $L$ meets $\partial S$ along a line segment. Let $a\in S$ be a point that maximizes the distance from $L$ among all points in $S$. This distance, say $w$, is the width of $S$. Let $b$ any point of $L\cap \partial S$ which is not the orthogonal projection of $a$ onto $L$. Then the distance from $a$ to $b$ is greater than $w$, a contradiction. (The projection onto the line through $a$ and $b$ will have length $>w$).