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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.

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  • $\begingroup$ What's the definition of constant width? $\endgroup$ – celtschk Jun 2 '13 at 1:30
  • $\begingroup$ The width of a convex curve in a given direction is the distance between a pair of supporting lines of the curve perpendicular to this direction. If the width of a curve is the same in all directions, then it is a curve of constant width. $\endgroup$ – user73564 Jun 2 '13 at 1:43
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    $\begingroup$ In that case I'd try to proof it in reverse (i.e. if there's a straight line segment in the boundary of the convex set, then S cannot be constant width). I'd start with the direction perpendicular to the straight line (in that direction, the one supporting line must certainly contain the complete line segment), and then try to prove that if you change the direction a little bit, you increase the width (it seems intuitively clear to me that the width must increase in that case). However I don't really know if this idea works. $\endgroup$ – celtschk Jun 2 '13 at 1:55
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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 $>w$).

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