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Let $C$ be a simple closed plane curve and let $D$ be its interior. Recall that the width of $C$ in a direction $\theta$ is the distance between two supporting lines for $D$ which are perpendicular to $\theta$. A curve is said to have constant width if its width is the same in every direction; see the Reuleaux polygons for nontrivial examples.

It is often stated (for instance in "The Enjoyment of Math" by Rademacher and Toeplitz) that curves of constant width are necessarily convex. Does anyone know how to prove this? It's not as easy as I thought it would be...

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1 Answer 1

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Here are some ideas:

Assume $C$ is a curve of constant width $\ell$; so $\bar D:=C\cup D$ is a domain of constant width $\ell$. The convex hull $H:={\rm conv}(\bar D)$ is a convex domain of the same constant width $\ell$. According to standard theorems about convex domains of constant width the radius of curvature of $\partial H$ cannot exceed $\ell$ at any point; so $\partial H$ cannot contain an open segment of positive length. This means that all points of $\partial H$ are extremal points. This should allow to prove that in fact $\partial H=\partial \bar D=C$.

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This sort of argument seems to prove that $\partial H$ is strictly convex, but I don't see how that helps us prove that $\partial H = C$. Also, I'm worried about making the curvature argument work without any regularity assumptions on $C$. –  Paul Siegel Feb 28 '13 at 14:16
    
Any point in $\partial H\setminus C$ is on a straight line segment. –  Kallus Mar 8 '13 at 23:57

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