Question

Cross-posted from math.stackexchange:

Let C be a simple closed plane curve and let D be its interior. Recall that the width of C in a direction θ is the distance between two supporting lines for D which are perpendicular to θ. 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? I can't find any references, and a proof eludes me.

Answer 1

First, observe that for a closed convex region $R$ of constant width $w$, there can be no interval $[a,b]\subset \partial R$. For take the support line $l$ to $R$ containing $[a,b]\subset l$, and let $c$ be a point in the other parallel support line for $R$ realizing the width $w$. Then the height of the triangle $abc$ is $w = d(c,l)$. However, the distance $w<\max\{d(a,c),d(b,c)\}$, and therefore the width of $R$ will be greater that the maximum of these two distances by choosing the support lines perpendicular to the longer side, contradicting constant width.

Now, suppose $K$ is a constant width region, and let $R$ be the convex hull of $K$. If $K \neq R$, then there exists an interval $[a,b]\subset \partial R-\partial K$, a contradiction.

Answer 2

Say that the compact set $K \subset \mathbb{R}^2$ of constant width $d$ is not convex; then there is some support line that touches two points, say $A$ and $B$ on $\partial K$. The parallel line that supports $K$ "from the other side" touches at least one point $C \in \partial K$. The distance between these two lines is the constant width $d$.

But either the distance $\overline{AC}$ or $\overline{BC}$ (or both) is strictly larger than $d$, so the width in the direction orthogonal to that line is larger than $d$ and $K$ has not constant width.