Barbier's theorem

Remarks

Apply the Crofton formula to $C$. A line that meets the interior of the convex body bounded by $C$ meets $C$ in exactly two points. The lines with normal direction $\theta$ that meet the body form a band of width $w(\theta)$. So the length of $C$ is $\frac{1}{2} \int_0^\pi 2 \, w(\theta) \, d\theta = \int_0^\pi w \, d\theta = \pi w$. Example: a circle of diameter $w$ has length $\pi w$, and the Reuleaux triangle of width $w$, made of three arcs of radius $w$ and angle $\pi/3$, also has length $\pi w$. The hypothesis of constant width is needed: the square of side $1$ has width $1$ in the direction of its sides, but its perimeter is $4$, not $\pi$.

Going deeper

For a student

The proof shows a more general fact: the perimeter of any plane convex body equals the integral of its width over the directions in $[0, \pi)$, which is Cauchy's formula. Barbier's theorem is the special case of a constant integrand. Together with the isoperimetric inequality it shows that the disc has the largest area among bodies of constant width $w$; the smallest area is the subject of the Blaschke–Lebesgue theorem.