Mean width and perimeter Does anyone know a simple, elementary and self-contained proof of the fact that the mean width of a convex two-dimensional body equals its perimeter divided by $\pi$?
 A: Here is Dan Klain's reply:
The only tricky step is that, instead of mean width, consider the functional that measures the number of crossings of a curve with any line in the plane.  In other words, integrate this number of crossings over all lines: over all families of parallel lines and over all directions for those families.  Call this function F(C), where C is the curve.
So F(C) = int_{unit vector u} int_{lines perp to u}  {# crossings  of C with line}  d-line du
If C is a line segment then F(C) is clearly proportional to the length |C| of C (by the same additivity argument as in Barbier's Buffon proof).    Moreover, F is also additive if you glue together two needles at an angle (this is why we use the function F instead of mean projection, which gets funky if you glue at the angle).  So by the same argument and continuity, F(C) is proportional to |C| for all curves C (from a reasonable class).
So we know F(C)  = k|C| for some constant k indep of C.  What is k?
If C happens to be the boundary of a convex region in the plane, then a line crosses C twice or not at all (the tangent cases have measure zero).  Moreover the 2 occurs when the line passes though a point of the projection of C to the perp of a line family.   So from the definition of C (integration over all line families and over the circle of directions) we have
F(C) = 2 * int_circle W_u(C) du
where W_u is the width in direction perp to unit vector u, so that
F(C) = 2 * 2pi * W(C) = 4pi * W(C)
where W(C) is the mean width of the region bounded by the convex curve C.
We now have
k|C| = F(C) =  4pi * W(C)
If we let S be the unit circle, then W_u is always 2 so
2 pi k = k |S| = F(S) = 4pi * W(S) = 8 pi
so k = 4.  This means
|C| = pi * W(C)
when C is a convex curve.
A: The statement you're looking for is the two-dimensional case of Cauchy's Surface Area Formula, since in two dimensions, the mean width of a convex body is equal to the average area of its projections onto 1-D subspaces. A simple proof can be found in this arXiv preprint by Tsukerman and Veomett.
