If one computes the average width of a triangle, then one gets $(s_1+s_2+s_3)/\pi$, where $s_1$, $s_2$, $s_3$ are the side lengths. I did this by brute force, using an integral which went through an interval of length $\pi$. For each angle $\theta$, the "width" $w_\theta$ of the triangle is the smallest distance one may place two parallel lines at that angle without penetrating the triangle. (Imagine using calipers for each angle.)

My questions: Is there an easier way (without using an integral) to see that the ratio of the perimeter $s_1+s_2+s_3$ to this average width is $\pi$? Does this generalize to any convex polygon (or polytope)?


The most basic statement along these lines is that the average width of any line segment is equal to $2L/\pi$, where $L$ is the length of the segment. This is highly related to the Buffon's needle problem, where the length of the needle is the same as the width of the strips. It is also related to the fact that the average value of $|x|$ on the unit circle is equal to $2/\pi$.

Now consider a convex polygon in the plane. No matter how you turn the polygon, its horizontal width is always equal to twice the sum of the horizontal widths of the edges. Specifically, the sum of the widths of the edges along the bottom is equal to the width of the polygon, and the same is true of the widths of the edges along the top. Thus, the average width of any convex polygon is equal to half the sum of the average widths of the side lengths: $$ \text{avg. width} \;=\; \frac{1}{2}\left(\frac{2s_1}{\pi} + \cdots + \frac{2s_n}{\pi}\right) \;=\; \frac{s_1+\cdots+s_n}{\pi}. $$

There is a similar formula in three dimensions involving area. Specifically, suppose we take a polyhedron and rotate it in all possible ways. For each rotation, we measure the area of the projection of the polyhedron onto the $xy$-plane. Then $$ \text{avg. $xy$-area} \;=\; \frac{A_1+\cdots+A_n}{4} $$ where $A_1,\ldots,A_n$ are the areas of the faces. Thus, if you take a unit cube outside on a sunny day and rotate it at random, the average area of its shadow will be $(1+1+1+1+1+1)/4 = 3/2$. Again, the factor of $4$ comes from the fact that a polyhedron has top faces and bottom faces, and the average $xy$-area of a flat object is $A/2$. The latter follows from the fact that the average value of $|z|$ on the unit sphere is $1/2$.

  • $\begingroup$ Jim: This is what I was looking for. I was suspicious that Buffon was behind this, but I couldn't see it. Thanks! (This still begs the question about the average width of a polytope....) $\endgroup$ – David Richter Feb 8 '11 at 16:45

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.