So far, only Veit Elser has solved the puzzle (or at least he's the only one who sent me a solution), so let me provide a few hints. But first, an obvious remark: if the answer is independent of A and B (as the wording of the puzzle implies), it can only be 4, since that's what you get when the sides of A are parallel to the respective sides of B (e.g., when A and B are rectangles whose sides are horizontal and vertical). But it's a priori not at all obvious that the answer should be the same for all pairs of parallelograms. I actually know two different proofs of the result. One follows the approach to such problems that I learned from George Hart (see his email to math-fun from October 18), and the other follows the approach I learned from Warren Smith (see my email to math-fun from October 19). If you look at their solutions to the random-slices-of-a-cube problem you might be inspired, as I was, to see how to solve the random-intersecting-parallelograms-puzzle. George's method can also be used to prove that if you take a random nonempty intersection of a regular hexagon with a nonempty translate of itself, the expected number of sides is exactly 5. Anyway, two days have passed since I posted the puzzle, so I think it's appropriate to have an unbridled conversation, with or without spoilers. Jim Propp On Sat, Nov 4, 2017 at 12:32 PM, James Propp <jamespropp@gmail.com> wrote:
Given two parallelograms A,B in the plane and a translation vector v, let P(v) be the intersection of A+v with B. If we choose a random v such that P(v) is nonempty, how many sides on average do we expect P(v) to have?
The set of such vectors v is a compact subset of R^2, so it's clear what probability measure to use: ordinary Lebesgue measure, rescaled.
There is a nice answer to my question and at least one nice proof. I'll be curious to see proofs different from the one I found.
I'm happy to answer requests for clarification, but no spoilers till Sunday, please.
Jim Propp