Here's Veit's solution (SPOILER): ^L ^L ^L (Remember the days when we'd use "^L" in situations like this? Are any of you using mail-readers that force you to hit the space-bar when it encounters this character?) ^L ^L ^L Veit wrote: I get 4. It doesn’t matter if you average over v, or v+a+b, where a are samples of the discrete lattice generated by A inside some large region R, and b are samples of the discrete lattice generated by B, also inside R. You can interpret this enlarged average in terms of a double covering of the plane, inside V, by the two parallelogram tilings. Every polygon formed by the edges in this “double tiling” in fact corresponds to an instance of one of your intersections. To answer the question all we want to know is the average number of sides per polygon in the double tiling. We forget about vertices of one tiling being coincident with vertices of the other because that has zero measure. That leaves only vertices of degree 4. Now we look at the polygons formed by the double tiling as a graph and use Euler. We make estimates inside a large R and therefore ignore boundary corrections. Let the number of vertices inside R be V. Because the degrees are all 4, the number of edges is E=2V.
From Euler, applied to a planar graph, we get F=E-V=V (ignoring boundary corrections and the Euler characteristic). The average number of polygon sides is 2E/F=4V/V=4.
There's a way to avoid the "boundary corrections from large regions are small" dodge. Specifically, one can show that if you superimpose the lattices associated with A and B, then as long as everything is generic (with no triple-intersections), each copy of A is dissected (by the B-lattice) into a finite number of polygons whose average "sidedness" is exactly 4, and vice versa. One can prove this by induction (adding one cut-line at a time), or prove it in one gulp by an appeal to Euler's formula V-E+F=2. This hinges on the assumption that each internal vertex in the dissection has degree 4, each vertex on the edge of the copy of A has degree 3, and the corners of the copy of A have degree 2. I'll provide more details if anyone wants them. Jim On Mon, Nov 6, 2017 at 9:13 AM, James Propp <jamespropp@gmail.com> wrote:
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