Nobody? ... Maybe these puzzles are less fun than I thought. (Or maybe "Fun is fun, but enough's enough.") I like the fact that the answer to this one isn't a whole number. One frustrating feature of this particular puzzle is that someone with no mathematical training might guess the correct answer using bogus reasoning. Jim On Tue, Nov 14, 2017 at 10:13 PM, James Propp <jamespropp@gmail.com> wrote:
Here's an easier puzzle of the same kind; it's probably the one I'd present to people first now that I know about it, since the calculations are simpler:
What's the expected number of vertices in the intersection of a rightside-up equilateral triangle and an upside-down equilateral triangle of the same size, conditioned on the event that the intersection is nonempty?
(The equilaterality of the triangle is a red herring; any triangle and its rotated-by-180-degrees twin will do.)
I'll let someone else have the fun of posting a solution.
Jim Propp
On Sat, Nov 11, 2017 at 11:06 AM, James Propp <jamespropp@gmail.com> wrote:
Let H be a regular hexagon and t a translation vector, so that H+t is a translate of H. Here's one way to prove that the average number of sides in the intersection of a H with H+t, conditioned on the event that the intersection is nonempty, is 5. (Veit found a nicer way, but I'll get to that later.)
A polygon P obtained by intersecting H with H+t can have three different sorts of vertices: a vertex of H, a vertex of H+t, and the intersection of an edge of H with an edge of H+t. So we can write the expected number of vertices of P as a sum of 48 probabilities (many of which are equal to one another, by symmetry): the probability that a particular vertex v of H belongs to H+t (which we sum over the 6 vertices v of H), the probability that a particular vertex v+t of H+t belongs to H (which we sum over the 6 vertices v of H), and the probability that a particular edge e of H intersects a particular edge e'+t of H+t (summed over all 24 pairs of nonparallel edges e,e' of H). But these probabilities are all easy to interpret geometrically and thus easy to compute.
First, let's find the common denominator of these probabilities: the area of the set of vectors {t: the intersection of H and H+t is nonempty}. This is the Minkowski difference H-H, and since H is centrally symmetric, it's just 2H (up to translation). Since H is made of 6 unit equilateral triangles, 2H has area equal to 24 unit equilateral triangles. (I'll just say "area 24" from now on.)
Now, on to the numerators and the probabilities! For each vertex of H, the set of vectors t for which that vertex of H belongs to H+t is just a translate of H, with area 6; so the probability that the vertex v belongs to H+t is just the area of H divided by the area of 2H, or 6/24=1/4. We get 6 such terms, for a total of 6/4 = 3/2. Likewise, for each vertex v of H, the probability that v+t belongs to H is 3/2. Finally, consider the pairs e,e'. The set of vectors t for which e intersects e'+t is just a parallelogram, namely, the parallelogram spanned by e and e' (treated as vectors), which consists of two unit triangles; so the probability measure of the set {t: e intersects e'+t} is 2/24, or 1/12. Summing over all 24 pairs e,e', we get 24/12, or 2. Finally, we add up all the probabilities, obtaining 3/2 + 3/2 + 2 = 5 as the expected number of vertices.
Veit's proof collapses the 24 edge-meets-edge terms into a single term whose value is the constant 2! What's more, his argument can be used to show that for any centrally symmetric 2n-gon, the expected number of vertices in the intersection of the 2n-gon with a translate of itself, conditioned on the event that the intersection is nonempty, is exactly n+2. The part of his argument that I still don't understand is this: how do we know that (generically) the boundaries of the two 2n-gons intersect exactly twice?
Jim Propp
On Mon, Nov 6, 2017 at 10:07 PM, James Propp <jamespropp@gmail.com> wrote:
I wrote:
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-parallelog rams-puzzle.
George's method can also be used to prove that if you take a random nonempty intersection of a regular hexagon with a translate of itself, the expected number of sides is exactly 5.
Hint: instead of counting sides, count vertices.
I'll sketch the proof tomorrow or Wednesday, if nobody beats me to it.
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.
Let me say, by way of motivation, that the "George-style" proof is really sweet. You'll feel good when you find it.
Jim