I've just been to a spectacularly good lecture given by

+Ravi Vakil. It was aimed at undergraduates, and therefore comprehensible (but more than that, it was completely absorbing and very entertaining). It included the following amazing gem of a proof. I'll give the problem and then leave some space, since I strongly recommend thinking about it for yourself. We didn't get the chance to do this, so I'm left feeling very curious about how it might be possible to come up with the argument.

The problem, he told us, originated in Russia. A train company stipulates that suitcases (which for the purposes of the problem are cuboids) must have height plus width plus depth equal to at most one metre. Is it possible to cheat the system by enclosing an illegal suitcase in a legal one? Equivalently, if d is the sum of the sidelengths, can a cuboid with larger d fit inside one with smaller d?

Now for the space.

First hint (which makes the problem massively easier). Define the r-neighbourhood of a suitcase to be the set of all points within distance r from that suitcase.

Second hint. OK, if you need it spelt out: consider the

*volume* of the r-neighbourhood.

Third hint. I didn't actually say this, but the volume in question is easy to compute, and once you've computed it, the answer to the question drops out.

OK, here's the answer (in sketch form). The r-neighbourhood can be chopped up naturally into pieces. One piece is the original cuboid. Six pieces are cuboids with one side of length r and a face in common with the original cuboid. Twelve pieces are quarters of cylinders of radius r that go round edges. And eight pieces are octants of a sphere of radius r.

When r is large, each kind of contribution contributes far more to the volume than the one before. But the spherical contributions are the same for the two suitcases, so the dominant term in the difference comes from the cylindrical contributions, which add up to pi r^2 times the sum of the height, width and depth. If one shape sits inside another, then so does its r-neighbourhood. Putting this together solves the problem.

So my question is, what should induce us to think of considering r-neighbourhoods? (I would be satisfied with an answer that said, "That's just a very nice way of expressing the proof, but here is an essentially equivalent argument that doesn't explicitly mention r-neighbourhoods," as long as the argument was clearly something one could have been expected to come up with.)