Couldn't follow the math, but the intuition is lovely. You use a very similar idea in the rather more spectacular sport of 'rock balancing' -- find three points of contact by feel, then rotate the rock (in just the same sense) until the centre of mass is between them. It's surprisingly easy to do, and can look fantastic.

Oh, the questions. Going by the intuition behind the proof, no, mismatched legs aren't covered. But you could stomp the Dylan CD into different-sized shards, use them to pad the legs out to perfect matching lengths, then apply the theorem.

I guess there's some probabilistic idea here that if a random partition doesn't give you the sizes you need, with each random refinement of the partition the probability of getting a matching set increases towards unity. Translation: if you can't do it with the first set of pieces, stomp them into smaller pieces and try again.

You know, I thought that was rather an old result, although this one might have slightly different conditions.

see, for instance,
http://www3.baylor.edu/Math/insumm05.pdf

which gives a walkthrough of two similar proofs, though at least one of them is attributed to an ancestor-professor.

Hee! Classic Clive. Who else in the world could make a post about something like this turn out really interesting and funny? Dissing Dylan, plastic surgery and cuddling Calvin -- it's been a good day.

Clive asks:

Would Martin's proof obtain for a table with uneven legs? Assuming the legs are off by a gradation of no more than 15%, would the bumps in the floor be able to compensate for the legs?

I think I can easily prove this is not the case. Imagine the floor is perfectly smooth. Certainly it meets the constraint that the "gradation" (is that the correct usage of that word?) is no more than 15%. And yet, the table will always be wobbly.

Here's my attempt at a proof the the original question, though. Say the table has four legs which we'll label going around as A, B, C, and D (note: A and C are on opposite sides as are B and D). Now, when the table is wobbly, either the A-C pair will be solid, and the table will wobble between B and D, or the roles will be flipped. As you rotate the table, eventually the B-D pair will be solid, and the table will wobble between A and C touching the ground. We know this will be eventually true, as once the table makes a quarter-turn, the legs themselves will have swapped ground bits. Now the table will not wobble at the point where the leg role-trading will take place. QED.

What do I win?

Brock, good quick logic on that one. You're right -- it's easy to posit conditions in which an uneven-legged table will always wobble.

Tikitu, I never thought of the rock-balancing comparison! Good one. I also strongly approve of the CD-smashing solution. Gotta try that some time.

Clew, great link! I love that they refer to it a "the wobbly table problem" ... is this question regularly spoken about by mathematicians?

Jason is way too nice.

Dude, that was written about by Martin Gardner in, like, 1970 (actually I have no idea what year, I just read about it in one of his book collections).

On the other hand, everyone could just buy 3 legged tables, which will always find a non wobbly purchase on any cafe terrain. Of course, the surface of the table may not be peerfectly level, but at least it would be stable...

Of course, the other solution might be telescoping legs, or legs which can extend out on a horizontal axis until they find an even spot on the ground further away than the radius provided by fixed legs.

Or maybe the cafe owner could get all bauhaus and suspend the tables from the ceiling on rigid struts (to keep the table from swinging).

I think the real question is whether the table user will have sufficient presence of mind to deal with any of these solutions until after they have had their coffee?

Bram, see, now this is like a Google version of the LazyWeb/LazyPhoto experiencer! I'm too lazy to Google the background to coffee-table math and do any, like, research ... so I just blog about it and wait for someone who knows more about it to come and correct the record! LazyGoogle!

Actually, in the pre-Internet days, this process would have been known by its lower-tech name: "Asking someone a question." Heh.

John, you ask questions only the philosophers can answer.

interesting posting. the matchbook trick is still probably the best solution for one of those tables with one pedestal from which sprout four "legs" at the floor level...

and as for the Dylan playing in the koffeehaus™:
better Dylan than The Gipsy Kings. (shudder)

it's easy to posit conditions in which an uneven-legged table will always wobble.

There's a reason for this. The 15% gradation requirement applies to the entire surface. That is, it holds for any two arbitrary points. However, the assertion that no table leg be inclined by more than 15% is much weaker, as the inclination is only defined here between the four points corresponding to the table legs. It is easy to come up with counterexamples to Clive's hypothesis because there are plenty of surfaces which satisfy the inclination requirement at the four points of some position of the table, but do not satisfy the requirement for every possible pair of points on the surface. Although the assertions sound similar, they impose very different mathematical constraints.

Bram, do you still remember in which of Martin Gardner's books you came across this trick?