Practical puzzle for math geeks

March 10, 2005 12:44 PM

I have a circular patio table with four legs. The paving stones are uneven, so it has always been difficult to set the table in place so it doesn't wobble. I move it regularly, since the shadows from the trees change, and sometimes I want shade, other times sun. I used to mark small dots on the stones so I could set it down quickly. Then I discovered something very interesting--no matter where I put the table, if I simply rotate it around its center, I can very quickly find a steady configuration. But move it from place to place, and it takes all day. This is like the four color map theorem. It works every time, but I haven't figured out why. It might even be tricky to put this into a mathematical format. The table doesn't need to be level, and all four legs needn't be the same height or "at" the same height. The table just can't wobble. Go geeks.
posted by weapons-grade pandemonium to science & nature (10 comments total)

Consider the leg heights and paving stone configuration as random. When you pick up and move the table, you're essentially taking two or three minutes to make another roll of the dice. When you rotate around the center, you're very quickly trying a number of different configurations, so it is much more efficient. IANAGFE (I Am Not A Garden Furniture Expert).
posted by brool at 12:54 PM on March 10, 2005

I saw virtually this exact situation presented as a puzzle some time ago--I don't remember where, so I can't cite it.

Anyway, first observe that no matter where you put down the table and in what orientation, three of the legs will touch the ground. (If you imagine just two legs touching, it will rotate along the line connecting the points where those two legs touch until a third leg touches.)

So let's call the feet of the four legs, going around clockwise, A, B, C, and D. Let's say in a certain orientation legs A, B, and C touch the ground, and D does not.

Now rotate the table 90° clockwise. A moves to the point where B was, B moves to where C was, and so forth. Since A, B, and C were touching before, that means B, C, and D are touching the ground now. (The table itself is even; it is the pavement that is not.)

Now, what happened as you rotated through that 90°? At some point D began touching the ground, and at some point A ceased to touch the ground. Which happened first? Well, we've already established that at least three legs touch the ground in any orientation, so A cannot have stopped touching before D started to touch! Thus, at some point within that 90° rotation both A and D (thus all four legs) were touching the ground.

I'm oversimplifying a bit by not considering the possibility that the ground is uneven enough that there's more than one change in which three legs are touching the ground during the 90° rotation, but the basic reasoning is the same there.
posted by DevilsAdvocate at 1:03 PM on March 10, 2005

An ancient proof from the newsgroups.
posted by smackfu at 1:03 PM on March 10, 2005

A moves to the point where B was, B moves to where C was, and so forth. Since A, B, and C were touching before, that means B, C, and D are touching the ground now.

Arrgh. I think I got that backward. D, A, and B are touching now. The general argument is still the same, though.
posted by DevilsAdvocate at 1:05 PM on March 10, 2005

The puzzle above can be considered to be a variant of the simpler puzzle: walking a circular route, will the walker ever be at the same altitude twice? Yes.

To show the similarity: At each point in the circle which passes through the four legs of the table, assign a value to the point that is the difference between the height (altitute) at that point and the average of heights of the two points that are 90 degrees from it.

Example: Given the table with legs A, B, C, and D, let's assume that leg A is positioned where the ground is at height 1", and leg C is positioned where the ground is at height 2". Then the average of the two is 1.5". Let's assume that the height (at the position of the bottom) of leg B is 1". Then the value assigned to the point where leg B is located is -.5". What this means is that leg B is 1/2 inch lower than the the midpoint of the line between legs A and C.

Continuing the example, let's assume that (the bottom of) leg D is at 1.5" of height. So the value of D is zero.

When the value of B and D add to zero, then the table is balanced. (If B and D happen to both be zero, then the table is also level.)

To return to the puzzle: in our hypothetical circle, every point has a calculated value, and it's easy to prove (I think) that there must exist at least two points that are opposite each other and also have values that are the negatives of each other (such as -.5" and +.5").

In summary: The simpler puzzle is about two points on a circle that are opposite each other and have the same value; this puzzle is about two points opposite each other that total to zero.
posted by WestCoaster at 1:39 PM on March 10, 2005

I knew someone would bring up a fixed point theorem argument.

Here's another cool variation which might help understand the concept involved:

You are in Iceland where it is cold and snowy. I am in South America laying on the beach.

We both have airplanes and I decide to go visit you while you, at the same time, decide to come and visit me. We take *entirely different paths* to exchange places but we end up taking the same amount of time.

Prove: At some point during our trips, we were someplace where the current temperature was exactly the same.

Proof: Say, our temperature difference in the beginning (mine minus yours) was 40 degrees. At the end of the trip, since we have exchanged places, the difference would be (mine minus yours) -40 degrees.
That difference varied wildly over the course of the trip as it moved from 40 to -40 but as long as it was continuous, at some point it passed through the 0 line, i.e. we both experienced the same temperature.
posted by vacapinta at 1:58 PM on March 10, 2005

Just wanted to say how much I enjoyed learning the solution to this, and reading the very clearly worded explanations. Thanks, math people.
posted by Miko at 5:45 PM on March 10, 2005

Uh, aren't these just versions of the Mean Value Theorem from algebra?
posted by notsnot at 6:37 PM on March 10, 2005

Great solutions. Smackfu's link, DevilsAdvocate's tripod, and WestCoaster's circular walk helped me visualize the geometry. Although the walk need not be circular, it has to do with returning to the point of departure. But here's the magic: since there are four identical legs, returning to the point of departure happens in a quarter of the expected time. In fact, it's faster than that, since you don't need to go that far. As the floating leg is moving to higher ground, the adjacent tripod leg is moving to lower ground. About an eighth rotation, more or less, will stabilize the table every time. Analogous problems and graphs linked by vacapinta are interesting, too. Thanks, everyone.

I think we could stabilize the world this way, if we put our minds to it.
posted by weapons-grade pandemonium at 8:18 PM on March 10, 2005

Just wanted to say how much I enjoyed learning the solution to this, and reading the very clearly worded explanations. Thanks, math people.

Hear, hear.
posted by carter at 8:39 PM on March 10, 2005