Deconstructing the Pentagon
No, it’s not about the latest Wikileaks scandal. Thursday at EPFL, the last Bernoulli guest lecture of the discrete/computational geometry theme semester was given by Miklos Laczkovich, on geometrical tilings. (I posted about another of the guest lectures; the remaining 2 talks, by Gil Kalai and Micha Sharir, were excellent too.)
Miklos’ talk was a survey with lots of pictures and just brief tastes of some proofs, which made it very entertaining. I liked that algebraic field isomorphisms came out of nowhere, as I’ll elaborate below.
The general theme of the talk was about decomposing shapes into smaller shapes. It started with the following classical puzzle. Note that this L-shaped figure can be decomposed into 3 identical pieces. Find a way to decompose it into 4 identical pieces. (Click here for the solution)
He moved on next to a hard result that sounds obvious, then another result which sounds obviously false, given the first one. (In the whole post, I only conider dissections into a finite number of pieces.)
Theorem (Max Dehn, 1903). A rectangle with side lengths P and Q can be dissected into squares (possibly of different sizes) if and only if P/Q is a rational number.
It’s obvious that when P/Q is rational, a tiling exists, like the chocolate bar shown at left (for P/Q = 3/2). But if P/Q is irrational, the proof that no tiling exists is quite tricky.
Given Dehn’s theorem, many people would immediately agree that the following dual theorem “should be” true:
A square can be dissected into rectangles (possibly of different sizes) with side lengths in ratio P/Q, if and only if P/Q is a rational number.
But this is false! There are irrational ratios P/Q yielding a dissection. For an example, we draw the diagram shown on the right-hand side (in blue), and hope that we can pick the side lengths to all be similar rectangles. Does this work? If the square has side length 1, we can deduce some side lengths of the rectangles (1/2 and 1/3). Then introduce x to stand for the ratio of the “long” side of the rectangle to the short one.
We can label all the lengths as shown in the figure below:
Then, we see that we must have x/2 + 1/3x = 1, which is a quadratic equation whose solution is x = 1±√(1/3), which is indeed irrational. The characterization of all ratios is the following:
Theorem (Laczkovich-Szekeres 1990): A square can be dissected into rectangles (possibly of different sizes) with side lengths in ratio P/Q, if and only if P/Q is an algebraic number, and all of its algebraic conjugates* have positive real part.
*: This means all the complex roots of the minimal rational polynomial.
If I understood the idea in the talk correctly, the proof uses a nice idea about morphisms between finite field extensions Q(x) of the rational numbers, roughly saying that we can replace x with any of its conjugates, and still have a rectangle tiling. The same methodology gives several other optimal results about dissections. He also mentioned the following nice open problem:
Open problem. Can the regular pentagon be dissected into 36°-36°-108° triangles (possibly of different size)?
The algebraic requirements that are always necessary and “often” sufficient to get a drawing, are true here. So if no drawing exists, proving this fact seems to require new ideas.
Filed under: b.k.a.t., math | 1 Comment