Recently I came up with an algebra problem with a nice geometric representation. Basically, I would like to know what happens if we repeatedly circumscribe a rectangle by another rectangle which is rotated by α∈(0,π4) radians. Use this picture as reference:

In particular, do the resulting rectangles converge to a square?

It is rather easy to show that the rectangles do converge to a square if α is constant throughout the process. However, if we make it more general by defineing a sequence (αn)∞n=1 of angles and use αi in the i‘th operation, then the answer seems to depend on the chosen sequence. So, for which sequences (αn)∞n=1 do the rectangles converge to a square?

Algebraically this problem can be defined like this:

Define two real sequences by A0=a,B0=b,a≠b,a,b∈R>0 and An+1=Bnsinαn+Ancosαn,Bn+1=Ansinαn+Bncosαn∀n∈N>0, where αi∈(0,π4)∀i∈N≥0. Is it true that limn→∞AnBn=1?

I tried out a few sequences in C++ to notice some patterns. Interestingly, rectangles seem to converge to a square if and only if limn→∞(∑ni=0αi)=∞. In particular, for αn=1n the convergence is really slow, however, it still seems to be converging.

Also, I believe that showing limn→∞(An−Bn)=1 would be an even stronger result for this problem. Does such a replacement have any influence on the result?

For the record, I am still a high school student, so I have no idea how hard this problem might actually be. Any help would be highly appreciated.

P.S. This is my first question on the site, so please don’t judge my wording and style too much. Feel free to ask questions if anything is unclear.

**Answer**

This is a particularly good first question!

Using a little linear algebra, we can actually write sufficiently explicit formulas for An and Bn to confirm your conjecture, namely that the divergence of ∑∞i=1αi is sufficient and necessary.

First, by writing the recurrence relations for An and Bn in matrix notation and applying a straightforward induction, we get that, for all nonnegative integers n,

\require{cancel}

\pmatrix{A_n \\ B_n}

=

\left(\prod_{i = 1}^n \underbrace{\pmatrix{\cos \alpha_i & \sin \alpha_i \\ \sin \alpha_i & \cos \alpha_i}}_{\Gamma_i}\right) \pmatrix{a \\ b} .

We can make this more explicit by employing the standard trick of diagonalization: The eigenvalues of \Gamma_i are \lambda_i^{\pm} := \cos \alpha_i \pm \sin \alpha_i, so for \alpha_i in the given range, we have that

0 < \lambda_i^- < \lambda_i^+ ,

and in particular the eigenvalues are distinct, guaranteeing that we can write

\Gamma_i = P_i \pmatrix{\lambda_i^- & 0 \\ 0 & \lambda_i^+} P_i^{-1} for some P_i. In fact, since the rotations \Gamma_i all commute with one another (and are diagonalizable) they are simultaneously diagonalizable, that is, we can choose all of the P_i to be the same matrix P. Computing eigenvectors corresponding to the eigenvalues shows that we may take P = \pmatrix{\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}}}, the rotation matrix for a clockwise rotation of \frac{\pi}{4} (see the remark below).

Our observation about being able to use the same P for all rotations pays off immediately, as the expression for the product \prod_{i = 1}^n \Gamma_i of rotations simplifies dramatically:

\prod_{i = 1}^n \Gamma_i

= \prod_{i = 1}^n \left[P \pmatrix{\lambda_i^- & 0 \\ 0 & \lambda_i^+} P^{-1} \right]

= P \pmatrix{\prod_{i = 1}^n \lambda_i^- & 0 \\ 0 & \prod_{i = 1}^n \lambda_i^+} P^{-1}.

This yields explicit formulas for A_n, B_n in terms of a, b, (\alpha_i):

\pmatrix{A_n \\ B_n} = \left(\prod_{i = 1}^n \Gamma_i\right) \pmatrix{a \\ b} = \pmatrix{\frac{1}{2}(a + b) \prod_{i = 1}^n \lambda_i^+ + \frac{1}{2} (a - b) \prod_{i = 1}^n \lambda_i^- \\ \frac{1}{2}(a + b) \prod_{i = 1}^n \lambda_i^+ + \frac{1}{2}(b - a) \prod_{i = 1}^n \lambda_i^-} .

Dividing and rewriting gives

\frac{A_n}{B_n} = \frac{\frac{1}{2}(a + b) \prod_{i = 1}^n \lambda_i^+ + \frac{1}{2} (a - b) \prod_{i = 1}^n \lambda_i^-}{\frac{1}{2}(a + b) \prod_{i = 1}^n \lambda_i^+ + \frac{1}{2}(b - a) \prod_{i = 1}^n \lambda_i^-} = \frac{1 + \mu_n}{1 - \mu_n},

where

\mu_n

:= \frac{a - b}{a + b} \prod_{i = 1}^n \frac{\lambda_i^-}{\lambda_i^+}

= \frac{a - b}{a + b} \prod_{i = 1}^n \frac{\cos \alpha_i - \sin \alpha_i}{\cos \alpha_i + \sin \alpha_i}

= \frac{a - b}{a + b} \prod_{i = 1}^n \tan \left(\frac{\pi}{4} - \alpha_i\right) ,

and 0 < \mu_n < 1.

So, if \limsup \alpha_i > 0, then \mu_n \to 0 and hence \frac{A_n}{B_n} \to 1. If, on the other hand, \limsup \alpha_i = 0, we need not have \mu_n \to 0. For small \alpha_i, \tan \left(\frac{\pi}{4} - \alpha_i\right) \sim 1 - 2 \alpha_i, so for \frac{A_n}{B_n} \to 1 I believe it's sufficient and necessary for \sum_{i = 1}^{\infty} \alpha_i to diverge.

Similarly, we get

\begin{align*}\lim_{n \to \infty} (A_n - B_n)

&= (a - b) \prod_{i = 1}^n (\cos \alpha_i - \sin \alpha_i) .

\end{align*}

Again, \limsup \alpha_i > 0 is sufficient to guarantee vanishing. For the case \limsup \alpha_i = 0 we can use \cos \alpha_i - \sin \alpha_i \sim 1 - \alpha_i to conclude that A_n - B_n \to 0 iff \sum_{i = 1}^{\infty} \alpha_i diverges.

**Remark** The Jordan decomposition \Gamma_i = P \pmatrix{\lambda_i^- & 0 \\ 0 & \lambda_i^+} P^{-1} is in this problem more than a convenient tool---it also gives some geometrical insight into why rotating and circumscribing makes rectangles more squarelike. The matrix P^{-1} rotates a vector in the ab-plane (which we can view as a vector from the rectangle to one of its corners) by \frac{\pi}{4}. Then \Gamma_i shears the rotated vector, lengthening its component in the direction that corresponds to the quantity a + b and shortening its component in the direction that corresponds to a - b, which we may view as the absolute defect from squareness. Finally, the matrix P rotates back the one-eighth of a turn, which we may view as returning back to the original ab-coordinates.

**Attribution***Source : Link , Question Author : samgiz , Answer Author : Travis Willse*