I recently saw a question here on mathoverflow: «For what n and t can a square be partitioned into n similar rectangles in t congruence classes?», where Joseph Gordon gave a proof that, indeed, a square can be partitioned into n non-congruent similar rectangles for any $n\ge3$. His method involves the use of Fibonacci-numbers. For n=3, the aspect ratio,r, of the similar rectangles is the square of the plastic number (also known as the Padovan constant).


Using his method I did some calclulations for a few n (stopped at n=16, where I found r approximately 1.6180355) and the results seem to suggest that as n grows larger, the aspect ratio of the similar rectangles tends towards the golden ratio. Can this be proved/disproved in any way?



Tiling a square with similar non-congruent rectangles. What is the aspect ratio of the rectangles as n grows large?
You are correct.


Let $\varphi$ be the golden ratio and $r\_n$ be the ratio you want (long side to short).


According to the answer you mention, for a certain positive constant $A\_n:$ $$r\_n=\frac{F\_{n-1} A\_n + F\_{n-2}}{F\_{n-2} A\_n + F\_{n-3}}. \tag{\*}$$ It follows that (for odd $n$) $$\frac{F\_{n-1}}{F\_{n-2}} <r\_n < \frac{F\_{n-2}}{F\_{n-3}}. $$ The same is true for even $n$ except the inequalities are reversed. Furthermore, the two rationals differ by exactly $\frac{1}{F\_{n-2}F\_{n-3}}.$


It is also well known that (again, up to reversing the inequalities) $$\frac{F\_{n-1}}{F\_{n-2}} < \varphi < \frac{F\_{n-2}}{F\_{n-3}}. \tag{\*\*}$$ This establishes the result and shows that the convergence is rapid.




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The exact values of $A\_n$ were not needed here, but in fact they converge quickly to $\frac{\sqrt 5}{\varphi}$ the root of the linear equation $$(A-1)\varphi^2-1.$$


$A\_n$ is the real root of the cubic polynomial $$f\_n(A)=(A-1)(F\_{n-1} A + F\_{n-2})^2 - (F\_{n-2} A + F\_{n-3})^2.$$ We see from $(\*\*)$ that $f\_n$ is almost exactly $$(A-1)(F\_{n-2}\varphi A + F\_{n-3}\varphi)^2 - (F\_{n-2} A + F\_{n-3})^2\\ =\left((A-1)\varphi^2-1\right)(F\_{n-2} A + F\_{n-3})^2.$$


