# What pair of rectangle shapes is exchanged by cutting off a square?

What pair of rectangle shapes is exchanged by cutting off a square?

According to https://oeis.org/A059966 the golden rectangle has period one when cutting off a square. What pair of rectangle (shapes) have period two?

Let's define a rectangle shape to be a rectangle normalised by scaling its smallest side to length one.

my attempt so far

we have sides $$1,b$$ then remove $$1,1$$ square to get $$1,b-1$$ then we scale by the smallest side to get $$1,1/(b-1)$$ then we remove the square $$1,1$$ again to get $$1,(b-1)^{-1}-1$$ and scale again to get $$1,((b-1)^{-1}-1)^{-1}$$ and this is our original rectangle so we have:

$$b=((b-1)^{-1}-1)^{-1}$$

So the solution to this is the first of our rectangle shapes. Am I on the right track?

Since "cutting off a square" is scale-invariant, we can postpone the scaling operation to the end.

Start with: $$(b,1)$$ where $$b > 1$$
The next rectangle is $$(b - 1, 1)$$

The next rectangle is either:
$$(b-2,1)$$ if $$(b-1) > 1$$ or
$$(b-1, 2-b)$$, if $$(b-1) \leq 1$$

Assume $$(b-2,1)$$ is the last rectangle, with $$b-2 > 1$$. Then we solve $$\frac{b-2}{1} = b$$, with no solutions. Assume $$b-2 < 1$$. Then we solve $$\frac{1}{b-2} = b$$ giving $$b\in\{1\pm \sqrt{2} \}$$, with valid solution $$b = 1 + \sqrt{2}$$ .

Assume $$(b-1,2-b)$$ is the last rectangle, with $$(b-1) > (2-b)$$ . We solve $$\frac{b-1}{2-b} = b$$ giving $$b= \frac{1+\sqrt{5}}{2}$$ as the only valid solution (matching $$b\geq 1$$). This is the ordinary golden rectangle. If instead we set $$(2-b) > (b-1)$$, solving $$\frac{2-b}{b-1} = b$$ gives $$b = \sqrt{2}$$.

The "cutting off a square" can be defined as an operation on the ratio of a rectangle, given by:

$$o(r) = \max(r-1, \frac{1}{r-1})$$

The solutions for $$o^{2}(r) = r$$ with $$r\geq 1$$ are: $$r\in\{\frac{1+\sqrt{5}}{2}, \sqrt{2}+1, \sqrt{2}\}$$

with the "cycles" going as: $$(\frac{1+\sqrt{5}}{2})\rightarrow (\frac{1+\sqrt{5}}{2}) \rightarrow (\frac{1+\sqrt{5}}{2})$$

$$(\sqrt{2}+1) \rightarrow (\sqrt{2}) \rightarrow (\sqrt{2}+1)$$ $$(\sqrt{2}) \rightarrow (\sqrt{2}+1) \rightarrow (\sqrt{2})$$

• Thanks for the answer... just a note out of courtesy as I'm short on time at the moment, I will digest these two answers, prune away this comment and accept one in due course. Rob Jun 13 at 13:35

Let $$[x,y]$$ represent a rectangle with sides of length $$x$$ and $$y$$. Let $$[x:y]$$ represent the equivalence class of rectangles $$[x,y]$$ with the same "shape" (i.e. ratio $$x/y$$). Define the transposition map $$T([x,y]):=[y,x]$$ which is an involution. Define the family of maps $$L_a([x,y]):=[x+a\,y,y]$$ which represents attaching a rectangle of shape $$[a:1]$$ to the given rectangle $$[x,y]$$ preserving the side $$y$$ if $$a>0$$. If $$a<0$$ then it represents cutting off instead of attaching.

These $$L_a$$ family maps satisfy

$$L_a(L_b([x,y]) =L_{a+b}([x,y]),\quad L_a(L_{-a}([x,y])) = [x,y].$$

Another family of maps $$R_a([x,y]):=[x,y+a\,x]$$ represents attaching a rectangle of shape $$[1:a]$$ to the given rectangle $$[x,y]$$ preserving the side $$x$$. These maps satisfy the same properties as those for the other family of maps because the transposition map $$T$$ conjugates the two families of maps.

Define special cases $$L = L_1$$ and $$R = R_1$$ whose inverse maps are $$L^{-1} = L_{-1}$$ and $$R^{-1} = R_{-1}$$.

The comment in the OEIS sequence A059966 entry states

The number of cycles of length n of rectangle shapes in the process of repeatedly cutting a square off the end of the rectangle. For example, the one cycle of length 1 is the golden rectangle.

The process of cutting off squares is inverse to the process of attaching squares which means the cycles of cutting off squares are exactly mirrored by the cycles of attaching squares.

The OEIS sequence enumerates irreducible cyclic compositions of the maps $$L,T$$ where $$n$$ is the number of $$L$$s and there is at least one $$T$$ no two of which are consecutive (up to cyclic equivalence).

For $$n=1$$ there is only the cycle $$TL = LT$$.

For $$n=2$$ there is only one cycle $$TLL = LLT = LTL$$ but $$TLTL$$ is excluded since it is the square of $$TL$$.

For $$n=3$$ there is $$TLLL$$ and $$TLLTL$$ but $$TLTLTL$$ is excluded since it is the cube of $$TL$$.

This rectangle shape must satisfy $$[x:y] = L(L(T([x:y]))) = [2x+y:x].$$ There is only one positive real solution to $$t := x/y = 2+1/t$$ which is $$\sqrt{2}+1$$. Transposing gives $$\sqrt{2}-1$$. Adding a unit square results in $$\sqrt{2}.$$ Adding another unit square results in $$\sqrt{2}+1$$ which is the original shape.