Paper folding and continued fractions This question is suggested by a prior question, which has not received a complete answer.  One corner of rectangular sheet of paper $P_1$ is folded down to make a trapezoid, and then a right triangle is cut off to make another rectangle $P_2.$  The process is repeated with $P_2$ to produce a rectangle $P_3$ and so on.  We assume that the ratio of the sides of $P_1$ is irrational, so none of the rectangles is a square, and the process goes on forever.

Let $S_n$ be the area of $P_n$ and suppose $$\lim_{n\to\infty}{S_{n+2}\over S_n}$$ exists.  What are the possible values of the limit.  
In the original question, the OP stated that he believed that, in order for the limit to exist, it must be the case that $P_n$ and $P_{n+2}$ are similar, for large $n,$ but that he couldn't prove it. 
In a partial answer, I corrected some calculation errors, and pointed out the relation of the problem to continued fractions, but I haven't been able to prove the OP's hypothesis either, though I think it's very likely true.
Can you prove that $P_n$ and $P_{n+2}$ must eventually be similar?  Alternatively, can you see how to modify the analysis so as not to use this hypothesis?
 A: There is another possible limit $\color{blue}{3-2\sqrt{2}=(1+\sqrt{2})^{-2}}$.  How?
Suppose rectangle $n$ has length $a$ and width $b$ with $a>2b$.  Then to get to rectangle $n+2$ you shorten the $a$ dimension twice leaving a rectangle with dimensions $a-2b$ and $b$.  If
$\dfrac{a-2b}{b}=\dfrac{b}{a}$
then the rectangles are similar forcing a fixed value of $S_{n+2}/S_n$.  Note that the intermediate rectangle would not be similar so we would not have $S_{n+1}/S_n=\sqrt{S_{n+2}/S_n}$.
The above equation for $a$ and $b$ is solved for a positive root $a/b=1+\sqrt{2}>2$.  Thereby
$\dfrac{S_{n+2}}{S_n}=\dfrac{a-2b}{a}=3-2\sqrt{2}$.
With $a/b=1+\sqrt{2}>2$ one can verify that also 
$\dfrac{S_{n+3}}{S_{n+1}}=3-2\sqrt{2}$.
A: The following are just some considerations, which might be of help.

We start with a rectangle of sides $a_0 \le b_0$ and
after ${\left\lfloor {{{b_{\,0} } \over {a_{\,0} }}} \right\rfloor }$ cuts we reach to a rectangle
of sides
$$
\left\{ {\matrix{
   {a_{\,0} \left\{ {{{b_{\,0} } \over {a_{\,0} }}} \right\} = b_{\,0} \bmod a_{\,0} } &  \to  & {a_{\,1} }  \cr 
   {a_{\,0} } &  \to  & {b_{\,1} }  \cr  } } \right.\;\;
$$
The ratio of the sides proceeds as
$$ \bbox[lightyellow] {  
\eqalign{
  & {{a_{\,0} } \over {b_{\,0} }} = {1 \over {{{b_{\,0} } \over {a_{\,0} }}}}
 = {1 \over {\left\lfloor {{{b_{\,0} } \over {a_{\,0} }}} \right\rfloor  + \left\{ {{{b_{\,0} } \over {a_{\,0} }}} \right\}}} =   \cr 
  &  = {1 \over {\left\lfloor {{{b_{\,0} } \over {a_{\,0} }}} \right\rfloor  + {{b_{\,0} \bmod a_{\,0} } \over {a_{\,0} }}}} =   \cr 
  &  = {1 \over {\left\lfloor {{{b_{\,0} } \over {a_{\,0} }}} \right\rfloor  + {{a_{\,1} } \over {b_{\,1} }}}}
 = {1 \over {\left\lfloor {{{b_{\,0} } \over {a_{\,0} }}} \right\rfloor  + {1 \over  \ddots }}} =  \cdots  \cr} 
 }\tag {1}$$
which is in fact the continued fraction development of $a_0 / b_0$, whose terms come from
the progressive steps in the Euclidean Algorithm for $gcd(a_0,b_0)$.
The sequence of the areas proceeds as follows
$$ \bbox[lightyellow] {  
\eqalign{
  & S_{\,0}  = a_{\,0} \,b_{\,0}  = a_{\,0} ^{\,2} \left( {{{b_{\,0} } \over {a_{\,0} }}} \right)=A_{\,0}  \cr 
  & S_{\,1}  = a_{\,0} \,\left( {b_{\,0}  - a_{\,0} } \right) = a_{\,0} ^{\,2} \,\left( {{{b_{\,0} } \over {a_{\,0} }} - 1} \right)  \cr 
  & \quad  \vdots   \cr 
  & S_{\,\left\lfloor {{{b_{\,0} } \over {a_{\,0} }}} \right\rfloor }
  = a_{\,0} ^{\,2} \,\left( {{{b_{\,0} } \over {a_{\,0} }} - \left\lfloor {{{b_{\,0} } \over {a_{\,0} }}} \right\rfloor } \right)
 = a_{\,0} ^{\,2} \,\left\{ {{{b_{\,0} } \over {a_{\,0} }}} \right\} = a_{\,1} b_{\,1} =A_{\,1}    \cr 
  & \quad  \vdots   \cr 
  & S_{\,n}  = a_{\,m} ^{\,2} \left( {{{b_{\,m} } \over {a_{\,m} }} - q} \right)\quad \left| \matrix{
  \;\sum\limits_{0\, \le \,k\, \le \,m - 1} {\left\lfloor {{{b_{\,k} } \over {a_{\,k} }}} \right\rfloor }  \le n < \sum\limits_{0\, \le \,k\, \le \,m}
 {\left\lfloor {{{b_{\,k} } \over {a_{\,k} }}} \right\rfloor }  \hfill \cr 
  \;0 \le n - \sum\limits_{0\, \le \,k\, \le \,m - 1} {\left\lfloor {{{b_{\,k} } \over {a_{\,k} }}} \right\rfloor }
  = q < \left\lfloor {{{b_{\,m} } \over {a_{\,\,m} }}} \right\rfloor  \hfill \cr}  \right. \cr} 
 }\tag {2}$$
where the meaning of the $A_n$ are obvious.
We can say that the $Sn$ are interpolating the $A_m$.
If we take $a_0=1,\;b_0=\varphi$ (golden ratio), we get
$$
\eqalign{
  & {1 \over {{{b_{\,0} } \over {a_{\,0} }}}} = {{\rm 1} \over \varphi } = {{\rm 1} \over {1 + \left( {\varphi  - 1} \right)}}
 = {{\rm 1} \over {1 + {1 \over \varphi }}}\quad  \Rightarrow \quad \left\{ \matrix{
  b_{\,m}  = \varphi ^{\,1 - \,m}  \hfill \cr 
  a_{\,m}  = \varphi ^{\, - \,m}  \hfill \cr 
  n = m\quad q = 0 \hfill \cr 
  S_{\,n}  = a_{\,n} \,b_{\,n}  = \varphi ^{\,2 - 2\,n}  \hfill \cr}  \right.  \cr 
  & {{S_{\,n + 2} } \over {S_{\,n} }} = \varphi ^{\, - \,4}  \cr} 
$$
But, apart from the golden ratio above, the Euclidean algorithm is in general not smooth,
and thus it is difficult to grasp the limit of the wanted ratio (to my knowledge).
In fact, if we take for example $(1,\pi)$ as the starting rectangle, it is known that the
the terms of the corresponding CF do not have any regular pattern.
For our problem that means that the side ratio $\left\lfloor {{{b_{\,n} } \over {a_{\,n} }}} \right\rfloor $
( and thus the ratio itself) is varying unpredictably.
And unpredictable as well will be the limit of the area ratio.
That is understandable in the model, as that many times we reach very near to a rational 
rectangle, after which we are left with a very thin and long stripe, which has an anomalously
low area wrt the parent sheet.
