Problem: For what real values of $k > 0$ is it possible to
dissect a $1 \times k$ rectangle into two similar but noncongruent polygons?

Solution: Answer: the dissection is possible for every $k > 0$ except for $k = 1$.

\paragraph{Construction.}
By symmetry it suffices to give a construction for $k > 1$
(since otherwise we replace $k$ by $k\inv$).
For every integer $n \ge 2$ and real number $r > 1$,
we define a shape $\mathcal R(n,r)$ as follows.
\begin{itemize}
  \ii We start with a rectangle of width $1$ and height $r$.
  To its left, we glue a rectangle of height $r$ and width $r^2$ to its left.
  \ii Then, we glue a rectangle of width $1+r^2$ and height $r^3$ below our figure,
  followed by a rectangle of height $r+r^3$ and width $r^4$ to the left of our figure.
  \ii Next, we glue a rectangle of width $1+r^2+r^4$ and height $r^5$ below our figure,
  followed by a rectangle of height $r+r^3+r^5$ and width $r^6$ to the left of our figure.
\end{itemize}
\dots and so on, up until we have put $2n$ rectangles together.
The picture $\mathcal R(3,r)$ is shown below as an example.
\begin{center}
\begin{asy}
size(8cm);
defaultpen(fontsize(10pt));

real r = 1.6;

pair A = (0,r^3+r);
pair B = (r^4,r^3+r);
pair C = (r^4+r^2,r^3+r);
pair D = (r^4+r^2+1,r^3+r);
pair E = (r^4,r^3);
pair F = (r^4+r^2, r^3);
pair G = (r^4+r^2+1, r^3);
pair O = (0,0);
pair H = (r^4,0);
pair I = (r^4+r^2+1,0);
pair W = (-r^6, A.y);
pair X = (W.x,-r^5);
pair Y = (0,-r^5);
pair Z = (I.x,-r^5);

pen p = black+1.3;
pen q = red;
fill(O--A--B--H--cycle, palecyan);
fill(B--C--F--E--cycle, palecyan);
fill(W--A--Y--X--cycle, palecyan);
fill(C--D--G--F--cycle, palegreen);
fill(E--G--I--H--cycle, palegreen);
fill(O--Y--Z--I--cycle, palegreen);

draw(O--A, q);
draw(B--E, q);
draw(F--G, q);
draw(H--I, q);
draw(W--D--Z--X--cycle, p);
draw(C--F--E--H--O--Y, p);

label("$1$", (C+D)/2, dir(90));
label("$r^2$", (B+C)/2, dir(90));
label("$r^4$", (A+B)/2, dir(90));
label("$r^6$", (W+A)/2, dir(90));

label("$r^1$", (D+G)/2, dir(0));
label("$r^3$", (G+I)/2, dir(0));
label("$r^5$", (I+Z)/2, dir(0));
\end{asy}
\end{center}
Observe that by construction,
the entire shape $\mathcal R(n,r)$ is a rectangle
which consists of two similar ``staircase'' polygons
(which are not congruent, since $r>1$).
Note that $\mathcal R(n,r)$ is similar to a $1 \times f_n(r)$ rectangle
where $f_n(r)$ is the aspect ratio of $\mathcal R(n,r)$, defined by
\[ f_n(r) = \frac{1+r^2+\dots+r^{2n}}{r+r^3+\dots+r^{2n-1}}
  = r + \frac{1}{r+r^3+\dots+r^{2n-1}}.  \]

We claim that this is enough.
Indeed for each fixed $n$, note that
\[ \lim_{r \to 1^+} f_n(r) = 1 + \frac1n
  \; \text{ and } \;
  \lim_{r \to \infty} f_n(r) = \infty. \]
Since $f_n$ is continuous, it achieves all values greater than $1+\frac1n$.
Thus by taking sufficiently large $n$ (such that $k > 1+\frac1n$),
we obtain a valid construction for any $k > 1$.

\paragraph{Proof of impossibility for a square.}
Now we show that $k = 1$ is impossible (the tricky part!).
Suppose we have a squared dissected into
two similar polygons $\mathcal P \sim \mathcal Q$.
Let $\Gamma$ be their common boundary.
By counting the number of sides of $\mathcal P$ and $\mathcal Q$
we see $\Gamma$ must run from one side of the square to an opposite side
(possibly ending at a corner of the square).
We orient the figure so $\Gamma$ runs from north to south,
with $\mathcal P$ to the west and $\mathcal Q$ to the east.
\begin{center}
\begin{asy}
size(4cm);
pair A = (0,0);
pair B = (1,0);
pair C = (1,1);
pair D = (0,1);
pair W = (0.4,0);
pair X = (0.3,0.7);
pair Y = (0.7,0.3);
pair Z = (0.6,1);
filldraw(A--W--X--Y--Z--D--cycle, palecyan, black);
filldraw(B--W--X--Y--Z--C--cycle, palegreen, black);
draw( W--X--Y--Z, black+1 );
label("$\Gamma$", Y, dir(15));
label("$\mathcal P$", (0.1,0.5), blue);
label("$\mathcal Q$", (0.9,0.5), deepgreen);
\end{asy}
\end{center}

Let $s$ be the longest length of a segment in $\Gamma$.
\begin{claim*}
  The longest side length of $\mathcal P$ is $\max(s,1)$.
  Similarly, the longest side length of $\mathcal Q$ is $\max(s,1)$ as well.
\end{claim*}
\begin{proof}
  The only edges of $\mathcal P$ not in $\Gamma$
  are the west edge of our original square, which has length $1$,
  and the north/south edges of $\mathcal P$ (if any),
  which have length at most $1$.
  An identical argument works for $\mathcal Q$.
\end{proof}
It follows the longest sides of $\mathcal P$ and $\mathcal Q$ have the same length!
Hence the two polygons are in fact congruent, ending the proof.