27.2.11 Find all real numbers $k>0$, such that a $1 \times k$ rectangle can be divided into two similar but non-congruent polygons.
The conclusion is: $k>0, k \neq 1$.
First, prove that a $1 \times 1$ square cannot be divided into two similar polygons.
Consider the endpoints of the dividing polyline. If these two endpoints belong to two opposite sides of the square (including the vertices of the square), then the two resulting polygons have the longest sides equal. Given that they are similar, these two polygons must be congruent. If one of the endpoints is a vertex of the square, then the two polygons have a difference of 1 in the number of sides (note that in this case, the other endpoint is on a side not adjacent to the vertex). If the two endpoints are on adjacent sides of the square, then the two polygons have a difference of 2 in the number of sides. Neither case can result in two similar polygons. Therefore, when $k=1$, there does not exist a valid division.
Next, prove that for $k>0, k \neq 1$, a $1 \times k$ rectangle can be divided into two similar polygons.
It suffices to prove the case when $k>1$ (when $k \in (0,1)$, treat $k$ as 1 unit, and consider the $\frac{1}{k} \times 1$ case). Take $n \in \mathbf{N}^{*}$ such that $k > \frac{n+1}{n}$. Consider the two similar step shapes, which can be combined to form a $1 \times k$ rectangle, where $\lambda (>1)$ and $x$ are to be determined.

Thus, we need to prove the existence of $\lambda$ and $x$ to show that a valid division exists. In fact, let $f(\lambda) = \frac{1 + \lambda^2 + \cdots + \lambda^{2n}}{\lambda + \lambda^3 + \cdots + \lambda^{2n-1}}$, then $f(1) = \frac{n+1}{n}$. There exists $\lambda > 1$ such that $f(\lambda) = k$. For this $\lambda$, choose $x$ such that $(\lambda + \lambda^3 + \cdots + \lambda^{2n-1}) x = 1$. Then the above division meets the requirements.
In summary, the required $k$ is any positive real number not equal to 1.