  The Euclidean Steiner problem is the problem of finding a set $St$, with the
shortest length, such that $St \cup A$ is connected, where $A$ is a given set
in a Euclidean space. The solutions $St$ to the Steiner problem will be called
Steiner sets while the set $A$ will be called input. Since every Steiner set is
acyclic we call it Steiner tree in the case when it is connected. We say that a
Steiner tree is indecomposable if it does not contain any Steiner tree for a
subset of the input.
  We are interested in finding the Steiner set when the input consists of
infinitely many points distributed on two lines. In particular we would like to
find a configuration which gives an indecomposable Steiner tree.
  We consider a self-similar input, namely the set $A_{\alpha,\lambda}$ of
points with coordinates $(\lambda^{k-1}\cos \alpha,$ $\pm \lambda^{k-1}\sin
\alpha)$, where $\lambda>0$ and $\alpha>0$ are small fixed values. These points
are distributed on the two sides of an angle of size $2\alpha$ in such a way
that the distances from the points to the vertex of the angle are in a
geometric progression.
  To our surprise, we show that in this case the solutions to the Steiner
problem for $A_{\alpha,\lambda}$, when $\alpha$ and $\lambda$ are small enough,
are always decomposable trees. More precisely, any Steiner tree for
$A_{\alpha,\lambda}$ is a countable union of Steiner trees, each one connecting
5 points from the input. By considering only a finite number of components we
obtain many solutions to the Steiner problem for finite sets composed of $4k+1$
points distributed on the two lines ($2k+1$ on a line and $2k$ on the other
line). These solutions are very similar to the ladders of Chung and Graham.
