Let $V$ be a 3-connected planar graph, $V^*$ be its dual graph.
Let $(C_i : i \in V)$ and $(D_p : p \in V^*)$ be a collection of circles so that if any edge $\{i,j\}$ borders any faces $a$ and $b$, the circles $C_i$ and $C_j$ are tangent at a point x, the circles $D_a$ and $D_b$ are tangent at the same point, these 4 circles intersect at this point and for which the C_i are disjoint, and any two $D_p$ not corresponding to the outer face are disjoint.
Suppose we have positions $x_i$ in R^2 for each vertex for each vertex at the center of the $C_i$; how can one show that this set of positions can also be obtained as a rubber band representation -- i.e. from the equilibrium of a graph with the vertices on the outer face $p_0$ nailed and ideal (natural length 0) rubber bands of appropriate strengths for each edge.
We could put the equations in variables $k_{ij}$: $\sum_{j \in N(i)} k_{ij}(x_i - x_j) = 0$ for every $i$ not on the outer face. This linear equations have a nontrivial solution since there are fewer equations than variables. But how can one show that we can then obtain $k_i$ all nonzero and positive?