Good morning,


I hope this question is not too far out of the scope of the forum. I am posting it here because this doesn't seem to be a very standard problem.


Yesterday we were calculating the equivalent resistances of various polyhedra between adjacent vertices and we noticed this:


Consider this bijection $f$ between edges of a polyhedron and its dual: if edge $AB$ on polyhedron $X$ separates faces $\alpha$ and $\beta$, $f(AB)$ on polyhedron $X\star$ connects the dual vertices of faces $\alpha$ and $\beta$.


Now construct a circuit of resistances all equal to $1$, each occupying an edge of polyhedron $X$ and do the same for $X\star$. We noticed that the equivalent resistance across $AB$ on $X$ and $f(AB)$ on $X\star$ sum to $1$ in all cases we calculated.


We could prove this works for platonic solids but the proof relies on too many symmetries to be easily generalised. However, we calculated this to work on a few other solids (EDIT: all pyramids and prisms and a few Archimedean and Catalan solids) which do not possess the same symmetries.


Is this true in general? If not, for what classes of solids is it true?



Dual polyhedra and electric circuits
Here is the [proof](https://www.ejgta.org/index.php/ejgta/article/view/657) for any planar graph.


