I feel sure this must be known, but can I find it??


Which connected plane graphs (graphs drawn in the plane without crossings) have a spanning tree such that at least one edge of each face is in the tree?


If multiple edges are allowed, there might be simply too many faces, and other obstructions are easy to find. But I don't know about simple graphs.



Spanning trees of plane graphs containing an edge of every face
A triangulation has a spanning tree with the required property if and only if its dual graph has a hamiltonian path (is *traceable*).


Zamfirescu constructed a 3-regular 3-connected planar non-traceable graph on 88 vertices. 
The dual of this graph is a triangulation with no spanning tree with required properties.


a reference:
[Tudor Zamfirescu, Three small cubic graphs with interesting Hamiltonian properties,
Journal of Graph Theory, Vol. 4 (1980), 287-292.](http://tzamfirescu.tricube.de/TZamfirescu-077.pdf)


