2-connected planar graph

Let G a simple 2-connected planar graph so that all vertices are incident with the infinite region. Suppose that every bounded region of G has length 3 (so is a cycle of length 3). Let k be the number of vertices of degree 2 in G, and let r be the number of regions of G sharing no edges with the infinite region.

If |V(G)| > 3, show that

k = r + 2

I'm trying to figure out how to go about this. So, I think that these could be of help:

• |E|−|V|=|R|−2
• every edge bounds two regions
• if the region shares no edge with the infinite region, then it only shares edges with other regions of length 3.
• ∑deg(v)=2|E(G)|
• 12≤∑[6−deg(v)] so ∑deg(v)≤6|V(G)|−12
• for 2-connected graphs, every vertex has deg(v)≥2

Any ideas?

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• $k=$ degree-1 nodes of the tree,
• $r=$ degree-3 nodes of the tree.
Contract all the degree one tree nodes. This gives almost a full binary tree, if you select one of the degree-3 nodes as root. "Almost", since the root has 3 children. For such a tree we have by induction #leaves=#interior nodes +2. And hence $k=r+2$.