Finding a vertex of degree 3 in a penny graph to prove that it can be 4 colored I need to prove that finite penny graphs can be 4-colored without using the 4 color theorem. It's obvious that the graph is planar and I know that I if I can always find a vertex of degree 3 then I can perform induction and complete the proof.
I know that since we have a finite penny graph, if there does not exist a vertex of degree 3 then the graph must be infinite in order to exist. I just don't know where to start to prove that I can always find a vertex of degree 3! It seems so obvious (since there should always be an 'outside' to the graph and the most compact way of having a vertex of degree 4 means I need 3 equilateral triangles which results in a non-convex boundary)  but I have no precise proof using contradiction of planarity or something else.... 
Any tips would be appreciated!
 A: You can get away with a vertex of degree four. First note there has to be one, since one can move a remote line parallel to itself until it first meets one of the pennies. Then all the pennies are on one side of that line, and it can be seen that penny has at most four neighboring pennies, which would be arranged around it as four of the total of six which would exactly fit around it making a complete hexagon. The limiting line makes more than four neighbors impossible.
Now the idea of finishing with a vertex of degree four is as follows: Suppose the colors around the vertex go B,R,W,G in circular order (black,red,white,green). Then one can start at the B, and alternately change everything reachable from there (as in a tree). If one gets to the W (so that W would become a B and the "black-white" change did no good), then there is a "black-white" chain connecting those two vertices. But then there cannot also be a "red-green" chain connecting the other two vertices at the same time, and one can start at the red vertex and alternately change reachable vertices between red and green. One will not reach the green vertex touching the starting penny because of the black-white chain connecting the B,W vertices around the starting penny.
I think you could also get three vertices touching a penny if you could show there must be an exterior penny having two different tangents for which all the pennies are on the same side of those tangents. This may be harder to show.
