TITLE: A property of circles
QUESTION [10 upvotes]: Consider a set of circles on a plane that don't overlap each other and any of them touches at least 6 other circles. Prove that this set has infinite number of circles. 
Well, there seems to be something special about number 6, but the circles are not even of the same radius. If they have the same radius the proof by contradiction would be easy. But this is really hard.

REPLY [4 votes]: Here is the argument @Rahul is hinting at.  Suppose that you have an arrangement of $n$ circles with disjoint interiors.  Now draw a line between the centers of circles that are tangent.  In the resulting figure $G$, none of the segments intersect except at endpoints, giving a planar graph (see http://en.wikipedia.org/wiki/Planar_graph) with the circle centers as vertices and tangencies as edges.  A fundamental fact about finite planar graphs is that if they have $n$ vertices there are at most $3n-6$ edges (this follows from Euler's formula http://en.wikipedia.org/wiki/Planar_graph#Euler.27s_formula), which is too few for every circle to be tangent to at least $6$ others.  (To the OP: this direction does lead to "something".  Thinking about why infinite graphs get around this is interesting.)
@user22705's is pretty much complete, though.  Consideration of angles says that if a circle in the kind of configuration you're interested in is tangent to one with strictly larger radius, it is also tangent to one with strictly smaller radius.  So if there is more than one radius, you end up with an infinite sequence of smaller and smaller circles.  The other case is all equal.  The tangency points on any circle are then determined: there are six of them, equally spaced.  Now check that this implies there must be circles arbitrarily far away from the origin.