Map-Coloring Problem When we are faced with map-coloring problem, why do we allow countries that meet at only one point to receive the same color? Is it because they do not share the same boundaries or common boundaries? Also if anyone if familiar with any source, can you direct me to a map that requires more than four colors if countries that meet only at one point that must get different colors?
 A: Historically, the map-coloring problem arose from (believe it or not) actually coloring maps.  There, if two countries share a common border that is a whole line or curve, then giving them the same color would make the map harder to read; the border would not be so clearly visible as if you used different colors.  This problem doesn't arise if the countries share only a single border point.  So it is reasonable to allow such countries to have the same color.  (Then, as pointed out in other answers and comments, this convention is necessary in order to have any chance for a non-trivial theorem.)
A: The reason that the four-color conjecture involves countries that share an edge has to do with the dual graph.  In planar graphs, a graph's dual treats the faces like vertices and adds an edge between two faces whenever those faces share an edge.  The dual graph is also planar.
A restatement of the four-color theorem says that:

Every planar graph is $4$-colorable.

Here, $4$-colorable refers to the fact that we can match the vertices to four distinct colors so that no two adjacent vertices are given the same color.  This shows the equivalence of the two theorems: since the dual graph can be $4$-colored, you can color the faces so that no two faces sharing an edge will be given the same color.
As noted in the comments, the wheel graph $W_6$ (or $W_7$, the "trivial pursuit pie") is a map where the faces cannot be $4$-colored so that faces sharing a vertex are given different colors.
