a light edge, then that edge must have been in the original graph. We give the charge to each vertex and the charge to each face , where denotes the degree of a vertex and the length of a face. (Since the graph is a triangulation, the charge on each face is 0.) Recall that the sum of all the degrees in the graph is equal to twice the number of edges; similarly, the sum of all the face lengths equals twice the number of edges. Using Euler's Formula, it's easy to see that the sum of all the charges is 12: We use only a single discharging rule: Each degree 5 vertex gives a charge of 1/5 to each neighbor. We consider which vertices could have positive final charge. The only vertices with positive initial charge are vertices of degree 5. Each degree 5 vertex gives a charge of 1/5 to each neighbor. So, each vertex is given a total charge of at most . The initial charge of each vertex v is . So, the final charge of each vertex is at most . Hence, a vertex can only have positive final charge if it has degree at most 7. Now we show that each vertex with positive final charge is adjacent to an endpoint of a light edge. If a vertex has degree 5 or 6 and has positive final charge, then received charge from an adjacent degree 5 vertex , so edge is light. If a vertex has degree 7 and has positive final charge, then received charge from at least 6 adjacent degree 5 vertices. Since the graph is a triangulation, the vertices adjacent to must form a cycle, and since it has only degree 7, the degree 5 neighbors cannot be all separated by vertices of higher degree; at least two of the degree 5 neighbors of must be adjacent to each other on this cycle. This yields the light edge. References . . . (Lecture text for Spring School on Combinatorics). . . Graph theory