364 (b) The planar graph shown at the right, which is isomorphic to the graph of the icosahedron (see Exercise 13, Section 10.2), has twelve vertices, each of degree 5. Solutions to Exercises 19. (a) [BB] Let 91,92, ... ,9n be the connected components of 9. Since 9i has at least three vertices, we have Ei :::; 3Vi -6. Hence, L: Ei :::; 3 L: Vi -6n, so E :::; 3V -6n as required. (b) Let 91, 92, ... ,9n be the connected components of 9. If 9i has at least three vertices, then Ei :::; 3Vi -6. If 9i has two vertices, then Ei = 1, so Ei = 3Vi -5. If 9i has one vertex, then Ei = 0 and Ei = 3Vi -3. In all cases, Ei :::; 3Vi -3, so E = L: Ei :::; 3 L: Vi -3n = 3V -3n. Finally, note that if n ~ 2, then 3V -3n :::; 3n -6 so Theorem 13.1.4 holds while, if n = 1, the graph is connected; we established E :::; 3V -6 for such a graph in the text. 20. We may assume in all parts of this question that 9 is connected. (a) [BB] Say there is only one vertex of degree at most 5. Then L: deg Vi ~ 6(V -1) = 6V -6, contradicting L: deg Vi = 2E :::; 6V -12. (b) Say there are only two vertices of degree at most 5.
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