Question 11 of 12: Euler's Formula for Planar Graphs and Polyhedra
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Examination, 04-BS-16 Discrete Mathematics, May 2013. Closed book, no aids, 3 hours, 12 questions of 10 marks each (100 marks); the exam instructs "answer 10 of 12" but every question is solved below as a complete study resource.
Reference texts: Rosen, Discrete Mathematics and Its Applications, 7th ed. (logic Ch.1, induction & recursion Ch.5, counting Ch.6, discrete probability Ch.7, relations Ch.9, graphs Ch.10-11); Epp, Discrete Mathematics with Applications.
Question 11: Euler's Formula for Planar Graphs and Polyhedra (10 marks)
Given. Polyhedron faces: 12 pentagons ($5$ edges each) + 20 hexagons ($6$ edges each); at every vertex exactly $2$ hexagons $+1$ pentagon meet (so every vertex has degree $3$).
Find. (a) Euler's formula relating $v,e,f$. (b) The vertex count $v$ and edge count $e$ of this polyhedron (a truncated icosahedron / "soccer-ball" solid).
(a) Euler's formula. For any finite, connected planar graph (equivalently, a convex polyhedron's surface graph):
$$\boxed{v - e + f = 2.}$$
(b) Edge count via face–edge incidences. Summing edges around every face counts each edge exactly twice (once from each adjacent face): $12\times5 + 20\times6 = 60+120=180$ face–edge incidences, so
$$e = \frac{180}{2} = \boxed{90.}$$
(b) Vertex count via vertex–edge incidences (degree sum). Every vertex has degree $3$ (2 hexagons + 1 pentagon meeting there means 3 edges emanate from it), so the sum of degrees is $3v$; the handshake lemma gives sum of degrees $=2e$:
$$3v = 2e = 2(90) = 180 \ \Rightarrow\ v = \boxed{60.}$$
Check against Euler's formula. $v-e+f = 60-90+32 = 2$ ✓, confirming the counts are consistent (this is exactly the truncated-icosahedron / soccer-ball polyhedron).
