Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Examination, 04-BS-16 Discrete Mathematics, May 2014. Closed book, no aids. The exam instructs "answer 10 of 12 questions"; every question is answered below as a complete study resource.
Reference texts: Rosen, Discrete Mathematics and Its Applications, 7th ed. (logic, induction, combinatorics, probability, relations, graph theory).
Given. Euler's polyhedron relation and the two structures above (a $K$-triangle mesh; the C$_{60}$ fullerene cage).
Find. (a) the Euler formula; (b) $V$ in terms of $K$; (c) the number of hexagonal and pentagonal faces on C$_{60}$.
Approach. (a) state the classical relation; (b) count edge-incidences from the triangle faces (each triangle contributes 3, each edge is counted twice) to get $E$, then invert Euler's formula for $V$; (c) combine Euler's formula with the vertex-degree-3 constraint (edge count) and the face-edge incidence count to solve a 2-equation linear system for hexagon/pentagon counts.
12a) Euler's polyhedron formula. For any convex polyhedron (or connected planar graph) with $V$ vertices, $E$ edges, and $F$ faces:
$$\boxed{V-E+F=2}$$
12b) Vertices in a $K$-triangle mesh. Each triangle has 3 edges, so summing over all $K$ triangles counts every edge exactly twice (each interior edge is shared by 2 triangles):
$$E=\frac{3K}{2}$$
Every triangle is one face, so $F=K$. Substituting into Euler's formula ($V=2-F+E$):
$$V=2-K+\frac{3K}{2}=2+\frac{K}{2}$$
$\boxed{V=\dfrac{K}{2}+2}$ (check: a tetrahedron has $K=4$ triangular faces $\Rightarrow V=2+2=4$ ✓; an icosahedron has $K=20$ $\Rightarrow V=2+10=12$ ✓).
12c) C$_{60}$ hexagon/pentagon split — set up two equations. Given $V=60$, $F=32$, and every vertex of a fullerene is 3-valent (each carbon bonds to exactly 3 neighbours), the edge count from vertex degrees is $E=\dfrac{3V}{2}=\dfrac{3\times60}{2}=90$. Let $p$ = number of pentagons, $h$ = number of hexagons. Two equations: total faces $p+h=32$, and total edge-incidences from faces $5p+6h=2E=180$ (each pentagon contributes 5 edge-sides, each hexagon 6, and every edge is shared by exactly 2 faces).
12c) Solve the system. From $h=32-p$, substitute into $5p+6h=180$:
$$5p+6(32-p)=180 \implies 5p+192-6p=180 \implies -p=-12 \implies p=12,\ h=20$$
Check: $V-E+F=60-90+32=2$ ✓ (Euler's formula holds). $\boxed{12 \text{ pentagons and } 20 \text{ hexagons}}$ — this is exactly the classical result: every fullerene has exactly 12 pentagons (a direct consequence of Euler's formula for any 3-valent polyhedron built from pentagons and hexagons only).
