04-BS-16 · December 2013

Question 11 of 12: Euler's Polyhedron Formula

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

National Examination, 04-BS-16 Discrete Mathematics, Dec 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, sets Ch.2, induction & pigeonhole Ch.5-6, relations Ch.9, counting Ch.6, discrete probability Ch.7, graphs Ch.10-11); Epp, Discrete Mathematics with Applications.

Question 11: Euler's Polyhedron Formula (10 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. (b) Truncated tetrahedron: 4 hexagons + 4 triangles. (c) Truncated octahedron: $v=24$, $e=36$, faces = hexagons + squares only.

Find. (a) Euler's formula. (b) $e,v$ for the truncated tetrahedron. (c) Number of hexagonal and square faces of the truncated octahedron.

(a) Euler's formula. For any finite, connected, planar graph drawn without crossings: $$\boxed{v - e + f = 2.}$$
(b) Count edges via face-edge incidences. Faces: $f=4+4=8$. Each hexagon contributes 6 edges, each triangle 3; every edge is shared by exactly 2 faces: $$e = \frac{4(6)+4(3)}{2} = \frac{24+12}{2} = \frac{36}{2} = \boxed{18.}$$
(b) Find vertices via Euler's formula. $$v = 2 - f + e = 2 - 8 + 18 = \boxed{12.}$$
(c) Find total faces via Euler's formula. $$f = 2 - v + e = 2 - 24 + 36 = 14.$$ Let $h$ = hexagons, $s$ = squares: $h+s=14$.
(c) Solve using the edge-incidence equation. Each hexagon has 6 edges, each square 4, each edge shared by 2 faces: $$6h+4s = 2e = 72.$$ Substituting $s=14-h$: $6h+4(14-h)=72\Rightarrow 6h+56-4h=72\Rightarrow2h=16\Rightarrow h=8$, so $s=14-8=6$: $$\boxed{8\text{ hexagonal faces and }6\text{ square faces.}}$$

Final results — Question 11
Part	Result
(a)	$v-e+f=2$
(b)	$e=18$, $v=12$
(c)	8 hexagons, 6 squares ($f=14$ total)

Topic: Euler's polyhedron formula combined with face-edge double-counting to solve for unknown vertex/edge/face counts.

Key relations: $v-e+f=2$ (Euler); for a polyhedron with face-degree sequence, $\sum(\text{edges per face}) = 2e$ (each edge borders exactly 2 faces) — this is the "handshake lemma" applied to faces instead of vertices.

Why this works: Two independent linear relations — Euler's formula and the face-edge double-count — are exactly enough to pin down two unknowns ($h$ and $s$, or $e$ and $v$) once the rest of the data is given, turning a solid-geometry counting problem into simple simultaneous equations.

Common pitfall: Forgetting the factor of 2 in the face-edge incidence sum (each edge is shared, not owned by one face); mixing up which quantity Euler's formula is being solved for.

Source: Rosen, Discrete Mathematics and Its Applications, §10.7 (planar graphs, Euler's formula and its applications to polyhedra).

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