Question 10 of 12

Question 10

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. (a) A connected planar simple graph. (b) 4 hexagons + 4 triangles. (c) 20 equilateral-triangle faces, a regular (vertex-transitive) polyhedron. (d) $E=48$, $V=32$, faces are squares and hexagons only.

Find. (a) Euler's formula. (b) V and E of the truncated tetrahedron. (c) triangles per vertex. (d) number of square faces and number of hexagon faces.

Approach. Apply Euler's polyhedron formula $v-e+f=2$ together with the "face-edge incidence" double count $\sum(\text{sides per face})=2e$ (each edge borders exactly two faces), and for vertex-regular solids the analogous $\sum(\text{sides per face})=(\text{faces meeting per vertex})\times v$.

1. Part (a) — Euler's formula. For any connected planar simple graph drawn in the plane, counting vertices v, edges e, and regions f (faces, including the unbounded outer region): $$\boxed{v - e + f = 2}$$
2. Part (b) — truncated tetrahedron (V, E). Total faces $f=4+4=8$. Count edge-face incidences (each face contributes one incidence per side, and every edge is shared by exactly 2 faces): $$\text{incidences} = 4(\text{triangle sides})\times3 + 4(\text{hexagon sides})\times6 = 12+24=36 \quad\Rightarrow\quad e=\frac{36}{2}=18.$$ Substitute into Euler's formula from part (a): $v = 2-f+e = 2-8+18=\boxed{12}$, and $e=\boxed{18}$.
3. Part (c) — icosahedron, triangles per vertex. Let s be the (constant, by regularity) number of triangles meeting at each vertex, $f=20$. The face-edge double count gives $e=\dfrac{3f}{2}=\dfrac{60}{2}=30$. Each triangle also contributes 3 vertex-incidences, and every vertex sees exactly s of them, so $3f = s\,v \Rightarrow v=\dfrac{3f}{s}=\dfrac{60}{s}$. Substituting v and e into Euler's formula: $$\frac{60}{s} - 30 + 20 = 2 \quad\Rightarrow\quad \frac{60}{s}=12 \quad\Rightarrow\quad \boxed{s=5}$$ giving $v=60/5=12$ vertices and $e=30$ edges — the familiar icosahedron $(V,E,F)=(12,30,20)$.
4. Part (d) — truncated rhombic dodecahedron faces. From Euler's formula with the given $V=32$, $E=48$: $$f = 2-v+e = 2-32+48=18 \text{ total faces.}$$ Let s = number of square faces, h = number of hexagon faces, so $s+h=18$. The face-edge double count gives $$4s+6h = 2e = 96.$$ Substitute $s=18-h$: $4(18-h)+6h=96 \Rightarrow 72+2h=96 \Rightarrow h=12$, so $s=18-12=6$: $$\boxed{6 \text{ square faces},\quad 12 \text{ hexagonal faces}}$$