average number of edges on the faces of the polyhedron is strictly less that 6

Question

V,E,F stands for vertices, edges, and faces respectively.

We have 3V ≤ 2E.

We want to show: 2E/F < 6 – (the average number of edges on the faces of the polyhedron is strictly less) that 6. use given fact and Euler formula V − E + F = 2.

I tried to solve this question straight using algebraic manipulation but I end up with 3F - E > 6. What i did essentially was :

1. V = 2 - F + E (1)
2. 3(2 - F + E) < 2E (plug 1 into the given fact)
3. 6 - 3F + 3E < 2E
4. 6- 3F < -E
5. E < 3F - 6

Answer 1

Denote $A$ to be the average number of edges per face.
It is also equal to the average number of vertices per face, as each face has an equal number of edges and vertices.

Each vertex is shared by at least $3$ faces. From that we get:

$$3V \leq A*F$$ That is, if you sum up for each face the number of vertices, you'll get no less than $3$ times the total vertices of the polyhedron, as each distinct vertex is counted at least trice (one for each face it is a part of). We rewrite this as: $$\frac{A*F}{3} \geq V$$

We also know each edge is shared by exactly $2$ faces. From that we get:

$$2 E = A*F$$

If we sum up the total number of edges of each face, we get twice the edges of the polyhedron. Let's substitute these into the Euler formula:

$$ V - E + F = 2 $$ $$\frac{A*F}{3} - \frac{A*F}{2} + F \geq 2$$ $$(1-\frac{A}{6})F \geq 2 $$ Since $F$ is positive, we get: $$1 > \frac{A}{6}$$