Discrete Mathematics- An Open Introduction - 3rd Edition, 2016a

264 4. Graph Theory
of edges is also kv/2. Putting this together gives
which says
e 3 f
2
k
k(2 + f /2)
,
2
6 f
4 + f .
Both k and f must be positive integers. Note that 6 f
4+ f
is an increasing
function for positive f , bounded above by a horizontal asymptote at k 6.
Thus the only possible values for k are 3, 4, and 5. Each of these are
possible. To get k 3, we need f 4 (this is the tetrahedron). For k 4
we take f 8 (the octahedron). For k 5 take f 20 (the icosahedron).
Thus there are exactly three regular polyhedra with triangles for faces.
Case 2: Each face is a square. Now we have e 4 f /2 2 f . Using
Euler’s formula we get v 2 + f , and counting edges using the degree k
of each vertex gives us
e 2 f k(2 + f ) .
2
Solving for k gives
k
4 f
2 + f
8 f
4 + 2 f .
This is again an increasing function, but this time the horizontal
asymptote is at k 4, so the only possible value that k could take is 3.
This produces 6 faces, and we have a cube. There is only one regular
polyhedron with square faces.
Case 3: Each face is a pentagon. We perform the same calculation as
above, this time getting e 5 f /2 so v 2 + 3 f /2. Then
so
e 5 f
2
k
k(2 + 3 f /2)
,
2
10 f
4 + 3 f .
Now the horizontal asymptote is at 10
3
. This is less than 4, so we can
only hope of making k 3. We can do so by using 12 pentagons, getting
the dodecahedron. This is the only regular polyhedron with pentagons as
faces.
Case 4: Each face is an n-gon with n ≥ 6. Following the same procedure
as above, we deduce that
k
2n f
4 + (n − 2) f ,