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# Euler's Characteristic, soccer balls, and golf balls

A typical soccer ball consists of 12 regular pentagons and 20 regular hexagons. There are also several golf balls on the market that have a mixture of pentagonal and hexagonal dimples. Both situations are examples of convex polyhedra. Loosely speaking, a polyhedron is a geometric solid in three dimensions with flat faces and straight edges. In this case, the faces are pentagons and hexagons. The adjective convex refers to the fact that a line segment joining any two points of the solid lies entirely inside or on the surface of the solid. For mathematicians, a natural question arises. Namely, what sorts of convex polyhedra can we build using only regular pentagons and regular hexagons? For example, is it possible to build a convex polyhedron using only regular pentagons? How about just hexagons? If we allow both, how many of each are possible? In this talk, we will explore these types questions by utilizing Euler's characteristic formula for polyhedra, which establishes a relationship between the number of vertices, edges, and faces of a polyhedron.

This talk was given at the Northern Arizona University Friday Afternoon Mathematics Undergraduate Seminar on Friday, November 2, 2012. ## Dana Ernst

November 02, 2012

## Transcript

1. Euler’s Characteristic, soccer
balls, & golf balls
Northern Arizona University
FAMUS, November 2, 2012

• Next week on Tuesday at 4-5pm in Room
164, I will be giving a department
which is a mathematical theorem that made
its debut in the TV show Futurama.

• Next week on Tuesday at 4-5pm in Room
164, I will be giving a department
which is a mathematical theorem that made
its debut in the TV show Futurama.
• I’m looking for a bright and motivated
sophomore or junior to work on a year-
long research project starting next fall.
Potential to get paid! Come talk to me if
you are interested.

5. 12 pentagons
20 hexagons

6. Lots of hexagons!

7. Lots of hexagons!
Anything else?

8. Lots of hexagons!
Anything else?
Pentagon!

9. Lots of hexagons!
Anything else?
It turns out that there are 12 pentagons.
Pentagon!

10. What conﬁgurations
are possible?

11. What conﬁgurations
are possible?
• Can I make a soccer or golf ball
(equivalently, tile a sphere) with only
hexagons?

12. What conﬁgurations
are possible?
• Can I make a soccer or golf ball
(equivalently, tile a sphere) with only
hexagons?

13. What conﬁgurations
are possible?
• Can I make a soccer or golf ball
(equivalently, tile a sphere) with only
hexagons?
• What combinations are possible if we allow
both hexagons and pentagons?

14. Platonic Solids

15. Platonic Solids
There are exactly 5 solids that one can form
using regular and congruent polygons (with
same number of faces meeting at each vertex).

16. Platonic Solids
There are exactly 5 solids that one can form
using regular and congruent polygons (with
same number of faces meeting at each vertex).

17. Platonic Solids
There are exactly 5 solids that one can form
using regular and congruent polygons (with
same number of faces meeting at each vertex).
It turns out that the dodecahedron is made up
of 12 pentagons!

18. Two down!

19. The classiﬁcation of the Platonic solids answers
two of our questions:
Two down!

20. The classiﬁcation of the Platonic solids answers
two of our questions:
• We can’t make a soccer or golf ball out of
just hexagons.
Two down!

21. The classiﬁcation of the Platonic solids answers
two of our questions:
• We can’t make a soccer or golf ball out of
just hexagons.
• We can make one out of only pentagons, but
must use exactly 12 of them.
Two down!

22. A side note
It turns out that the soccer ball is a truncated
icosahedron, which is an example of an
Archimedean solid.

23. Euler’s Characteristic

24. Euler’s Characteristic
• Consider any convex polyhedron
(equivalently, any planar discrete graph drawn
on a sphere).

25. Euler’s Characteristic
• Consider any convex polyhedron
(equivalently, any planar discrete graph drawn
on a sphere).
• Let V be the number of vertices.

26. Euler’s Characteristic
• Consider any convex polyhedron
(equivalently, any planar discrete graph drawn
on a sphere).
• Let V be the number of vertices.
• Let E be the number of edges.

27. Euler’s Characteristic
• Consider any convex polyhedron
(equivalently, any planar discrete graph drawn
on a sphere).
• Let V be the number of vertices.
• Let E be the number of edges.
• Let F be the number of faces.

28. Euler’s Characteristic
• Consider any convex polyhedron
(equivalently, any planar discrete graph drawn
on a sphere).
• Let V be the number of vertices.
• Let E be the number of edges.
• Let F be the number of faces.
• Euler discovered the following formula:

29. Euler’s Characteristic
• Consider any convex polyhedron
(equivalently, any planar discrete graph drawn
on a sphere).
• Let V be the number of vertices.
• Let E be the number of edges.
• Let F be the number of faces.
• Euler discovered the following formula:
V-E+F = 2

30. Euler’s Characteristic

31. Euler’s Characteristic
Let’s look at some data:

32. Euler’s Characteristic
Let’s look at some data:
Polyhedron V E F Characteristic
Tetrahedron 4 6 4 2
Cube 8 12 6 2
Octahedron 6 12 8 2
Dodecahedron 20 30 12 2
Icosahedron 12 30 20 2
Soccer Ball 12⋅5 = 60 30 + 12⋅5 = 90 20 + 12 = 32 2

33. Proof of Euler’s Formula

34. Proof of Euler’s Formula
Let’s sketch the proof of Euler’s characteristic
for polyhedra (Cauchy, 1811).

35. Proof of Euler’s Formula
Let’s sketch the proof of Euler’s characteristic
for polyhedra (Cauchy, 1811).
• Pick a random face of polyhedron and
remove it.

36. Proof of Euler’s Formula
Let’s sketch the proof of Euler’s characteristic
for polyhedra (Cauchy, 1811).
• Pick a random face of polyhedron and
remove it.
• By pulling the edges of the missing face away
from each other, deform all the rest into a
planar graph.

37. Proof of Euler’s Formula
Let’s sketch the proof of Euler’s characteristic
for polyhedra (Cauchy, 1811).
• Pick a random face of polyhedron and
remove it.
• By pulling the edges of the missing face away
from each other, deform all the rest into a
planar graph.
• We just removed one face, but number of
vertices and edges is the same.

38. Proof of Euler’s Formula
Let’s sketch the proof of Euler’s characteristic
for polyhedra (Cauchy, 1811).
• Pick a random face of polyhedron and
remove it.
• By pulling the edges of the missing face away
from each other, deform all the rest into a
planar graph.
• We just removed one face, but number of
vertices and edges is the same.
• It’s now enough to prove V-E+F = 1.

39. Proof (Continued)

40. • If there is a face with more than 3 sides, draw
a new edge across this face joining two non-
connected).
Proof (Continued)

41. • If there is a face with more than 3 sides, draw
a new edge across this face joining two non-
connected).
• This adds one face and one edge, but does
not change the quantity V-E+F.
Proof (Continued)

42. • If there is a face with more than 3 sides, draw
a new edge across this face joining two non-
connected).
• This adds one face and one edge, but does
not change the quantity V-E+F.
• Continue adding new edges as above until all
faces have 3 sides.
Proof (Continued)

43. • If there is a face with more than 3 sides, draw
a new edge across this face joining two non-
connected).
• This adds one face and one edge, but does
not change the quantity V-E+F.
• Continue adding new edges as above until all
faces have 3 sides.
• At the end of this process, V-E+F has not
changed.
Proof (Continued)

44. Proof (Continued)

45. • Apply repeatedly either of the following two
steps, making sure that exterior boundary is
always a simple cycle:
Proof (Continued)

46. • Apply repeatedly either of the following two
steps, making sure that exterior boundary is
always a simple cycle:
1. Remove a triangle with only one edge
Proof (Continued)

47. • Apply repeatedly either of the following two
steps, making sure that exterior boundary is
always a simple cycle:
1. Remove a triangle with only one edge
2. Remove a triangle with two edges on
exterior.
Proof (Continued)

48. • Apply repeatedly either of the following two
steps, making sure that exterior boundary is
always a simple cycle:
1. Remove a triangle with only one edge
2. Remove a triangle with two edges on
exterior.
• In ﬁrst case, we remove one edge & one face.
Proof (Continued)

49. • Apply repeatedly either of the following two
steps, making sure that exterior boundary is
always a simple cycle:
1. Remove a triangle with only one edge
2. Remove a triangle with two edges on
exterior.
• In ﬁrst case, we remove one edge & one face.
• In second case, we remove one vertex, two
edges, and one face.
Proof (Continued)

50. Proof (Continued)

51. • In both cases, we maintain V-E+F.
Proof (Continued)

52. • In both cases, we maintain V-E+F.
• Continue removing triangles following the
steps above until we are left with a single
triangle.
Proof (Continued)

53. • In both cases, we maintain V-E+F.
• Continue removing triangles following the
steps above until we are left with a single
triangle.
• We have reduced the problem to V = 3,
E = 3, F = 1, which yields V-E+F = 1, as
desired.
Proof (Continued)

pentagons & hexagons?

55. • Let p be the number of pentagons & let h be
the number of hexagons.
pentagons & hexagons?

56. • Let p be the number of pentagons & let h be
the number of hexagons.
• Claim: Each vertex has degree 3. Vague
justiﬁcation: There’s not enough room for
more than 3 pentagons and/or hexagons to
meet at a point, and 2 is not enough.
pentagons & hexagons?

57. • Let p be the number of pentagons & let h be
the number of hexagons.
• Claim: Each vertex has degree 3. Vague
justiﬁcation: There’s not enough room for
more than 3 pentagons and/or hexagons to
meet at a point, and 2 is not enough.
• Then V = (5p+6h)/3, E = (5p+6h)/2, and
F = p+h.
pentagons & hexagons?

58. Using both?

59. • Then we see that
2 = V-E+F = (5p+6h)/3 - (5p+6h)/2 + (p+h).
Using both?

60. • Then we see that
2 = V-E+F = (5p+6h)/3 - (5p+6h)/2 + (p+h).
• This implies
12 = 10p+12h-15p-18h+6p+6h = p.
Using both?

61. • Then we see that
2 = V-E+F = (5p+6h)/3 - (5p+6h)/2 + (p+h).
• This implies
12 = 10p+12h-15p-18h+6p+6h = p.
• That is, the number of pentagons is always 12.
Using both?

62. • Then we see that
2 = V-E+F = (5p+6h)/3 - (5p+6h)/2 + (p+h).
• This implies
12 = 10p+12h-15p-18h+6p+6h = p.
• That is, the number of pentagons is always 12.
• It turns out that the number of hexagons that
is possible is unbounded.
Using both?

63. • Then we see that
2 = V-E+F = (5p+6h)/3 - (5p+6h)/2 + (p+h).
• This implies
12 = 10p+12h-15p-18h+6p+6h = p.
• That is, the number of pentagons is always 12.
• It turns out that the number of hexagons that
is possible is unbounded.
• Question: What numbers of hexagons are
allowed? I’ll let you think about it.
Using both?