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.

Transcript

1. Euler’s Characteristic, soccer balls, & golf balls Northern Arizona University FAMUS, November 2, 2012
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3. Advertisements • Next week on Tuesday at 4-5pm in Room 164, I will be giving a department colloquium about the Futurama theorem, which is a mathematical theorem that made its debut in the TV show Futurama.
4. Advertisements • Next week on Tuesday at 4-5pm in Room 164, I will be giving a department colloquium about the Futurama theorem, 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.
6. 12 pentagons 20 hexagons
8. Lots of hexagons!
9. Lots of hexagons! Anything else?
10. Lots of hexagons! Anything else? Pentagon!
11. Lots of hexagons! Anything else? It turns out that there are 12 pentagons. Pentagon!
12. What configurations are possible?
13. What configurations are possible? • Can I make a soccer or golf ball (equivalently, tile a sphere) with only hexagons?
14. What configurations are possible? • Can I make a soccer or golf ball (equivalently, tile a sphere) with only hexagons? • What about only pentagons?
15. What configurations are possible? • Can I make a soccer or golf ball (equivalently, tile a sphere) with only hexagons? • What about only pentagons? • What combinations are possible if we allow both hexagons and pentagons?
16. Platonic Solids
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).
18. 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).
19. 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!
20. Two down!
21. The classification of the Platonic solids answers two of our questions: Two down!
22. 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!
23. 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!
24. A side note It turns out that the soccer ball is a truncated icosahedron, which is an example of an Archimedean solid.
25. Euler’s Characteristic
26. Euler’s Characteristic • Consider any convex polyhedron (equivalently, any planar discrete graph drawn on a sphere).
27. Euler’s Characteristic • Consider any convex polyhedron (equivalently, any planar discrete graph drawn on a sphere). • Let V be the number of vertices.
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.
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.
30. 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:
31. 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
32. Euler’s Characteristic
33. Euler’s Characteristic Let’s look at some data:
34. 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
35. Proof of Euler’s Formula
36. Proof of Euler’s Formula Let’s sketch the proof of Euler’s characteristic for polyhedra (Cauchy, 1811).
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.
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.
39. 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.
40. 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.
41. Proof (Continued)
42. • If there is a face with more than 3 sides, draw a new edge across this face joining two non- adjacent vertices (that aren’t already connected). Proof (Continued)
43. • If there is a face with more than 3 sides, draw a new edge across this face joining two non- adjacent vertices (that aren’t already connected). • This adds one face and one edge, but does not change the quantity V-E+F. Proof (Continued)
44. • If there is a face with more than 3 sides, draw a new edge across this face joining two non- adjacent vertices (that aren’t already 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)
45. • If there is a face with more than 3 sides, draw a new edge across this face joining two non- adjacent vertices (that aren’t already 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)
46. Proof (Continued)
47. • Apply repeatedly either of the following two steps, making sure that exterior boundary is always a simple cycle: 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 adjacent to exterior. 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 adjacent to exterior. 2. Remove a triangle with two edges on exterior. Proof (Continued)
50. • 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 adjacent to exterior. 2. Remove a triangle with two edges on exterior. • In first case, we remove one edge & one face. Proof (Continued)
51. • 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 adjacent to exterior. 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)
52. Proof (Continued)
53. • In both cases, we maintain V-E+F. Proof (Continued)
54. • 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)
55. • 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)
56. What about using both pentagons & hexagons?
57. • Let p be the number of pentagons & let h be the number of hexagons. What about using both pentagons & hexagons?
58. • 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. What about using both pentagons & hexagons?
59. • 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. What about using both pentagons & hexagons?
60. Using both?
61. • Then we see that 2 = V-E+F = (5p+6h)/3 - (5p+6h)/2 + (p+h). 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. 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. Using both?
64. • 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?
65. • 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?