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
   

   View Slide

2. Advertisements
   

   View Slide

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.
   

   View Slide

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.
   

   View Slide

5. View Slide

6. 12 pentagons
   20 hexagons
   

   View Slide

7. View Slide

8. Lots of hexagons!
   

   View Slide

9. Lots of hexagons!
   Anything else?
   

   View Slide

10. Lots of hexagons!
    Anything else?
    Pentagon!
    

    View Slide

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

    View Slide

12. What configurations
    are possible?
    

    View Slide

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

    View Slide

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

    View Slide

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?
    

    View Slide

16. Platonic Solids
    

    View Slide

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).
    

    View Slide

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).
    

    View Slide

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!
    

    View Slide

20. Two down!
    

    View Slide

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

    View Slide

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!
    

    View Slide

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!
    

    View Slide

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

    View Slide

25. Euler’s Characteristic
    

    View Slide

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

    View Slide

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

    View Slide

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.
    

    View Slide

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.
    

    View Slide

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:
    

    View Slide

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
    

    View Slide

32. Euler’s Characteristic
    

    View Slide

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

    View Slide

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
    

    View Slide

35. Proof of Euler’s Formula
    

    View Slide

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

    View Slide

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.
    

    View Slide

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.
    

    View Slide

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.
    

    View Slide

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.
    

    View Slide

41. Proof (Continued)
    

    View Slide

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)
    

    View Slide

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)
    

    View Slide

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)
    

    View Slide

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)
    

    View Slide

46. Proof (Continued)
    

    View Slide

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

    View Slide

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)
    

    View Slide

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)
    

    View Slide

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)
    

    View Slide

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)
    

    View Slide

52. Proof (Continued)
    

    View Slide

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

    View Slide

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)
    

    View Slide

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)
    

    View Slide

56. What about using both
    pentagons & hexagons?
    

    View Slide

57. • Let p be the number of pentagons & let h be
    the number of hexagons.
    What about using both
    pentagons & hexagons?
    

    View Slide

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?
    

    View Slide

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?
    

    View Slide

60. Using both?
    

    View Slide

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

    View Slide

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?
    

    View Slide

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?
    

    View Slide

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?
    

    View Slide

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?
    

    View Slide