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Number of edges of a convex polytope with n vertices

  1. Dec 11, 2008 #1
    1. The problem statement, all variables and given/known data

    What is the number of edges of a convex polytope with n vertices all of whose faces are triangles.


    2. Relevant equations

    # of faces + #of vertecies = # of edges + 2

    3. The attempt at a solution

    My reasoning is as follows:

    n/3 + n = # of edges + 2

    4n/3 - 2 = #of edges

    This doesn't seem right though. Am I using the correct formula?
     
    Last edited: Dec 11, 2008
  2. jcsd
  3. Dec 11, 2008 #2

    Dick

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    Right formula. Wrong logic. Each edge meets two faces. Each face meets three edges. What's the relation between edges and faces?
     
  4. Dec 11, 2008 #3
    3/2 * faces = edges?

    Is the result

    3/2*n - 2 = # of edges by any chance?

    I just did

    3/2*1/3*n + n = e + 2
     
    Last edited: Dec 11, 2008
  5. Dec 11, 2008 #4

    Dick

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    Yes. Three times the number of faces equals twice the number of edges. If you have to do this again pick some example like tetrahedron E=6, V=4, F=4. Octohedron E=12, V=6 and F=8. It's a good sanity check.
     
  6. Dec 11, 2008 #5
    That's actually how I found it. I drew a little tetrahedron with 4 vertecies, 6 edges and 4 faces. I don't know why but i just couldn't wrap my mind around it without a drawing. Thanks again for your time.
     
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