Number of edges of a convex polytope with n vertices

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In summary, the number of edges of a convex polytope with n vertices, where all faces are triangles, can be calculated using the formula 3/2 * n - 2. This is derived from the fact that three times the number of faces equals twice the number of edges. A good way to understand this is by using a simple example, such as a tetrahedron with 4 vertices, 6 edges, and 4 faces.
  • #1
squaremeplz
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Homework Statement



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


Homework Equations



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

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?
 
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  • #2
Right formula. Wrong logic. Each edge meets two faces. Each face meets three edges. What's the relation between edges and faces?
 
  • #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:
  • #4
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.
 
  • #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.
 

FAQ: Number of edges of a convex polytope with n vertices

How do you calculate the number of edges in a convex polytope with n vertices?

To calculate the number of edges in a convex polytope with n vertices, you can use the formula: E = (n * (n-1))/2. This formula is based on the fact that each vertex is connected to n-1 other vertices, and since each edge connects two vertices, we divide by 2 to avoid counting each edge twice.

Can you give an example of a convex polytope and its corresponding number of edges?

Yes, a regular tetrahedron is an example of a convex polytope with 4 vertices. Using the formula mentioned above, we can calculate the number of edges as: E = (4 * (4-1))/2 = 6. Therefore, a regular tetrahedron has 6 edges.

How does the number of edges change when the number of vertices in a convex polytope increases?

The number of edges in a convex polytope increases as the number of vertices increases. This is because each new vertex adds n-1 new edges. For example, if we have a convex polytope with 5 vertices, adding 1 more vertex will result in 5 new edges, adding 2 more vertices will result in 10 new edges, and so on.

Is the number of edges always an even number in a convex polytope?

Yes, the number of edges in a convex polytope is always an even number. This is because the formula for calculating the number of edges involves dividing by 2, and any whole number divided by 2 will result in an even number.

Are there any exceptions to the formula for calculating the number of edges in a convex polytope?

Yes, the formula for calculating the number of edges in a convex polytope only applies to regular convex polytopes. For irregular convex polytopes, the number of edges can vary and cannot be calculated using a simple formula. In such cases, the number of edges must be determined by counting them individually.

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