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This is a practice question from a practice test I am working on.

An icosahedron is a regular polyhedron with 20 faces, each of which is a triangle.
Determine the number of edges and the number of vertices in an icosahedron. 
(Hint: Remember that an icosahedron can be thought of as a planar graph with
20 triangular faces).

I have been looking at this problem and cannot determine how to approach it. Given that it can be thought of as a planar graph, I assume I must use something regarding Euler's formula:

|V|-|E|+|F|=2

And potentially Euler's formula regarding planar graphs:

|E|<=3|V|-6

But I am unsure how to connect the two. Given either edges or vertices I can solve for the other, but I don't know how to find these. Apparently the number of vertices is 12, but why?

I am also unsure if the number of faces should be 20 or 21. The hint given says 20, but that's 20 triangular faces, leaving me to assume that they did not count the face surrounding the graph, thus making a total of 21 faces.

Thanks for any help.

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So you have $20$ faces. Here If we were just counting the total edges given $20$ triangles we get $20 \times 3 =60$ edges, but each triangular face shares edges with adjacent triangles so we've actually double counted our set of actual vertices for Icosa. Thus, $E = 30$. Then by Euler's formula we have that $V-E+F=2 \Rightarrow V-30 +20 =2 \Rightarrow V =12$

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