Consider a simple convex polyhedron, determine the vertices

Question

Consider a simple convex polyhedron Δ in ℝ with 2013 faces.

How many vertices are there in Δ ? How many of the vertices have index one with respect to linear function?(not equal to a constant on each edge of Δ )

Answer 1

The most common definition of a simple polyhedron is one all of whose vertices are at the intersection of exactly three facets.

The questioner is suggesting that you have enough information to determine the number of vertices, which isn't obvious at first glance.

You can approach this by setting up equations that include the number of vertices V, and then solving for V. One equation you know going in is Euler's equation:

F + V - E = $2$;

That is, the number of facets (i.e., sides) plus the number of vertices, minus the number of edges, equals two.

You know the number of facets, so you have two unknowns, V and E. You need one more linear equation, and you'll have two equations with two unknowns that you can solve.

Say you were to count all the edges to get E. In doing so, also count the two endpoints of each edge. When you're done counting the edges, you also have a list of twice as many vertices. In the process of counting edges and vertices, you've counted each vertex exactly three times, once for each edge ending at that vertex. Combining this information, you conclude that V = $2/3$ E.

You've got your second equation, so you can solve for V. Also, you know that you were given enough information to determine V, even though there are a lot of different polyhedra that are described by the question.

As an alternative to doing the algebra (since you know V is determined), you could think of an specific example of a simple polyhedron with 2,013 sides, such as a prism, and just count the number of vertices.