Projective Geometry: Combinatorially, but not projectively equivalent polytopes I have a hard time understanding Projective Geometry.
My task is to 

Find two polytopes, that are combinatorially, but not projectively
  equivalent.

What combinatorially equivalent means is clear to me, however I do not really understand the term "projectively equivalent". Can someone explain it to me?
How do I solve this task? I'd appreciate if someone could give me a hint towards the solution.
Thanks a lot!
 A: Projective equivalence means that there exists a projective transformation transforming one into the other.
A projective transformation in $d$-dimensional projective space is uniquely defined by $d+2$ points and their images, both in general position. So you can choose any polytope with more than $d+2$ vertices, and if you move one vertex a bit then they have to be projectively non-equivalent since the unmodified vertices would imply an identity transformation but the polytope is not identical.
A: Consider a pyramid over an arbitrary rectangular base and cut it with a plane parallel to the base to obtain a frustum. The polytope thus obtained is equivalent to a cube. However, it will not be projectively equivalent to the cube in general. Indeed, the diagonals joining opposite vertices from top and bottom are not concurrent in general ( the opposite top and bottom sides will not be in the same plane, if the bottom quadrilateral has opposite sides non-parallel). 
Note: a quadruple of points in space lying in the same plane is not projectively equivalent to a quadruple of points not lying in the same plane.
