6. REVIEW
    

1. Sketch the vertices and edges of a transparent cube viewed from near the
centre of a face. Find V, E and F. Does Euler’s Formula apply to the vertices, edges and faces of a cube?

Yes. V – E + F = 2 (in this case V = 8, E = 12, F = 6), and similarly for all convex polyhedra. Note that the planar ‘exterior’ corresponds to the face being looked through. It also holds for most non-convex regular polyhedra, but curiously fails for the great dodecahedron {5, 5/2} and the small stellated dodecahedron {5/2, 5}.
2. In fact, for (convex) polyhedra, V – E + F = 2; for polygons V – E = 0. Can you find an analogue for the line segment? There is a trivial analogue. For a line segment, V = 2. Notice the pattern of the three formulae.
3. Continuing Q2, would there be an analogue for a convex 4-dimensional ‘polytope’? Yes. We need to let C denote the number of ‘cells’ (3-dimenional faces). In this case it can be shown that V – E + F + C = 0.
4. Investigate Pick’s Theorem for this lattice polygon. This polygon has an edge intersection which is not a lattice point, so we have no analogue of Pick’s Theorem in this case.
5. And for this one ... For this polygon we have V = 8, E = 12, F = 4, A = 4, B = 12, I = 0, P   =  0, and  =  – 4 . Pick’s (Extended) Formula is satisfied for these values.
6. What about this one?
Can you relate your answer to the Extended Pick’s Theorem?
The Extended Pick’s Theorem fails in this case because the polygon is not ‘all in one piece’. In fact, V = 8, E = 8, F = 1, A = 8, B = 16, I = 0, P  =  1, and     = 0. Pick’s Formula now gives 7 instead of 8. We can see what is happening here by joining an outside vertex to an inside vertex. This now makes E = 10 (the new edge is traversed twice), P = –1, and
= – 2. Pick’s Formula now gives the correct value 8 for A.