Euler Characteristics Solutions

Solution to Question 1 : Yes, it is the dodecahedron, having 12 pentagons and no other faces. But it is not really round enough.

Solution to Question 2 : No, the calculation showed that if you are only using pentagons and hexagons, you need to have 12 pentagons.So many hexagons and 0 pentagons doesn't work.

Solution to Question 3 : F = 528+18+1 = 547
E = 3*528 + 4*18 + 24)/2 = 840
V = 2 - F + E = 2 - 547 + 840 = 295.

Solution to Question 4 and 5 : Drawing polyhedra on the `flat' manifolds we get

torus	'pretzel'	Klein bottle	projective plane
F = 2	F = 4	F = 2	F = 3
E = 6	E = 8	E = 6	E = 6
V = 4	V = 2	V = 4	V = 4
χ = 0	χ = -2	χ = 0	χ = 1

About the calculation for the pretzel: Note that the eight vertices on the edge of the octagon are all the same vertex, and the boundary of the octagon does not contain any edge of the drawn in polyhedron.

Solution to Question 6 : It depends how you cut through the Klein bottle. If you cut along edge A, then you get a cylinder; if you cut along edge B then it will be a Möbius band. For the projective plain it doesn't matter: cutting along edge A or edge B both gives you a Möbius band.

Solution to the problem : There are six houses and utilities (V = 6) and nine leads (E = 9). The number of `faces' is slightly more tricky. Each face is bounded by leads, and since house goes to utility and vice versa, there cannot be any triangular face. The least number of edges a single face has is four. Every edge is (part of) the common boundary of two faces, so there at at most 2*E/4 = 4.5 faces. This is not an integer, but four faces (say three quadrilaterals and one hexagon) is possible. But then we get χ = 4-9+6 = 1 < 2. So it is impossible to connect the houses and utilities on Earth, but I invite you to do it on the projective plane.