What is wrong with the following argument for the existence of a regular dodecahedron, an argument which is supposed to describe what one actually does when making one out of cardboard? Draw a regular pentagon in the plane, and surround it by five further regular pentagons of the same size, each one sharing a different edge with the original pentagon. Now fold these upwards, all by the same angle, until they just touch each other, so that you have a sort of cup. The top of this cup consists of ten edges, two for each of the five further pentagons, that zigzag round roughly in a circle.
Let us label the original pentagon A and the five subsequent ones B,C,D,E and F, ordered cyclically.
The upper corners of the zigzag make an angle of exactly 108, the angle of a regular pentagon, since each one is formed by two edges of one of B,C,D,E or F. I claim that the same is true for the lower corners. This can be seen as follows. Consider the corner of the zigzag at the top of the edge e shared by B and C. If you reflect in the plane P that bisects the edge e, then the angle in question becomes one of the five angles of the base pentagon A. Therefore it is 108.
Hence, there is a regular pentagon G, of the same size as all the other ones, which shares an edge with B and an edge with C. This argument works for the four other lower corners of the zigzag, giving pentagons H, I, J and K (again, let us say, in cyclic order and going round the same way as B, C, D, E and F).
It is important to show that these pentagons fit together in the sense that G shares an edge with H, which shares an edge with I, and so on. This can again be done by reflecting in the plane P. It is not a bad idea to draw a picture at this point, but if you reflect in P, then G maps to A, and H maps to a pentagon that shares an edge with A and C, which means that it must be B or D. Since B maps to itself, H maps to D, which shares an edge with A. Thus, the reflected images of G and H share an edge, which implies that G and H share an edge (which can easily be checked by a more careful version of the above argument to be the right edge).
By symmetry, we conclude that the pentagons G, H, I, J and K all fit together as they should. Every edge is now shared by two pentagons except for one edge each of G, H, I, J and K. By symmetry once again (rotating through 108 about a line through the bottom pentagon and perpendicular to it) these lines form a regular pentagon.
To see that the symmetry group of the resulting shape is transitive in all the ways one wants, notice that, once we had chosen the bottom pentagon, there was no choice about how to choose all the rest, given the rule that two neighbours of a given pentagon sharing adjacent edges were required to share a further edge with each other. Hence, we could have started the process at any of the other pentagons and would have obtained the same shape. Therefore, any isometry of R^3 that maps one of the faces to another, with the outer and inner sides mapping to the outer and inner sides respectively, can be extended to a symmetry of the entire shape.
Of course, I have not written the above proof in a totally formal way. My question is, where would the difficulty arise if I tried to do so?
Click here to return to the main geometry page.