On Tim Gowers' webpage he has an example "proof" of the regular dodecahedron's existence which he claims contains a flaw.
He writes
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?
which suggests to me that he believes the proof contains a serious flaw that can't be fixed by simply adding more detail.
However I can't detect any such error. So how does the proof fail?
To prevent the possibility of link-rot, the entire argument is as follows.
Why isn't it obvious that a regular dodecahedron exists?
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?
Best Answer
To find the flaw, tile the plane with regular hexagons and apply the “proof” to the tiling. There's nothing specific to pentagons in it, so it should go through with hexagons, too – but of course it doesn't, as there's no Platonic solid with hexagonal facets.
Everything works out up to the point “if you reflect in $P$, then $G$ maps to $A$”; but the following statement “$H$ maps to a pentagon that shares an edge with $A$ and $C$” is false for hexagons, and there's no justification why it should be true for pentagons.