Sue Liu of Madras College, St Andrews sent an excellent solution to
this question.

Suppose that $a$ is the radius of the axle, $b$ is the radius of
each ball-bearing, and $c$ is the radius of the hub (see the
figure).

The solution is based on the following figure.

The angle subtended by each ball-bearing at the centre of the axle
is $2\pi/n$, so that in the triangle $OPQ$ we have $$OQ= a+b, \quad
PQ = b, \quad \angle POQ = \pi/n$$ and hence

- $$(a+b)\sin (\pi/n) = b.$$

We also know that

- $$c = OP' = OP +PP' = a+2b.$$

The rest of the solution is applying simple algebra to (1) and (2),
and to simplify this we will (temporarily) write $s$ for $\sin
(\pi/n)$. First, (1) gives $$ as = b(1-s).$$ Next, (2) gives $$c =
a+2b = a + {2as\over 1-s} = a\left({1+s\over 1-s}\right),$$ so
that, finally, we have $${a \over b} = \left({1 - \sin (\pi/n)\over
\sin (\pi /n)}\right), \quad {b \over c} = \left({\sin (\pi/n)\over
1+\sin (\pi /n)}\right), \quad {c\over a} = \left({1+\sin
(\pi/n)\over 1-\sin (\pi /n)}\right).$$ When there are 3
ballbearings, $n = 3$ and $\sin (\pi /3) = \sqrt 3 /2$. Hence$${a
\over b}= {2\sqrt 3 \over 3} - 1, \quad {b \over c} = \sqrt 3(2 -
\sqrt 3), \quad {c \over a} = 7 + 4\sqrt 3.$$ When there are 4
ballbearings, $n = 4$ and $\sin (\pi /4) = \sqrt 1 /\sqrt 2$. Hence
$${a \over b}= \sqrt 2 - 1, \quad {b \over c} = \sqrt 2 - 1, \quad
{c \over a} = 3 + 2\sqrt 2.$$ {\bf If $n=6$ then} $b = a$ and $c =
3a$. This is a very special case for suppose, in the general case,
that the internal radius $c$ of the hub is an integer multiple of
the radius $b$ of each ball-bearing. Then $c/b = N$, say, where $N$
is an integer, and this gives $$N\sin {\pi \over n} = 1+\sin
{\pi\over n},$$ or $$\sin {\pi\over n} = {1\over N-1}.$$ Now it is
known (although this is NOT an elementary result) that if $\sin x$
is rational, then $\sin x$ is $0$, $1/2$ or $1$. Thus we must have
$${1\over N-1} = {1\over 2},$$ so that $N=3$. This means that $\sin
(\pi/n) = 1/2$ so that $n=6$.

The same conclusion can be made if we know that in the general case
the internal radius of the hub is an integer multiple of the radius
of each ball-bearing.

Finally, as this same conclusion, namely $n=6$, can be drawn
whenever one of the ratios $a/b$, $b/c$, $c/a$ are rational numbers
and, as the case $n=6$ is not practical (many more ballbearings are
needed) it follows that in all cases of, for example, bicycle
wheels, these ratios are irrational. It therefore follows that it
is impossible to manufacture a perfectly fitting set of
ball-bearings.