### Baby Circle

A small circle fits between two touching circles so that all three circles touch each other and have a common tangent? What is the exact radius of the smallest circle?

### Pericut

Two semicircle sit on the diameter of a semicircle centre O of twice their radius. Lines through O divide the perimeter into two parts. What can you say about the lengths of these two parts?

### Kissing

Two perpendicular lines are tangential to two identical circles that touch. What is the largest circle that can be placed in between the two lines and the two circles and how would you construct it?

# Ball Bearings

##### Stage: 5 Challenge Level:

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
1. $$(a+b)\sin (\pi/n) = b.$$
We also know that
1. $$c = OP' = OP +PP' = a+2b.$$
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.