In the following animation press '+' to zoom in to see an infinite
sequence of smaller and smaller circles, all sitting on the same
line, and just touching the parent circles immediately above them.
This problem is about finding a mathematical explanation for what
you see.
This text is usually replaced by the Flash movie.
Twelve such circles are shown in the diagram to the right. They
touch the $x$axis at the points marked in the diagram. We will
need the following two definitions in this problem:
1) The Farey
sequence $F_n$ is the list written in increasing order of all
the rational numbers between $0$ and $1$ that have only the numbers
$1, 2, 3, ..., n$ as denominators.
2) For the two rational numbers $\frac{b}{d}$ and
$\frac{a}{c}$ the mediant is $\frac{a+b}{c+d}$. It can be proved
that $\frac{b}{d}$ and $\frac{a}{c}$ are consecutive terms in a
Farey sequence if and only if $adbc=1$. (See the problem Farey Neighbours )
Notice that the $x$coordinates of the points where these
circles touch the axis form a Farey sequence.


Imagine drawing two circles of radius $\frac{1}{2}$ with centres
$(0,\frac{1}{2})$ and $(1, \frac{1}{2})$. We could describe these
as sitting on a horizontal axis (the $x$axis) and touching each
other. Next imagine drawing a circle which sits on the same
horizontal axis and touches both circles. Can you see why such a
circle is unique? Can you see how we could repeat this process
indefinitely?
The $x$coordinates of the points where the circles touch the axis
are always Farey sequences and, as you zoom in, these contact
points give an infinite sequence of Farey sequences. To prove this
result you first need to show that two circles which touch the
$x$axis at $(\frac{b}{d},0)$ and $(\frac{a}{c},0)$ and have radii
$\frac{1}{2d^2}$ and $\frac{1}{2c^2}$ respectively will touch each
other if and only if $\frac{b}{d}$ and $\frac{a}{c}$ are Farey
neighbours, that is if and only if $adbc=1$.
Finally you need to show that, given two such circles, the circle
which touches the x axis at the mediant point $\frac{a+b}{c+d}$,
and has radius $\frac{1}{2(c+d)^2}$, is tangent to both these
circles.