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'Ford Circles' printed from http://nrich.maths.org/

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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.

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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 $|ad-bc|=1$. (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 $|ad-bc|=1$.

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.