Exercise 6.1.9 (Ford circles)

Proof. (of Lemma.)

• Suppose that $0<a∕b<c∕d$ are adjacent Farey fractions of order $n$. By section 6.1, $\mathit{bc}-\mathit{ad}=1$. Similarly, if $0<c∕d<a∕b$, then $\mathit{ad}-\mathit{bc}=1$. In both cases $|\mathit{bc}-\mathit{ad}|=1$
• Conversely, assume that $|\mathit{bc}-\mathit{ad}|=1$. Put $n=\max <!--\; FUNCTION\; APPLICATION\; -->(b,d)$. By Problem 5, $a∕b$ and $c∕d$ are adjacent Farey fractions of order $n$.

Proof. (of Problem 6.1.9)

See figures on

 https://en.wikipedia.org/wiki/Ford_circle

(a)

Consider two fractions $a∕b\ne c∕d$ and the two associate Farey circle $𝒞(a∕b)$ and $𝒞(c∕d)$, with respective radius $R={(2b^2)}^{-1}$ and $r={(2d^2)}^{-1}$ and centers $A=((a∕b,{(2b^2)}^{-1}))$, $B=(c∕d,{(2d^2)}^{-1})$.

We start from the formula

$$\begin{array}{lll}\hfill {(R+r)}^2-{(R-r)}^2=4\mathit{Rr}=\frac{1}{b^2{d}^2}.& & \hfill \text{(1)}\end{array}$$

By the Pythagorean Theorem, the distance $\mathit{BC}$ between the two centers satisfy

$$\begin{array}{lllll}\hfill B{C}^2& ={\left(\frac{c}{d}-\frac{a}{b}\right)}^2+{(R-r)}^2& \hfill & & \hfill \text{(2)}\\ \hfill & =\frac{{(\mathit{bc}-\mathit{ad})}^2}{b^2{d}^2}+{(R-r)}^2& \hfill & & \hfill \text{(3)}\\ \hfill & \ge \frac{1}{b^2{d}^2}+{(R-r)}^2& \hfill (\text{since }|\mathit{bc}-\mathit{ad}|\ge 1)& & \hfill \text{(4)}\\ \hfill & ={(R+r)}^2& \hfill (\text{by (1)}).& & \hfill \text{(5)}\end{array}$$

Therefore $\mathit{BC}\ge R+r$. This shows that the interior of $𝒞(a∕b)$ contains no point $𝒞(c∕d)$.

The interior of a Ford circle contains no point of other Ford circle.

(b)

• Suppose that $a∕b$ and $c∕d$ are adjacent Farey fractions of some order. By the Lemma, $|\mathit{bc}-\mathit{ad}|=1$. Then the inequality (4) becomes an equality, thus $\mathit{BC}=R+r$. This shows that the circles $𝒞(a∕b)$ and $𝒞(c∕d)$ are tangent.

• Conversely, suppose that $𝒞(a∕b)$ and $𝒞(c∕d)$ are tangent. Then $\mathit{BC}=R+r$. Using part (a),

  $$\begin{array}{llll}\hfill {(R+r)}^2& =B{C}^2& \hfill & \\ \hfill & =\frac{{(\mathit{bc}-\mathit{ad})}^2}{b^2{d}^2}+{(R-r)}^2& \hfill & \\ \hfill & \ge \frac{1}{b^2{d}^2}+{(R-r)}^2& \hfill & \\ \hfill & ={(R+r)}^2.& \hfill & \end{array}$$

  Therefore

  $$\frac{{(\mathit{bc}-\mathit{ad})}^2}{b^2{d}^2}+{(R-r)}^2=\frac{1}{b^2{d}^2}+{(R-r)}^2,$$

  thus ${(\mathit{bc}-\mathit{ad})}^2=1$, so $|\mathit{bc}-\mathit{ad}|=1$. The Lemma shows that $a∕b$ and $c∕d$ are adjacent Farey fractions of some order.

In conclusion, if $a∕b>0,c∕d>0$, the two Ford circles $𝒞(a∕b),𝒞(c∕d)$ are tangent if and only if $a∕b$ and $c∕d$ are adjacent Farey fractions of some order.