MATHS BITE: Ford Circles

A ford circle is a circle with centre (p/q,1/(2q^{2})), and radius 1/(2q^{2}), where p and q are coprime integers.

File:Ford circles colour.svg
Source: Wikipedia

Notice that each Ford Circle is tangent to to the horizontal axis and any two Ford circles are either tangent or disjoint. The latter statement can be proven by finding the squared distance d^2 between the centres of the circles with (p,q) and (p’,q’) as the pairs of coprime integers.

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Let s be the sum of the radii:

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Then

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However, we have that latex-image-4.png and so latex-image-5.png, thus the distance between circles is greater or equal to the sum of the radii of the circles. There is equality iff

latex-image-6.png

In this case, the circles are tangent to one another.

Total area of Ford Circles

(taken from Wikipedia)

As no two ford circles intersect, it follows immediately that the total area of the Ford circles:

\left\{C[p,q]:0\leq {\frac  {p}{q}}\leq 1\right\}

is less than 1.

From the definition, the area is

A=\sum _{{q\geq 1}}\sum _{{(p,q)=1 \atop 1\leq p<q}}\pi \left({\frac  {1}{2q^{2}}}\right)^{2}.

Simplifying this expression gives us

A={\frac  {\pi }{4}}\sum _{{q\geq 1}}{\frac  {1}{q^{4}}}\sum _{{(p,q)=1 \atop 1\leq p<q}}1={\frac  {\pi }{4}}\sum _{{q\geq 1}}{\frac  {\varphi (q)}{q^{4}}}={\frac  {\pi }{4}}{\frac  {\zeta (3)}{\zeta (4)}},

noting that the last equality is given by considering the Dirichlet generating function for Euler’s totient function φ(q).

Given that ζ(4) = π^4/90, we get

A={\frac  {45}{2}}{\frac  {\zeta (3)}{\pi ^{3}}}\approx 0.872284041.

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