Diving hard fractions?

Hard Math Proof Involving Farey Fractions?

• For each Farey fraction a/b let C(a/b) denote the circle in the plane of radius (2b^2)^-1 and center (a/b,(2b^2)^-2). These circles, called the Ford circles, lie in the half-plane 0≤y and are tangent to the x-axis at the point a/b. Show that the interior of a Ford circle contains no point of any other Ford circle and two Ford circles C(a/b), C(a'/b') are tangent if and only iff a/b and a'/b' are adjacent Farey fractions of same order

• Answer:

  C(a/b) and C(a'/b') have a point in their common interior iff distance between centers is less than sum of their radii 1/(2b^2) + 1/(2b'^2) you are given their centers, so calculate their differences, noting the fractions are reduced. tangent means the distance is exactly the value above and adjacent Farey fractions of same order means ab' - a'b = +/-1, which must lead quickly to the equality.