Show that we can find a set of disjoint circles such that given any rational point on the x-axis, there is a circle touching the x-axis at that point. Show that we cannot find such a set for the irrational points.
Solution
Two disjoint circles touching the x-axis at A and B and each with radius r cannot have AB < 2r. Now suppose a circle radius R ≥ r touches at A and a disjoint circle radius r touches at B. The circle radius r touching at A does not extend outside the circle radius R, so the circle at B must also be disjoint from it. Hence we still have AB ≥ 2r. Hence there can only be countably many disjoint circles radius r or more touching the x-axis. But a countable set of countable sets is still countable, so there can only be countably many disjoint circles touching the x-axis (any such circle has radius > 1/n for some n). There are uncountably many irrational points, so we cannot have disjoint circles touching at all the irrational points.
For the rational points take the circle touching at m/n (in lowest terms) to have radius 1/(3n2). Now suppose the circles at m/n and a/b have centres P and Q. We have PQ2 = (m/n - a/b)2 + (1/(3n2) - 1/(3b2) )2. If they intersect, then PQ ≤ (1/(3n2) + 1/(3b2) ) and hence (m/n - a/b)2 ≤ (1/(3n2) + 1/(3b2) )2 - (1/(3n2) - 1/(3b2) )2 = 4/9 1/(n2b2), so (mb - an)2 < 4/9. But (mb - an) must be integral, so mb - an = 0 and hence m/n = a/b. So distinct circles do not intersect.
© John Scholes
jscholes@kalva.demon.co.uk
5 Mar 2002