26th Putnam 1965

Problem B6

Four distinct points A1, A2, B1, B2 have the property that any circle through A1 and A2 has at least one point in common with any circle through B1 and B2. Show that the four points are collinear or lie on a circle.

Solution

The trick is to take concentric circles. If A1A2 is not parallel to B1B2 then their perpendicular bisectors must intersect at some point O. Take circles centre O through A1, A2 and through B1, B2. These must either coincide, in which case the 4 points lie on a circle, or have no points in common.

In the case where A1A2 is parallel to B1B2. Assume they do not coincide (otherwise the 4 points would be collinear). Then take a point O on the perpendicular bisector of A1A2 on the opposite side of A1A2 to B1B2 and sufficiently distant from A1A2 that the circle through A1, A2 centre O only extends less than halfway towards the line B1B2. Similarly, take a circle through B1, B2 which extends less than halfway towards A1A2. Then these circles will not meet.

© John Scholes
jscholes@kalva.demon.co.uk
25 Jan 2002