Two circles C and C' have centers O and O' and meet at M and N. The common tangent closer to M touches C at A and C' at B. The line through B perpendicular to AM meets the line OO' at D. BO'B' is a diameter of C'. Show that M, D and B' are collinear.
Solution
A neat coordinate solution by Massaki Yamamoto (a competitor) is as follows.
Take AB as the x-axis and the perpendicular line through M as the y-axis. Choose the unit of length so that M has coordinates (0, 1). Let A be (-m, 0) and B be (n, 0). Then considering the right-angled triangle O'MK, where K is (n, 1) we find that O' is (n, (n2+1)/2 ). Similarly, O is (-m, (m2+1)/2 ) ).
The gradient of the lie AM is 1/m, so the gradient of the line BD is -m and hence its equation is mx + y = mn. The gradient of the line OO' is (n-m)/2, so its equation is 2y - x(n-m) = mn+1. These intersect at ( (mn-1)/(m+n), (mn2+m)/(m+n) ). B' is (n, n2+1). It is now easy to check that the lines MB' and MD both have gradient n, so M, D, B' are collinear.
This looks easy, but choosing the right coordinate system is critical.
© John Scholes
jscholes@kalva.demon.co.uk
20 Oct 2000