C is a circle and L a line not meeting it. M and N are variable points on L such that the circle diameter MN touches C but does not contain it. Show that there is a fixed point P such that the ∠MPN is constant.
Solution
Solution by Demetres Christofides
Answer: let O be the center of C, let C have radius 1, let D be the foot of the perpendicular from O to L and let OD = a. Then P is the point on the segment OD a distance √(a2 - 1) from D.
Let MN = 2x. Let Q be the midpoint of MN. The circles touch, so OQ = x + 1. Suppose first that D is between M and N and that N is closer to D. We have QD2 = (x+1)2 - a2, QP2 = QD2 + DP2 = (x+1)2 - a2 + a2 - 1. Now tan MPD = (x + QD)/PD, tan NPD = (x - QD)/PD, so tan MPN = tan(MPD + NPD) = 2x PD/(PD2 + QD2 - x2) = √(a2 - 1), which is independent of x.
Similarly, if D is not between M and N, but is closer to M, then we have tan MPD = (QD - x)/PD, tan NPD = (QD + x)/PD, tan MPD = tan(NPD - MPD) and we get the same result.
© John Scholes
jscholes@kalva.demon.co.uk
11 Sep 2002