Prove that there is a point D on the side AB of the triangle ABC, such that CD is the geometric mean of AD and DB if and only if sin A sin B ≤ sin2(C/2)
Solution
Extend CD to meet the circumcircle of ABC at E. Then CD·DE = AD·DB, so CD is the geometric mean of AD and DB iff CD = DE. So we can find such a point iff the distance of C from AB is less than the distance of AB from the furthest point of the arc AB on the opposite side of AB to C. The furthest point F is evidently the midpoint of the arc AB. F lies on the angle bisector of C. So ∠FAB = ∠FAC = ∠C/2. Hence distance of F from AB is c/2 tan C/2 (as usual we set c = AB, b = CA, a = BC). The distance of C from AB is a sin B. So a necessary and sufficient condition is c/2 tan C/2 ≥ a sin B. But by the sine rule, a = c sin A/sin C, so the condition becomes (sin C/2 sin C)/(2 cos C/2) ≥ sin A sin B. But sin C = 2 sin C/2 cos C/2, so we obtain the condition quoted in the question.
Solutions are also available in: Samuel L Greitzer, International Mathematical Olympiads 1959-1977, MAA 1978, and in István Reiman, International Mathematical Olympiad 1959-1999, ISBN 189-8855-48-X.
© John Scholes
jscholes@kalva.demon.co.uk
10 Oct 1998