In a given right triangle ABC, the hypoteneuse BC, length a, is divided into n equal parts with n an odd integer. The central part subtends an angle α at A. h is the perpendicular distance from A to BC. Prove that:
tan α = 4nh/(an2 - a).
Solution
Let M be the midpoint of BC, and P and Q the two points a/2n either side of it, with P nearer B. Then α = ∠PAQ = ∠QAH - ∠PAH (taking angles as negative if P (or Q) lies to the left of H). So tan α = (QH - PH)/(AH2 + QH·PH) = AH·PQ/(AH2 + (MH - a/2n)(MH + a/2n)) = (ah/n)/(a2/4 - a2/(4n2)) = 4nh/(an2 - a).
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.
2nd IMO 1960
© John Scholes
jscholes@kalva.demon.co.uk
18 Sep 1998
Last corrected/updated 24 Sep 2003