20th BMO 1984 Problem 1 In the triangle ABC, ∠C = 90 o . Find all points D such that AD·BC = AC·BD = AB·CD/√2.
Take rectangular coordinates with C as (0,0), A as (a,0), B as (0,b). Take D as (x,y). Then ((x-a) 2 +y 2 )b 2 = a 2 (x 2 +(y-b) 2 ) = (a 2 +b 2 )(x 2 +y 2 )/2. Hence (a 2 -b 2 )(x 2 +y 2 ) + 2ab(bx-ay) = 0 and (a 2 -b 2 )(x 2 +y 2 ) - 4a 2 by + 2a 2 b 2 = 0. Hence bx + ay = ab. So D lies on the line AB. Substituting back, we find (x,y) = (ab/(a+b),ab/(a+b)) or (ab/(a-b),ab/(a-b)). Geometrically, the two points are the intersections of the internal and external angle bisectors of angle C with the line AB. 20th BMO 1984 © John Scholes jscholes@kalva.demon.co.uk 10 February 2004 Last corrected/updated 10 Feb 04