IMO 1968 Problem A1 Find all triangles whose side lengths are consecutive integers, and one of whose angles is twice another.
Let the sides be a, a+1, a+2, the angle oppose a be A, the angle opposite a+1 be B, and the angle opposite a+2 be C. Using the cosine rule, we find cos A = (a+5)/(2a+4), cos B = (a+1)/2a, cos C = (a-3)/2a. Finally, using cos 2x = 2 cos 2 x - 1, we find solutions a = 4 for C = 2A, a = 1 for B = 2A, and no solutions for C = 2B. a = 1 is a degenerate solution (the triangle has the three vertices collinear). The other solution is 4, 5, 6. s 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. 10th IMO 1968 © John Scholes jscholes@kalva.demon.co.uk 27 Sep 1998 Last corrected/updated 27 Sep 1998