Problem A1
Find all triangles whose side lengths are consecutive integers, and one of whose angles is twice another.
Solution
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 cos2x - 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.
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
27 Sep 1998
Last corrected/updated 27 Sep 1998