1st Irish 1988 Problem 4 The triangle ABC (with sidelengths a, b, c as usual) satisfies log(a 2 ) = log(b 2 ) + log(c 2 ) - log(2bc cos A). What can we say about the triangle?
Isosceles (specifically, BC equals one of the other two sides). Solution We have log(2bc cos A) = log(b 2 ) + log(c 2 ) - log(a 2 ). Taking exponentials gives 2bc cos A = b 2 c 2 /a 2 . By the cosine formula, 2bc cos A = b 2 + c 2 - a 2 . Hence b 2 + c 2 - a 2 = b 2 c 2 /a 2 . Multiplying by a 2 and factorising gives (a 2 - b 2 ) (a 2 - c 2 ) = 0. Hence a = b or a = c. Conversely it is easy to see that if a = b or c, then b 2 + c 2 - a 2 = b 2 c 2 /a 2 and hence 2bc cos A = b 2 c 2 /a 2 . Taking logs gives the relation given. 1st Irish 1988 © John Scholes jscholes@kalva.demon.co.uk 11 June 2003 Last updated/corrected 11 Jun 03