IMO 1998 SL 4

Origin: ARM

Problem

Let M and N be points inside triangle ABC such that \angleMAB = \angleNAC and \angleMBA = \angleNBC. Prove that AM \cdot AN AB \cdot AC + BM \cdot BN BA \cdot BC + CM \cdot CN CA \cdot CB = 1.

Solution

Let K be the point on the ray BN with \angleBCK = \angleBMA. Since \angleKBC = \angleABM, we get \triangleBCK ∼\triangleBMA. It follows that BC/BM = BK/BA, which implies that also \triangleBAK ∼\triangleBMC. The quadrilat- eral ANCK is cyclic, because \angleBKC = \angleBAM = \angleNAC. Then by Ptolemy’s theorem we obtain AC \cdot BK = AC \cdot BN + AN \cdot CK + CN \cdot AK. (1) On the other hand, from the similarities noted above we get

CK = BC \cdot AM BM , AK = AB \cdot CM BM and BK = AB \cdot BC BM . After substitution of these values, the equality (1) becomes AB \cdot BC \cdot AC BM = AC \cdot BN + BC \cdot AM \cdot AN BM

• AB \cdot CM \cdot CN BM , which is exactly the equality we must prove multiplied by AB\cdotBC\cdotCA BM .