A. 468. We are given two triangles. Their side lengths are a,b,c and A,B,C respectively, the areas are t and T, respectively. Prove that -a2A2+a2B2+a2C2+b2A2-b2B2+b2C2+c2A2+c2B2-c2C216tT.
Deadline expired on 15 January 2009.
Solution. Let and the two angles opposite to sides c and C, respectively. From the cosine law we have a2+b2-c2=2abcos and A2+B2-C2=2ABsin ; the two areas are and .
Substituing these quantities,
=2a2B2+2A2b2-4abABcos cos -4abABsin sin =