Mathematical and Physical Journal
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Problem A. 468. (December 2008)

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-c2C2\ge16tT.

(5 pont)

Deadline expired on January 15, 2009.


Solution. Let \gamma and \Gamma the two angles opposite to sides c and C, respectively. From the cosine law we have a2+b2-c2=2abcos \gamma and A2+B2-C2=2ABsin \Gamma; the two areas are t=\frac12ab\sin\gamma and T=\frac12AB\sin\Gamma.

Substituing these quantities,

-a2A2+a2B2+a2C2+b2A2-b2B2+b2C2+c2A2+c2B2-c2C2-16tT=

=2a2B2+2A2b2-(a2+b2-c2)(A2+B2-C2)-16tT=

=2a2B2+2A2b2-4abABcos \gammacos \Gamma-4abABsin \gammasin \Gamma=

=2(aB-Ab)2+4abAB(1-cos (\gamma-\Gamma))\ge0.


Statistics:

12 students sent a solution.
5 points:Blázsik Zoltán, Bodor Bertalan, Éles András, Márkus Bence, Nagy 235 János, Nagy 648 Donát, Somogyi Ákos, Tomon István, Tossenberger Anna, Varga 171 László, Wolosz János.
4 points:Backhausz Tibor.

Problems in Mathematics of KöMaL, December 2008