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C. 859. Prove that the equality


c^2+2ab\sin\, (\gamma+30^{\circ}) =b^2+2ac\sin\, (\beta+30^{\circ}) =a^2+2bc\sin\,
(\alpha+30^{\circ}).

holds in every triangle.

Suggested by G. Holló, Budapest

(5 points)

Deadline expired on 15 June 2006.


Google Translation (Sorry, the solution is published in Hungarian only.)

Megoldás.

c^2+2ab\sin\left(\gamma+30^{\circ}\right)=
c^2+2ab\left(\sin\gamma\cdot{\sqrt3\over2}+\cos\gamma\cdot{1\over2}\right)=
c^2+ab\sin\gamma\cdot\sqrt3+ab\cos\gamma=

=c^2+2t\cdot\sqrt3+ab\cdot{a^2+b^2-c^2\over2ab}=2\sqrt3\cdot t+{a^2+b^2+c^2\over2}.

Ez konstans. Vagyis az egyenlőség valóban teljesül, mert mindhárom ezzel a konstanssal egyenlő.


Statistics on problem C. 859.
100 students sent a solution.
5 points:88 students.
4 points:7 students.
3 points:1 student.
2 points:2 students.
1 point:1 student.
Unfair, not evaluated:1 solution.


  • Problems in Mathematics of KöMaL, May 2006

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