Problem C. 859. (May 2006)

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 pont)

Deadline expired on June 15, 2006.

Sorry, the solution is available only in Hungarian. Google translation

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:

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 solutions.

Problems in Mathematics of KöMaL, May 2006

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 solutions.

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