Problem: △ A B C △ A B C cos ( 3 A ) + cos ( 3 B ) + cos ( 3 C ) = 1 cos ( 3 A ) + cos ( 3 B ) + cos ( 3 C ) = 1 10 1 0 13 1 3 m m △ A B C △ A B C m m m m
Solution:
The condition cos ( 3 A ) + cos ( 3 B ) + cos ( 3 C ) = 1 cos ( 3 A ) + cos ( 3 B ) + cos ( 3 C ) = 1
0 = 1 − cos 3 A − ( cos 3 B + cos 3 C ) = 2 2 ( 3 2 A ) − 2 cos ( 3 2 ( B + C ) ) cos ( 3 2 ( B − C ) ) = 2 2 ( 3 2 A ) + 2 sin ( 3 2 A ) cos ( 3 2 ( B − C ) ) = 2 sin ( 3 2 A ) ( sin ( 3 2 A ) + cos ( 3 2 ( B − C ) ) ) = 2 sin ( 3 2 A ) ( − cos ( 3 2 ( B + C ) ) + cos ( 3 2 ( B − C ) ) ) = 4 sin ( 3 2 A ) sin ( 3 2 B ) sin ( 3 2 C ) . 0 = 1 − cos 3 A − ( cos 3 B + cos 3 C ) = 2 sin 2 ( 2 3 A ) − 2 cos ( 2 3 ( B + C ) ) cos ( 2 3 ( B − C ) ) = 2 sin 2 ( 2 3 A ) + 2 sin ( 2 3 A ) cos ( 2 3 ( B − C ) ) = 2 sin ( 2 3 A ) ( sin ( 2 3 A ) + cos ( 2 3 ( B − C ) ) ) = 2 sin ( 2 3 A ) ( − cos ( 2 3 ( B + C ) ) + cos ( 2 3 ( B − C ) ) ) = 4 sin ( 2 3 A ) sin ( 2 3 B ) sin ( 2 3 C ) .
Therefore one of ∠ A , ∠ B ∠ A , ∠ B ∠ C ∠ C 12 0 ∘ 1 2 0 ∘ △ A B C △ A B C 12 0 ∘ 1 2 0 ∘ 10 1 0 13 1 3 1 0 2 + 1 3 2 + 10 ⋅ 13 = 399 1 0 2 + 1 3 2 + 1 0 ⋅ 1 3 = 3 9 9
The problems on this page are the property of the MAA's American Mathematics Competitions