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Ukrainian Mathematical Olympiad, Final Round

Algebra Difficulty 5.4 AIME, harder Prove it Ukraine

Let α,β,γ∈[0;π2]\alpha, \beta, \gamma \in [0; \frac{\pi}{2}] satisfy the conditions
sin⁡α+sin⁡β+sin⁡γ=1,sin⁡αcos⁡2α+sin⁡βcos⁡2β+sin⁡γcos⁡2γ=−1. \sin \alpha + \sin \beta + \sin \gamma = 1, \sin \alpha \cos 2\alpha + \sin \beta \cos 2\beta + \sin \gamma \cos 2\gamma = -1.
Find all the possible values of the expression sin⁡2α+sin⁡2β+sin⁡2γ\sin^2 \alpha + \sin^2 \beta + \sin^2 \gamma.

Solution

З умови задачі випливають рівності
sin⁡α−2sin⁡3α+sin⁡β−2sin⁡3β+sin⁡γ−2sin⁡3γ=−1,sin⁡3α+sin⁡3β+sin⁡3γ=1. \sin \alpha - 2\sin^3 \alpha + \sin \beta - 2\sin^3 \beta + \sin \gamma - 2\sin^3 \gamma = -1, \\ \sin^3 \alpha + \sin^3 \beta + \sin^3 \gamma = 1.
Оскільки для φ∈[0;π2]\varphi \in [0; \frac{\pi}{2}] справджуються нерівності sin⁡3φ≤sin⁡2φ≤sin⁡φ\sin^3 \varphi \le \sin^2 \varphi \le \sin \varphi, то
1=sin⁡3α+sin⁡3β+sin⁡3γ≤sin⁡2α+sin⁡2β+sin⁡2γ≤sin⁡α+sin⁡β+sin⁡γ=1. 1 = \sin^3 \alpha + \sin^3 \beta + \sin^3 \gamma \le \sin^2 \alpha + \sin^2 \beta + \sin^2 \gamma \le \sin \alpha + \sin \beta + \sin \gamma = 1.
Залишається зауважити, що, наприклад, при α=β=0\alpha = \beta = 0, γ=π2\gamma = \frac{\pi}{2} маємо рівність
sin⁡2α+sin⁡2β+sin⁡2γ=1. \sin^2 \alpha + \sin^2 \beta + \sin^2 \gamma = 1.

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