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Macedonian Mathematical Competitions

Geometry Difficulty 5.5 AIME, harder Prove it North Macedonia

Prove that there is an angle xx such that
sinx=sin⁡β⋅sin⁡γ1−cos⁡α⋅cos⁡β⋅cos⁡γ, sin x = \frac{\sin \beta \cdot \sin \gamma}{1 - \cos \alpha \cdot \cos \beta \cdot \cos \gamma},
for every angle α\alpha, and β\beta and γ\gamma acute angles.

Solution

We will prove that
sin⁡β⋅sin⁡γ1−cos⁡α⋅cos⁡β⋅cos⁡γ \frac{\sin \beta \cdot \sin \gamma}{1 - \cos \alpha \cdot \cos \beta \cdot \cos \gamma}
belongs in [−1,1][-1, 1].

Since β\beta and γ\gamma are acute angles we have cos⁡β>0\cos \beta > 0, cos⁡γ>0\cos \gamma > 0, so cos⁡β⋅cos⁡γ>0\cos \beta \cdot \cos \gamma > 0. Because cos⁡α≤1\cos \alpha \le 1, we have that cos⁡α⋅cos⁡β⋅cos⁡γ≤cos⁡β⋅cos⁡γ\cos \alpha \cdot \cos \beta \cdot \cos \gamma \le \cos \beta \cdot \cos \gamma, i.e. −cos⁡α⋅cos⁡β⋅cos⁡γ≥−cos⁡β⋅cos⁡γ-\cos \alpha \cdot \cos \beta \cdot \cos \gamma \ge -\cos \beta \cdot \cos \gamma.

From the sum identities, we have
sin⁡β⋅sin⁡γ+cos⁡β⋅cos⁡γ=cos⁡(β−γ)≤1. \sin \beta \cdot \sin \gamma + \cos \beta \cdot \cos \gamma = \cos(\beta - \gamma) \le 1.
Then 0<sin⁡β⋅sin⁡γ≤1−cos⁡β⋅cos⁡γ≤1−cos⁡α⋅cos⁡β⋅cos⁡γ. \text{Then } 0 < \sin \beta \cdot \sin \gamma \le 1 - \cos \beta \cdot \cos \gamma \le 1 - \cos \alpha \cdot \cos \beta \cdot \cos \gamma.
So
0<sin⁡β⋅sin⁡γ1−cos⁡α⋅cos⁡β⋅cos⁡γ≤1. 0 < \frac{\sin \beta \cdot \sin \gamma}{1 - \cos \alpha \cdot \cos \beta \cdot \cos \gamma} \le 1.
There is an angle xx such that
sin⁡x=sin⁡β⋅sin⁡γ1−cos⁡α⋅cos⁡β⋅cos⁡γ. \sin x = \frac{\sin \beta \cdot \sin \gamma}{1 - \cos \alpha \cdot \cos \beta \cdot \cos \gamma}.

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