Maths Olympiad Prep

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Geometry Difficulty 5.1 AIME, harder Prove it Russia

Let α,β,γ\alpha, \beta, \gamma be the angles in triangle Δ\Delta. Suppose that sin⁡α>cos⁡β\sin \alpha > \cos \beta, sin⁡β>cos⁡γ\sin \beta > \cos \gamma, and sin⁡γ>cos⁡α\sin \gamma > \cos \alpha. Prove that Δ\Delta is acute-angled.

Углы треугольника α,β,γ\alpha, \beta, \gamma удовлетворяют неравенствам sin⁡α>cos⁡β,sin⁡β>cos⁡γ,sin⁡γ>cos⁡α\sin \alpha > \cos \beta, \sin \beta > \cos \gamma, \sin \gamma > \cos \alpha. Докажите, что треугольник остроугольный.

Solution

Предположим противное; пусть для определённости γ≥90∘\gamma \ge 90^\circ. Тогда α+β≤90∘\alpha + \beta \le 90^\circ, и углы α\alpha и β\beta острые. Поэтому 0<β≤90∘−α<90∘0 < \beta \le 90^\circ - \alpha < 90^\circ, откуда cos⁡β≥cos⁡(90∘−α)=sin⁡α\cos \beta \ge \cos(90^\circ - \alpha) = \sin \alpha, что противоречит условию.

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Source: MathNet, licensed CC-BY-4.0. Statement and solution reproduced as published; topic and difficulty added by this site.