Maths Olympiad Prep

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Geometry Difficulty 5.8 AIME, harder Prove it

11.5. The angles of a triangle α,β,γ\alpha, \beta, \gamma satisfy the inequalities sin⁡α>cos⁡β,sin⁡β>cos⁡γ,sin⁡γ>cos⁡α\sin \alpha > \cos \beta, \sin \beta > \cos \gamma, \sin \gamma > \cos \alpha. Prove that the triangle is acute-angled. (I. Bogdanov)

Solution

Solution. Suppose the opposite; let γ⩾90∘\gamma \geqslant 90^{\circ} for definiteness. Then α+β⩽90∘\alpha+\beta \leqslant 90^{\circ}, and the angles α\alpha and β\beta are acute. Therefore, 0<β⩽90∘−α<90∘0<\beta \leqslant 90^{\circ}-\alpha<90^{\circ}, from which cos⁡β⩾cos⁡(90∘−α)=sin⁡α\cos \beta \geqslant \cos \left(90^{\circ}-\alpha\right)=\sin \alpha, which contradicts the condition.

Comment. There may be solutions in which a participant uses the monotonicity of a trigonometric function on an interval where it is actually non-monotonic (for example, the functions sin⁡x\sin x on the interval (0∘,180∘)\left(0^{\circ}, 180^{\circ}\right) or cos⁡x\cos x on (−90∘,90∘))\left.\left(-90^{\circ}, 90^{\circ}\right)\right). Such solutions are scored 0 points.

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.