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Geometry Difficulty 5.8 AIME, harder Prove it

5. (France) Let PP be a point inside △ABC\triangle A B C. Prove that ∠PAB,∠PBC,∠PCA\angle P A B, \angle P B C, \angle P C A is at least one less than or equal to 30∘30^{\circ}.

Solution

Let α=∠PAB,β=∠PBC\alpha=\angle P A B, \beta=\angle P B C, γ=∠PCA\gamma=\angle P C A. Then,
the distance from PP to ABA B
=PAsin⁡α=PBsin⁡(B−β), \begin{array}{c} =P A \sin \alpha \\ =P B \sin (B-\beta), \end{array}
the distance from PP to BCB C
=PBsin⁡β=PCsin⁡(C−γ),=PAsin⁡(A−α). \begin{array}{l} =P B \sin \beta \\ =P C \sin (C-\gamma), \\ =P A \sin (A-\alpha). \\ \end{array}
the distance from PP to CAC A =PCsin⁡γ=P C \sin \gamma

Thus,
sin⁡α⋅sin⁡β⋅sin⁡γ=sin⁡(A−α)⋅sin⁡(B−β)⋅sin⁡(C−γ). \begin{array}{l} \sin \alpha \cdot \sin \beta \cdot \sin \gamma \\ =\sin (A-\alpha) \cdot \sin (B-\beta) \\ \cdot \sin (C-\gamma) . \end{array}

By the convexity of ln⁡sin⁡x\ln \sin x in (0,π)(0, \pi), we have
sin⁡2α⋅sin⁡2β⋅sin⁡2γ=sin⁡α⋅sin⁡(A−α)⋅sin⁡β⋅sin⁡(B−β)⋅sin⁡γ⋅sin⁡(C−γ)⩽sin⁡0α+Aα+β+B−β+γ+C⋅γ=164. \begin{array}{l} \sin ^{2} \alpha \cdot \sin ^{2} \beta \cdot \sin ^{2} \gamma \\ =\sin \alpha \cdot \sin (A-\alpha) \cdot \sin \beta \cdot \sin (B-\beta) \\ \quad \cdot \sin \gamma \cdot \sin (C-\gamma) \\ \leqslant \sin ^{0} \alpha+A \quad \alpha+\beta+B-\beta+\gamma+C \cdot \gamma \\ =\frac{1}{64} . \end{array}

Therefore, sin⁡α⋅sin⁡β⋅sin⁡γ⩽18\sin \alpha \cdot \sin \beta \cdot \sin \gamma \leqslant \frac{1}{8}.
This implies that among α,β,γ\alpha, \beta, \gamma, there exists one, say α\alpha, such that
sin⁡α⩽12. \sin \alpha \leqslant \frac{1}{2}.

Thus, either α⩽30∘\alpha \leqslant 30^{\circ} or α⩾150∘\alpha \geqslant 150^{\circ}. In the latter case, β,γ\beta, \gamma must both be less than 30∘30^{\circ}.

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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.