Arie maxima
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Arie maxima
Fie P un punct in interiorul triunghiului \( ABC \) astfel incat \( PA=\sqrt{2} \), \( PB=2 \) si \( PC=\sqrt{3}-1 \). Gasiti aria maxima a triunghiului \( ABC \).
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Marius Mainea
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Maximul se atinge cand triunghiul ABC o latura de lungime \( \sqrt{2}+\sqrt{3}-1 \), inaltimea corespunzatoare de lungime 2 iar punctul P este piciorul acestei inaltimi.
\( S_{ABC}=S_{APB}+S_{BPC}+S_{CPA}=\frac{\sqrt{2}\cdot 2\cdot \sin\alpha}{2}+\frac{(\sqrt{3}-1)\cdot 2\cdot \sin\beta}{2}-\frac{(\sqrt{3}-1)\cdot \sqrt{2}\cdot \sin(\alpha+\beta)}{2}\le\frac{\sqrt{2}\cdot 2}{2}+\frac{(\sqrt{3}-1)\cdot 2}{2} \)
cu egalitate cand \( \alpha=\beta=\frac{\pi}{2} \)
\( S_{ABC}=S_{APB}+S_{BPC}+S_{CPA}=\frac{\sqrt{2}\cdot 2\cdot \sin\alpha}{2}+\frac{(\sqrt{3}-1)\cdot 2\cdot \sin\beta}{2}-\frac{(\sqrt{3}-1)\cdot \sqrt{2}\cdot \sin(\alpha+\beta)}{2}\le\frac{\sqrt{2}\cdot 2}{2}+\frac{(\sqrt{3}-1)\cdot 2}{2} \)
cu egalitate cand \( \alpha=\beta=\frac{\pi}{2} \)
Last edited by Marius Mainea on Wed Jan 27, 2010 10:35 pm, edited 1 time in total.