已知三棱锥 $S-ABC$ 中侧棱 $SA$、$SB$、$SC$ 互相垂直，$M$ 是底面三角形 $ABC$ 内一动点，直线 $MS$ 与 $SA$、$SB$、$SC$ 所成的角分别是 $\alpha$、$\beta$、$\gamma$．
【难度】
【出处】
2018年全国高中数学联赛河北省预赛（高三）
【标注】
-
证明：$\alpha$、$\beta$、$\gamma$ 不可能是锐角三角形的三个内角标注答案略解析以线段 $MS$ 为体对角线构造长方体，则 $\alpha$、$\beta$、$\gamma$ 恰好为长方体的体对角线与从一个顶点出发的三条棱所成的角，因此 $\cos^{2}\alpha+\cos^{2}\beta+\cos^{2}\gamma=1$．因为 ${{\cos}^{2}}\alpha +{{\cos }^{2}}\beta \text{=}1-{{\cos }^{2}}\gamma $，所以 $\dfrac{1}{2}\left(\cos 2\alpha +\cos 2\beta \right)\text{=}-{{\cos }^{2}}\gamma $，所以 $\cos\left( \alpha +\beta \right)\times \cos\left( \alpha -\beta \right)\text{=}-{{\cos }^{2}}\gamma <0$，故 $cos(\alpha+\beta)<0$．所以 $\dfrac{\pi}{2}<\alpha+\beta<\pi$．下面证明 $\alpha+\beta+\gamma<\pi$．要证 $\alpha+\beta+\gamma<\pi$，只需证 $\alpha+\beta<\pi-\gamma$，只需证 $\cos(\pi-y)<\cos(\alpha+\beta)<0$，只需证 $\cos^{2}(\pi-y)>\cos^{2}(\alpha+\beta)$．因为 $\begin{align}
& {{\cos }^{2}}\left( \alpha +\beta \right)-{{\cos }^{2}}\gamma \text{=}{{\cos }^{2}}\left( \alpha+\beta \right)+\frac{1}{2}\left( \cos2\alpha +\cos 2\beta \right) \\
& \text{=}{{\cos }^{2}}\left( \alpha+\beta \right)+\cos \left( \alpha+\beta \right)\cdot \cos \left( \alpha-\beta \right) \\
& =2\cos \left( \alpha +\beta \right)\cdot \cos \alpha \cdot \cos\beta <0\\
\end{align}$，所以 $\alpha+\beta+\gamma<\pi$，故 $\alpha$、$\beta$、$\gamma$ 不可能是锐角三角形的三个内角． -
设 $S=\dfrac{1}{\cos^{2}\alpha}+\dfrac{1}{\cos^{2}\beta}+\dfrac{1}{\cos^{2}\gamma}-\dfrac{2(\cos^{3}\alpha+\cos^{3}\beta+\cos^{3}\gamma)}{\cos\alpha\cdot\cos\beta\cdot\cos\gamma}$，证明：$S\geqslant 3$标注答案略解析因为\[\begin{align}
& S-3 \\
& \text{=}\dfrac{1}{{{\cos }^{2}}\alpha}+\dfrac{1}{{{\cos }^{2}}\beta }+\dfrac{1}{{{\cos }^{2}}\gamma }-\dfrac{2\left({{\cos }^{3}}\alpha +{{\cos }^{3}}\beta +{{\cos }^{3}}\gamma \right)}{\cos \alpha \cdot \cos \beta \cdot\cos \gamma }-3 \\
& =\dfrac{{{\cos }^{2}}\alpha +{{\cos}^{2}}\beta +{{\cos }^{2}}\gamma }{{{\cos }^{2}}\alpha }+\dfrac{{{\cos}^{2}}\alpha +{{\cos }^{2}}\beta +{{\cos }^{2}}\gamma }{{{\cos }^{2}}\beta }+\dfrac{{{\cos}^{2}}\alpha +{{\cos }^{2}}\beta +{{\cos }^{2}}\gamma }{{{\cos }^{2}}\gamma}-2\left( \dfrac{{{\cos }^{2}}\alpha }{\cos \beta \cdot \cos \gamma }+\dfrac{{{\cos}^{2}}\beta }{\cos \alpha \cdot \cos \gamma }+\dfrac{{{\cos }^{2}}\gamma }{\cos\alpha \cdot \cos \beta } \right)-3 \\
& ={{\cos }^{2}}\alpha \left( \dfrac{1}{{{\cos}^{2}}\beta }+\dfrac{1}{{{\cos }^{2}}\gamma } \right)+{{\cos }^{2}}\beta \left(\dfrac{1}{{{\cos }^{2}}\alpha }+\dfrac{1}{{{\cos }^{2}}\gamma } \right)+{{\cos}^{2}}\gamma \left( \dfrac{1}{{{\cos }^{2}}\alpha }+\dfrac{1}{{{\cos}^{2}}\beta } \right)-2\left( \dfrac{{{\cos }^{2}}\alpha }{\cos \beta \cdot\cos \gamma }+\dfrac{{{\cos }^{2}}\beta }{\cos \alpha \cdot \cos \gamma }+\dfrac{{{\cos}^{2}}\gamma }{\cos \alpha \cdot \cos \beta } \right) \\
& ={{\cos }^{2}}\alpha {{\left( \dfrac{1}{\cos\beta }-\dfrac{1}{\cos \gamma } \right)}^{2}}+{{\cos }^{2}}\beta {{\left( \dfrac{1}{\cos\alpha }-\dfrac{1}{\cos \gamma } \right)}^{2}}+{{\cos }^{2}}\gamma {{\left( \dfrac{1}{\cos\alpha }-\dfrac{1}{\cos \beta } \right)}^{2}}\geqslant 0 \\
\end{align}\]所以 $S\geqslant 3$．
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