25~26高三上·山东菏泽期中·第14题

$\sqrt{3} + 1$

【题目】在$\triangle ABC$中，$P$为边$AB$上一点，$CP = 1$，$\angle ACP = 30^{\circ}$，$\angle BCP = 45^{\circ}$，$AP = \lambda BP$，$\angle CPB = \theta$．当$\triangle ABC$面积最小时，$\tan\theta = $________．

【答案】$\sqrt{3} + 1$
【解析】$S_{\triangle BCP} = \frac{1}{2}BC \cdot CP\sin45^{\circ} $$= \frac{\sqrt{2}}{4}BC$，$S_{\triangle ACP} = \frac{1}{2}AC \cdot CP\sin30^{\circ} = \frac{1}{4}AC$，
所以$S_{\triangle ABC} = \frac{\sqrt{2}}{4}BC + \frac{1}{4}AC$．

在$\triangle ACP$中，由正弦定理得$\frac{AC}{\sin(180^{\circ} – \theta)} = \frac{CP}{\sin(\theta – 30^{\circ})}$，
化简得$AC = \frac{2\sin\theta}{\sqrt{3}\sin\theta – \cos\theta}$．

在$\triangle BCP$中，由正弦定理得$\frac{BC}{\sin\theta} = \frac{CP}{\sin(135^{\circ} – \theta)}$，
化简得$BC = \frac{\sqrt{2}\sin\theta}{\cos\theta + \sin\theta}$．

故$S_{\triangle ABC} = \frac{1}{2}\left( \frac{\sin\theta}{\cos\theta + \sin\theta} + \frac{\sin\theta}{\sqrt{3}\sin\theta – \cos\theta} \right) $$= \frac{1}{2}\left( \frac{1}{\tan\theta + 1} + \frac{1}{\sqrt{3} – \frac{1}{\tan\theta}} \right)$，

令$\frac{1}{\tan\theta} = m$，

则$S_{\triangle ABC} = \frac{1}{2}\left( \frac{1}{m + 1} + \frac{1}{\sqrt{3} – m} \right) $$= \frac{\sqrt{3} + 1}{2} \cdot \frac{1}{(m + 1)(\sqrt{3} – m)}$，

由二次函数性质可知，$f(m) = (m + 1)(\sqrt{3} – m) $$= -m^2 + (\sqrt{3} – 1)m + \sqrt{3}$，

函数开口向下，对称轴为$m = \frac{\sqrt{3} – 1}{2}$，

所以当$m = \frac{\sqrt{3} – 1}{2}$时，$f(m)$取得最大值$\left( \frac{\sqrt{3} – 1}{2} + 1 \right)\left( \sqrt{3} – \frac{\sqrt{3} – 1}{2} \right) $$= \left( \frac{\sqrt{3} + 1}{2} \right)^2 = \frac{2 + \sqrt{3}}{2}$，

即当$m = \frac{\sqrt{3} – 1}{2}$时，$S_{\triangle ABC}$取最小值，

此时$\tan\theta = \frac{1}{m} = \frac{2}{\sqrt{3} – 1} = \sqrt{3} + 1$．

故答案为：$\sqrt{3} + 1$.