25~26高三上·南昌二中周测·第14题

$\frac{\sqrt{6}}{5}$

【题目】在三棱锥$P – ABC$中，$AB = 2\sqrt{6}$，$PC = 1$，$PA + PB = 6$，$CA – CB = 4$，且$PC \perp AB$，则二面角$P – AB – C$的余弦值的最小值为_______.

【答案】$\frac{\sqrt{6}}{5}$

【解析】因为$PA + PB = 6 = 2a_1$，所以$a_1 = 3$，且$2c = 2\sqrt{6}$，可得$c = \sqrt{6}$，

所以$b_1 = \sqrt{a_1^2 – c^2} = \sqrt{3}$，点$P$的轨迹方程为$\frac{x^2}{9} + \frac{y^2}{3} = 1$，

又因为$CA – CB = 4 = 2a_2$，则$a_2 = 2$，且$c = \sqrt{6}$，则$b_2 = \sqrt{c^2 – a_2^2} = \sqrt{2}$，
所以点$C$的轨迹方程为$\frac{x^2}{4} – \frac{y^2}{2} = 1$（双曲线的一支），

过点$P$在平面$PAB$内作$PH \perp AB$，垂足为点$H$，连接$CH$，
因为$AB \perp PC$，而$PH \cap PC = P$，$PH$、$PC \subset$面$PHC$，所以$AB \perp$面$PHC$，

因为$CH \subset$平面$PHC$，则$CH \perp AB$，则二面角$P – AB – C$的平面角为$\angle PHC$，

设$O$为$AB$中点，因为$P$点在椭圆上，设$P(3\cos\theta, \sqrt{3}\sin\theta)(\theta \in \left( 0, \frac{\pi}{2} \right])$，所以$OH = 3\cos\theta (\theta \in \left( 0, \frac{\pi}{2} \right])$，$PH = \sqrt{3}\sin\theta$，
因为$C$在双曲线上，$x_C = 3\cos\theta$，

所以$y_C = CH = \sqrt{\frac{9}{2}\cos^2\theta – 2}$，

所以$\cos \angle PHC = \frac{PH^2 + CH^2 – PC^2}{2PH \cdot CH} $$= \frac{3\sin^2\theta + \frac{9}{2}\cos^2\theta – 2 – 1}{2\sqrt{3}\sin\theta \sqrt{\frac{9}{2}\cos^2\theta – 2}}$
$= \frac{\frac{3}{2}\cos^2\theta}{2\sqrt{3}\sin\theta \sqrt{\frac{9}{2}\cos^2\theta – 2}} $$= \frac{\sqrt{6}}{4} \cdot \frac{\cos^2\theta}{\sin\theta \sqrt{9\cos^2\theta – 4}} $$= \frac{\sqrt{6}}{4} \cdot \frac{\cos^2\theta}{\sqrt{(9\cos^2\theta – 4)(1 – \cos^2\theta)}}$
$= \frac{\sqrt{6}\cos^2\theta}{2\sqrt{(9\cos^2\theta – 4)(4 – 4\cos^2\theta)}} $$\geq \frac{\sqrt{6}\cos^2\theta}{(9\cos^2\theta – 4) + (4 – 4\cos^2\theta)} = \frac{\sqrt{6}}{5}$，

当且仅当$9\cos^2\theta – 4 = 4 – 4\cos^2\theta$时，即当$\cos\theta = \frac{2\sqrt{26}}{13}$时，等号成立，

因此二面角$P – AB – C$的余弦值的最小值为$\frac{\sqrt{6}}{5}$．

故答案为：$\frac{\sqrt{6}}{5}$.