990

如图所示，用一个不平行于圆柱底面的平面，截该圆柱所得的截面为椭圆面，得到的几何体称之为 "斜截圆柱". ${\displaystyle AB}$ 是底面圆 ${\displaystyle O}$ 的直径, ${\displaystyle AB=2BC=2}$, 椭圆所在平面垂直于平面 ${\displaystyle ABCD}$, 且与底面所成二面角为 ${\displaystyle {45}^{\circ }}$. 图一中，点 ${\displaystyle P}$ 是椭圆上的动点，点 ${\displaystyle P}$ 在底面上的投影为点 ${\displaystyle P_1}$; 图二中，椭圆上的点 ${\displaystyle E_i(i=1,2,3,\cdots ,n)}$ 在底面上的投影分别为 ${\displaystyle F_i}$, 且 ${\displaystyle F_i}$ 均在直径 ${\displaystyle AB}$ 的同一侧.

1. 当 ${\displaystyle \mathrm{\angle }AO{P}_1={\displaystyle \frac{2\pi }{3}}}$ 时，求 ${\displaystyle P{P}_1}$ 的长度;
2. 1. 当 ${\displaystyle n=6}$ 时，若图二中, 点 ${\displaystyle F_1,F_2,\cdots ,F_6}$ 将半圆平均分为 7 等份，求 ${\displaystyle (E_1{F}_1-2)\cdot (E_2{F}_2-2)\cdot (E_3{F}_3-2)}$;
   2. 证明: $$\stackrel{⌢}{A{F}_1}\cdot E_1{F}_1+\stackrel{⌢}{A{F}_1}\cdot E_2{F}_2+\cdots +\stackrel{⌢}{A{F}_1}\cdot E_n{F}_n+\stackrel{⌢}{A{F}_1}\cdot BC<2\pi .$$

1. 解 如图，取截面 ${\displaystyle ABCD}$, 设 ${\displaystyle P_1,P}$ 在 ${\displaystyle AB,DC}$ 上的投影分别为 ${\displaystyle P_1^{\prime },P^{\prime }}$, 则易得 ${\displaystyle P^{\prime }{P}_1^{\prime }=P{P}_1}$. 因此在梯形 ${\displaystyle ABCD}$ 中，$$\begin{array}{r}{P}^{\prime }{P}_1^{\prime }=\frac{A{P}_1^{\prime }}{AB}\cdot BC+\frac{B{P}_1^{\prime }}{AB}\cdot AD=\frac{1-\cos \theta }{2}+3\frac{1+\cos \theta }{2}=2+\cos \theta ,\end{array}$$
   代入 ${\displaystyle \theta ={\displaystyle \frac{2\pi }{3}}}$, 得 ${\displaystyle P{P}_1={\displaystyle \frac{3}{2}}}$.
2. 1. 解 由 1 得 ${\displaystyle {\displaystyle E_i{F}_i-2=\cos \frac{i\pi }{7}}}$.
      从而利用点鞭炮法:$$\begin{array}{rl}\text{原式}=& \cos \frac{\pi }{7}\cos \frac{2\pi }{7}\cos \frac{3\pi }{7}=\frac{8\sin {\displaystyle \frac{\pi }{7}}\cos {\displaystyle \frac{\pi }{7}}\cos {\displaystyle \frac{2\pi }{7}}\cos {\displaystyle \frac{3\pi }{7}}}{8\sin {\displaystyle \frac{\pi }{7}}}\\ =& -\frac{4\sin {\displaystyle \frac{2\pi }{7}}\cos {\displaystyle \frac{2\pi }{7}}\cos {\displaystyle \frac{4\pi }{7}}}{8\sin {\displaystyle \frac{\pi }{7}}}=-\frac{2\sin {\displaystyle \frac{4\pi }{7}}\cos {\displaystyle \frac{4\pi }{7}}}{8\sin {\displaystyle \frac{\pi }{7}}}=-\frac{\sin {\displaystyle \frac{8\pi }{7}}}{8\sin {\displaystyle \frac{\pi }{7}}}=\frac{1}{8}.\end{array}$$
   2. 证明 设 ${\displaystyle {\alpha }_0=0,{\alpha }_i=\mathrm{\angle }AO{F}_i,1\le i\le n,{\alpha }_{n+1}=\pi }$. 则$$\begin{array}{rl}\text{左边}=& \sum _{i=1}^{n+1}({\alpha }_i-{\alpha }_{i-1})(2+\cos {\alpha }_i)=2\pi +\sum _{i=1}^{n+1}({\alpha }_i-{\alpha }_{i-1})\cos {\alpha }_i\\ <& 2\pi +\sum _{i=1}^{n+1}{\int }_{{\alpha }_{i-1}}^{{\alpha }_i}\cos x\mathrm{d}x=2\pi +{\int }_0^{\pi }\cos x\mathrm{d}x=2\pi =\text{右边}.\end{array}$$