Subfield: Solid Geometry


Question: $《瀑布》(图1)是最为人所知的作品之一，图中的瀑布会源源不断地落下，落下的水又逆流而上，荒唐至极，但又会让你百看不腻，画面下方还有一位饶有兴致地观察者，似乎她没发现什么不对劲.画面两座高塔各有一个几何体，左塔上方是著名的“三立方体合体”，它由三个正方体构成，右塔上的几何体是首次出现，后称“埃舍尔多面体”.$![Image](https://cdntk.53zaixian.com/tq/1693878120023072770/img/compression/M_a4dd6db2e8bc965684cd1e3d3aec4410.jpg)$图1埃舍尔多面体可以用两两垂直且中心重合的三个正方形构造，设边长均为2，定义正方形A_{n}B_{n}C_{n}D_{n}，n=1,2,3的顶点为“框架点”，定义两正方形交线为“极轴”，其端点为“极点”，记为P_{n},Q_{n}，将极点P_{1},Q_{1}分别与正方形A_{2}B_{2}C_{2}D_{2}的顶点连线，取其中点记为E_{m}，F_{m}，m=1,2,3,4(如图2).埃舍尔多面体可视部分是由12个四棱锥构成的，这些四棱锥顶点均为“框架点”，底面四边形由两个“极点”与两个“中点”构成，为了便于理解，图3中我们构造了其中两个四棱锥A_{1}-P_{1}E_{1}P_{2}E_{2}和A_{2}-P_{2}E_{1}P_{3}F_{1}.$![Image](https://cdntk.53zaixian.com/tq/1693878120023072770/img/compression/M_e6e18ec43d5e54b1307eb23fd57d1fc7.jpg)$图2$![Image](https://cdntk.53zaixian.com/tq/1693878120023072770/img/compression/M_2300d48d68978bd9afab366e8380361f.jpg)$图3$
$(3)求埃舍尔多面体的表面积与体积(直接写出答案).$

Solution:
$由(1)可得，\overrightarrow{P_1P_2}=(1,0,-1)，\overrightarrow{E_1E_2}=(0,1,0)，\overrightarrow{P_1E_1}=\left(\frac{1}{2},-\frac{1}{2},-\frac{1}{2}\right)，\overrightarrow{P_2E_2}=\left(-\frac{1}{2},\frac{1}{2},\frac{1}{2}\right)，\overrightarrow{A_1P_1}=(-1,0,0)，\overrightarrow{P_1E_2}=\left(\frac{1}{2},\frac{1}{2},-\frac{1}{2}\right)，所以\overrightarrow{P_2E_2}=-\overrightarrow{P_1E_1}，所以P_{2}E_{2}\parallel P_{1}E_{1}，且P_{2}E_{2}=P_{1}E_{1}，所以四边形P_{1}E_{1}P_{2}E_{2}为平行四边形.又\overrightarrow{P_1P_2}\cdot \overrightarrow{E_1E_2}=0，所以\overrightarrow{P_1P_2}\perp \overrightarrow{E_1E_2}，即P_{1}P_{2}\perp E_{1}E_{2}，所以四边形P_{1}E_{1}P_{2}E_{2}为菱形，又|\overrightarrow{P_1P_2}|=\sqrt{2}，|\overrightarrow{E_1E_2}|=1，所以S_{四边形P_{1}E_{1}P_{2}E_{2}}=\frac{1}{2}\times |\overrightarrow{P_1P_2}|\times |\overrightarrow{E_1E_2}|=\frac{\sqrt{2}}{2}.设n_{3}=(x_{3},y_{3},z_{3})是平面P_{1}E_{1}P_{2}E_{2}的一个法向量，则\left\{\begin{matrix}\boldsymbol{n}_3\cdot \overrightarrow{P_1P_2}=0,\\ \boldsymbol{n}_3\cdot \overrightarrow{P_1E_1}=0,\end{matrix}\right.即\left\{\begin{matrix}x_3-z_3=0,\\ \frac{1}{2}x_3-\frac{1}{2}y_3-\frac{1}{2}z_3=0,\end{matrix}\right.令x_{3}=1，则y_{3}=0,z_{3}=1,所以n_{3}=(1,0,1)是平面P_{1}E_{1}P_{2}E_{2}的一个法向量.又\overrightarrow{A_1P_1}=(-1,0,0)，所以点A_{1}到平面P_{1}E_{1}P_{2}E_{2}的距离d=\frac{|\overrightarrow{A_1P_1}\cdot \boldsymbol{n}_3|}{|\boldsymbol{n}_3|}=\frac{1}{\sqrt{2}}=\frac{\sqrt{2}}{2}，所以四棱锥A_{1}-P_{1}E_{1}P_{2}E_{2}的体积V_{1}=\frac{1}{3}\times S_{四边形P_{1}E_{1}P_{2}E_{2}}\times d=\frac{1}{3}\times \frac{\sqrt{2}}{2}\times \frac{\sqrt{2}}{2}=\frac{1}{6}.因为\overrightarrow{A_1P_1}=(-1,0,0)，\overrightarrow{P_1E_2}=\left(\frac{1}{2},\frac{1}{2},-\frac{1}{2}\right)，\overrightarrow{P_1E_1}=\left(\frac{1}{2},-\frac{1}{2},-\frac{1}{2}\right)，所以\overrightarrow{A_1P_1}在\overrightarrow{P_1E_2}方向上的投影为\frac{\overrightarrow{A_1P_1}\cdot \overrightarrow{P_1E_2}}{|\overrightarrow{P_1E_2}|}=\frac{-\frac{1}{2}}{\sqrt{\frac{3}{4}}}=-\frac{\sqrt{3}}{3}，所以点A_{1}到直线P_{1}E_{2}的距离h_{1}=\sqrt{{|\overrightarrow{A_1P_1}|}^2-{\left|\frac{\overrightarrow{A_1P_1}\cdot \overrightarrow{P_1E_2}}{|\overrightarrow{P_1E_2}|}\right|}^2}=\frac{\sqrt{6}}{3}.同理可得，点A_{1}到直线P_{1}E_{1}的距离h_{2}=\frac{\sqrt{6}}{3}.所以四棱锥A_{1}-P_{1}E_{1}P_{2}E_{2}的侧面积S_{1}=\frac{1}{2}\times |\overrightarrow{P_1E_2}|\times h_{1}\times 4=\frac{1}{2}\times \frac{\sqrt{3}}{2}\times \frac{\sqrt{6}}{3}\times 4=\sqrt{2}.所以埃舍尔多面体的表面积为12S_{1}=12\sqrt{2}，体积为12V_{1}=2.$

Final answer: $12\sqrt 2,2$