di-zhang-fdu/OlympiadExtended huggingface dataset:
Subfield: Solid Geometry


Question: $如图，P-ABCD是一个四棱锥，已知四边形ABCD是梯形，PD\perp 平面ABCD，AD\perp CD，AB\parallel CD，PD＝AD＝AB＝1，CD＝2，点E是棱PC的中点，点F在棱PB上，PF=\frac{1}{2}FB.$<img_1930>
$(1)证明：直线BE\parallel 平面PAD；$

Solution:
$证明：取PD的中点G，连接AG，GE，因为G，E分别是PD，PC的中点，所以GE\parallel DC，GE=\frac{1}{2}DC，因为AB\parallel DC,AB=\frac{1}{2}DC，所以GE\parallel AB,GE=AB，所以四边形AGEB为平行四边形，所以BE\parallel AG，因为BE\not\subset 平面PAD，AG\subset 平面PAD，所以直线BE\parallel 平面PAD.$
Subfield: Solid Geometry


Question: $如图，P-ABCD是一个四棱锥，已知四边形ABCD是梯形，PD\perp 平面ABCD，AD\perp CD，AB\parallel CD，PD＝AD＝AB＝1，CD＝2，点E是棱PC的中点，点F在棱PB上，PF=\frac{1}{2}FB.$<img_1930>
$(2)求直线BE与平面PBD所成角的正弦值；$

Solution:
$因为PD\perp 平面ABCD，AD\subset 平面ABCD，DC\subset 平面ABCD，所以PD\perp AD,PD\perp DC，又AD\perp CD，所以以D为原点，分别以DA，DC，DP所在直线为x，y，z轴建立空间直角坐标系，如图，$<img_2081>$则B(1,1,0)，E\left(0,1,\frac{1}{2}\right)，D(0,0,0)，P(0,0,1),所以\overrightarrow{BE}=\left(-1,0,\frac{1}{2}\right)，\overrightarrow{DP}=(0,0，1)，\overrightarrow{DB}=(1,1，0).设平面PBD的法向量为m=(x_{1}，y_{1}，z_{1})，则\left\{\begin{matrix}\boldsymbol{m}\cdot \overrightarrow{DP}=z_1=0,\\ \boldsymbol{m}\cdot \overrightarrow{DB}=x_1+y_1=0,\end{matrix}\right.取x_{1}=1，则m=(1,-1,0)， 设直线BE与平面PBD所成角为\alpha ，则\\sin   \alpha =|\\cos  \langle \overrightarrow{BE}，m\rangle |=\frac{|\overrightarrow{BE}\cdot \boldsymbol{m}|}{|\overrightarrow{BE}|\cdot |\boldsymbol{m}|}=\frac{\sqrt{10}}{5}，所以直线BE与平面PBD所成角的正弦值为\frac{\sqrt{10}}{5}.$

Final answer: $\frac{\sqrt{10}}{5}$
Subfield: Solid Geometry


Question: $如图，P-ABCD是一个四棱锥，已知四边形ABCD是梯形，PD\perp 平面ABCD，AD\perp CD，AB\parallel CD，PD＝AD＝AB＝1，CD＝2，点E是棱PC的中点，点F在棱PB上，PF=\frac{1}{2}FB.$![Image](https://cdntk.53zaixian.com/tq/1701760590998343681/img/compression/M_937e644cda087b8e38c5443fae39af17.jpg)
$(2)求直线BE与平面PBD所成角的正弦值；$

Solution:
$因为PD\perp 平面ABCD，AD\subset 平面ABCD，DC\subset 平面ABCD，所以PD\perp AD,PD\perp DC，又AD\perp CD，所以以D为原点，分别以DA，DC，DP所在直线为x，y，z轴建立空间直角坐标系，如图，$![Image](https://cdntk.53zaixian.com/tq/1701760591342276610/img/compression/M_c6cbcf8a8f33dbb787b677cbd98c712b.jpg)$则B(1,1,0)，E\left(0,1,\frac{1}{2}\right)，D(0,0,0)，P(0,0,1),所以\overrightarrow{BE}=\left(-1,0,\frac{1}{2}\right)，\overrightarrow{DP}=(0,0，1)，\overrightarrow{DB}=(1,1，0).设平面PBD的法向量为m=(x_{1}，y_{1}，z_{1})，则\left\{\begin{matrix}\boldsymbol{m}\cdot \overrightarrow{DP}=z_1=0,\\ \boldsymbol{m}\cdot \overrightarrow{DB}=x_1+y_1=0,\end{matrix}\right.取x_{1}=1，则m=(1,-1,0)， 设直线BE与平面PBD所成角为\alpha ，则\\sin   \alpha =|\\cos  \langle \overrightarrow{BE}，m\rangle |=\frac{|\overrightarrow{BE}\cdot \boldsymbol{m}|}{|\overrightarrow{BE}|\cdot |\boldsymbol{m}|}=\frac{\sqrt{10}}{5}，所以直线BE与平面PBD所成角的正弦值为\frac{\sqrt{10}}{5}.$

Final answer: $\frac{\sqrt{10}}{5}$
Subfield: Solid Geometry


Question: $如图，P-ABCD是一个四棱锥，已知四边形ABCD是梯形，PD\perp 平面ABCD，AD\perp CD，AB\parallel CD，PD＝AD＝AB＝1，CD＝2，点E是棱PC的中点，点F在棱PB上，PF=\frac{1}{2}FB.$![Image](https://cdntk.53zaixian.com/tq/1701760590998343681/img/compression/M_937e644cda087b8e38c5443fae39af17.jpg)
$(3)求平面DEF与平面ABCD的夹角的余弦值．$

Solution:
$由(2)知，\overrightarrow{DE}=\left(0,1,\frac{1}{2}\right)，又PF=\frac{1}{2}FB，所以\overrightarrow{PF}=\frac{1}{2}\overrightarrow{FB}，所以\overrightarrow{DF}=\frac{2}{3}\overrightarrow{DP}+\frac{1}{3}\overrightarrow{DB}=\frac{2}{3}(0，0,1)+\frac{1}{3}(1，1，0)=\left(\frac{1}{3},\frac{1}{3},\frac{2}{3}\right)，设平面DEF的法向量为n=(x_{2}，y_{2}，z_{2})，则\left\{\begin{matrix}\boldsymbol{n}\cdot \overrightarrow{DF}=\frac{1}{3}x_2+\frac{1}{3}y_2+\frac{2}{3}z_2=0,\\ \boldsymbol{n}\cdot \overrightarrow{DE}=y_2+\frac{1}{2}z_2=0,\end{matrix}\right.取z_{2}=2，则y_{2}=-1，x_{2}=-3，则n=(-3,-1，2)， 易知平面ABCD的一个法向量为\overrightarrow{DP}=(0,0，1)，设平面DEF与平面ABCD的夹角为\beta ，则\\cos   \beta =|\\cos  \langle n，\overrightarrow{DP}\rangle |=\frac{|\boldsymbol{n}\cdot \overrightarrow{DP}|}{|\boldsymbol{n}|\cdot |\overrightarrow{DP}|}=\frac{\sqrt{14}}{7}，所以平面DEF与平面ABCD的夹角的余弦值为\frac{\sqrt{14}}{7}.$

Final answer: $\frac{\sqrt{14}}{7}$
Subfield: Solid Geometry


Question: $如图，P-ABCD是一个四棱锥，已知四边形ABCD是梯形，PD\perp 平面ABCD，AD\perp CD，AB\parallel CD，PD＝AD＝AB＝1，CD＝2，点E是棱PC的中点，点F在棱PB上，PF=\frac{1}{2}FB.$<img_1930>
$(3)求平面DEF与平面ABCD的夹角的余弦值．$

Solution:
$由(2)知，\overrightarrow{DE}=\left(0,1,\frac{1}{2}\right)，又PF=\frac{1}{2}FB，所以\overrightarrow{PF}=\frac{1}{2}\overrightarrow{FB}，所以\overrightarrow{DF}=\frac{2}{3}\overrightarrow{DP}+\frac{1}{3}\overrightarrow{DB}=\frac{2}{3}(0，0,1)+\frac{1}{3}(1，1，0)=\left(\frac{1}{3},\frac{1}{3},\frac{2}{3}\right)，设平面DEF的法向量为n=(x_{2}，y_{2}，z_{2})，则\left\{\begin{matrix}\boldsymbol{n}\cdot \overrightarrow{DF}=\frac{1}{3}x_2+\frac{1}{3}y_2+\frac{2}{3}z_2=0,\\ \boldsymbol{n}\cdot \overrightarrow{DE}=y_2+\frac{1}{2}z_2=0,\end{matrix}\right.取z_{2}=2，则y_{2}=-1，x_{2}=-3，则n=(-3,-1，2)， 易知平面ABCD的一个法向量为\overrightarrow{DP}=(0,0，1)，设平面DEF与平面ABCD的夹角为\beta ，则\\cos   \beta =|\\cos  \langle n，\overrightarrow{DP}\rangle |=\frac{|\boldsymbol{n}\cdot \overrightarrow{DP}|}{|\boldsymbol{n}|\cdot |\overrightarrow{DP}|}=\frac{\sqrt{14}}{7}，所以平面DEF与平面ABCD的夹角的余弦值为\frac{\sqrt{14}}{7}.$

Final answer: $\frac{\sqrt{14}}{7}$
Subfield: Solid Geometry


Question: $已知四棱锥P-ABCD中，DA\perp 平面PAB，DA\parallel BC，\angle BAP=120^\circ ,DA=AP=2AB=2BC=2,E为线段DP的中点.$![Image](https://cdntk.53zaixian.com/tq/1701760591996588033/img/compression/M_f834ea2761201614408810a631930602.jpg)
$(1)求证:直线CE\parallel 平面PAB;$

Solution:
$以点A为坐标原点，AB,AD所在直线分别为x轴,z轴，与AB垂直的直线为y轴，建立如图所示的空间直角坐标系.$![Image](https://cdntk.53zaixian.com/tq/1701760591761707010/img/compression/M_4897845fceb0ebef5f4a306b4c917f74.jpg)$则A(0,0,0),B(1,0,0),C(1,0,1),D(0,0,2),P(-1,\sqrt{3},0),E\left(-\frac{1}{2},\frac{\sqrt{3}}{2},1\right).
(2分)
证明:\overrightarrow{CE}=\left(-\frac{3}{2},\frac{\sqrt{3}}{2},0\right)，易知平面PAB的一个法向量为n=(0,0,1)，因为\overrightarrow{CE}\cdot n=0，所以\overrightarrow{CE}\perp n,又因为CE\not\subset 平面PAB，所以直线CE\parallel 平面PAB.
(6分)$

Final answer: 证明题
Subfield: Solid Geometry


Question: $已知四棱锥P-ABCD中，DA\perp 平面PAB，DA\parallel BC，\angle BAP=120^\circ ,DA=AP=2AB=2BC=2,E为线段DP的中点.$<img_2113>
$(1)求证:直线CE\parallel 平面PAB;$

Solution:
$以点A为坐标原点，AB,AD所在直线分别为x轴,z轴，与AB垂直的直线为y轴，建立如图所示的空间直角坐标系.$<img_3021>$则A(0,0,0),B(1,0,0),C(1,0,1),D(0,0,2),P(-1,\sqrt{3},0),E\left(-\frac{1}{2},\frac{\sqrt{3}}{2},1\right).
(2分)
证明:\overrightarrow{CE}=\left(-\frac{3}{2},\frac{\sqrt{3}}{2},0\right)，易知平面PAB的一个法向量为n=(0,0,1)，因为\overrightarrow{CE}\cdot n=0，所以\overrightarrow{CE}\perp n,又因为CE\not\subset 平面PAB，所以直线CE\parallel 平面PAB.
(6分)$
Subfield: Solid Geometry


Question: $已知四棱锥P-ABCD中，DA\perp 平面PAB，DA\parallel BC，\angle BAP=120^\circ ,DA=AP=2AB=2BC=2,E为线段DP的中点.$![Image](https://cdntk.53zaixian.com/tq/1701760591996588033/img/compression/M_f834ea2761201614408810a631930602.jpg)
$(2)求直线BE与平面PCD所成角的正弦值;$

Solution:
$\overrightarrow{CE}=\left(-\frac{3}{2},\frac{\sqrt{3}}{2},0\right)，\overrightarrow{CD}=(-1,0,1),\overrightarrow{BE}=\left(-\frac{3}{2},\frac{\sqrt{3}}{2},1\right)，设m=(x,y,z)为平面PCD的法向量，则\left\{\begin{matrix}\boldsymbol{m}\cdot \overrightarrow{CE}=-\frac{3}{2}x+\frac{\sqrt{3}}{2}y=0,\\ \boldsymbol{m}\cdot \overrightarrow{CD}=-x+z=0,\end{matrix}\right.令x=1，则y=\sqrt{3}，z=1，所以m=(1,\sqrt{3},1)，设直线BE与平面PCD所成角为\theta ，则\\sin   \theta =|\\cos  \langle m,\overrightarrow{BE}\rangle |=\frac{\left|-\frac{3}{2}+\frac{3}{2}+1\right|}{\sqrt{5}\times 2}=\frac{\sqrt{5}}{10}，所以直线BE与平面PCD所成角的正弦值为\frac{\sqrt{5}}{10}.
(10分)$

Final answer: $\frac{\sqrt{5}}{10}$
Subfield: Solid Geometry


Question: $已知四棱锥P-ABCD中，DA\perp 平面PAB，DA\parallel BC，\angle BAP=120^\circ ,DA=AP=2AB=2BC=2,E为线段DP的中点.$<img_2113>
$(2)求直线BE与平面PCD所成角的正弦值;$

Solution:
$\overrightarrow{CE}=\left(-\frac{3}{2},\frac{\sqrt{3}}{2},0\right)，\overrightarrow{CD}=(-1,0,1),\overrightarrow{BE}=\left(-\frac{3}{2},\frac{\sqrt{3}}{2},1\right)，设m=(x,y,z)为平面PCD的法向量，则\left\{\begin{matrix}\boldsymbol{m}\cdot \overrightarrow{CE}=-\frac{3}{2}x+\frac{\sqrt{3}}{2}y=0,\\ \boldsymbol{m}\cdot \overrightarrow{CD}=-x+z=0,\end{matrix}\right.令x=1，则y=\sqrt{3}，z=1，所以m=(1,\sqrt{3},1)，设直线BE与平面PCD所成角为\theta ，则\\sin   \theta =|\\cos  \langle m,\overrightarrow{BE}\rangle |=\frac{\left|-\frac{3}{2}+\frac{3}{2}+1\right|}{\sqrt{5}\times 2}=\frac{\sqrt{5}}{10}，所以直线BE与平面PCD所成角的正弦值为\frac{\sqrt{5}}{10}.
(10分)$

Final answer: $\frac{\sqrt{5}}{10}$
Subfield: Solid Geometry


Question: $已知四棱锥P-ABCD中，DA\perp 平面PAB，DA\parallel BC，\angle BAP=120^\circ ,DA=AP=2AB=2BC=2,E为线段DP的中点.$![Image](https://cdntk.53zaixian.com/tq/1701760591996588033/img/compression/M_f834ea2761201614408810a631930602.jpg)
$(3)求平面PCD与平面PAD夹角的余弦值.$

Solution:
$\overrightarrow{AD}=(0，0，2)，\overrightarrow{AP}=(-1，\sqrt{3}，0)，设平面PAD的法向量为n_{0}=(x_{0},y_{0},z_{0})，则\left\{\begin{matrix}\boldsymbol{n}_0\cdot \overrightarrow{AD}=2z_0=0,\\ \boldsymbol{n}_0\cdot \overrightarrow{AP}=-x_0+\sqrt{3}y_0=0,\end{matrix}\right.令x_{0}=3，则y_{0}=\sqrt{3}，z_{0}=0，所以n_{0}=(3,\sqrt{3},0)，由(2)知，平面PCD的一个法向量为m=(1,\sqrt{3},1)，设平面PCD与平面PAD的夹角为\phi ，则\\cos   \phi =|\\cos  \langle m,n_{0}\rangle |=\frac{|3+3+0|}{\sqrt{5}\times 2\sqrt{3}}=\frac{\sqrt{15}}{5}，所以平面PCD与平面PAD夹角的余弦值为\frac{\sqrt{15}}{5}.
(15分)$

Final answer: $\frac{\sqrt{15}}{5}$
Subfield: Solid Geometry


Question: $已知四棱锥P-ABCD中，DA\perp 平面PAB，DA\parallel BC，\angle BAP=120^\circ ,DA=AP=2AB=2BC=2,E为线段DP的中点.$<img_2113>
$(3)求平面PCD与平面PAD夹角的余弦值.$

Solution:
$\overrightarrow{AD}=(0，0，2)，\overrightarrow{AP}=(-1，\sqrt{3}，0)，设平面PAD的法向量为n_{0}=(x_{0},y_{0},z_{0})，则\left\{\begin{matrix}\boldsymbol{n}_0\cdot \overrightarrow{AD}=2z_0=0,\\ \boldsymbol{n}_0\cdot \overrightarrow{AP}=-x_0+\sqrt{3}y_0=0,\end{matrix}\right.令x_{0}=3，则y_{0}=\sqrt{3}，z_{0}=0，所以n_{0}=(3,\sqrt{3},0)，由(2)知，平面PCD的一个法向量为m=(1,\sqrt{3},1)，设平面PCD与平面PAD的夹角为\phi ，则\\cos   \phi =|\\cos  \langle m,n_{0}\rangle |=\frac{|3+3+0|}{\sqrt{5}\times 2\sqrt{3}}=\frac{\sqrt{15}}{5}，所以平面PCD与平面PAD夹角的余弦值为\frac{\sqrt{15}}{5}.
(15分)$

Final answer: $\frac{\sqrt{15}}{5}$
Subfield: Solid Geometry


Question: $在苏州博物馆有一类典型建筑八角亭(如图1)，既美观又利于采光，其中一角(如图2所示)为多面体ABCDE-A_{1}B_{1}C_{1}D_{1}，AB\perp AE，AE\parallel BC，AB\parallel ED,AA_{1}\perp 底面ABCDE,四边形A_{1}B_{1}C_{1}D_{1}是边长为2的正方形且平行于底面，AB\parallel A_{1}B_{1}，D_{1}E，B_{1}B的中点分别为F,G，AB=AE=2DE=2BC=4，AA_{1}=1.
$![Image](https://cdntk.53zaixian.com/tq/1701760592671870977/img/compression/M_8695bf8aa58575451fe2f6ef027ed3c3.jpg)$
图1
　
$![Image](https://cdntk.53zaixian.com/tq/1701760592671870977/img/compression/M_42faacc072c7519fa4036c02eee9946c.jpg)$
图2
$
$(1)证明:FG\parallel 平面C_{1}CD;$

Solution:
$证明：如图，以E为原点建立空间直角坐标系E-xyz，则A(0,4,0),C(4,2，0)，D(2,0,0)，C_{1}(2,2,1)，F\left(0,1,\frac{1}{2}\right)，G\left(3,4,\frac{1}{2}\right)，所以\overrightarrow{C_1C}=(2,0,-1),\overrightarrow{C_1D}=(0,-2,-1)，\overrightarrow{FG}=(3,3,0).$![Image](https://cdntk.53zaixian.com/tq/1701760592332132354/img/compression/M_7aff65e8ba588e2249b6dc889c066549.jpg)$设平面C_{1}CD的法向量为n_{1}=(x,y,z),则\left\{\begin{matrix}\boldsymbol{n}_1\cdot \overrightarrow{C_1C}=0,\\ \boldsymbol{n}_1\cdot \overrightarrow{C_1D}=0,\end{matrix}\right.即\left\{\begin{matrix}2x-z=0,\\ -2y-z=0,\end{matrix}\right.令x=1,则n_{1}=(1,-1,2),\therefore n_{1}\cdot \overrightarrow{FG}=3-3+0=0，\therefore n_{1}\perp \overrightarrow{FG}，又FG\not\subset 平面C_{1}CD，\therefore FG\parallel 平面C_{1}CD.$

Final answer: 证明题
Subfield: Solid Geometry


Question: $在苏州博物馆有一类典型建筑八角亭(如图1)，既美观又利于采光，其中一角(如图2所示)为多面体ABCDE-A_{1}B_{1}C_{1}D_{1}，AB\perp AE，AE\parallel BC，AB\parallel ED,AA_{1}\perp 底面ABCDE,四边形A_{1}B_{1}C_{1}D_{1}是边长为2的正方形且平行于底面，AB\parallel A_{1}B_{1}，D_{1}E，B_{1}B的中点分别为F,G，AB=AE=2DE=2BC=4，AA_{1}=1.
$<img_2278>$
图1
　
$<img_2224>$
图2
$
$(1)证明:FG\parallel 平面C_{1}CD;$

Solution:
$证明：如图，以E为原点建立空间直角坐标系E-xyz，则A(0,4,0),C(4,2，0)，D(2,0,0)，C_{1}(2,2,1)，F\left(0,1,\frac{1}{2}\right)，G\left(3,4,\frac{1}{2}\right)，所以\overrightarrow{C_1C}=(2,0,-1),\overrightarrow{C_1D}=(0,-2,-1)，\overrightarrow{FG}=(3,3,0).$<img_1548>$设平面C_{1}CD的法向量为n_{1}=(x,y,z),则\left\{\begin{matrix}\boldsymbol{n}_1\cdot \overrightarrow{C_1C}=0,\\ \boldsymbol{n}_1\cdot \overrightarrow{C_1D}=0,\end{matrix}\right.即\left\{\begin{matrix}2x-z=0,\\ -2y-z=0,\end{matrix}\right.令x=1,则n_{1}=(1,-1,2),\therefore n_{1}\cdot \overrightarrow{FG}=3-3+0=0，\therefore n_{1}\perp \overrightarrow{FG}，又FG\not\subset 平面C_{1}CD，\therefore FG\parallel 平面C_{1}CD.$
Subfield: Solid Geometry


Question: $在苏州博物馆有一类典型建筑八角亭(如图1)，既美观又利于采光，其中一角(如图2所示)为多面体ABCDE-A_{1}B_{1}C_{1}D_{1}，AB\perp AE，AE\parallel BC，AB\parallel ED,AA_{1}\perp 底面ABCDE,四边形A_{1}B_{1}C_{1}D_{1}是边长为2的正方形且平行于底面，AB\parallel A_{1}B_{1}，D_{1}E，B_{1}B的中点分别为F,G，AB=AE=2DE=2BC=4，AA_{1}=1.
$![Image](https://cdntk.53zaixian.com/tq/1701760592671870977/img/compression/M_8695bf8aa58575451fe2f6ef027ed3c3.jpg)$
图1
　
$![Image](https://cdntk.53zaixian.com/tq/1701760592671870977/img/compression/M_42faacc072c7519fa4036c02eee9946c.jpg)$
图2
$
$(2)求平面C_{1}CD与平面AA_{1}B_{1}B夹角的余弦值;$

Solution:
$由(1)知，平面C_{1}CD的一个法向量为n_{1}=(1，-1，2)，易知平面AA_{1}B_{1}B的一个法向量为n_{2}=(0,1,0)，设平面C_{1}CD与平面AA_{1}B_{1}B的夹角为\phi ，\therefore \\cos   \phi =|\\cos  \langle n_{1},n_{2}\rangle |=\frac{|\boldsymbol{n}_1\cdot \boldsymbol{n}_2|}{|\boldsymbol{n}_1||\boldsymbol{n}_2|}=\frac{\sqrt{6}}{6}.$

Final answer: $\frac{\sqrt{6}}{6}$
Subfield: Solid Geometry


Question: $在苏州博物馆有一类典型建筑八角亭(如图1)，既美观又利于采光，其中一角(如图2所示)为多面体ABCDE-A_{1}B_{1}C_{1}D_{1}，AB\perp AE，AE\parallel BC，AB\parallel ED,AA_{1}\perp 底面ABCDE,四边形A_{1}B_{1}C_{1}D_{1}是边长为2的正方形且平行于底面，AB\parallel A_{1}B_{1}，D_{1}E，B_{1}B的中点分别为F,G，AB=AE=2DE=2BC=4，AA_{1}=1.
$<img_2278>$
图1
　
$<img_2224>$
图2
$
$(2)求平面C_{1}CD与平面AA_{1}B_{1}B夹角的余弦值;$

Solution:
$由(1)知，平面C_{1}CD的一个法向量为n_{1}=(1，-1，2)，易知平面AA_{1}B_{1}B的一个法向量为n_{2}=(0,1,0)，设平面C_{1}CD与平面AA_{1}B_{1}B的夹角为\phi ，\therefore \\cos   \phi =|\\cos  \langle n_{1},n_{2}\rangle |=\frac{|\boldsymbol{n}_1\cdot \boldsymbol{n}_2|}{|\boldsymbol{n}_1||\boldsymbol{n}_2|}=\frac{\sqrt{6}}{6}.$

Final answer: $\frac{\sqrt{6}}{6}$
Subfield: Solid Geometry


Question: $在苏州博物馆有一类典型建筑八角亭(如图1)，既美观又利于采光，其中一角(如图2所示)为多面体ABCDE-A_{1}B_{1}C_{1}D_{1}，AB\perp AE，AE\parallel BC，AB\parallel ED,AA_{1}\perp 底面ABCDE,四边形A_{1}B_{1}C_{1}D_{1}是边长为2的正方形且平行于底面，AB\parallel A_{1}B_{1}，D_{1}E，B_{1}B的中点分别为F,G，AB=AE=2DE=2BC=4，AA_{1}=1.
$<img_2278>$
图1
　
$<img_2224>$
图2
$
$(3)一束光从玻璃窗面C_{1}CD上点C_{1}射入恰经过点A(假设此时光经过玻璃为直射)，求这束光在玻璃窗C_{1}CD上的入射角的正切值.$

Solution:
$由(1)知，\overrightarrow{AC_1}=(2,-2,1),平面C_{1}CD的一个法向量为n_{1}=(1，-1，2)，设入射角为\theta ,则\\cos   \theta =|\\cos  \langle n_{1},\overrightarrow{AC_1}\rangle |=\frac{|\boldsymbol{n}_1\cdot \overrightarrow{AC_1}|}{|\boldsymbol{n}_1||\overrightarrow{AC_1}|}=\frac{\sqrt{6}}{3}，则\\sin   \theta =\frac{\sqrt{3}}{3}，\therefore \\tan   \theta =\frac{\sqrt{2}}{2}.$

Final answer: $\frac{\sqrt{2}}{2}$
Subfield: Solid Geometry


Question: $在苏州博物馆有一类典型建筑八角亭(如图1)，既美观又利于采光，其中一角(如图2所示)为多面体ABCDE-A_{1}B_{1}C_{1}D_{1}，AB\perp AE，AE\parallel BC，AB\parallel ED,AA_{1}\perp 底面ABCDE,四边形A_{1}B_{1}C_{1}D_{1}是边长为2的正方形且平行于底面，AB\parallel A_{1}B_{1}，D_{1}E，B_{1}B的中点分别为F,G，AB=AE=2DE=2BC=4，AA_{1}=1.
$![Image](https://cdntk.53zaixian.com/tq/1701760592671870977/img/compression/M_8695bf8aa58575451fe2f6ef027ed3c3.jpg)$
图1
　
$![Image](https://cdntk.53zaixian.com/tq/1701760592671870977/img/compression/M_42faacc072c7519fa4036c02eee9946c.jpg)$
图2
$
$(3)一束光从玻璃窗面C_{1}CD上点C_{1}射入恰经过点A(假设此时光经过玻璃为直射)，求这束光在玻璃窗C_{1}CD上的入射角的正切值.$

Solution:
$由(1)知，\overrightarrow{AC_1}=(2,-2,1),平面C_{1}CD的一个法向量为n_{1}=(1，-1，2)，设入射角为\theta ,则\\cos   \theta =|\\cos  \langle n_{1},\overrightarrow{AC_1}\rangle |=\frac{|\boldsymbol{n}_1\cdot \overrightarrow{AC_1}|}{|\boldsymbol{n}_1||\overrightarrow{AC_1}|}=\frac{\sqrt{6}}{3}，则\\sin   \theta =\frac{\sqrt{3}}{3}，\therefore \\tan   \theta =\frac{\sqrt{2}}{2}.$

Final answer: $\frac{\sqrt{2}}{2}$
Subfield: Solid Geometry


Question: $如图，在四棱柱ABCD-A_{1}B_{1}C_{1}D_{1}中，已知侧棱AA_{1}\perp 底面ABCD，侧面ABB_{1}A_{1}是正方形，AB_{1}与A_{1}B交于点O，AB\perp BC，AB\parallel CD，AB=2，BC=CD=1.$![Image](https://cdntk.53zaixian.com/tq/1701760594039214083/img/compression/M_d0eb5821223e4ab862b67c2c21d16de9.jpg)
$(1)求证：AD\parallel 平面COC_{1}；$

Solution:
$证明：分别取线段AB，CC_{1}的中点E，F，连接CE，OE，OF，则AE=CD,OE=\frac{1}{2}AA_{1}=CF，又AE\parallel CD,OE\parallel AA_{1}\parallel CF，所以四边形AECD和四边形OECF均为平行四边形，所以AD\parallel CE\parallel OF，又AD\not\subset 平面COC_{1}，OF\subset 平面COC_{1}，所以AD\parallel 平面COC_{1}.$

Final answer: 证明题
Subfield: Solid Geometry


Question: $如图，在四棱柱ABCD-A_{1}B_{1}C_{1}D_{1}中，已知侧棱AA_{1}\perp 底面ABCD，侧面ABB_{1}A_{1}是正方形，AB_{1}与A_{1}B交于点O，AB\perp BC，AB\parallel CD，AB=2，BC=CD=1.$<img_2898>
$(1)求证：AD\parallel 平面COC_{1}；$

Solution:
$证明：分别取线段AB，CC_{1}的中点E，F，连接CE，OE，OF，则AE=CD,OE=\frac{1}{2}AA_{1}=CF，又AE\parallel CD,OE\parallel AA_{1}\parallel CF，所以四边形AECD和四边形OECF均为平行四边形，所以AD\parallel CE\parallel OF，又AD\not\subset 平面COC_{1}，OF\subset 平面COC_{1}，所以AD\parallel 平面COC_{1}.$
Subfield: Solid Geometry


Question: $如图，在四棱柱ABCD-A_{1}B_{1}C_{1}D_{1}中，已知侧棱AA_{1}\perp 底面ABCD，侧面ABB_{1}A_{1}是正方形，AB_{1}与A_{1}B交于点O，AB\perp BC，AB\parallel CD，AB=2，BC=CD=1.$![Image](https://cdntk.53zaixian.com/tq/1701760594039214083/img/compression/M_d0eb5821223e4ab862b67c2c21d16de9.jpg)
$(2)求直线OC_{1}与平面AB_{1}C所成角的正弦值；$

Solution:
$依题意，以B为原点，分别以\overrightarrow{BA}，\overrightarrow{BB_1}，\overrightarrow{BC}的方向为x轴，y轴，z轴的正方向建立空间直角坐标系(如图)，$![Image](https://cdntk.53zaixian.com/tq/1701760594374758402/img/compression/M_bd9274e9558adfabcdd434d61aa827dc.jpg)$则A(2，0，0)，C(0，0，1)，B_{1}(0，2，0)，C_{1}(0，2，1)，O(1，1，0)，所以\overrightarrow{OC_1}=(-1，1，1)，\overrightarrow{AC}=(-2，0，1)，\overrightarrow{AB_1}=(-2，2，0)，设平面AB_{1}C的法向量为m=(x_{1},y_{1},z_{1}),则\left\{\begin{matrix}\boldsymbol{m}\cdot \overrightarrow{AC}=-2x_1+z_1=0,\\ \boldsymbol{m}\cdot \overrightarrow{AB_1}=-2x_1+2y_1=0,\end{matrix}\right.不妨令x_{1}=1,所以m=(1,1,2),于是\\cos  \langle m,\overrightarrow{OC_1}\rangle =\frac{\boldsymbol{m}\cdot \overrightarrow{OC_1}}{|\boldsymbol{m}||\overrightarrow{OC_1}|}=\frac{2}{\sqrt{6}\times \sqrt{3}}=\frac{\sqrt{2}}{3},所以直线OC_{1}与平面AB_{1}C所成角的正弦值为\frac{\sqrt{2}}{3}.$

Final answer: $\frac{\sqrt{2}}{3}$
Subfield: Solid Geometry


Question: $如图，在四棱柱ABCD-A_{1}B_{1}C_{1}D_{1}中，已知侧棱AA_{1}\perp 底面ABCD，侧面ABB_{1}A_{1}是正方形，AB_{1}与A_{1}B交于点O，AB\perp BC，AB\parallel CD，AB=2，BC=CD=1.$<img_2898>
$(2)求直线OC_{1}与平面AB_{1}C所成角的正弦值；$

Solution:
$依题意，以B为原点，分别以\overrightarrow{BA}，\overrightarrow{BB_1}，\overrightarrow{BC}的方向为x轴，y轴，z轴的正方向建立空间直角坐标系(如图)，$<img_2111>$则A(2，0，0)，C(0，0，1)，B_{1}(0，2，0)，C_{1}(0，2，1)，O(1，1，0)，所以\overrightarrow{OC_1}=(-1，1，1)，\overrightarrow{AC}=(-2，0，1)，\overrightarrow{AB_1}=(-2，2，0)，设平面AB_{1}C的法向量为m=(x_{1},y_{1},z_{1}),则\left\{\begin{matrix}\boldsymbol{m}\cdot \overrightarrow{AC}=-2x_1+z_1=0,\\ \boldsymbol{m}\cdot \overrightarrow{AB_1}=-2x_1+2y_1=0,\end{matrix}\right.不妨令x_{1}=1,所以m=(1,1,2),于是\\cos  \langle m,\overrightarrow{OC_1}\rangle =\frac{\boldsymbol{m}\cdot \overrightarrow{OC_1}}{|\boldsymbol{m}||\overrightarrow{OC_1}|}=\frac{2}{\sqrt{6}\times \sqrt{3}}=\frac{\sqrt{2}}{3},所以直线OC_{1}与平面AB_{1}C所成角的正弦值为\frac{\sqrt{2}}{3}.$

Final answer: $\frac{\sqrt{2}}{3}$
Subfield: Solid Geometry


Question: $如图，在四棱柱ABCD-A_{1}B_{1}C_{1}D_{1}中，已知侧棱AA_{1}\perp 底面ABCD，侧面ABB_{1}A_{1}是正方形，AB_{1}与A_{1}B交于点O，AB\perp BC，AB\parallel CD，AB=2，BC=CD=1.$![Image](https://cdntk.53zaixian.com/tq/1701760594039214083/img/compression/M_d0eb5821223e4ab862b67c2c21d16de9.jpg)
$(3)若点P在线段A_{1}D_{1}上，且A_{1}P=\frac{2}{3}A_{1}D_{1}，求二面角C-AB_{1}-P的正弦值.$

Solution:
$由(1)知A_{1}(2，2，0)，D_{1}(1，2，1)，则\overrightarrow{A_1D_1}=(-1，0，1)，设P(x_{P},y_{P},z_{P}),则\overrightarrow{A_1P}=(x_{P}-2,y_{P}-2,z_{P}),由\overrightarrow{A_1P}=\frac{2}{3}\overrightarrow{A_1D_1},得x_{P}-2=-\frac{2}{3},y_{P}-2=0,z_{P}=\frac{2}{3},从而P\left(\frac{4}{3},2,\frac{2}{3}\right)，所以\overrightarrow{AP}=\left(-\frac{2}{3},2,\frac{2}{3}\right).设平面AB_{1}P的法向量为n=(x_{2},y_{2},z_{2}),则\left\{\begin{matrix}\boldsymbol{n}\cdot \overrightarrow{AP}=-\frac{2}{3}x_2+2y_2+\frac{2}{3}z_2=0,\\ \boldsymbol{n}\cdot \overrightarrow{AB_1}=-2x_2+2y_2=0,\end{matrix}\right.不妨令x_{2}=1,所以n=(1,1,-2),因此有\\cos  \langle m,n\rangle =\frac{\boldsymbol{m}\cdot \boldsymbol{n}}{|\boldsymbol{m}||\boldsymbol{n}|}=\frac{-2}{\sqrt{6}\times \sqrt{6}}=-\frac{1}{3}.于是\\sin  \langle m,n\rangle =\sqrt{1-\\\cos   ^2\langle \boldsymbol{m},\boldsymbol{n}\rangle }=\frac{2\sqrt{2}}{3},所以二面角C-AB_{1}-P的正弦值为\frac{2\sqrt{2}}{3}.$

Final answer: $\frac{2\sqrt{2}}{3}$
Subfield: Solid Geometry


Question: $如图，在四棱柱ABCD-A_{1}B_{1}C_{1}D_{1}中，已知侧棱AA_{1}\perp 底面ABCD，侧面ABB_{1}A_{1}是正方形，AB_{1}与A_{1}B交于点O，AB\perp BC，AB\parallel CD，AB=2，BC=CD=1.$<img_2898>
$(3)若点P在线段A_{1}D_{1}上，且A_{1}P=\frac{2}{3}A_{1}D_{1}，求二面角C-AB_{1}-P的正弦值.$

Solution:
$由(1)知A_{1}(2，2，0)，D_{1}(1，2，1)，则\overrightarrow{A_1D_1}=(-1，0，1)，设P(x_{P},y_{P},z_{P}),则\overrightarrow{A_1P}=(x_{P}-2,y_{P}-2,z_{P}),由\overrightarrow{A_1P}=\frac{2}{3}\overrightarrow{A_1D_1},得x_{P}-2=-\frac{2}{3},y_{P}-2=0,z_{P}=\frac{2}{3},从而P\left(\frac{4}{3},2,\frac{2}{3}\right)，所以\overrightarrow{AP}=\left(-\frac{2}{3},2,\frac{2}{3}\right).设平面AB_{1}P的法向量为n=(x_{2},y_{2},z_{2}),则\left\{\begin{matrix}\boldsymbol{n}\cdot \overrightarrow{AP}=-\frac{2}{3}x_2+2y_2+\frac{2}{3}z_2=0,\\ \boldsymbol{n}\cdot \overrightarrow{AB_1}=-2x_2+2y_2=0,\end{matrix}\right.不妨令x_{2}=1,所以n=(1,1,-2),因此有\\cos  \langle m,n\rangle =\frac{\boldsymbol{m}\cdot \boldsymbol{n}}{|\boldsymbol{m}||\boldsymbol{n}|}=\frac{-2}{\sqrt{6}\times \sqrt{6}}=-\frac{1}{3}.于是\\sin  \langle m,n\rangle =\sqrt{1-\\\cos   ^2\langle \boldsymbol{m},\boldsymbol{n}\rangle }=\frac{2\sqrt{2}}{3},所以二面角C-AB_{1}-P的正弦值为\frac{2\sqrt{2}}{3}.$

Final answer: $\frac{2\sqrt{2}}{3}$
Subfield: Solid Geometry


Question: $如图,在四棱柱ABCD-A_{1}B_{1}C_{1}D_{1}中,底面ABCD是正方形,平面A_{1}ADD_{1}\perp 平面ABCD,AD=AA_{1}=A_{1}D=2.$![Image](https://cdntk.53zaixian.com/tq/1701760609782321153/img/compression/M_bafb81f07cd6d372ad69314fd0614640.jpg)
$(1)求证：A_{1}D\perp AB；$

Solution:
$证明:因为四边形ABCD为正方形，所以AB\perp AD,又平面A_{1}ADD_{1}\perp 平面ABCD，平面A_{1}ADD_{1}\cap 平面ABCD=AD，AB\subset 平面ABCD,所以AB\perp 平面AA_{1}D_{1}D,因为A_{1}D\subset 平面AA_{1}D_{1}D,所以AB\perp A_{1}D.(3分)$

Final answer: 证明题
Subfield: Solid Geometry


Question: $如图,在四棱柱ABCD-A_{1}B_{1}C_{1}D_{1}中,底面ABCD是正方形,平面A_{1}ADD_{1}\perp 平面ABCD,AD=AA_{1}=A_{1}D=2.$<img_2998>
$(1)求证：A_{1}D\perp AB；$

Solution:
$证明:因为四边形ABCD为正方形，所以AB\perp AD,又平面A_{1}ADD_{1}\perp 平面ABCD，平面A_{1}ADD_{1}\cap 平面ABCD=AD，AB\subset 平面ABCD,所以AB\perp 平面AA_{1}D_{1}D,因为A_{1}D\subset 平面AA_{1}D_{1}D,所以AB\perp A_{1}D.(3分)$
Subfield: Solid Geometry


Question: $如图,在四棱柱ABCD-A_{1}B_{1}C_{1}D_{1}中,底面ABCD是正方形,平面A_{1}ADD_{1}\perp 平面ABCD,AD=AA_{1}=A_{1}D=2.$![Image](https://cdntk.53zaixian.com/tq/1701760609782321153/img/compression/M_bafb81f07cd6d372ad69314fd0614640.jpg)
$(2)求直线AB与平面A_{1}DC_{1}所成角的余弦值；$

Solution:
$取AD的中点O,连接A_{1}O,因为AA_{1}=A_{1}D，O为AD的中点，所以A_{1}O\perp AD，又平面AA_{1}D_{1}D\perp 平面ABCD,平面AA_{1}D_{1}D\cap 平面ABCD=AD，A_{1}O\subset 平面AA_{1}D_{1}D, 所以A_{1}O\perp 平面ABCD,以点O为坐标原点，\overrightarrow{OD},\overrightarrow{OA_1}的方向分别为y轴，z轴的正方向，建立如图所示的空间直角坐标系.(5分)$![Image](https://cdntk.53zaixian.com/tq/1701760610180780034/img/compression/M_1f459bb6d7dd1dafde5287ec649fac22.jpg)$则A(0,-1,0)，B(2,-1,0)， A_{1}(0,0,\sqrt{3}),C_{1}(2,2,\sqrt{3})，D(0,1,0),C(2，1，0)，所以\overrightarrow{AB}=(2，0,0)，\overrightarrow{A_1C_1}=(2,2,0)，\overrightarrow{A_1D}=(0,1,-\sqrt{3})，设平面A_{1}C_{1}D的法向量为n_{1}=(x_{1},y_{1},z_{1}),由\left\{\begin{matrix}\boldsymbol{n}_1\cdot \overrightarrow{A_1C_1}=2x_1+2y_1=0,\\ \boldsymbol{n}_1\cdot \overrightarrow{A_1D}=y_1-\sqrt{3}z_1=0,\end{matrix}\right.令z_{1}=1,则n_{1}=(-\sqrt{3}，\sqrt{3}，1)，(7分)设直线AB与平面A_{1}DC_{1}所成角为\theta _{1}，则\\sin   \theta _{1}=|\\cos  \langle \overrightarrow{AB},n_{1}\rangle |=\frac{|\overrightarrow{AB}\cdot \boldsymbol{n}_1|}{|\overrightarrow{AB}||\boldsymbol{n}_1|}=\frac{\sqrt{21}}{7}，所以\\cos   \theta _{1}=\sqrt{1-\\\sin   ^2\theta _1}=\frac{2\sqrt{7}}{7},则直线AB与平面A_{1}DC_{1}所成角的余弦值为\frac{2\sqrt{7}}{7}.(10分)$

Final answer: $\frac{2\sqrt{7}}{7}$