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


Question: $如图，四棱锥P－ABCD中，底面ABCD是梯形，AD\parallel BC，AD\perp 平面PAB，\triangle PAB是等腰三角形，PA=PB，AB=BC=2AD=2，E是AB的中点.$<img_2687>
$(2)设PA与CD所成的角为\theta _{1}，直线PD与平面ABCD所成的角为\theta _{2}，二面角P－BC－A为\theta _{3}，从以下所给的三个条件中选出其中一个作为已知条件，求四棱锥P－ABCD的体积.①\\cos   \theta _{1}=\frac{2}{5}；②\\sin   \theta _{2}=\frac{\sqrt{6}}{3}；③\\cos   \theta _{3}=\frac{\sqrt{5}}{5}.$

Solution:
$选择条件①.如图1，取BC的中点F，连接AF，PF，
$<img_2052>$
图1
因为AD\parallel CF，AD=CF，所以四边形ADCF为平行四边形，所以AF\parallel CD，AF=CD.所以\angle PAF就是PA与CD所成的角，\angle PAF=\theta _{1}.易得AF=\sqrt{5}，设PE=x(x>0)，则PA=PB=\sqrt{x^2+1}，PF=\sqrt{x^2+2}.在\triangle PAF中，由余弦定理得PF^{2}=PA^{2}+AF^{2}－2PA\cdot AF\cdot \\cos   \theta _{1}，所以x^{2}+2=x^{2}+1+5－2\sqrt{x^2+1}\times \sqrt{5}\times \frac{2}{5}，解得x=2(舍负).所以四棱锥P－ABCD的体积为\frac{1}{3}\cdot S_{梯形ABCD}\cdot PE=\frac{1}{3}\times \frac{1}{2}\cdot (AD+BC)\cdot AB\cdot PE=\frac{1}{3}\times \frac{1}{2}\times (1+2)\times 2\times 2=2.选择条件②.如图2，连接DE.
$<img_1936>$
图2
因为PE\perp 平面ABCD，所以ED是PD在平面ABCD内的射影.所以\angle PDE就是PD与平面ABCD所成的角，\angle PDE=\theta _{2}，且\theta _{2}\in (0^\circ ，90^\circ )，因为AE=AD=1，所以DE=\sqrt{2}.因为\\sin   \theta _{2}=\frac{\sqrt{6}}{3}，所以\\cos   \theta _{2}=\frac{\sqrt{3}}{3}，则\\tan   \theta _{2}=\sqrt{2}.所以\frac{PE}{DE}=\sqrt{2}，故PE=2.所以四棱锥P－ABCD的体积为\frac{1}{3}\cdot S_{梯形ABCD}\cdot PE=\frac{1}{3}\times \frac{1}{2}\cdot (AD+BC)\cdot AB\cdot PE=\frac{1}{3}\times \frac{1}{2}\times (1+2)\times 2\times 2=2.选择条件③.因为AD\parallel BC，AD\perp 平面PAB，所以BC\perp 平面PAB，又AB,PB\subset 平面PAB,所以BC\perp AB，BC\perp PB.所以\angle PBA就是二面角P－BC－A的平面角，\angle PBA=\theta _{3}.因为\\cos   \theta _{3}=\frac{BE}{PB}=\frac{1}{PB}=\frac{\sqrt{5}}{5}，所以PB=\sqrt{5}，所以PE=2.所以四棱锥P－ABCD的体积为\frac{1}{3}\cdot S_{梯形ABCD}\cdot PE=\frac{1}{3}\times \frac{1}{2}\cdot (AD+BC)\cdot AB\cdot PE=\frac{1}{3}\times \frac{1}{2}\times (1+2)\times 2\times 2=2.$

Final answer: 2
Subfield: Solid Geometry


Question: $如图，四棱锥P－ABCD中，底面ABCD是梯形，AD\parallel BC，AD\perp 平面PAB，\triangle PAB是等腰三角形，PA=PB，AB=BC=2AD=2，E是AB的中点.$![Image](https://cdntk.53zaixian.com/tq/1716268087906070530/img/compression/M_2ecfa69fcaa04b2c4bc957fe6af5cc0f.jpg)
$(2)设PA与CD所成的角为\theta _{1}，直线PD与平面ABCD所成的角为\theta _{2}，二面角P－BC－A为\theta _{3}，从以下所给的三个条件中选出其中一个作为已知条件，求四棱锥P－ABCD的体积.①\\cos   \theta _{1}=\frac{2}{5}；②\\sin   \theta _{2}=\frac{\sqrt{6}}{3}；③\\cos   \theta _{3}=\frac{\sqrt{5}}{5}.$

Solution:
$选择条件①.如图1，取BC的中点F，连接AF，PF，
$![Image](https://cdntk.53zaixian.com/tq/1716268088229031937/img/compression/M_939629676dced3f56b60013c15888e5e.jpg)$
图1
因为AD\parallel CF，AD=CF，所以四边形ADCF为平行四边形，所以AF\parallel CD，AF=CD.所以\angle PAF就是PA与CD所成的角，\angle PAF=\theta _{1}.易得AF=\sqrt{5}，设PE=x(x>0)，则PA=PB=\sqrt{x^2+1}，PF=\sqrt{x^2+2}.在\triangle PAF中，由余弦定理得PF^{2}=PA^{2}+AF^{2}－2PA\cdot AF\cdot \\cos   \theta _{1}，所以x^{2}+2=x^{2}+1+5－2\sqrt{x^2+1}\times \sqrt{5}\times \frac{2}{5}，解得x=2(舍负).所以四棱锥P－ABCD的体积为\frac{1}{3}\cdot S_{梯形ABCD}\cdot PE=\frac{1}{3}\times \frac{1}{2}\cdot (AD+BC)\cdot AB\cdot PE=\frac{1}{3}\times \frac{1}{2}\times (1+2)\times 2\times 2=2.选择条件②.如图2，连接DE.
$![Image](https://cdntk.53zaixian.com/tq/1716268088229031937/img/compression/M_a6cfde096d79dac1937c127bb3619ddd.jpg)$
图2
因为PE\perp 平面ABCD，所以ED是PD在平面ABCD内的射影.所以\angle PDE就是PD与平面ABCD所成的角，\angle PDE=\theta _{2}，且\theta _{2}\in (0^\circ ，90^\circ )，因为AE=AD=1，所以DE=\sqrt{2}.因为\\sin   \theta _{2}=\frac{\sqrt{6}}{3}，所以\\cos   \theta _{2}=\frac{\sqrt{3}}{3}，则\\tan   \theta _{2}=\sqrt{2}.所以\frac{PE}{DE}=\sqrt{2}，故PE=2.所以四棱锥P－ABCD的体积为\frac{1}{3}\cdot S_{梯形ABCD}\cdot PE=\frac{1}{3}\times \frac{1}{2}\cdot (AD+BC)\cdot AB\cdot PE=\frac{1}{3}\times \frac{1}{2}\times (1+2)\times 2\times 2=2.选择条件③.因为AD\parallel BC，AD\perp 平面PAB，所以BC\perp 平面PAB，又AB,PB\subset 平面PAB,所以BC\perp AB，BC\perp PB.所以\angle PBA就是二面角P－BC－A的平面角，\angle PBA=\theta _{3}.因为\\cos   \theta _{3}=\frac{BE}{PB}=\frac{1}{PB}=\frac{\sqrt{5}}{5}，所以PB=\sqrt{5}，所以PE=2.所以四棱锥P－ABCD的体积为\frac{1}{3}\cdot S_{梯形ABCD}\cdot PE=\frac{1}{3}\times \frac{1}{2}\cdot (AD+BC)\cdot AB\cdot PE=\frac{1}{3}\times \frac{1}{2}\times (1+2)\times 2\times 2=2.$

Final answer: 2
Subfield: Solid Geometry


Question: $如图，在三棱柱ABC-A_{1}B_{1}C_{1}中，\triangle ABC为等边三角形，四边形BCC_{1}B_{1}是边长为2的正方形，AC_{1}=2\sqrt{2}，D_{1}为B_{1}C_{1}的中点，D为棱BC上一点，BD_{1}\parallel 平面ADC_{1}.$<img_2966>
$(1)求证：D为BC中点；$

Solution:
$证明：\because BD_{1}\parallel 平面ADC_{1}，BD_{1}\subset 平面BB_{1}C_{1}C，平面ADC_{1}\cap 平面BB_{1}C_{1}C=DC_{1}，\therefore BD_{1}\parallel DC_{1}，又BD\parallel D_{1}C_{1}，\therefore 四边形BD_{1}C_{1}D为平行四边形，又D_{1}为B_{1}C_{1}的中点，\therefore BD=D_{1}C_{1}=\frac{1}{2}B_{1}C_{1}=\frac{1}{2}BC，\therefore D为BC的中点.$
Subfield: Solid Geometry


Question: $如图，在三棱柱ABC-A_{1}B_{1}C_{1}中，\triangle ABC为等边三角形，四边形BCC_{1}B_{1}是边长为2的正方形，AC_{1}=2\sqrt{2}，D_{1}为B_{1}C_{1}的中点，D为棱BC上一点，BD_{1}\parallel 平面ADC_{1}.$![Image](https://cdntk.53zaixian.com/tq/1716268088812040194/img/compression/M_2cb30fa4161c4f58d3b075da7f665e11.jpg)
$(1)求证：D为BC中点；$

Solution:
$证明：\because BD_{1}\parallel 平面ADC_{1}，BD_{1}\subset 平面BB_{1}C_{1}C，平面ADC_{1}\cap 平面BB_{1}C_{1}C=DC_{1}，\therefore BD_{1}\parallel DC_{1}，又BD\parallel D_{1}C_{1}，\therefore 四边形BD_{1}C_{1}D为平行四边形，又D_{1}为B_{1}C_{1}的中点，\therefore BD=D_{1}C_{1}=\frac{1}{2}B_{1}C_{1}=\frac{1}{2}BC，\therefore D为BC的中点.$

Final answer: 证明题
Subfield: Solid Geometry


Question: $如图，在三棱柱ABC-A_{1}B_{1}C_{1}中，\triangle ABC为等边三角形，四边形BCC_{1}B_{1}是边长为2的正方形，AC_{1}=2\sqrt{2}，D_{1}为B_{1}C_{1}的中点，D为棱BC上一点，BD_{1}\parallel 平面ADC_{1}.$![Image](https://cdntk.53zaixian.com/tq/1716268088812040194/img/compression/M_2cb30fa4161c4f58d3b075da7f665e11.jpg)
$(2)求直线BC与平面ADC_{1}所成角的正弦值.$

Solution:
$连接DD_{1},可知DD_{1}\perp BC,由(1)得BD=1,又\triangle ABC为等边三角形，\therefore AD=\sqrt{3}，\because BD_{1}=\sqrt{BB^2_1+B_1D^2_1}=\sqrt{5},\therefore DC_{1}=BD_{1}=\sqrt{5}，又AC_{1}=2\sqrt{2}，\therefore AD^{2}+DC^2_1=AC^2_1，\therefore AD\perp DC_{1}，又AD\perp BC，BC\cap DC_{1}=D，BC，DC_{1}\subset 平面BB_{1}C_{1}C，\therefore AD\perp 平面BB_{1}C_{1}C，又DD_{1}\subset 平面BB_{1}C_{1}C,\therefore AD\perp DD_{1}.如图建立以D为坐标原点，DB所在直线为x轴，DD_{1}所在直线为y轴，DA所在直线为z轴的空间直角坐标系，$![Image](https://cdntk.53zaixian.com/tq/1716268089072087041/img/compression/M_6d1771f1f6e537686dcdb45f97b2bdd2.jpg)$A(0，0，\sqrt{3})，B(1，0，0)，C(－1，0，0)，D(0，0，0)，C_{1}(－1，2，0)，\therefore \overrightarrow{CB}=(2，0，0)，\overrightarrow{DA}=(0，0，\sqrt{3})，\overrightarrow{DC_1}=(－1，2，0)，设平面ADC_{1}的法向量为m=(x，y，z)，则\left\{\begin{matrix}\boldsymbol{m}\cdot \overrightarrow{DA}=\sqrt{3}z=0,\\ \boldsymbol{m}\cdot \overrightarrow{DC_1}=-x+2y=0,\end{matrix}\right.令y=1，得x=2，\therefore m=(2，1，0)，设直线BC与平面ADC_{1}所成角为\alpha ，\\sin   \alpha =|\\cos  \langle m，\overrightarrow{CB}\rangle |=\frac{|\boldsymbol{m}\cdot \overrightarrow{CB}|}{|\boldsymbol{m}||\overrightarrow{CB}|}=\frac{4}{\sqrt{5}\times 2}=\frac{2\sqrt{5}}{5}，故直线BC与平面ADC_{1}所成角的正弦值为\frac{2\sqrt{5}}{5}.$

Final answer: $\frac{2\sqrt{5}}{5}$
Subfield: Solid Geometry


Question: $如图，在三棱柱ABC-A_{1}B_{1}C_{1}中，\triangle ABC为等边三角形，四边形BCC_{1}B_{1}是边长为2的正方形，AC_{1}=2\sqrt{2}，D_{1}为B_{1}C_{1}的中点，D为棱BC上一点，BD_{1}\parallel 平面ADC_{1}.$<img_2966>
$(2)求直线BC与平面ADC_{1}所成角的正弦值.$

Solution:
$连接DD_{1},可知DD_{1}\perp BC,由(1)得BD=1,又\triangle ABC为等边三角形，\therefore AD=\sqrt{3}，\because BD_{1}=\sqrt{BB^2_1+B_1D^2_1}=\sqrt{5},\therefore DC_{1}=BD_{1}=\sqrt{5}，又AC_{1}=2\sqrt{2}，\therefore AD^{2}+DC^2_1=AC^2_1，\therefore AD\perp DC_{1}，又AD\perp BC，BC\cap DC_{1}=D，BC，DC_{1}\subset 平面BB_{1}C_{1}C，\therefore AD\perp 平面BB_{1}C_{1}C，又DD_{1}\subset 平面BB_{1}C_{1}C,\therefore AD\perp DD_{1}.如图建立以D为坐标原点，DB所在直线为x轴，DD_{1}所在直线为y轴，DA所在直线为z轴的空间直角坐标系，$<img_1546>$A(0，0，\sqrt{3})，B(1，0，0)，C(－1，0，0)，D(0，0，0)，C_{1}(－1，2，0)，\therefore \overrightarrow{CB}=(2，0，0)，\overrightarrow{DA}=(0，0，\sqrt{3})，\overrightarrow{DC_1}=(－1，2，0)，设平面ADC_{1}的法向量为m=(x，y，z)，则\left\{\begin{matrix}\boldsymbol{m}\cdot \overrightarrow{DA}=\sqrt{3}z=0,\\ \boldsymbol{m}\cdot \overrightarrow{DC_1}=-x+2y=0,\end{matrix}\right.令y=1，得x=2，\therefore m=(2，1，0)，设直线BC与平面ADC_{1}所成角为\alpha ，\\sin   \alpha =|\\cos  \langle m，\overrightarrow{CB}\rangle |=\frac{|\boldsymbol{m}\cdot \overrightarrow{CB}|}{|\boldsymbol{m}||\overrightarrow{CB}|}=\frac{4}{\sqrt{5}\times 2}=\frac{2\sqrt{5}}{5}，故直线BC与平面ADC_{1}所成角的正弦值为\frac{2\sqrt{5}}{5}.$

Final answer: $\frac{2\sqrt{5}}{5}$
Subfield: Solid Geometry


Question: $如图，已知直三棱柱ABC-A_{1}B_{1}C_{1}中，AB=AC=2，D为BC中点，AA_{1}=2，再从条件①，条件②这两个条件中选择一个作为已知，完成以下问题：$<img_2447>
$(1)证明：AB_{1}\perp BC_{1}；$

Solution:
$选择条件②
证明：在\triangle ABC中，AB=AC=2，BC=2\sqrt{2}，\therefore AB^{2}+AC^{2}=BC^{2}.\therefore \angle BAC=90^\circ ，即AB\perp AC.在直三棱柱ABC-A_{1}B_{1}C_{1}中，侧棱AA_{1}\perp 底面ABC，AB,AC\subset 平面ABC,\therefore AA_{1}\perp AB，AA_{1}\perp AC.以点A为原点，分别以AC，AB，AA_{1}所在直线为x轴，y轴，z轴建立空间直角坐标系，如图1，
$<img_2713>$
图1
则A(0,0,0),B(0，2，0)，C_{1}(2，0，2)，B_{1}(0，2，2).\therefore \overrightarrow{AB_1}=(0，2，2)，\overrightarrow{BC_1}=(2，-2，2).\therefore \overrightarrow{AB_1}\cdot \overrightarrow{BC_1}=0\times 2+2\times (-2)+2\times 2=0.\therefore AB_{1}\perp BC_{1}.
选择条件①
证明：连接AD.在直三棱柱ABC-A_{1}B_{1}C_{1}中，CC_{1}\perp 底面ABC，AD\subset 平面ABC，\therefore CC_{1}\perp AD.在\triangle ABC中，AB=AC，D为BC中点，\therefore AD\perp BC.又CC_{1}\cap BC=C，CC_{1},BC\subset 平面BB_{1}C_{1}C,\therefore AD\perp 平面BB_{1}C_{1}C.又BC_{1}\subset 平面BB_{1}C_{1}C，\therefore AD\perp BC_{1}.又B_{1}D\perp BC_{1}，AD\cap B_{1}D=D，AD,B_{1}D\subset 平面AB_{1}D,\therefore BC_{1}\perp 平面AB_{1}D.\because AB_{1}\subset 平面AB_{1}D，\therefore AB_{1}\perp BC_{1}.$
Subfield: Solid Geometry


Question: $如图，已知直三棱柱ABC-A_{1}B_{1}C_{1}中，AB=AC=2，D为BC中点，AA_{1}=2，再从条件①，条件②这两个条件中选择一个作为已知，完成以下问题：$![Image](https://cdntk.53zaixian.com/tq/1716268090556870659/img/compression/M_612990b9a585408063fd48a80490b7e6.jpg)
$(1)证明：AB_{1}\perp BC_{1}；$

Solution:
$选择条件②
证明：在\triangle ABC中，AB=AC=2，BC=2\sqrt{2}，\therefore AB^{2}+AC^{2}=BC^{2}.\therefore \angle BAC=90^\circ ，即AB\perp AC.在直三棱柱ABC-A_{1}B_{1}C_{1}中，侧棱AA_{1}\perp 底面ABC，AB,AC\subset 平面ABC,\therefore AA_{1}\perp AB，AA_{1}\perp AC.以点A为原点，分别以AC，AB，AA_{1}所在直线为x轴，y轴，z轴建立空间直角坐标系，如图1，
$![Image](https://cdntk.53zaixian.com/tq/1716268090556870659/img/compression/M_211532afd4e9faa3f2199901bd65be1e.jpg)$
图1
则A(0,0,0),B(0，2，0)，C_{1}(2，0，2)，B_{1}(0，2，2).\therefore \overrightarrow{AB_1}=(0，2，2)，\overrightarrow{BC_1}=(2，-2，2).\therefore \overrightarrow{AB_1}\cdot \overrightarrow{BC_1}=0\times 2+2\times (-2)+2\times 2=0.\therefore AB_{1}\perp BC_{1}.
选择条件①
证明：连接AD.在直三棱柱ABC-A_{1}B_{1}C_{1}中，CC_{1}\perp 底面ABC，AD\subset 平面ABC，\therefore CC_{1}\perp AD.在\triangle ABC中，AB=AC，D为BC中点，\therefore AD\perp BC.又CC_{1}\cap BC=C，CC_{1},BC\subset 平面BB_{1}C_{1}C,\therefore AD\perp 平面BB_{1}C_{1}C.又BC_{1}\subset 平面BB_{1}C_{1}C，\therefore AD\perp BC_{1}.又B_{1}D\perp BC_{1}，AD\cap B_{1}D=D，AD,B_{1}D\subset 平面AB_{1}D,\therefore BC_{1}\perp 平面AB_{1}D.\because AB_{1}\subset 平面AB_{1}D，\therefore AB_{1}\perp BC_{1}.$

Final answer: 证明题
Subfield: Solid Geometry


Question: $如图，已知直三棱柱ABC-A_{1}B_{1}C_{1}中，AB=AC=2，D为BC中点，AA_{1}=2，再从条件①，条件②这两个条件中选择一个作为已知，完成以下问题：$![Image](https://cdntk.53zaixian.com/tq/1716268090556870659/img/compression/M_612990b9a585408063fd48a80490b7e6.jpg)
$(2)求直线BC_{1}与平面A_{1}B_{1}D所成角的正弦值.条件①：B_{1}D\perp BC_{1}；条件②：BC=2\sqrt{2}.注：如果选择条件①和条件②分别解答，按第一个解答计分.$

Solution:
$选择条件②
B(0，2，0)，D(1，1，0)，A_{1}(0，0，2)，B_{1}(0，2，2)，C_{1}(2，0，2)，\therefore \overrightarrow{A_1B_1}=(0，2，0)，\overrightarrow{A_1D}=(1，1，-2)，\overrightarrow{BC_1}=(2，-2，2).设平面A_{1}B_{1}D的法向量为m=(x，y，z)，则\left\{\begin{matrix}\boldsymbol{m}\cdot \overrightarrow{A_1B_1}=2y=0,\\ \boldsymbol{m}\cdot \overrightarrow{A_1D}=x+y-2z=0,\end{matrix}\right. 令z=1，得x=2,则m=(2，0，1).设直线BC_{1}与平面A_{1}B_{1}D所成角为\alpha ，\\sin   \alpha =|\\cos  \langle \overrightarrow{BC_1},m\rangle |=\frac{|\overrightarrow{BC_1}\cdot \boldsymbol{m}|}{|\boldsymbol{m}|\cdot |\overrightarrow{BC_1}|}=\frac{6}{\sqrt{5}\times 2\sqrt{3}}=\frac{\sqrt{15}}{5}.\therefore 直线BC_{1}与平面A_{1}B_{1}D所成角的正弦值为\frac{\sqrt{15}}{5}.
选择条件①
设B_{1}D\cap BC_{1}=O,在直三棱柱ABC-A_{1}B_{1}C_{1}中可得BB_{1}\perp B_{1}C_{1},在Rt\triangle BB_{1}C_{1}中，由B_{1}D\perp BC_{1}，得B_{1}B^{2}=BO\cdot BC_{1},\because BC\parallel B_{1}C_{1},D为BC中点，\therefore \frac{BO}{OC_1}=\frac{BD}{B_1C_1}=\frac{1}{2},\therefore BO=\frac{1}{3}BC_{1}，\therefore B_{1}B^{2}=\frac{1}{3}BC^2_1,又B_{1}B=AA_{1}=2,\therefore BC_{1}=2\sqrt{3}(舍负)，\therefore BC=B_{1}C_{1}=\sqrt{BC^2_1-BB^2_1}=2\sqrt{2}.又在\triangle ABC中，AB=AC=2,\therefore AB^{2}+AC^{2}=BC^{2}.\therefore \angle BAC=90^\circ ，即AB\perp AC.在直三棱柱ABC-A_{1}B_{1}C_{1}中，侧棱AA_{1}\perp 底面ABC,AB,AC\subset 平面ABC,\therefore AA_{1}\perp AB,AA_{1}\perp AC.以点A为原点，分别以AC,AB,AA_{1}所在直线为x轴，y轴，z轴建立空间直角坐标系，如图2，
$![Image](https://cdntk.53zaixian.com/tq/1716268090556870659/img/compression/M_d3721534d853e7ea4e8443bae69e36eb.jpg)$
图2
则B(0,2,0),D(1,1,0),A_{1}(0,0,2),B_{1}(0,2,2),C_{1}(2,0,2),\therefore \overrightarrow{A_1B_1}=(0，2，0)，\overrightarrow{A_1D}=(1,1,-2),\overrightarrow{BC_1}=(2,-2,2).设平面A_{1}B_{1}D的法向量为n=(x_{1},y_{1},z_{1}),则\left\{\begin{matrix}\boldsymbol{n}\cdot \overrightarrow{A_1B_1}=2y_1=0,\\ \boldsymbol{n}\cdot \overrightarrow{A_1D}=x_1+y_1-2z_1=0,\end{matrix}\right.令z_{1}=1,得x_{1}=2,则n=(2,0,1).设直线BC_{1}与平面A_{1}B_{1}D所成角为\theta ,\\sin   \theta =|\\cos  \langle \overrightarrow{BC_1},n\rangle |=\frac{|\overrightarrow{BC_1}\cdot \boldsymbol{n}|}{|\boldsymbol{n}|\cdot |\overrightarrow{BC_1}|}=\frac{6}{\sqrt{5}\times 2\sqrt{3}}=\frac{\sqrt{15}}{5}.\therefore 直线BC_{1}与平面A_{1}B_{1}D所成角的正弦值为\frac{\sqrt{15}}{5}.$

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


Question: $如图，已知直三棱柱ABC-A_{1}B_{1}C_{1}中，AB=AC=2，D为BC中点，AA_{1}=2，再从条件①，条件②这两个条件中选择一个作为已知，完成以下问题：$<img_2447>
$(2)求直线BC_{1}与平面A_{1}B_{1}D所成角的正弦值.条件①：B_{1}D\perp BC_{1}；条件②：BC=2\sqrt{2}.注：如果选择条件①和条件②分别解答，按第一个解答计分.$

Solution:
$选择条件②
B(0，2，0)，D(1，1，0)，A_{1}(0，0，2)，B_{1}(0，2，2)，C_{1}(2，0，2)，\therefore \overrightarrow{A_1B_1}=(0，2，0)，\overrightarrow{A_1D}=(1，1，-2)，\overrightarrow{BC_1}=(2，-2，2).设平面A_{1}B_{1}D的法向量为m=(x，y，z)，则\left\{\begin{matrix}\boldsymbol{m}\cdot \overrightarrow{A_1B_1}=2y=0,\\ \boldsymbol{m}\cdot \overrightarrow{A_1D}=x+y-2z=0,\end{matrix}\right. 令z=1，得x=2,则m=(2，0，1).设直线BC_{1}与平面A_{1}B_{1}D所成角为\alpha ，\\sin   \alpha =|\\cos  \langle \overrightarrow{BC_1},m\rangle |=\frac{|\overrightarrow{BC_1}\cdot \boldsymbol{m}|}{|\boldsymbol{m}|\cdot |\overrightarrow{BC_1}|}=\frac{6}{\sqrt{5}\times 2\sqrt{3}}=\frac{\sqrt{15}}{5}.\therefore 直线BC_{1}与平面A_{1}B_{1}D所成角的正弦值为\frac{\sqrt{15}}{5}.
选择条件①
设B_{1}D\cap BC_{1}=O,在直三棱柱ABC-A_{1}B_{1}C_{1}中可得BB_{1}\perp B_{1}C_{1},在Rt\triangle BB_{1}C_{1}中，由B_{1}D\perp BC_{1}，得B_{1}B^{2}=BO\cdot BC_{1},\because BC\parallel B_{1}C_{1},D为BC中点，\therefore \frac{BO}{OC_1}=\frac{BD}{B_1C_1}=\frac{1}{2},\therefore BO=\frac{1}{3}BC_{1}，\therefore B_{1}B^{2}=\frac{1}{3}BC^2_1,又B_{1}B=AA_{1}=2,\therefore BC_{1}=2\sqrt{3}(舍负)，\therefore BC=B_{1}C_{1}=\sqrt{BC^2_1-BB^2_1}=2\sqrt{2}.又在\triangle ABC中，AB=AC=2,\therefore AB^{2}+AC^{2}=BC^{2}.\therefore \angle BAC=90^\circ ，即AB\perp AC.在直三棱柱ABC-A_{1}B_{1}C_{1}中，侧棱AA_{1}\perp 底面ABC,AB,AC\subset 平面ABC,\therefore AA_{1}\perp AB,AA_{1}\perp AC.以点A为原点，分别以AC,AB,AA_{1}所在直线为x轴，y轴，z轴建立空间直角坐标系，如图2，
$<img_1540>$
图2
则B(0,2,0),D(1,1,0),A_{1}(0,0,2),B_{1}(0,2,2),C_{1}(2,0,2),\therefore \overrightarrow{A_1B_1}=(0，2，0)，\overrightarrow{A_1D}=(1,1,-2),\overrightarrow{BC_1}=(2,-2,2).设平面A_{1}B_{1}D的法向量为n=(x_{1},y_{1},z_{1}),则\left\{\begin{matrix}\boldsymbol{n}\cdot \overrightarrow{A_1B_1}=2y_1=0,\\ \boldsymbol{n}\cdot \overrightarrow{A_1D}=x_1+y_1-2z_1=0,\end{matrix}\right.令z_{1}=1,得x_{1}=2,则n=(2,0,1).设直线BC_{1}与平面A_{1}B_{1}D所成角为\theta ,\\sin   \theta =|\\cos  \langle \overrightarrow{BC_1},n\rangle |=\frac{|\overrightarrow{BC_1}\cdot \boldsymbol{n}|}{|\boldsymbol{n}|\cdot |\overrightarrow{BC_1}|}=\frac{6}{\sqrt{5}\times 2\sqrt{3}}=\frac{\sqrt{15}}{5}.\therefore 直线BC_{1}与平面A_{1}B_{1}D所成角的正弦值为\frac{\sqrt{15}}{5}.$

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


Question: $如图，平面PAC\perp 平面ABC，AB\perp BC，AB=BC，D，O分别为PA，AC的中点，AC=8，PA=PC=5.$![Image](https://cdntk.53zaixian.com/tq/1716268093820039169/img/compression/M_d187d648d509685520f8f8113549dcc3.jpg)
$(1)设平面PBC\cap 平面BOD=l，判断直线l与PC的位置关系，并证明；$

Solution:
$直线l\parallel PC，证明如下：因为D，O分别为PA，AC的中点，所以DO\parallel PC.又因为DO\subset 平面BOD，PC\not\subset 平面BOD，所以PC\parallel 平面BOD.因为PC\subset 平面PBC，平面PBC\cap 平面BOD=l，所以l\parallel PC.$

Final answer: 证明题
Subfield: Solid Geometry


Question: $如图，平面PAC\perp 平面ABC，AB\perp BC，AB=BC，D，O分别为PA，AC的中点，AC=8，PA=PC=5.$<img_2200>
$(1)设平面PBC\cap 平面BOD=l，判断直线l与PC的位置关系，并证明；$

Solution:
$直线l\parallel PC，证明如下：因为D，O分别为PA，AC的中点，所以DO\parallel PC.又因为DO\subset 平面BOD，PC\not\subset 平面BOD，所以PC\parallel 平面BOD.因为PC\subset 平面PBC，平面PBC\cap 平面BOD=l，所以l\parallel PC.$
Subfield: Solid Geometry


Question: $如图，平面PAC\perp 平面ABC，AB\perp BC，AB=BC，D，O分别为PA，AC的中点，AC=8，PA=PC=5.$<img_2200>
$(2)求直线PB与平面BOD所成角的正弦值.$

Solution:
$连接PO.因为PA=PC，O为AC中点，所以PO\perp AC.因为平面PAC\perp 平面ABC，平面PAC\cap 平面ABC=AC,所以PO\perp 平面ABC.因为OB\subset 平面ABC,所以PO\perp OB.因为AB=BC，O为AC中点，所以OB\perp AC.所以OB、OC、OP两两垂直.如图，建立空间直角坐标系O－xyz，则O(0，0，0)，A(0，－4，0)，B(4，0，0),P(0，0，3).所以\overrightarrow{PB}=(4，0，-3),\overrightarrow{OB}=(4，0，0).因为点D为PA中点，所以D\left(0,-2,\frac{3}{2}\right).所以\overrightarrow{OD}=\left(0,-2,\frac{3}{2}\right)，设平面BOD的法向量为n=(x,y,z)，则\left\{\begin{matrix}\boldsymbol{n}\cdot \overrightarrow{OB}=0,\\ \boldsymbol{n}\cdot \overrightarrow{OD}=0,\end{matrix}\right. 即\left\{\begin{matrix}4x=0,\\ -2y+\frac{3}{2}z=0.\end{matrix}\right.令z=4，则y=3，得n=(0,3,4).设直线PB与平面BOD所成角为\alpha ，所以\\sin   \alpha =|\\cos  \langle \overrightarrow{PB}，n\rangle |=\frac{|\overrightarrow{PB}\cdot \boldsymbol{n}|}{|\overrightarrow{PB}|\cdot |\boldsymbol{n}|}=\frac{12}{25}.所以直线PB与平面BOD所成角的正弦值为\frac{12}{25}.$<img_2443>

Final answer: $\frac{12}{25}$
Subfield: Solid Geometry


Question: $如图，平面PAC\perp 平面ABC，AB\perp BC，AB=BC，D，O分别为PA，AC的中点，AC=8，PA=PC=5.$![Image](https://cdntk.53zaixian.com/tq/1716268093820039169/img/compression/M_d187d648d509685520f8f8113549dcc3.jpg)
$(2)求直线PB与平面BOD所成角的正弦值.$

Solution:
$连接PO.因为PA=PC，O为AC中点，所以PO\perp AC.因为平面PAC\perp 平面ABC，平面PAC\cap 平面ABC=AC,所以PO\perp 平面ABC.因为OB\subset 平面ABC,所以PO\perp OB.因为AB=BC，O为AC中点，所以OB\perp AC.所以OB、OC、OP两两垂直.如图，建立空间直角坐标系O－xyz，则O(0，0，0)，A(0，－4，0)，B(4，0，0),P(0，0，3).所以\overrightarrow{PB}=(4，0，-3),\overrightarrow{OB}=(4，0，0).因为点D为PA中点，所以D\left(0,-2,\frac{3}{2}\right).所以\overrightarrow{OD}=\left(0,-2,\frac{3}{2}\right)，设平面BOD的法向量为n=(x,y,z)，则\left\{\begin{matrix}\boldsymbol{n}\cdot \overrightarrow{OB}=0,\\ \boldsymbol{n}\cdot \overrightarrow{OD}=0,\end{matrix}\right. 即\left\{\begin{matrix}4x=0,\\ -2y+\frac{3}{2}z=0.\end{matrix}\right.令z=4，则y=3，得n=(0,3,4).设直线PB与平面BOD所成角为\alpha ，所以\\sin   \alpha =|\\cos  \langle \overrightarrow{PB}，n\rangle |=\frac{|\overrightarrow{PB}\cdot \boldsymbol{n}|}{|\overrightarrow{PB}|\cdot |\boldsymbol{n}|}=\frac{12}{25}.所以直线PB与平面BOD所成角的正弦值为\frac{12}{25}.$![Image](https://cdntk.53zaixian.com/tq/1716268094956695553/img/compression/M_441c1f98e5aa1ee8b0cf3feced383ed8.jpg)

Final answer: $\frac{12}{25}$
Subfield: Solid Geometry


Question: $如图,在长方体ABCD－A_{1}B_{1}C_{1}D_{1}中,底面ABCD是边长为2的正方形,DD_{1}=4,E,F分别是CC_{1},B_{1}C_{1}的中点.$![Image](https://cdntk.53zaixian.com/tq/1716268095271268354/img/compression/M_4b6add02f0706778992248e7fe4690ff.jpg)
$(1)求证：A_{1}F\parallel 平面AED_{1}；$

Solution:
$证明：连接A_{1}D，设A_{1}D\cap AD_{1}=O，连接OE,EF,B_{1}C.在长方体ABCD-A_{1}B_{1}C_{1}D_{1}中，因为A_{1}B_{1}\parallel CD，且A_{1}B_{1}=CD,所以四边形A_{1}B_{1}CD是平行四边形．所以A_{1}D\parallel B_{1}C,且A_{1}D=B_{1}C.因为E，F分别是CC_{1}，B_{1}C_{1}的中点，所以FE\parallel B_{1}C,且FE=\frac{1}{2}B_{1}C.在矩形A_{1}ADD_{1}中,O是A_{1}D的中点，所以A_{1}O\parallel FE,且A_{1}O=FE.所以四边形A_{1}OEF是平行四边形.所以A_{1}F\parallel OE.因为A_{1}F\not\subset 平面AED_{1},OE\subset 平面AED_{1}，所以A_{1}F\parallel 平面AED_{1}.$

Final answer: 证明题
Subfield: Solid Geometry


Question: $如图,在长方体ABCD－A_{1}B_{1}C_{1}D_{1}中,底面ABCD是边长为2的正方形,DD_{1}=4,E,F分别是CC_{1},B_{1}C_{1}的中点.$<img_1942>
$(1)求证：A_{1}F\parallel 平面AED_{1}；$

Solution:
$证明：连接A_{1}D，设A_{1}D\cap AD_{1}=O，连接OE,EF,B_{1}C.在长方体ABCD-A_{1}B_{1}C_{1}D_{1}中，因为A_{1}B_{1}\parallel CD，且A_{1}B_{1}=CD,所以四边形A_{1}B_{1}CD是平行四边形．所以A_{1}D\parallel B_{1}C,且A_{1}D=B_{1}C.因为E，F分别是CC_{1}，B_{1}C_{1}的中点，所以FE\parallel B_{1}C,且FE=\frac{1}{2}B_{1}C.在矩形A_{1}ADD_{1}中,O是A_{1}D的中点，所以A_{1}O\parallel FE,且A_{1}O=FE.所以四边形A_{1}OEF是平行四边形.所以A_{1}F\parallel OE.因为A_{1}F\not\subset 平面AED_{1},OE\subset 平面AED_{1}，所以A_{1}F\parallel 平面AED_{1}.$
Subfield: Solid Geometry


Question: $如图,在长方体ABCD－A_{1}B_{1}C_{1}D_{1}中,底面ABCD是边长为2的正方形,DD_{1}=4,E,F分别是CC_{1},B_{1}C_{1}的中点.$![Image](https://cdntk.53zaixian.com/tq/1716268095271268354/img/compression/M_4b6add02f0706778992248e7fe4690ff.jpg)
$(2)设H在棱BB_{1}上,且BH=\frac{1}{4}BB_{1},N为CD的中点,求证：NH\perp 平面AED_{1},并求直线AN与平面AED_{1}所成角的正弦值．$

Solution:
$建立如图所示的空间直角坐标系D-xyz,
$![Image](https://cdntk.53zaixian.com/tq/1716268095543898113/img/compression/M_060cb1119f01251c4d8664681640fffb.jpg)$则D(0,0,0),A(2,0,0),D_{1}(0,0,4),E(0,2,2),H(2,2,1),N(0,1,0).所以\overrightarrow{AD_1}=(-2，0，4)，\overrightarrow{D_1E}=(0,2,-2)，\overrightarrow{NH}=(2,1,1),\overrightarrow{AN}=(-2,1,0).设平面AED_{1}的法向量为m=(x,y,z),则\left\{\begin{matrix}\boldsymbol{m}\cdot \overrightarrow{AD_1}=0,\\ \boldsymbol{m}\cdot \overrightarrow{D_1E}=0,\end{matrix}\right.即\left\{\begin{matrix}-2x+4z=0,\\ 2y-2z=0.\end{matrix}\right.令z=1,则x=2,y=1.所以m=(2,1,1).所以\overrightarrow{NH}=m.所以NH\perp 平面AED_{1}.设AN与平面AED_{1}所成角为\theta ,则\\sin   \theta =|\\cos  \langle \overrightarrow{AN},m\rangle |=\frac{|\overrightarrow{AN}\cdot \boldsymbol{m}|}{|\overrightarrow{AN}|\cdot |\boldsymbol{m}|}=\frac{|-4+1+0｜}{\sqrt{4+1+0}\times \sqrt{4+1+1}}=\frac{\sqrt{30}}{10}.$

Final answer: 证明题
Subfield: Solid Geometry


Question: $如图,在长方体ABCD－A_{1}B_{1}C_{1}D_{1}中,底面ABCD是边长为2的正方形,DD_{1}=4,E,F分别是CC_{1},B_{1}C_{1}的中点.$<img_1942>
$(2)设H在棱BB_{1}上,且BH=\frac{1}{4}BB_{1},N为CD的中点,求证：NH\perp 平面AED_{1},并求直线AN与平面AED_{1}所成角的正弦值．$

Solution:
$建立如图所示的空间直角坐标系D-xyz,
$<img_1975>$则D(0,0,0),A(2,0,0),D_{1}(0,0,4),E(0,2,2),H(2,2,1),N(0,1,0).所以\overrightarrow{AD_1}=(-2，0，4)，\overrightarrow{D_1E}=(0,2,-2)，\overrightarrow{NH}=(2,1,1),\overrightarrow{AN}=(-2,1,0).设平面AED_{1}的法向量为m=(x,y,z),则\left\{\begin{matrix}\boldsymbol{m}\cdot \overrightarrow{AD_1}=0,\\ \boldsymbol{m}\cdot \overrightarrow{D_1E}=0,\end{matrix}\right.即\left\{\begin{matrix}-2x+4z=0,\\ 2y-2z=0.\end{matrix}\right.令z=1,则x=2,y=1.所以m=(2,1,1).所以\overrightarrow{NH}=m.所以NH\perp 平面AED_{1}.设AN与平面AED_{1}所成角为\theta ,则\\sin   \theta =|\\cos  \langle \overrightarrow{AN},m\rangle |=\frac{|\overrightarrow{AN}\cdot \boldsymbol{m}|}{|\overrightarrow{AN}|\cdot |\boldsymbol{m}|}=\frac{|-4+1+0｜}{\sqrt{4+1+0}\times \sqrt{4+1+1}}=\frac{\sqrt{30}}{10}.$
Subfield: Solid Geometry


Question: $如图，在三棱柱ABC－A_{1}B_{1}C_{1}中，四边形AA_{1}C_{1}C是边长为4的菱形，AB=BC=\sqrt{13}，点D为棱AC上的动点(不与A，C重合)，平面B_{1}BD与棱A_{1}C_{1}交于点E.$![Image](https://cdntk.53zaixian.com/tq/1716268096479227906/img/compression/M_02777938469333898802ea27cbea54b0.jpg)
$(1)求证：BB_{1}\parallel DE；$

Solution:
$证明：在三棱柱ABC－A_{1}B_{1}C_{1}中，AA_{1}\parallel BB_{1}，因为BB_{1}\not\subset 平面ACC_{1}A_{1}，AA_{1}\subset 平面ACC_{1}A_{1}，所以BB_{1}\parallel 平面ACC_{1}A_{1}，又因为平面B_{1}BDE\cap 平面ACC_{1}A_{1}=DE，所以BB_{1}\parallel DE.$

Final answer: 证明题
Subfield: Solid Geometry


Question: $如图，在三棱柱ABC－A_{1}B_{1}C_{1}中，四边形AA_{1}C_{1}C是边长为4的菱形，AB=BC=\sqrt{13}，点D为棱AC上的动点(不与A，C重合)，平面B_{1}BD与棱A_{1}C_{1}交于点E.$<img_2041>
$(1)求证：BB_{1}\parallel DE；$

Solution:
$证明：在三棱柱ABC－A_{1}B_{1}C_{1}中，AA_{1}\parallel BB_{1}，因为BB_{1}\not\subset 平面ACC_{1}A_{1}，AA_{1}\subset 平面ACC_{1}A_{1}，所以BB_{1}\parallel 平面ACC_{1}A_{1}，又因为平面B_{1}BDE\cap 平面ACC_{1}A_{1}=DE，所以BB_{1}\parallel DE.$