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


Question: $如图，在三棱锥P-ABC中，AB=BC=2\sqrt{2}，PA=PB=PC=AC=4，O为AC的中点.$<img_2067>
$(2)若点M在棱BC上，BM=\lambda BC，\lambda \in (0，1)，且二面角M-PA-C的大小为30^\circ ，求\lambda 的值.$

Solution:
$如图，以O为原点，OB,OC,OP所在直线分别为x轴、y轴、z轴建立空间直角坐标系，$<img_3024>$则A(0,-2,0),B(2,0,0),C(0,2,0),P(0,0,2\sqrt{3}),\therefore \overrightarrow{BC}=(-2,2,0),\overrightarrow{AP}=(0,2,2\sqrt{3}),\overrightarrow{AB}=(2,2,0).\because BM=\lambda BC，点M在棱BC上，\therefore \overrightarrow{BM}=\lambda \overrightarrow{BC}=(-2\lambda ,2\lambda ,0),\lambda \in (0,1)，\therefore \overrightarrow{AM}=\overrightarrow{AB}+\overrightarrow{BM}=(2-2\lambda ,2+2\lambda ,0).设平面PAM的一个法向量为m=(x,y,z)，则\left\{\begin{matrix}\boldsymbol{m} \cdot \overrightarrow{AP}=2y+2\sqrt{3}z=0,\\ \boldsymbol{m} \cdot \overrightarrow{AM}=(2-2\lambda )x+(2+2\lambda )y=0,\end{matrix}\right.即\left\{\begin{matrix}y=-\sqrt{3}z,\\ x=\frac{1+\lambda }{\lambda -1}y,\end{matrix}\right.令y=\sqrt{3}(\lambda -1)，则m=(\sqrt{3}(1+\lambda ),\sqrt{3}(\lambda -1),1-\lambda )，易知平面PAC的一个法向量为\overrightarrow{OB}=(2,0,0).\because 二面角M-PA-C的大小为30^\circ ，\therefore |\\cos  \langle m,\overrightarrow{OB}\rangle |=\frac{|2\sqrt{3}(1+\lambda )|}{2\sqrt{7\lambda ^2-2\lambda +7}}=\\cos   30^\circ =\frac{\sqrt{3}}{2}，结合二面角的平面角的余弦值，构建关于参数的方程，从而达到求解的目的.解得\lambda =\frac{1}{3}或\lambda =3(舍去)，故\lambda =\frac{1}{3}.$

Final answer: $\frac{1}{3}$
Subfield: Solid Geometry


Question: $如图，在三棱锥P-ABC中，AB=BC=2\sqrt{2}，PA=PB=PC=AC=4，O为AC的中点.$![Image](https://cdntk.53zaixian.com/tq/1658661599331414018/img/compression/M_29e3530672de434f2dc818fd6ef0b08f.jpg)
$(2)若点M在棱BC上，BM=\lambda BC，\lambda \in (0，1)，且二面角M-PA-C的大小为30^\circ ，求\lambda 的值.$

Solution:
$如图，以O为原点，OB,OC,OP所在直线分别为x轴、y轴、z轴建立空间直角坐标系，$![Image](https://cdntk.53zaixian.com/tq/1658661599616626689/img/compression/M_ea12205c0b1a45d9131aa9302808a177.jpg)$则A(0,-2,0),B(2,0,0),C(0,2,0),P(0,0,2\sqrt{3}),\therefore \overrightarrow{BC}=(-2,2,0),\overrightarrow{AP}=(0,2,2\sqrt{3}),\overrightarrow{AB}=(2,2,0).\because BM=\lambda BC，点M在棱BC上，\therefore \overrightarrow{BM}=\lambda \overrightarrow{BC}=(-2\lambda ,2\lambda ,0),\lambda \in (0,1)，\therefore \overrightarrow{AM}=\overrightarrow{AB}+\overrightarrow{BM}=(2-2\lambda ,2+2\lambda ,0).设平面PAM的一个法向量为m=(x,y,z)，则\left\{\begin{matrix}\boldsymbol{m} \cdot \overrightarrow{AP}=2y+2\sqrt{3}z=0,\\ \boldsymbol{m} \cdot \overrightarrow{AM}=(2-2\lambda )x+(2+2\lambda )y=0,\end{matrix}\right.即\left\{\begin{matrix}y=-\sqrt{3}z,\\ x=\frac{1+\lambda }{\lambda -1}y,\end{matrix}\right.令y=\sqrt{3}(\lambda -1)，则m=(\sqrt{3}(1+\lambda ),\sqrt{3}(\lambda -1),1-\lambda )，易知平面PAC的一个法向量为\overrightarrow{OB}=(2,0,0).\because 二面角M-PA-C的大小为30^\circ ，\therefore |\\cos  \langle m,\overrightarrow{OB}\rangle |=\frac{|2\sqrt{3}(1+\lambda )|}{2\sqrt{7\lambda ^2-2\lambda +7}}=\\cos   30^\circ =\frac{\sqrt{3}}{2}，结合二面角的平面角的余弦值，构建关于参数的方程，从而达到求解的目的.解得\lambda =\frac{1}{3}或\lambda =3(舍去)，故\lambda =\frac{1}{3}.$

Final answer: $\frac{1}{3}$
Subfield: Solid Geometry


Question: $如图，在四棱锥P-ABCD中，底面ABCD是矩形，且AD=2，AB=1，PA\perp 平面ABCD，E，F分别是AB，BC的中点.$![Image](https://cdntk.53zaixian.com/tq/1658661554733256706/img/compression/M_941c7ac17c41e1b9541687af1a15b406.jpg)
$(1)证明：PF\perp DF；$

Solution:
$如图，建立空间直角坐标系A-xyz，则F(1,1,0)，D(0,2,0).$![Image](https://cdntk.53zaixian.com/tq/1658661554733256706/img/compression/M_bfd12a90d96da5e6776e4d2b158b9ee0.jpg)$证明：不妨令P(0,0,t)，则\overrightarrow{PF}=(1,1,-t)，\overrightarrow{DF}=(1,-1,0)，\because \overrightarrow{PF}\cdot \overrightarrow{DF}=1\times 1+1\times (-1)+(-t)\times 0=0，\therefore \overrightarrow{PF}\perp \overrightarrow{DF}，即PF\perp DF.$

Final answer: 证明题
Subfield: Solid Geometry


Question: $如图，在四棱锥P-ABCD中，底面ABCD是矩形，且AD=2，AB=1，PA\perp 平面ABCD，E，F分别是AB，BC的中点.$<img_2437>
$(1)证明：PF\perp DF；$

Solution:
$如图，建立空间直角坐标系A-xyz，则F(1,1,0)，D(0,2,0).$<img_1484>$证明：不妨令P(0,0,t)，则\overrightarrow{PF}=(1,1,-t)，\overrightarrow{DF}=(1,-1,0)，\because \overrightarrow{PF}\cdot \overrightarrow{DF}=1\times 1+1\times (-1)+(-t)\times 0=0，\therefore \overrightarrow{PF}\perp \overrightarrow{DF}，即PF\perp DF.$
Subfield: Solid Geometry


Question: $如图，在四棱锥P-ABCD中，底面ABCD是矩形，且AD=2，AB=1，PA\perp 平面ABCD，E，F分别是AB，BC的中点.$![Image](https://cdntk.53zaixian.com/tq/1658661554733256706/img/compression/M_941c7ac17c41e1b9541687af1a15b406.jpg)
$(2)判断并说明PA上是否存在点G，使得EG\parallel 平面PFD.$

Solution:
$设平面PFD的一个法向量为n=(x,y,z)，则\left\{\begin{matrix}\boldsymbol{n} \cdot \overrightarrow{PF}=0,\\ \boldsymbol{n} \cdot \overrightarrow{DF}=0,\end{matrix}\right.即\left\{\begin{matrix}x+y-tz=0,\\ x-y=0,\end{matrix}\right.令z=1，得x=y=\frac{t}{2},\therefore n=\left(\frac{t}{2},\frac{t}{2},1\right).设G(0,0,m)，又E\left(\frac{1}{2},0,0\right)，\therefore \overrightarrow{EG}=\left(-\frac{1}{2},0,m\right).要使EG\parallel 平面PFD，只需\overrightarrow{EG}\cdot n=0，即-\frac{1}{2}\times \frac{t}{2}+0\times \frac{t}{2}+m\times 1=0，解得m=\frac{1}{4}t.\therefore 当点G满足AG=\frac{1}{4}AP时，EG\parallel 平面PFD.$

Final answer: $当点G满足AG=1/4AP时，EG\parallel 平面PFD.$
Subfield: Solid Geometry


Question: $如图1，四边形ABCD是梯形，AB\parallel CD,AD=DC=CB=\frac{1}{2}AB=4,M是AB的中点，将\triangle ADM沿DM折起至\triangle A'DM的位置，如图2，点N在线段A'C上.$![Image](https://cdntk.53zaixian.com/tq/1658661615395475459/img/compression/M_7db3000fc64c156f4eb28e0cf44c17b0.jpg)$图1$![Image](https://cdntk.53zaixian.com/tq/1658661615395475459/img/compression/M_9640f0523c7687891322aa7d68f70688.jpg)$图2$
$(1)若N是A'C的中点，求证：平面DNM\perp 平面A'BC;$

Solution:
$证明：取DM的中点O，连接A'O,CO，易证DM\perp 平面A'CO，所以DM\perp A'C.因为DC=A'D，N为A'C的中点，所以DN\perp A'C，又DN\cap DM=D，所以A'C\perp 平面DNM，又A'C\subset 平面A'BC，所以平面DNM\perp 平面A'BC.$

Final answer: 证明题
Subfield: Solid Geometry


Question: $如图1，四边形ABCD是梯形，AB\parallel CD,AD=DC=CB=\frac{1}{2}AB=4,M是AB的中点，将\triangle ADM沿DM折起至\triangle A'DM的位置，如图2，点N在线段A'C上.$<img_2882>$图1$<img_1478>$图2$
$(1)若N是A'C的中点，求证：平面DNM\perp 平面A'BC;$

Solution:
$证明：取DM的中点O，连接A'O,CO，易证DM\perp 平面A'CO，所以DM\perp A'C.因为DC=A'D，N为A'C的中点，所以DN\perp A'C，又DN\cap DM=D，所以A'C\perp 平面DNM，又A'C\subset 平面A'BC，所以平面DNM\perp 平面A'BC.$
Subfield: Solid Geometry


Question: $如图1，四边形ABCD是梯形，AB\parallel CD,AD=DC=CB=\frac{1}{2}AB=4,M是AB的中点，将\triangle ADM沿DM折起至\triangle A'DM的位置，如图2，点N在线段A'C上.$<img_2882>$图1$<img_1478>$图2$
$(2)若A'C=2\sqrt{6}，平面DNM与平面CDM夹角的余弦值为\frac{2\sqrt{5}}{5}，求\frac{A'N}{NC}的值.$

Solution:
$易求得OC=OA'=2\sqrt{3}，又A'C=2\sqrt{6}，所以OC^{2}+OA'^{2}=A'C^{2}，可得OC\perp A'O，又A'O\perp OD,OC\perp OD，所以以O为坐标原点，OD,OC,OA'所在直线分别为x轴、y轴、z轴，建立如图所示的空间直角坐标系O-xyz，$<img_1639>$则D(2,0,0),M(-2,0,0),C(0,2\sqrt{3},0),A'(0,0,2\sqrt{3})，所以A'C=(0,2\sqrt{3},-2\sqrt{3}),\overrightarrow{MD}=(4，0，0).设\overrightarrow{A'N}=\lambda \overrightarrow{A'C}(0\leq \lambda \leq 1)，则\overrightarrow{A'N}=(0,2\sqrt{3}\lambda ,-2\sqrt{3}\lambda )，得N(0,2\sqrt{3}\lambda ,2\sqrt{3}-2\sqrt{3}\lambda )，所以\overrightarrow{DN}=(-2,2\sqrt{3}\lambda ,2\sqrt{3}-2\sqrt{3}\lambda ).设平面DNM的一个法向量为n_{1}=(x,y,z)，由\left\{\begin{matrix}\overrightarrow{MD} \cdot \boldsymbol{n}_1=0,\\ \overrightarrow{DN} \cdot \boldsymbol{n}_1=0,\end{matrix}\right.得\left\{\begin{matrix}4x=0,\\ -2x+2\sqrt{3}\lambda y+(2\sqrt{3}-2\sqrt{3}\lambda )z=0,\end{matrix}\right.令y=\lambda -1，则n_{1}=(0,\lambda -1,\lambda ).易得平面CDM的一个法向量为n_{2}=(0,0,1).设平面DNM与平面CDM的夹角为\theta ，则\\cos   \theta =\left|\frac{\boldsymbol{n}_1 \cdot \boldsymbol{n}_2}{|\boldsymbol{n}_1||\boldsymbol{n}_2|}\right|=\frac{2\sqrt{5}}{5}，即\left|\frac{\lambda }{\sqrt{(\lambda -1)^2+\lambda ^2}}\right|=\frac{2\sqrt{5}}{5}，解得\lambda =\frac{2}{3}或\lambda =2(舍去).所以\frac{A'N}{NC}=2.$

Final answer: $2$
Subfield: Solid Geometry


Question: $如图1，四边形ABCD是梯形，AB\parallel CD,AD=DC=CB=\frac{1}{2}AB=4,M是AB的中点，将\triangle ADM沿DM折起至\triangle A'DM的位置，如图2，点N在线段A'C上.$![Image](https://cdntk.53zaixian.com/tq/1658661615395475459/img/compression/M_7db3000fc64c156f4eb28e0cf44c17b0.jpg)$图1$![Image](https://cdntk.53zaixian.com/tq/1658661615395475459/img/compression/M_9640f0523c7687891322aa7d68f70688.jpg)$图2$
$(2)若A'C=2\sqrt{6}，平面DNM与平面CDM夹角的余弦值为\frac{2\sqrt{5}}{5}，求\frac{A'N}{NC}的值.$

Solution:
$易求得OC=OA'=2\sqrt{3}，又A'C=2\sqrt{6}，所以OC^{2}+OA'^{2}=A'C^{2}，可得OC\perp A'O，又A'O\perp OD,OC\perp OD，所以以O为坐标原点，OD,OC,OA'所在直线分别为x轴、y轴、z轴，建立如图所示的空间直角坐标系O-xyz，$![Image](https://cdntk.53zaixian.com/tq/1658661615852654593/img/compression/M_0c84d1eaab172ddb4d31246a990c0138.jpg)$则D(2,0,0),M(-2,0,0),C(0,2\sqrt{3},0),A'(0,0,2\sqrt{3})，所以A'C=(0,2\sqrt{3},-2\sqrt{3}),\overrightarrow{MD}=(4，0，0).设\overrightarrow{A'N}=\lambda \overrightarrow{A'C}(0\leq \lambda \leq 1)，则\overrightarrow{A'N}=(0,2\sqrt{3}\lambda ,-2\sqrt{3}\lambda )，得N(0,2\sqrt{3}\lambda ,2\sqrt{3}-2\sqrt{3}\lambda )，所以\overrightarrow{DN}=(-2,2\sqrt{3}\lambda ,2\sqrt{3}-2\sqrt{3}\lambda ).设平面DNM的一个法向量为n_{1}=(x,y,z)，由\left\{\begin{matrix}\overrightarrow{MD} \cdot \boldsymbol{n}_1=0,\\ \overrightarrow{DN} \cdot \boldsymbol{n}_1=0,\end{matrix}\right.得\left\{\begin{matrix}4x=0,\\ -2x+2\sqrt{3}\lambda y+(2\sqrt{3}-2\sqrt{3}\lambda )z=0,\end{matrix}\right.令y=\lambda -1，则n_{1}=(0,\lambda -1,\lambda ).易得平面CDM的一个法向量为n_{2}=(0,0,1).设平面DNM与平面CDM的夹角为\theta ，则\\cos   \theta =\left|\frac{\boldsymbol{n}_1 \cdot \boldsymbol{n}_2}{|\boldsymbol{n}_1||\boldsymbol{n}_2|}\right|=\frac{2\sqrt{5}}{5}，即\left|\frac{\lambda }{\sqrt{(\lambda -1)^2+\lambda ^2}}\right|=\frac{2\sqrt{5}}{5}，解得\lambda =\frac{2}{3}或\lambda =2(舍去).所以\frac{A'N}{NC}=2.$

Final answer: $2$
Subfield: Solid Geometry


Question: $如图，在四棱锥P-ABCD中，底面ABCD是等腰梯形，CD=2AB=2BC=2AD=4，平面ADP\perp 平面ABCD,E是PC的中点，且\triangle ADP为等边三角形，平面ADP\cap 平面PBC=m.$<img_2436>
$(1)设m\cap 直线BC=M，求点M到平面PDC的距离；$

Solution:
$延长DA,CB，交于点F，则直线PF即为m，则点F即为点M.取AD的中点O,CD的中点为Q，连接OQ，AQ，OP，易得\triangle MCD，\triangle AQD均为等边三角形，所以OQ\perp AD.因为平面ADP\perp 平面ABCD,平面ADP\cap 平面ABCD=AD，OP\perp AD，OP\subset 平面ADP，所以OP\perp 平面ABCD,所以OP\perp OQ,OP\perp OA,以O为原点建立如图所示的空间直角坐标系，$<img_2465>$由题意得B(2,\sqrt{3},0),P(0,0,\sqrt{3}),D(-1,0,0),C(1,2\sqrt{3},0),M(3,0,0),E\left(\frac{1}{2},\sqrt{3},\frac{\sqrt{3}}{2}\right)，则\overrightarrow{PD}=(-1,0,-\sqrt{3}),\overrightarrow{PC}=(1,2\sqrt{3},-\sqrt{3})，\overrightarrow{MC}=(-2，2\sqrt{3}，0)，设平面PDC的一个法向量为n_{1}=(x_{1},y_{1},z_{1})，由\left\{\begin{matrix}\boldsymbol{n}_1 \cdot \overrightarrow{PD}=0,\\ \boldsymbol{n}_1 \cdot \overrightarrow{PC}=0,\end{matrix}\right.得\left\{\begin{matrix}-x_1-\sqrt{3}z_1=0,\\ x_1+2\sqrt{3}y_1-\sqrt{3}z_1=0,\end{matrix}\right.令x_{1}=\sqrt{3}，则y_{1}=-1,z_{1}=-1，所以n_{1}=(\sqrt{3},-1,-1).所以点M到平面PDC的距离为\frac{|\boldsymbol{n}_1 \cdot \overrightarrow{MC}|}{|\boldsymbol{n}_1|}=\frac{4\sqrt{15}}{5}.$

Final answer: $\frac{4\sqrt 15}{5}$
Subfield: Solid Geometry


Question: $如图，在四棱锥P-ABCD中，底面ABCD是等腰梯形，CD=2AB=2BC=2AD=4，平面ADP\perp 平面ABCD,E是PC的中点，且\triangle ADP为等边三角形，平面ADP\cap 平面PBC=m.$![Image](https://cdntk.53zaixian.com/tq/1658661616108507140/img/compression/M_d437c7404c7a5c04180328ca49052a1c.jpg)
$(1)设m\cap 直线BC=M，求点M到平面PDC的距离；$

Solution:
$延长DA,CB，交于点F，则直线PF即为m，则点F即为点M.取AD的中点O,CD的中点为Q，连接OQ，AQ，OP，易得\triangle MCD，\triangle AQD均为等边三角形，所以OQ\perp AD.因为平面ADP\perp 平面ABCD,平面ADP\cap 平面ABCD=AD，OP\perp AD，OP\subset 平面ADP，所以OP\perp 平面ABCD,所以OP\perp OQ,OP\perp OA,以O为原点建立如图所示的空间直角坐标系，$![Image](https://cdntk.53zaixian.com/tq/1658661616599240706/img/compression/M_797918cae04bb8b29c1dcf33d20168fa.jpg)$由题意得B(2,\sqrt{3},0),P(0,0,\sqrt{3}),D(-1,0,0),C(1,2\sqrt{3},0),M(3,0,0),E\left(\frac{1}{2},\sqrt{3},\frac{\sqrt{3}}{2}\right)，则\overrightarrow{PD}=(-1,0,-\sqrt{3}),\overrightarrow{PC}=(1,2\sqrt{3},-\sqrt{3})，\overrightarrow{MC}=(-2，2\sqrt{3}，0)，设平面PDC的一个法向量为n_{1}=(x_{1},y_{1},z_{1})，由\left\{\begin{matrix}\boldsymbol{n}_1 \cdot \overrightarrow{PD}=0,\\ \boldsymbol{n}_1 \cdot \overrightarrow{PC}=0,\end{matrix}\right.得\left\{\begin{matrix}-x_1-\sqrt{3}z_1=0,\\ x_1+2\sqrt{3}y_1-\sqrt{3}z_1=0,\end{matrix}\right.令x_{1}=\sqrt{3}，则y_{1}=-1,z_{1}=-1，所以n_{1}=(\sqrt{3},-1,-1).所以点M到平面PDC的距离为\frac{|\boldsymbol{n}_1 \cdot \overrightarrow{MC}|}{|\boldsymbol{n}_1|}=\frac{4\sqrt{15}}{5}.$

Final answer: $\frac{4\sqrt 15}{5}$
Subfield: Solid Geometry


Question: $如图，在四棱锥P-ABCD中，底面ABCD是等腰梯形，CD=2AB=2BC=2AD=4，平面ADP\perp 平面ABCD,E是PC的中点，且\triangle ADP为等边三角形，平面ADP\cap 平面PBC=m.$![Image](https://cdntk.53zaixian.com/tq/1658661616108507140/img/compression/M_d437c7404c7a5c04180328ca49052a1c.jpg)
$(2)求二面角P-BE-D的平面角的正弦值.$

Solution:
$由(1)得\overrightarrow{BP}=(-2,-\sqrt{3},\sqrt{3}),\overrightarrow{BD}=(-3,-\sqrt{3},0),\overrightarrow{BE}=\left(-\frac{3}{2},0,\frac{\sqrt{3}}{2}\right).设平面BDE的一个法向量为n_{2}=(x_{2},y_{2},z_{2})，由\left\{\begin{matrix}\boldsymbol{n}_2 \cdot \overrightarrow{BD}=0,\\ \boldsymbol{n}_2 \cdot \overrightarrow{BE}=0,\end{matrix}\right.得\left\{\begin{matrix}-3x_2-\sqrt{3}y_2=0,\\ -\frac{3}{2}x_2+\frac{\sqrt{3}}{2}z_2=0,\end{matrix}\right.令x_{2}=1，得y_{2}=-\sqrt{3},z_{2}=\sqrt{3}，则n_{2}=(1,-\sqrt{3},\sqrt{3}).设平面PBE的一个法向量为n_{3}=(x_{3},y_{3},z_{3})，由\left\{\begin{matrix}\boldsymbol{n}_3 \cdot \overrightarrow{BP}=0,\\ \boldsymbol{n}_3 \cdot \overrightarrow{BE}=0,\end{matrix}\right.得\left\{\begin{matrix}-2x_3-\sqrt{3}y_3+\sqrt{3}z_3=0,\\ -\frac{3}{2}x_3+\frac{\sqrt{3}}{2}z_3=0,\end{matrix}\right.令x_{3}=3，得y_{3}=\sqrt{3},z_{3}=3\sqrt{3}，则n_{3}=(3,\sqrt{3},3\sqrt{3}).设二面角P-BE-D的平面角为\theta ,则|\\cos   \theta |=|\\cos  \langle n_{2},n_{3}\rangle |=\frac{|\boldsymbol{n}_2 \cdot \boldsymbol{n}_3|}{|\boldsymbol{n}_2||\boldsymbol{n}_3|}=\frac{3\sqrt{273}}{91}，因此\\sin   \theta =\sqrt{1-\\\cos   ^2\theta }=\frac{8\sqrt{91}}{91}，所以二面角P-BE-D的平面角的正弦值是\frac{8\sqrt{91}}{91}$

Final answer: $\frac{8\sqrt{91}}{91}$
Subfield: Solid Geometry


Question: $如图，在四棱锥P-ABCD中，底面ABCD是等腰梯形，CD=2AB=2BC=2AD=4，平面ADP\perp 平面ABCD,E是PC的中点，且\triangle ADP为等边三角形，平面ADP\cap 平面PBC=m.$<img_2436>
$(2)求二面角P-BE-D的平面角的正弦值.$

Solution:
$由(1)得\overrightarrow{BP}=(-2,-\sqrt{3},\sqrt{3}),\overrightarrow{BD}=(-3,-\sqrt{3},0),\overrightarrow{BE}=\left(-\frac{3}{2},0,\frac{\sqrt{3}}{2}\right).设平面BDE的一个法向量为n_{2}=(x_{2},y_{2},z_{2})，由\left\{\begin{matrix}\boldsymbol{n}_2 \cdot \overrightarrow{BD}=0,\\ \boldsymbol{n}_2 \cdot \overrightarrow{BE}=0,\end{matrix}\right.得\left\{\begin{matrix}-3x_2-\sqrt{3}y_2=0,\\ -\frac{3}{2}x_2+\frac{\sqrt{3}}{2}z_2=0,\end{matrix}\right.令x_{2}=1，得y_{2}=-\sqrt{3},z_{2}=\sqrt{3}，则n_{2}=(1,-\sqrt{3},\sqrt{3}).设平面PBE的一个法向量为n_{3}=(x_{3},y_{3},z_{3})，由\left\{\begin{matrix}\boldsymbol{n}_3 \cdot \overrightarrow{BP}=0,\\ \boldsymbol{n}_3 \cdot \overrightarrow{BE}=0,\end{matrix}\right.得\left\{\begin{matrix}-2x_3-\sqrt{3}y_3+\sqrt{3}z_3=0,\\ -\frac{3}{2}x_3+\frac{\sqrt{3}}{2}z_3=0,\end{matrix}\right.令x_{3}=3，得y_{3}=\sqrt{3},z_{3}=3\sqrt{3}，则n_{3}=(3,\sqrt{3},3\sqrt{3}).设二面角P-BE-D的平面角为\theta ,则|\\cos   \theta |=|\\cos  \langle n_{2},n_{3}\rangle |=\frac{|\boldsymbol{n}_2 \cdot \boldsymbol{n}_3|}{|\boldsymbol{n}_2||\boldsymbol{n}_3|}=\frac{3\sqrt{273}}{91}，因此\\sin   \theta =\sqrt{1-\\\cos   ^2\theta }=\frac{8\sqrt{91}}{91}，所以二面角P-BE-D的平面角的正弦值是\frac{8\sqrt{91}}{91}$

Final answer: $\frac{8\sqrt{91}}{91}$
Subfield: Solid Geometry


Question: $如图，在直三棱柱ABC-A_{1}B_{1}C_{1}中，AC\perp BC，AC=BC=AA_{1}=2.$![Image](https://cdntk.53zaixian.com/tq/1658661561549000707/img/compression/M_64fa35664be5580b12fd319b38695824.jpg)
$(1)求证：A_{1}C\perp BC；$

Solution:
$证明：因为三棱柱ABC-A_{1}B_{1}C_{1}是直三棱柱，所以CC_{1}\perp 平面ABC，所以CC_{1}\perp AC,CC_{1}\perp BC，又AC\perp BC，AC\cap CC_{1}=C，所以BC\perp 平面ACC_{1}A_{1}，所以A_{1}C\perp BC.$

Final answer: 证明题
Subfield: Solid Geometry


Question: $如图，在直三棱柱ABC-A_{1}B_{1}C_{1}中，AC\perp BC，AC=BC=AA_{1}=2.$<img_2245>
$(1)求证：A_{1}C\perp BC；$

Solution:
$证明：因为三棱柱ABC-A_{1}B_{1}C_{1}是直三棱柱，所以CC_{1}\perp 平面ABC，所以CC_{1}\perp AC,CC_{1}\perp BC，又AC\perp BC，AC\cap CC_{1}=C，所以BC\perp 平面ACC_{1}A_{1}，所以A_{1}C\perp BC.$
Subfield: Solid Geometry


Question: $如图，在直三棱柱ABC-A_{1}B_{1}C_{1}中，AC\perp BC，AC=BC=AA_{1}=2.$![Image](https://cdntk.53zaixian.com/tq/1658661561549000707/img/compression/M_64fa35664be5580b12fd319b38695824.jpg)
$(2)求直线AC_{1}和A_{1}B_{1}夹角的大小.$

Solution:
$以C为原点建立如图所示的空间直角坐标系，$![Image](https://cdntk.53zaixian.com/tq/1658661562043928577/img/compression/M_f19d75d12e524ab81432adfd637454ab.jpg)$则A(2,0,0),A_{1}(2，0，2)，C_{1}(0,0,2),B_{1}(0,2,2)，则\overrightarrow{AC_1}=(-2,0,2),\overrightarrow{A_1B_1}=(-2,2,0)，设直线AC_{1}和A_{1}B_{1}的夹角为\alpha ，则\\cos   \alpha =\left|\frac{\overrightarrow{AC_1} \cdot \overrightarrow{A_1B_1}}{|\overrightarrow{AC_1}||\overrightarrow{A_1B_1}|}\right|=\frac{4}{2\sqrt{2} \times 2\sqrt{2}}=\frac{1}{2}，所以直线AC_{1}与A_{1}B_{1}的夹角为\frac{\pi }{3}.$

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


Question: $如图，在直三棱柱ABC-A_{1}B_{1}C_{1}中，AC\perp BC，AC=BC=AA_{1}=2.$<img_2245>
$(2)求直线AC_{1}和A_{1}B_{1}夹角的大小.$

Solution:
$以C为原点建立如图所示的空间直角坐标系，$<img_3090>$则A(2,0,0),A_{1}(2，0，2)，C_{1}(0,0,2),B_{1}(0,2,2)，则\overrightarrow{AC_1}=(-2,0,2),\overrightarrow{A_1B_1}=(-2,2,0)，设直线AC_{1}和A_{1}B_{1}的夹角为\alpha ，则\\cos   \alpha =\left|\frac{\overrightarrow{AC_1} \cdot \overrightarrow{A_1B_1}}{|\overrightarrow{AC_1}||\overrightarrow{A_1B_1}|}\right|=\frac{4}{2\sqrt{2} \times 2\sqrt{2}}=\frac{1}{2}，所以直线AC_{1}与A_{1}B_{1}的夹角为\frac{\pi }{3}.$

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


Question: $在三棱柱ABC-A_{1}B_{1}C_{1}中，AA_{1}\perp 底面ABC，AB=AC=AA_{1}，AB\perp AC，P为线段BC_{1}上一点.$![Image](https://cdntk.53zaixian.com/tq/1658661571749548033/img/compression/M_7f5a0dd4deed62c8c73473e7e6786aeb.jpg)
$(1)若BP=PC_{1}，求PC与AA_{1}夹角的余弦值；$

Solution:
$以A为原点，AB，AC，AA_{1}所在直线分别为x轴、y轴、z轴建立空间直角坐标系，如图所示.$![Image](https://cdntk.53zaixian.com/tq/1658661571485306882/img/compression/M_0ad0445e89554a86ac3c49ca3c2ae01d.jpg)$设AB=1，则A(0,0,0),B(1,0,0),C(0,1,0),A_{1}(0,0,1),C_{1}(0,1,1),\therefore \overrightarrow{AA_1}=(0,0,1).\because BP=PC_{1}，\therefore P\left(\frac{1}{2},\frac{1}{2},\frac{1}{2}\right)，\therefore \overrightarrow{CP}=\left(\frac{1}{2},-\frac{1}{2},\frac{1}{2}\right).设PC与AA_{1}的夹角为\theta ，则\\cos   \theta =\frac{|\overrightarrow{CP} \cdot \overrightarrow{AA_1}|}{|\overrightarrow{CP}||\overrightarrow{AA_1}|}=\frac{\sqrt{3}}{3}.$

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


Question: $在三棱柱ABC-A_{1}B_{1}C_{1}中，AA_{1}\perp 底面ABC，AB=AC=AA_{1}，AB\perp AC，P为线段BC_{1}上一点.$<img_1754>
$(1)若BP=PC_{1}，求PC与AA_{1}夹角的余弦值；$

Solution:
$以A为原点，AB，AC，AA_{1}所在直线分别为x轴、y轴、z轴建立空间直角坐标系，如图所示.$<img_3099>$设AB=1，则A(0,0,0),B(1,0,0),C(0,1,0),A_{1}(0,0,1),C_{1}(0,1,1),\therefore \overrightarrow{AA_1}=(0,0,1).\because BP=PC_{1}，\therefore P\left(\frac{1}{2},\frac{1}{2},\frac{1}{2}\right)，\therefore \overrightarrow{CP}=\left(\frac{1}{2},-\frac{1}{2},\frac{1}{2}\right).设PC与AA_{1}的夹角为\theta ，则\\cos   \theta =\frac{|\overrightarrow{CP} \cdot \overrightarrow{AA_1}|}{|\overrightarrow{CP}||\overrightarrow{AA_1}|}=\frac{\sqrt{3}}{3}.$

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


Question: $在三棱柱ABC-A_{1}B_{1}C_{1}中，AA_{1}\perp 底面ABC，AB=AC=AA_{1}，AB\perp AC，P为线段BC_{1}上一点.$<img_1754>
$(2)若BP=\sqrt{2}PC_{1}，求PC与平面ABB_{1}A_{1}的夹角；$

Solution:
$设P(a,b,c)，由BP=\sqrt{2}PC_{1}，即\overrightarrow{BP}=\sqrt{2}\overrightarrow{PC_1}得(a-1,b,c)=\sqrt{2}(-a,1-b,1-c),可得P(\sqrt{2}-1,2-\sqrt{2},2-\sqrt{2})，\therefore \overrightarrow{CP}=(\sqrt{2}-1,1-\sqrt{2},2-\sqrt{2}).设PC与平面ABB_{1}A_{1}的夹角为\alpha ，易知平面ABB_{1}A_{1}的一个法向量为n=(0,1,0),则\\sin   \alpha =\frac{|\boldsymbol{n} \cdot \overrightarrow{CP}|}{|\boldsymbol{n}||\overrightarrow{CP}|}=\frac{|1-\sqrt{2}|}{2\sqrt{2}-2}=\frac{1}{2},\therefore PC与平面ABB_{1}A_{1}的夹角为30^\circ .$

Final answer: $30^\circ $
Subfield: Solid Geometry


Question: $在三棱柱ABC-A_{1}B_{1}C_{1}中，AA_{1}\perp 底面ABC，AB=AC=AA_{1}，AB\perp AC，P为线段BC_{1}上一点.$![Image](https://cdntk.53zaixian.com/tq/1658661571749548033/img/compression/M_7f5a0dd4deed62c8c73473e7e6786aeb.jpg)
$(2)若BP=\sqrt{2}PC_{1}，求PC与平面ABB_{1}A_{1}的夹角；$

Solution:
$设P(a,b,c)，由BP=\sqrt{2}PC_{1}，即\overrightarrow{BP}=\sqrt{2}\overrightarrow{PC_1}得(a-1,b,c)=\sqrt{2}(-a,1-b,1-c),可得P(\sqrt{2}-1,2-\sqrt{2},2-\sqrt{2})，\therefore \overrightarrow{CP}=(\sqrt{2}-1,1-\sqrt{2},2-\sqrt{2}).设PC与平面ABB_{1}A_{1}的夹角为\alpha ，易知平面ABB_{1}A_{1}的一个法向量为n=(0,1,0),则\\sin   \alpha =\frac{|\boldsymbol{n} \cdot \overrightarrow{CP}|}{|\boldsymbol{n}||\overrightarrow{CP}|}=\frac{|1-\sqrt{2}|}{2\sqrt{2}-2}=\frac{1}{2},\therefore PC与平面ABB_{1}A_{1}的夹角为30^\circ .$

Final answer: $30^\circ $
Subfield: Solid Geometry


Question: $在三棱柱ABC-A_{1}B_{1}C_{1}中，AA_{1}\perp 底面ABC，AB=AC=AA_{1}，AB\perp AC，P为线段BC_{1}上一点.$<img_1754>
$(3)若平面A_{1}ACC_{1}与平面ACP的夹角为45^\circ ，求\frac{BP}{PC_1}的值.$

Solution:
$设\overrightarrow{BP}=\lambda \overrightarrow{BC_1}=(-\lambda ,\lambda ,\lambda )，0\leq \lambda \leq 1,则\overrightarrow{CP}=\overrightarrow{CB}+\overrightarrow{BP}=(1,-1,0)+(-\lambda ,\lambda ,\lambda )=(1-\lambda ,\lambda -1,\lambda ).易得\overrightarrow{AC}=(0,1,0)，设平面ACP的一个法向量为m=(x,y,z),则\left\{\begin{matrix}\boldsymbol{m} \cdot \overrightarrow{AC}=0,\\ \boldsymbol{m} \cdot \overrightarrow{CP}=0,\end{matrix}\right.即\left\{\begin{matrix}y=0,\\ (1-\lambda )x+(\lambda -1)y+\lambda z=0,\end{matrix}\right.取z=\lambda -1，得m=(\lambda ,0,\lambda -1)，易知平面A_{1}ACC_{1}的一个法向量为p=(1,0,0),\because 平面A_{1}ACC_{1}与平面ACP的夹角为45^\circ ，\therefore \\cos   45^\circ =\frac{|\boldsymbol{m} \cdot \boldsymbol{p}|}{|\boldsymbol{m}||\boldsymbol{p}|}=\frac{|\lambda |}{\sqrt{\lambda ^2+{(\lambda -1)}^2}}=\frac{\sqrt{2}}{2},解得\lambda =\frac{1}{2}，则\overrightarrow{BP}=\frac{1}{2}\overrightarrow{BC_1}，\therefore P为BC_{1}的中点，\therefore \frac{BP}{PC_1}=1,\therefore 当平面A_{1}ACC_{1}与平面ACP的夹角为45^\circ 时，\frac{BP}{PC_1}=1.$

Final answer: $1$
Subfield: Solid Geometry


Question: $在三棱柱ABC-A_{1}B_{1}C_{1}中，AA_{1}\perp 底面ABC，AB=AC=AA_{1}，AB\perp AC，P为线段BC_{1}上一点.$![Image](https://cdntk.53zaixian.com/tq/1658661571749548033/img/compression/M_7f5a0dd4deed62c8c73473e7e6786aeb.jpg)
$(3)若平面A_{1}ACC_{1}与平面ACP的夹角为45^\circ ，求\frac{BP}{PC_1}的值.$

Solution:
$设\overrightarrow{BP}=\lambda \overrightarrow{BC_1}=(-\lambda ,\lambda ,\lambda )，0\leq \lambda \leq 1,则\overrightarrow{CP}=\overrightarrow{CB}+\overrightarrow{BP}=(1,-1,0)+(-\lambda ,\lambda ,\lambda )=(1-\lambda ,\lambda -1,\lambda ).易得\overrightarrow{AC}=(0,1,0)，设平面ACP的一个法向量为m=(x,y,z),则\left\{\begin{matrix}\boldsymbol{m} \cdot \overrightarrow{AC}=0,\\ \boldsymbol{m} \cdot \overrightarrow{CP}=0,\end{matrix}\right.即\left\{\begin{matrix}y=0,\\ (1-\lambda )x+(\lambda -1)y+\lambda z=0,\end{matrix}\right.取z=\lambda -1，得m=(\lambda ,0,\lambda -1)，易知平面A_{1}ACC_{1}的一个法向量为p=(1,0,0),\because 平面A_{1}ACC_{1}与平面ACP的夹角为45^\circ ，\therefore \\cos   45^\circ =\frac{|\boldsymbol{m} \cdot \boldsymbol{p}|}{|\boldsymbol{m}||\boldsymbol{p}|}=\frac{|\lambda |}{\sqrt{\lambda ^2+{(\lambda -1)}^2}}=\frac{\sqrt{2}}{2},解得\lambda =\frac{1}{2}，则\overrightarrow{BP}=\frac{1}{2}\overrightarrow{BC_1}，\therefore P为BC_{1}的中点，\therefore \frac{BP}{PC_1}=1,\therefore 当平面A_{1}ACC_{1}与平面ACP的夹角为45^\circ 时，\frac{BP}{PC_1}=1.$

Final answer: $1$
Subfield: Solid Geometry


Question: $如图所示，已知空间四边形ABCD的每条边和对角线都等于1，E，F分别是AB，AD的中点，设{a，b，c}为空间向量的一组基，\overrightarrow{AB}=a，\overrightarrow{AC}=b，\overrightarrow{AD}=c，试用基向量法求解以下各题.$<img_2967>
$(1)求\overrightarrow{EF}\cdot \overrightarrow{AB}的值；$

Solution:
$由题意得|a|=|b|=|c|=1，\langle a,b\rangle =\langle b,c\rangle =\langle c,a\rangle =60^\circ ，故a\cdot b=b\cdot c=c\cdot a=1\times 1\times \\cos   60^\circ =\frac{1}{2}.
(3分)
\because \overrightarrow{EF}=\frac{1}{2}\overrightarrow{BD}=\frac{1}{2}(\overrightarrow{AD}-\overrightarrow{AB})=\frac{1}{2}(c-a)，\therefore \overrightarrow{EF}\cdot \overrightarrow{AB}=\frac{1}{2}(c-a)\cdot a=\frac{1}{2}c\cdot a-\frac{1}{2}a^{2}=\frac{1}{2}\times \frac{1}{2}-\frac{1}{2}=-\frac{1}{4}.
(6分)$

Final answer: $-1/4$
Subfield: Solid Geometry


Question: $如图所示，已知空间四边形ABCD的每条边和对角线都等于1，E，F分别是AB，AD的中点，设{a，b，c}为空间向量的一组基，\overrightarrow{AB}=a，\overrightarrow{AC}=b，\overrightarrow{AD}=c，试用基向量法求解以下各题.$![Image](https://cdntk.53zaixian.com/tq/1658663329313918977/img/compression/M_80b124012f7f2378c9b99900c76a97f1.jpg)
$(1)求\overrightarrow{EF}\cdot \overrightarrow{AB}的值；$

Solution:
$由题意得|a|=|b|=|c|=1，\langle a,b\rangle =\langle b,c\rangle =\langle c,a\rangle =60^\circ ，故a\cdot b=b\cdot c=c\cdot a=1\times 1\times \\cos   60^\circ =\frac{1}{2}.
(3分)
\because \overrightarrow{EF}=\frac{1}{2}\overrightarrow{BD}=\frac{1}{2}(\overrightarrow{AD}-\overrightarrow{AB})=\frac{1}{2}(c-a)，\therefore \overrightarrow{EF}\cdot \overrightarrow{AB}=\frac{1}{2}(c-a)\cdot a=\frac{1}{2}c\cdot a-\frac{1}{2}a^{2}=\frac{1}{2}\times \frac{1}{2}-\frac{1}{2}=-\frac{1}{4}.
(6分)$

Final answer: $-1/4$