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


Question: $如图，已知平行六面体ABCD-A'B'C'D'.$![Image](https://cdntk.53zaixian.com/tq/1693855834103803906/img/compression/M_7deee49da7f2ee1cfd7bb65cd9e773e0.jpg)
$(3)设M是底面ABCD的中心，N是侧面BCC'B'对角线BC'上靠近C'的四等分点，设\overrightarrow{MN}=\alpha \overrightarrow{AB}+\beta \overrightarrow{AD}+\gamma \overrightarrow{AA'}，试求\alpha ，\beta ，\gamma 的值.$

Solution:
$因为\overrightarrow{MN}=\overrightarrow{MB}+\overrightarrow{BN}=\frac{1}{2}\overrightarrow{DB}+\frac{3}{4}\overrightarrow{BC'}=\frac{1}{2}(\overrightarrow{DA}+\overrightarrow{AB})+\frac{3}{4}(\overrightarrow{BC}+\overrightarrow{CC'})=\frac{1}{2}\overrightarrow{AB}+\frac{1}{4}\overrightarrow{AD}+\frac{3}{4}\overrightarrow{AA'}，所以\alpha =\frac{1}{2}，\beta =\frac{1}{4}，\gamma =\frac{3}{4}.$

Final answer: $\frac{1}{2},\frac{1}{4},\frac{3}{4}$
Subfield: Solid Geometry


Question: $如图，已知平行六面体ABCD-A'B'C'D'.$<img_1483>
$(3)设M是底面ABCD的中心，N是侧面BCC'B'对角线BC'上靠近C'的四等分点，设\overrightarrow{MN}=\alpha \overrightarrow{AB}+\beta \overrightarrow{AD}+\gamma \overrightarrow{AA'}，试求\alpha ，\beta ，\gamma 的值.$

Solution:
$因为\overrightarrow{MN}=\overrightarrow{MB}+\overrightarrow{BN}=\frac{1}{2}\overrightarrow{DB}+\frac{3}{4}\overrightarrow{BC'}=\frac{1}{2}(\overrightarrow{DA}+\overrightarrow{AB})+\frac{3}{4}(\overrightarrow{BC}+\overrightarrow{CC'})=\frac{1}{2}\overrightarrow{AB}+\frac{1}{4}\overrightarrow{AD}+\frac{3}{4}\overrightarrow{AA'}，所以\alpha =\frac{1}{2}，\beta =\frac{1}{4}，\gamma =\frac{3}{4}.$

Final answer: $\frac{1}{2},\frac{1}{4},\frac{3}{4}$
Subfield: Solid Geometry


Question: $如图所示，已知正四面体OABC的棱长为1，E，F分别是OA，OC的中点,则:$<img_2650>
$(1)\overrightarrow{OA}\cdot \overrightarrow{OB}=　　　　；$

Solution:
$因为正四面体OABC的棱长为1，所以|\overrightarrow{OA}|=|\overrightarrow{OB}|=1，\triangle OAB为等边三角形，所以\angle AOB=60^\circ ，故\overrightarrow{OA}\cdot \overrightarrow{OB}=|\overrightarrow{OA}||\overrightarrow{OB}|\\cos  \langle \overrightarrow{OA}，\overrightarrow{OB}\rangle =|\overrightarrow{OA}||\overrightarrow{OB}|\cdot \\cos  \angle AOB=1\times 1\times \\cos   60^\circ =\frac{1}{2}.$

Final answer: $\frac{1}{2}$
Subfield: Solid Geometry


Question: $如图所示，已知正四面体OABC的棱长为1，E，F分别是OA，OC的中点,则:$![Image](https://cdntk.53zaixian.com/tq/1693855840068104194/img/compression/M_458b047e5a960906db2b10afedc5811e.jpg)
$(1)\overrightarrow{OA}\cdot \overrightarrow{OB}=　　　　；$

Solution:
$因为正四面体OABC的棱长为1，所以|\overrightarrow{OA}|=|\overrightarrow{OB}|=1，\triangle OAB为等边三角形，所以\angle AOB=60^\circ ，故\overrightarrow{OA}\cdot \overrightarrow{OB}=|\overrightarrow{OA}||\overrightarrow{OB}|\\cos  \langle \overrightarrow{OA}，\overrightarrow{OB}\rangle =|\overrightarrow{OA}||\overrightarrow{OB}|\cdot \\cos  \angle AOB=1\times 1\times \\cos   60^\circ =\frac{1}{2}.$

Final answer: $\frac{1}{2}$
Subfield: Solid Geometry


Question: $如图所示，已知正四面体OABC的棱长为1，E，F分别是OA，OC的中点,则:$![Image](https://cdntk.53zaixian.com/tq/1693855840068104194/img/compression/M_458b047e5a960906db2b10afedc5811e.jpg)
$(2)\overrightarrow{EF}\cdot \overrightarrow{CB}=　　　　；$

Solution:
$因为E，F分别是OA，OC的中点，所以EF\parallel AC，EF=\frac{1}{2}AC，故\overrightarrow{EF}\cdot \overrightarrow{CB}=|\overrightarrow{EF}||\overrightarrow{CB}|\\cos  \langle \overrightarrow{EF}，\overrightarrow{CB}\rangle =\frac{1}{2}|\overrightarrow{CA}||\overrightarrow{CB}|\cdot \\cos  \langle \overrightarrow{AC}，\overrightarrow{CB}\rangle =\frac{1}{2}\times 1\times 1\times \\cos   120^\circ =-\frac{1}{4}.$

Final answer: $-\frac{1}{4}$
Subfield: Solid Geometry


Question: $如图所示，已知正四面体OABC的棱长为1，E，F分别是OA，OC的中点,则:$<img_2650>
$(2)\overrightarrow{EF}\cdot \overrightarrow{CB}=　　　　；$

Solution:
$因为E，F分别是OA，OC的中点，所以EF\parallel AC，EF=\frac{1}{2}AC，故\overrightarrow{EF}\cdot \overrightarrow{CB}=|\overrightarrow{EF}||\overrightarrow{CB}|\\cos  \langle \overrightarrow{EF}，\overrightarrow{CB}\rangle =\frac{1}{2}|\overrightarrow{CA}||\overrightarrow{CB}|\cdot \\cos  \langle \overrightarrow{AC}，\overrightarrow{CB}\rangle =\frac{1}{2}\times 1\times 1\times \\cos   120^\circ =-\frac{1}{4}.$

Final answer: $-\frac{1}{4}$
Subfield: Solid Geometry


Question: $如图所示，已知正四面体OABC的棱长为1，E，F分别是OA，OC的中点,则:$<img_2650>
$(3)(\overrightarrow{OA}+\overrightarrow{OB})\cdot (\overrightarrow{CA}+\overrightarrow{CB})=　　　　.$

Solution:
$(\overrightarrow{OA}+\overrightarrow{OB})\cdot (\overrightarrow{CA}+\overrightarrow{CB})=(\overrightarrow{OA}+\overrightarrow{OB})\cdot (\overrightarrow{OA}-\overrightarrow{OC}+\overrightarrow{OB}-\overrightarrow{OC})=(\overrightarrow{OA}+\overrightarrow{OB})\cdot (\overrightarrow{OA}+\overrightarrow{OB}-2\overrightarrow{OC})=\overrightarrow{OA}^{2}+\overrightarrow{OA}\cdot \overrightarrow{OB}-2\overrightarrow{OA}\cdot \overrightarrow{OC}+\overrightarrow{OB}\cdot \overrightarrow{OA}+\overrightarrow{OB}^{2}-2\overrightarrow{OB}\cdot \overrightarrow{OC}=1+\frac{1}{2}-2\times \frac{1}{2}+\frac{1}{2}+1-2\times \frac{1}{2}=1.$

Final answer: $1$
Subfield: Solid Geometry


Question: $如图所示，已知正四面体OABC的棱长为1，E，F分别是OA，OC的中点,则:$![Image](https://cdntk.53zaixian.com/tq/1693855840068104194/img/compression/M_458b047e5a960906db2b10afedc5811e.jpg)
$(3)(\overrightarrow{OA}+\overrightarrow{OB})\cdot (\overrightarrow{CA}+\overrightarrow{CB})=　　　　.$

Solution:
$(\overrightarrow{OA}+\overrightarrow{OB})\cdot (\overrightarrow{CA}+\overrightarrow{CB})=(\overrightarrow{OA}+\overrightarrow{OB})\cdot (\overrightarrow{OA}-\overrightarrow{OC}+\overrightarrow{OB}-\overrightarrow{OC})=(\overrightarrow{OA}+\overrightarrow{OB})\cdot (\overrightarrow{OA}+\overrightarrow{OB}-2\overrightarrow{OC})=\overrightarrow{OA}^{2}+\overrightarrow{OA}\cdot \overrightarrow{OB}-2\overrightarrow{OA}\cdot \overrightarrow{OC}+\overrightarrow{OB}\cdot \overrightarrow{OA}+\overrightarrow{OB}^{2}-2\overrightarrow{OB}\cdot \overrightarrow{OC}=1+\frac{1}{2}-2\times \frac{1}{2}+\frac{1}{2}+1-2\times \frac{1}{2}=1.$

Final answer: $1$
Subfield: Solid Geometry


Question: $在直三棱柱ABC-A_{1}B_{1}C_{1}中，AA_{1}=AB=BC=3，AC=2，D是AC的中点.$<img_2848>
$(1)求证：B_{1}C\parallel 平面A_{1}BD；$

Solution:
$证明：连接AB_{1}交A_{1}B于点E，连接DE，易得DE\parallel B_{1}C，因为DE\subset 平面A_{1}BD，B_{1}C\not\subset 平面A_{1}BD，所以B_{1}C\parallel 平面A_{1}BD.$
Subfield: Solid Geometry


Question: $在直三棱柱ABC-A_{1}B_{1}C_{1}中，AA_{1}=AB=BC=3，AC=2，D是AC的中点.$![Image](https://cdntk.53zaixian.com/tq/1693855913519620097/img/compression/M_bd13a21e5a8c7804dc308fb1bb686e58.jpg)
$(1)求证：B_{1}C\parallel 平面A_{1}BD；$

Solution:
$证明：连接AB_{1}交A_{1}B于点E，连接DE，易得DE\parallel B_{1}C，因为DE\subset 平面A_{1}BD，B_{1}C\not\subset 平面A_{1}BD，所以B_{1}C\parallel 平面A_{1}BD.$

Final answer: 证明题
Subfield: Solid Geometry


Question: $在直三棱柱ABC-A_{1}B_{1}C_{1}中，AA_{1}=AB=BC=3，AC=2，D是AC的中点.$![Image](https://cdntk.53zaixian.com/tq/1693855913519620097/img/compression/M_bd13a21e5a8c7804dc308fb1bb686e58.jpg)
$(2)求直线B_{1}C到平面A_{1}BD的距离.$

Solution:
$由(1)知B_{1}C\parallel 平面A_{1}BD，所以B_{1}C到平面A_{1}BD的距离即为点B_{1}到平面A_{1}BD的距离，建立如图所示的空间直角坐标系D-xyz，$![Image](https://cdntk.53zaixian.com/tq/1693855913519620097/img/compression/M_48217433406d95cece2ef9cdec238e5f.jpg)$则D(0,0,0),B_{1}(0,2\sqrt{2}，3)，B(0,2\sqrt{2}，0)，A_{1}(-1,0,3)，所以\overrightarrow{DB_1}=(0,2\sqrt{2}，3)，\overrightarrow{DB}=(0,2\sqrt{2}，0)，\overrightarrow{DA_1}=(-1,0,3).设平面A_{1}BD的一个法向量为n=(x，y，z)，则\left\{\begin{matrix}\boldsymbol{n}\cdot \overrightarrow{DB}=0,\\ \boldsymbol{n}\cdot \overrightarrow{DA_1}=0,\end{matrix}\right.即\left\{\begin{matrix}2\sqrt{2}y=0,\\ -x+3z=0,\end{matrix}\right.令z=1,则y=0,x=3,所以n=(3,0,1)，故直线B_{1}C到平面A_{1}BD的距离d=\frac{|\boldsymbol{n}\cdot \overrightarrow{DB_1}|}{|\boldsymbol{n}|}=\frac{3\sqrt{10}}{10}.$

Final answer: $\frac{3\sqrt{10}}{10}$
Subfield: Solid Geometry


Question: $在直三棱柱ABC-A_{1}B_{1}C_{1}中，AA_{1}=AB=BC=3，AC=2，D是AC的中点.$<img_2848>
$(2)求直线B_{1}C到平面A_{1}BD的距离.$

Solution:
$由(1)知B_{1}C\parallel 平面A_{1}BD，所以B_{1}C到平面A_{1}BD的距离即为点B_{1}到平面A_{1}BD的距离，建立如图所示的空间直角坐标系D-xyz，$<img_2560>$则D(0,0,0),B_{1}(0,2\sqrt{2}，3)，B(0,2\sqrt{2}，0)，A_{1}(-1,0,3)，所以\overrightarrow{DB_1}=(0,2\sqrt{2}，3)，\overrightarrow{DB}=(0,2\sqrt{2}，0)，\overrightarrow{DA_1}=(-1,0,3).设平面A_{1}BD的一个法向量为n=(x，y，z)，则\left\{\begin{matrix}\boldsymbol{n}\cdot \overrightarrow{DB}=0,\\ \boldsymbol{n}\cdot \overrightarrow{DA_1}=0,\end{matrix}\right.即\left\{\begin{matrix}2\sqrt{2}y=0,\\ -x+3z=0,\end{matrix}\right.令z=1,则y=0,x=3,所以n=(3,0,1)，故直线B_{1}C到平面A_{1}BD的距离d=\frac{|\boldsymbol{n}\cdot \overrightarrow{DB_1}|}{|\boldsymbol{n}|}=\frac{3\sqrt{10}}{10}.$

Final answer: $\frac{3\sqrt{10}}{10}$
Subfield: Solid Geometry


Question: $如图所示，在四棱锥P-ABCD中，底面ABCD为正方形，PA\perp 底面ABCD，点E在线段PC上(不含端点).$![Image](https://cdntk.53zaixian.com/tq/1693855885307740163/img/compression/M_faae8356c78862bcff2aac37864a0ad5.jpg)
$(2)是否存在点E，使得平面PCD\perp 平面AED？$

Solution:
$\overrightarrow{PC}=(a,a,-c)，\overrightarrow{DC}=(a,0,0)，\overrightarrow{PA}=(0，0，-c)，设平面PCD的一个法向量为n_{1}=(x_{1},y_{1},z_{1}),则\left\{\begin{matrix}\boldsymbol{n}_1\cdot \overrightarrow{PC}=ax_1+ay_1-cz_1=0,\\ \boldsymbol{n}_1\cdot \overrightarrow{DC}=ax_1=0,\end{matrix}\right.令y_{1}=c，则x_{1}=0，z_{1}=a，\therefore n_{1}=(0,c,a).设\overrightarrow{PE}=\mu  \overrightarrow{PC}，0<\mu <1,则\overrightarrow{AE}=\overrightarrow{PE}-\overrightarrow{PA}=\mu \overrightarrow{PC}-\overrightarrow{PA}=(\mu a,\mu a,c-\mu c)，易知\overrightarrow{AD}=(0,a,0).设平面AED的一个法向量为n_{2}=(x_{2},y_{2},z_{2})，则\left\{\begin{matrix}\boldsymbol{n}_2\cdot \overrightarrow{AE}=\mu ax_2+\mu ay_2+cz_2-\mu cz_2=0,\\ \boldsymbol{n}_2\cdot \overrightarrow{AD}=ay_2=0,\end{matrix}\right.令x_{2}=-c，则y_{2}=0,z_{2}=\frac{\mu a}{1-\mu }，\therefore n_{2}=\left(-c,0,\frac{\mu a}{1-\mu }\right).若平面PCD\perp 平面AED，则n_{1}\cdot n_{2}=0，即\frac{\mu a^2}{1-\mu }=0，此方程无解，\therefore 不存在点E，使得n_{1}\perp n_{2}，即不存在点E，使得平面PCD\perp 平面AED.$

Final answer: $不存在点E，使得平面PCD\perp 平面AED。$
Subfield: Solid Geometry


Question: $如图所示，在四棱锥P-ABCD中，底面ABCD为正方形，PA\perp 底面ABCD，点E在线段PC上(不含端点).$<img_2616>
$(2)是否存在点E，使得平面PCD\perp 平面AED？$

Solution:
$\overrightarrow{PC}=(a,a,-c)，\overrightarrow{DC}=(a,0,0)，\overrightarrow{PA}=(0，0，-c)，设平面PCD的一个法向量为n_{1}=(x_{1},y_{1},z_{1}),则\left\{\begin{matrix}\boldsymbol{n}_1\cdot \overrightarrow{PC}=ax_1+ay_1-cz_1=0,\\ \boldsymbol{n}_1\cdot \overrightarrow{DC}=ax_1=0,\end{matrix}\right.令y_{1}=c，则x_{1}=0，z_{1}=a，\therefore n_{1}=(0,c,a).设\overrightarrow{PE}=\mu  \overrightarrow{PC}，0<\mu <1,则\overrightarrow{AE}=\overrightarrow{PE}-\overrightarrow{PA}=\mu \overrightarrow{PC}-\overrightarrow{PA}=(\mu a,\mu a,c-\mu c)，易知\overrightarrow{AD}=(0,a,0).设平面AED的一个法向量为n_{2}=(x_{2},y_{2},z_{2})，则\left\{\begin{matrix}\boldsymbol{n}_2\cdot \overrightarrow{AE}=\mu ax_2+\mu ay_2+cz_2-\mu cz_2=0,\\ \boldsymbol{n}_2\cdot \overrightarrow{AD}=ay_2=0,\end{matrix}\right.令x_{2}=-c，则y_{2}=0,z_{2}=\frac{\mu a}{1-\mu }，\therefore n_{2}=\left(-c,0,\frac{\mu a}{1-\mu }\right).若平面PCD\perp 平面AED，则n_{1}\cdot n_{2}=0，即\frac{\mu a^2}{1-\mu }=0，此方程无解，\therefore 不存在点E，使得n_{1}\perp n_{2}，即不存在点E，使得平面PCD\perp 平面AED.$



Final answer: $不存在点E，使得平面PCD\perp 平面AED。$
Subfield: Solid Geometry


Question: $等边\triangle ABC的边长为3，点D,E分别是AB，AC上的点，且满足\frac{AD}{DB}=\frac{CE}{EA}=\frac{1}{2}(如图①)，将\triangle ADE沿DE折起到\triangle A_{1}DE的位置，使平面A_{1}DE\perp 平面BCED，连接A_{1}B,A_{1}C(如图②).
$![Image](https://cdntk.53zaixian.com/tq/1693855897134235649/img/compression/M_ad109c585827f318072e59750b070b38.jpg)![Image](https://cdntk.53zaixian.com/tq/1693855897134235649/img/compression/M_8fca8f89be66def10310552e359b4124.jpg)
$(1)求证：A_{1}D\perp 平面BCED；$

Solution:
$证明：根据题意，知AE=2,AD=1,A=60^\circ ，从而DE=\sqrt{1^2+2^2-2\times 1\times 2\times \\\cos    60^{\circ}}=\sqrt{3}，故AD^{2}+DE^{2}=AE^{2}，\therefore AD\perp DE，BD\perp DE，\therefore A_{1}D\perp DE.\because 平面A_{1}DE\perp 平面BCED,平面A_{1}DE\cap 平面BCED=DE,A_{1}D\subset 平面A_{1}DE,\therefore A_{1}D\perp 平面BCED.$

Final answer: 证明题
Subfield: Solid Geometry


Question: $等边\triangle ABC的边长为3，点D,E分别是AB，AC上的点，且满足\frac{AD}{DB}=\frac{CE}{EA}=\frac{1}{2}(如图①)，将\triangle ADE沿DE折起到\triangle A_{1}DE的位置，使平面A_{1}DE\perp 平面BCED，连接A_{1}B,A_{1}C(如图②).
$<img_2619><img_2840>
$(1)求证：A_{1}D\perp 平面BCED；$

Solution:
$证明：根据题意，知AE=2,AD=1,A=60^\circ ，从而DE=\sqrt{1^2+2^2-2\times 1\times 2\times \\\cos    60^{\circ}}=\sqrt{3}，故AD^{2}+DE^{2}=AE^{2}，\therefore AD\perp DE，BD\perp DE，\therefore A_{1}D\perp DE.\because 平面A_{1}DE\perp 平面BCED,平面A_{1}DE\cap 平面BCED=DE,A_{1}D\subset 平面A_{1}DE,\therefore A_{1}D\perp 平面BCED.$
Subfield: Solid Geometry


Question: $等边\triangle ABC的边长为3，点D,E分别是AB，AC上的点，且满足\frac{AD}{DB}=\frac{CE}{EA}=\frac{1}{2}(如图①)，将\triangle ADE沿DE折起到\triangle A_{1}DE的位置，使平面A_{1}DE\perp 平面BCED，连接A_{1}B,A_{1}C(如图②).
$![Image](https://cdntk.53zaixian.com/tq/1693855897134235649/img/compression/M_ad109c585827f318072e59750b070b38.jpg)![Image](https://cdntk.53zaixian.com/tq/1693855897134235649/img/compression/M_8fca8f89be66def10310552e359b4124.jpg)
$(2)在线段A_{1}B上是否存在点P，使直线DP与直线EA_{1}所成角的余弦值为\frac{\sqrt{5}}{10}？若存在，求出\frac{A_1P}{A_1B}的值；若不存在，请说明理由.$

Solution:
$存在.由(1)知A_{1}D，BD，DE两两垂直，则以D为坐标原点，DB，DE，DA_{1}所在直线分别为x轴，y轴，z轴建立空间直角坐标系D-xyz，如图，$![Image](https://cdntk.53zaixian.com/tq/1693855897901793282/img/compression/M_8dcc813ccfee4687ef3acbd2ecb3c908.jpg)$则D(0,0,0),A_{1}(0,0,1),B(2,0,0),E(0,\sqrt{3},0),\therefore \overrightarrow{A_1B}=(2,0,-1),\overrightarrow{A_1E}=(0，\sqrt{3}，-1)，设\overrightarrow{A_1P}=\lambda \overrightarrow{A_1B}=(2\lambda ,0,-\lambda ),0\leq \lambda \leq 1，\therefore P(2\lambda ,0,1-\lambda ),则\overrightarrow{DP}=(2\lambda ,0,1-\lambda ),\therefore ｜\\cos  \langle \overrightarrow{DP},\overrightarrow{A_1E}\rangle ｜=\left |  \frac{\overrightarrow{DP}\cdot \overrightarrow{A_1E}}{|\overrightarrow{DP}||\overrightarrow{A_1E}|}                     \right | 
=\frac{｜\lambda -1｜}{2\sqrt{4\lambda ^2+(1-\lambda )^2}}=\frac{\sqrt{5}}{10},\therefore \lambda =\frac{1}{2},即\frac{A_1P}{A_1B}=\frac{1}{2}.$

Final answer: $存在，\frac{A_1P}{A_1B}=\frac{1}{2}。$
Subfield: Solid Geometry


Question: $等边\triangle ABC的边长为3，点D,E分别是AB，AC上的点，且满足\frac{AD}{DB}=\frac{CE}{EA}=\frac{1}{2}(如图①)，将\triangle ADE沿DE折起到\triangle A_{1}DE的位置，使平面A_{1}DE\perp 平面BCED，连接A_{1}B,A_{1}C(如图②).
$<img_2619><img_2840>
$(2)在线段A_{1}B上是否存在点P，使直线DP与直线EA_{1}所成角的余弦值为\frac{\sqrt{5}}{10}？若存在，求出\frac{A_1P}{A_1B}的值；若不存在，请说明理由.$

Solution:
$存在.由(1)知A_{1}D，BD，DE两两垂直，则以D为坐标原点，DB，DE，DA_{1}所在直线分别为x轴，y轴，z轴建立空间直角坐标系D-xyz，如图，$<img_2573>$则D(0,0,0),A_{1}(0,0,1),B(2,0,0),E(0,\sqrt{3},0),\therefore \overrightarrow{A_1B}=(2,0,-1),\overrightarrow{A_1E}=(0，\sqrt{3}，-1)，设\overrightarrow{A_1P}=\lambda \overrightarrow{A_1B}=(2\lambda ,0,-\lambda ),0\leq \lambda \leq 1，\therefore P(2\lambda ,0,1-\lambda ),则\overrightarrow{DP}=(2\lambda ,0,1-\lambda ),\therefore ｜\\cos  \langle \overrightarrow{DP},\overrightarrow{A_1E}\rangle ｜=\left |  \frac{\overrightarrow{DP}\cdot \overrightarrow{A_1E}}{|\overrightarrow{DP}||\overrightarrow{A_1E}|}                     \right | 
=\frac{｜\lambda -1｜}{2\sqrt{4\lambda ^2+(1-\lambda )^2}}=\frac{\sqrt{5}}{10},\therefore \lambda =\frac{1}{2},即\frac{A_1P}{A_1B}=\frac{1}{2}.$





Final answer: $\frac{A_1P}{A_1B}=\frac{1}{2}$
Subfield: Solid Geometry


Question: $在多面体ABCDE中，平面ACDE\perp 平面ABC，四边形ACDE为直角梯形，CD\parallel AE，AC\perp AE，AB\perp BC，CD=1，AE=AC=2，F为DE的中点，且点E满足\overrightarrow{EB}=4\overrightarrow{EG}.$<img_1967>
$(1)证明：GF\parallel 平面ABC；$

Solution:
$证明：分别取AB,EB的中点M,N，连接CM,MN,ND，则MN\parallel AE，MN=\frac{1}{2}AE.在梯形ACDE中，CD\parallel AE且CD=\frac{1}{2}AE，\therefore MN\parallel CD，MN=CD，\therefore 四边形CDNM是平行四边形，\therefore CM\parallel DN.\because \overrightarrow{EG}=\frac{1}{4}\overrightarrow{EB}，N为EB的中点，\therefore G为EN的中点.又F为DE的中点，\therefore GF\parallel DN，\therefore GF\parallel CM，又CM\subset 平面ABC，GF\not\subset 平面ABC，\therefore GF\parallel 平面ABC.$
Subfield: Solid Geometry


Question: $在多面体ABCDE中，平面ACDE\perp 平面ABC，四边形ACDE为直角梯形，CD\parallel AE，AC\perp AE，AB\perp BC，CD=1，AE=AC=2，F为DE的中点，且点E满足\overrightarrow{EB}=4\overrightarrow{EG}.$![Image](https://cdntk.53zaixian.com/tq/1693855901001383938/img/compression/M_1ea9444473df09695e26821ff5978f65.jpg)
$(1)证明：GF\parallel 平面ABC；$

Solution:
$证明：分别取AB,EB的中点M,N，连接CM,MN,ND，则MN\parallel AE，MN=\frac{1}{2}AE.在梯形ACDE中，CD\parallel AE且CD=\frac{1}{2}AE，\therefore MN\parallel CD，MN=CD，\therefore 四边形CDNM是平行四边形，\therefore CM\parallel DN.\because \overrightarrow{EG}=\frac{1}{4}\overrightarrow{EB}，N为EB的中点，\therefore G为EN的中点.又F为DE的中点，\therefore GF\parallel DN，\therefore GF\parallel CM，又CM\subset 平面ABC，GF\not\subset 平面ABC，\therefore GF\parallel 平面ABC.$

Final answer: 证明题
Subfield: Solid Geometry


Question: $在多面体ABCDE中，平面ACDE\perp 平面ABC，四边形ACDE为直角梯形，CD\parallel AE，AC\perp AE，AB\perp BC，CD=1，AE=AC=2，F为DE的中点，且点E满足\overrightarrow{EB}=4\overrightarrow{EG}.$![Image](https://cdntk.53zaixian.com/tq/1693855901001383938/img/compression/M_1ea9444473df09695e26821ff5978f65.jpg)
$(2)当多面体ABCDE的体积最大时，求二面角A-BE-D的余弦值.$

Solution:
$过B作BH\perp AC于H.\because 平面ACDE\perp 平面ABC，平面ACDE\cap 平面ABC=AC，BH\subset 平面ABC，\therefore BH\perp 平面ACDE，\therefore BH即为四棱锥B-ACDE的高，又底面ACDE的面积确定，\therefore 要使多面体ABCDE体积最大，应使BH的长度最大，此时AB=BC=\sqrt{2}，H为AC的中点,易知HB，HC，HF两两垂直.以H为原点，HB，HC，HF所在直线分别为x轴，y轴，z轴建立如图所示的空间直角坐标系H-xyz,$![Image](https://cdntk.53zaixian.com/tq/1693855901001383938/img/compression/M_79626d1c479c2d4a9f14fa3662ae837f.jpg)$则A(0,-1,0)，B(1,0,0)，E(0,-1,2),D(0,1,1)，\therefore \overrightarrow{AB}=(1,1,0),\overrightarrow{BE}=(-1,-1,2),\overrightarrow{DE}=(0,-2,1).设n_{1}=(x_{1},y_{1},z_{1})为平面ABE的一个法向量，则\left\{\begin{matrix}\boldsymbol{n}_1\cdot \overrightarrow{AB}=0,\\ \boldsymbol{n}_1\cdot \overrightarrow{BE}=0,\end{matrix}\right.\therefore \left\{\begin{matrix}x_1+y_1=0,\\ -x_1-y_1+2z_1=0,\end{matrix}\right.令x_{1}=1,则y_{1}=-1,z_{1}=0,\therefore n_{1}=(1,-1,0).设n_{2}=(x_{2},y_{2},z_{2})为平面DBE的一个法向量，则\left\{\begin{matrix}\boldsymbol{n}_2\cdot \overrightarrow{DE}=0,\\ \boldsymbol{n}_2\cdot \overrightarrow{BE}=0,\end{matrix}\right.\therefore \left\{\begin{matrix}-2y_2+z_2=0,\\ -x_2-y_2+2z_2=0,\end{matrix}\right.令y_{2}=1,则x_{2}=3,z_{2}=2,\therefore n_{2}=(3,1,2),\therefore \\cos  \langle n_{1},n_{2}\rangle =\frac{\boldsymbol{n}_1\cdot \boldsymbol{n}_2}{|\boldsymbol{n}_1|\cdot |\boldsymbol{n}_2|}=\frac{\sqrt{7}}{7}，由图知，二面角A-BE-D为钝二面角，\therefore 二面角A-BE-D的余弦值为-\frac{\sqrt{7}}{7}.$

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


Question: $在多面体ABCDE中，平面ACDE\perp 平面ABC，四边形ACDE为直角梯形，CD\parallel AE，AC\perp AE，AB\perp BC，CD=1，AE=AC=2，F为DE的中点，且点E满足\overrightarrow{EB}=4\overrightarrow{EG}.$<img_1967>
$(2)当多面体ABCDE的体积最大时，求二面角A-BE-D的余弦值.$

Solution:
$过B作BH\perp AC于H.\because 平面ACDE\perp 平面ABC，平面ACDE\cap 平面ABC=AC，BH\subset 平面ABC，\therefore BH\perp 平面ACDE，\therefore BH即为四棱锥B-ACDE的高，又底面ACDE的面积确定，\therefore 要使多面体ABCDE体积最大，应使BH的长度最大，此时AB=BC=\sqrt{2}，H为AC的中点,易知HB，HC，HF两两垂直.以H为原点，HB，HC，HF所在直线分别为x轴，y轴，z轴建立如图所示的空间直角坐标系H-xyz,$<img_2374>$则A(0,-1,0)，B(1,0,0)，E(0,-1,2),D(0,1,1)，\therefore \overrightarrow{AB}=(1,1,0),\overrightarrow{BE}=(-1,-1,2),\overrightarrow{DE}=(0,-2,1).设n_{1}=(x_{1},y_{1},z_{1})为平面ABE的一个法向量，则\left\{\begin{matrix}\boldsymbol{n}_1\cdot \overrightarrow{AB}=0,\\ \boldsymbol{n}_1\cdot \overrightarrow{BE}=0,\end{matrix}\right.\therefore \left\{\begin{matrix}x_1+y_1=0,\\ -x_1-y_1+2z_1=0,\end{matrix}\right.令x_{1}=1,则y_{1}=-1,z_{1}=0,\therefore n_{1}=(1,-1,0).设n_{2}=(x_{2},y_{2},z_{2})为平面DBE的一个法向量，则\left\{\begin{matrix}\boldsymbol{n}_2\cdot \overrightarrow{DE}=0,\\ \boldsymbol{n}_2\cdot \overrightarrow{BE}=0,\end{matrix}\right.\therefore \left\{\begin{matrix}-2y_2+z_2=0,\\ -x_2-y_2+2z_2=0,\end{matrix}\right.令y_{2}=1,则x_{2}=3,z_{2}=2,\therefore n_{2}=(3,1,2),\therefore \\cos  \langle n_{1},n_{2}\rangle =\frac{\boldsymbol{n}_1\cdot \boldsymbol{n}_2}{|\boldsymbol{n}_1|\cdot |\boldsymbol{n}_2|}=\frac{\sqrt{7}}{7}，由图知，二面角A-BE-D为钝二面角，\therefore 二面角A-BE-D的余弦值为-\frac{\sqrt{7}}{7}.$

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


Question: $如图，四棱锥P-ABCD的底面是等腰梯形，AD\parallel BC，BC=2AD，\angle ABC=60^\circ ，E是PB的中点，F是棱PC上的点，且A，D，E，F四点共面.$![Image](https://cdntk.53zaixian.com/tq/1693855901982851073/img/compression/M_a5fd4cb35d1e11b577a1ac2d6f0241a1.jpg)
$(1)求证：F是PC的中点；$

Solution:
$证明：\because AD\parallel BC，BC\subset 平面PBC，AD\not\subset 平面PBC，\therefore AD\parallel 平面PBC.\because A,D,E,F四点共面，平面ADFE\cap 平面PBC=EF，\therefore AD\parallel EF，而AD\parallel BC，\therefore EF\parallel BC，\because E是PB的中点，\therefore F为PC的中点.$

Final answer: 证明题
Subfield: Solid Geometry


Question: $如图，四棱锥P-ABCD的底面是等腰梯形，AD\parallel BC，BC=2AD，\angle ABC=60^\circ ，E是PB的中点，F是棱PC上的点，且A，D，E，F四点共面.$<img_2982>
$(1)求证：F是PC的中点；$

Solution:
$证明：\because AD\parallel BC，BC\subset 平面PBC，AD\not\subset 平面PBC，\therefore AD\parallel 平面PBC.\because A,D,E,F四点共面，平面ADFE\cap 平面PBC=EF，\therefore AD\parallel EF，而AD\parallel BC，\therefore EF\parallel BC，\because E是PB的中点，\therefore F为PC的中点.$
Subfield: Solid Geometry


Question: $如图，四棱锥P-ABCD的底面是等腰梯形，AD\parallel BC，BC=2AD，\angle ABC=60^\circ ，E是PB的中点，F是棱PC上的点，且A，D，E，F四点共面.$![Image](https://cdntk.53zaixian.com/tq/1693855901982851073/img/compression/M_a5fd4cb35d1e11b577a1ac2d6f0241a1.jpg)
$(2)若\triangle PAD为等边三角形，二面角P-AD-B的大小为120^\circ ，求直线BD与平面ADFE所成角的正弦值.$

Solution:
$如图，设AD，BC的中点分别为G，O，连接OG，PG，以O为坐标原点，OC，OG所在直线分别为x轴，y轴，过点O作垂直于平面ABCD的直线为z轴，建立空间直角坐标系O-xyz，$![Image](https://cdntk.53zaixian.com/tq/1693855902469390338/img/compression/M_534a35b65ea82286bc5a2058724abb31.jpg)$设AD=a，\because AB=CD，BC=2AD，\angle ABC=60^\circ ,\therefore BC=2a，AB=CD=a，\therefore OG=\frac{\sqrt{3}}{2}a,则A\left(-\frac{a}{2},\frac{\sqrt{3}}{2}a,0\right),B(-a,0,0),D\left(\frac{a}{2},\frac{\sqrt{3}}{2}a,0\right)，\therefore \overrightarrow{BD}=\left(\frac{3}{2}a,\frac{\sqrt{3}}{2}a,0\right),\overrightarrow{AD}=(a,0,0).\because \triangle PAD为等边三角形，\therefore PG\perp AD，由题意知OG\perp AD,\therefore \angle PGO为二面角P-AD-B的平面角，又二面角P-AD-B的大小为120^\circ ，\therefore \angle PGO=120^\circ ，\because PG\perp AD，GO\perp AD，又PG\cap GO=G,PG,GO\subset 平面PGO，\therefore AD\perp 平面PGO，过P作PH垂直于y轴于点H，\because PH\subset 平面PGO，\therefore AD\perp PH，又PH\perp GH，GH,AD\subset 平面ABCD，GH\cap AD=G，\therefore PH\perp 平面ABCD，且\angle PGH=60^\circ ,\therefore PG=\frac{\sqrt{3}}{2}a,PH=\frac{\sqrt{3}}{2}a\times \frac{\sqrt{3}}{2}=\frac{3}{4}a,GH=\frac{\sqrt{3}}{4}a，\therefore OH=OG+GH=\frac{\sqrt{3}}{2}a+\frac{\sqrt{3}}{4}a=\frac{3\sqrt{3}}{4}a，\therefore P\left(0,\frac{3\sqrt{3}}{4}a,\frac{3}{4}a\right)，\because E，F分别为PB，PC的中点，\therefore E\left(-\frac{a}{2},\frac{3\sqrt{3}}{8}a,\frac{3}{8}a\right),F\left(\frac{a}{2},\frac{3\sqrt{3}}{8}a,\frac{3}{8}a\right),\therefore \overrightarrow{AE}=\left(0,-\frac{\sqrt{3}}{8}a,\frac{3}{8}a\right)，设平面ADFE的一个法向量为n=(x,y,z)，则\left\{\begin{matrix}\boldsymbol{n}\cdot \overrightarrow{AD}=0,\\ \boldsymbol{n}\cdot \overrightarrow{AE}=0,\end{matrix}\right.即\left\{\begin{matrix}ax=0,\\ -\frac{\sqrt{3}}{8}ay+\frac{3}{8}az=0,\end{matrix}\right.取z=1，则x=0,y=\sqrt{3},\therefore n=(0,\sqrt{3},1)，设直线BD与平面ADFE所成的角为\theta ，则\\sin   \theta =|\\cos  \langle \overrightarrow{BD},n\rangle |=\frac{\frac{3}{2}a}{\sqrt{3}a\times 2}=\frac{\sqrt{3}}{4}，即直线BD与平面ADFE所成角的正弦值为\frac{\sqrt{3}}{4}.$

Final answer: $\frac{\sqrt{3}}{4}.$