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


Question: $如图，在四棱锥P-ABCD中，PA\perp 底面ABCD，AB\perp AD，BC\parallel AD，PA=AB=BC=2，AD=4，E为棱PD的中点，\overrightarrow{PF}=\lambda \overrightarrow{PC}(\lambda 为常数，且0<\lambda <1).$![Image](https://cdntk.53zaixian.com/tq/1715283959261249539/img/compression/M_2204568acb0be54936034d86506013b5.jpg)
$(1)若直线BF\parallel 平面ACE，求实数\lambda 的值；$

Solution:
$因为PA\perp 底面ABCD，AB，AD\subset 平面ABCD，所以PA\perp AB，PA\perp AD.由题意可知，AB，AD，AP两两垂直，建立如图所示的空间直角坐标系O-xyz，则A(0，0，0)，B(2，0，0)，C(2，2，0)，D(0，4，0)，P(0，0，2)，E(0，2，1)，所以\overrightarrow{AC}=(2，2，0)，\overrightarrow{AE}=(0，2，1)，\overrightarrow{BP}=(-2，0，2)，\overrightarrow{PC}=(2，2，-2)，则\overrightarrow{PF}=\lambda \overrightarrow{PC}=(2\lambda ，2\lambda ，-2\lambda )，所以\overrightarrow{BF}=\overrightarrow{BP}+\overrightarrow{PF}=(2\lambda -2，2\lambda ，2-2\lambda ).设平面ACE的法向量为m=(x,y,z).由\left\{\begin{matrix}\overrightarrow{AC}\cdot \boldsymbol{m}=0,\\ \overrightarrow{AE}\cdot \boldsymbol{m}=0,\end{matrix}\right.得\left\{\begin{matrix}2x+2y=0,\\ 2y+z=0,\end{matrix}\right.令x=1，得m=(1,-1,2).因为BF\parallel 平面ACE，所以\overrightarrow{BF}\cdot m=2\lambda -2-2\lambda +4-4\lambda =0，解得\lambda =\frac{1}{2}.$![Image](https://cdntk.53zaixian.com/tq/1715283959538073601/img/compression/M_7ef0944d6897207535df5d208c76e45b.jpg)

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


Question: $如图，在四棱锥P-ABCD中，PA\perp 底面ABCD，AB\perp AD，BC\parallel AD，PA=AB=BC=2，AD=4，E为棱PD的中点，\overrightarrow{PF}=\lambda \overrightarrow{PC}(\lambda 为常数，且0<\lambda <1).$<img_2149>
$(2)当\lambda =\frac{1}{4}时，求二面角C-AE-F的大小.$

Solution:
$由(1)知\overrightarrow{AP}=(0，0，2)，\overrightarrow{PF}=(2\lambda ,2\lambda ,-2\lambda )，\overrightarrow{AE}=(0，2，1)，平面ACE的一个法向量为m=(1，-1，2).当\lambda =\frac{1}{4}时，\overrightarrow{AF}=\overrightarrow{AP}+\overrightarrow{PF}=(2\lambda ,2\lambda ,2-2\lambda )=\left(\frac{1}{2},\frac{1}{2},\frac{3}{2}\right).设平面AEF的法向量为n=(x_{0},y_{0},z_{0}).由\left\{\begin{matrix}\overrightarrow{AE}\cdot \boldsymbol{n}=0,\\ \overrightarrow{AF}\cdot \boldsymbol{n}=0,\end{matrix}\right.得\left\{\begin{matrix}2y_0+z_0=0,\\ \frac{1}{2}x_0+\frac{1}{2}y_0+\frac{3}{2}z_0=0.\end{matrix}\right.令y_{0}=1，得n=(5，1，-2)，所以\\cos  \langle m，n\rangle =\frac{\boldsymbol{m}\cdot \boldsymbol{n}}{｜\boldsymbol{m}｜｜\boldsymbol{n}｜}=0，所以二面角C-AE-F的大小为\frac{\pi }{2}.$

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


Question: $如图，在四棱锥P-ABCD中，PA\perp 底面ABCD，AB\perp AD，BC\parallel AD，PA=AB=BC=2，AD=4，E为棱PD的中点，\overrightarrow{PF}=\lambda \overrightarrow{PC}(\lambda 为常数，且0<\lambda <1).$![Image](https://cdntk.53zaixian.com/tq/1715283959261249539/img/compression/M_2204568acb0be54936034d86506013b5.jpg)
$(2)当\lambda =\frac{1}{4}时，求二面角C-AE-F的大小.$

Solution:
$由(1)知\overrightarrow{AP}=(0，0，2)，\overrightarrow{PF}=(2\lambda ,2\lambda ,-2\lambda )，\overrightarrow{AE}=(0，2，1)，平面ACE的一个法向量为m=(1，-1，2).当\lambda =\frac{1}{4}时，\overrightarrow{AF}=\overrightarrow{AP}+\overrightarrow{PF}=(2\lambda ,2\lambda ,2-2\lambda )=\left(\frac{1}{2},\frac{1}{2},\frac{3}{2}\right).设平面AEF的法向量为n=(x_{0},y_{0},z_{0}).由\left\{\begin{matrix}\overrightarrow{AE}\cdot \boldsymbol{n}=0,\\ \overrightarrow{AF}\cdot \boldsymbol{n}=0,\end{matrix}\right.得\left\{\begin{matrix}2y_0+z_0=0,\\ \frac{1}{2}x_0+\frac{1}{2}y_0+\frac{3}{2}z_0=0.\end{matrix}\right.令y_{0}=1，得n=(5，1，-2)，所以\\cos  \langle m，n\rangle =\frac{\boldsymbol{m}\cdot \boldsymbol{n}}{｜\boldsymbol{m}｜｜\boldsymbol{n}｜}=0，所以二面角C-AE-F的大小为\frac{\pi }{2}.$

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


Question: $如图，四棱锥P-ABCD的底面为直角梯形，BC\parallel AD，\angle BAD=90^\circ ，AD=PD=2AB=2BC=2，M为PA的中点.$<img_2104>
$(1)求证：BM\parallel 平面PCD；$

Solution:
$证明：取PD的中点N，连接CN，MN，因为M为PA的中点，所以MN\parallel AD，MN=\frac{1}{2}AD，又BC\parallel AD，BC=\frac{1}{2}AD，所以MN\parallel BC，MN=BC，所以四边形BMNC为平行四边形，所以BM\parallel CN，又CN\subset 平面PCD，BM\not\subset 平面PCD，所以BM\parallel 平面PCD.$<img_1506>
Subfield: Solid Geometry


Question: $如图，四棱锥P-ABCD的底面为直角梯形，BC\parallel AD，\angle BAD=90^\circ ，AD=PD=2AB=2BC=2，M为PA的中点.$![Image](https://cdntk.53zaixian.com/tq/1715283959890395139/img/compression/M_77310709ff52b6c505dd3dc7499d5609.jpg)
$(1)求证：BM\parallel 平面PCD；$

Solution:
$证明：取PD的中点N，连接CN，MN，因为M为PA的中点，所以MN\parallel AD，MN=\frac{1}{2}AD，又BC\parallel AD，BC=\frac{1}{2}AD，所以MN\parallel BC，MN=BC，所以四边形BMNC为平行四边形，所以BM\parallel CN，又CN\subset 平面PCD，BM\not\subset 平面PCD，所以BM\parallel 平面PCD.$![Image](https://cdntk.53zaixian.com/tq/1715283959890395139/img/compression/M_37853db918c7e9f336ce3f6c646400a8.jpg)

Final answer: 证明题
Subfield: Solid Geometry


Question: $如图，四棱锥P-ABCD的底面为直角梯形，BC\parallel AD，\angle BAD=90^\circ ，AD=PD=2AB=2BC=2，M为PA的中点.$<img_2104>
$(2)若平面ABCD\perp 平面PAD，异面直线BC与PD所成角为60^\circ ，且\triangle PAD是钝角三角形，求二面角B-PC-D的正弦值.$

Solution:
$因为BC\parallel AD，所以\angle ADP或其补角为异面直线BC与PD所成角，又AD=PD，\triangle PAD为钝角三角形，所以\angle ADP=120^\circ ，又平面ABCD\perp 平面PAD，平面ABCD\cap 平面PAD=AD，AB\perp AD，所以AB\perp 平面PAD，以A为坐标原点，AD，AB所在直线分别为y，z轴建立空间直角坐标系，则B(0，0，1)，D(0，2，0)，C(0，1，1)，P(\sqrt{3}，3，0)，所以\overrightarrow{PC}=(-\sqrt{3}，-2，1)，\overrightarrow{PB}=(-\sqrt{3}，-3，1).设平面PBC的法向量为n=(x,y,z).由\left\{\begin{matrix}\boldsymbol{n}\cdot \overrightarrow{PC}=0,\\ \boldsymbol{n}\cdot \overrightarrow{PB}=0\end{matrix}\right.得\left\{\begin{matrix}-\sqrt{3}x-2y+z=0,\\ -\sqrt{3}x-3y+z=0,\end{matrix}\right.令x=1，得平面PBC的一个法向量为n=(1,0,\sqrt{3})，同理可得平面PCD的一个法向量为m=(-1，\sqrt{3}，\sqrt{3}).则\\cos  \langle m,n\rangle =\frac{\boldsymbol{m}\cdot \boldsymbol{n}}{｜\boldsymbol{m}｜｜\boldsymbol{n}｜}=\frac{\sqrt{7}}{7}，设二面角B-PC-D的平面角为\theta ，则\\sin   \theta =\sqrt{1-\left(\frac{\sqrt{7}}{7}\right)^2}=\frac{\sqrt{42}}{7}.故二面角B-PC-D的正弦值为\frac{\sqrt{42}}{7}.$

Final answer: $\frac{\sqrt{42}}{7}$
Subfield: Solid Geometry


Question: $如图，四棱锥P-ABCD的底面为直角梯形，BC\parallel AD，\angle BAD=90^\circ ，AD=PD=2AB=2BC=2，M为PA的中点.$![Image](https://cdntk.53zaixian.com/tq/1715283959890395139/img/compression/M_77310709ff52b6c505dd3dc7499d5609.jpg)
$(2)若平面ABCD\perp 平面PAD，异面直线BC与PD所成角为60^\circ ，且\triangle PAD是钝角三角形，求二面角B-PC-D的正弦值.$

Solution:
$因为BC\parallel AD，所以\angle ADP或其补角为异面直线BC与PD所成角，又AD=PD，\triangle PAD为钝角三角形，所以\angle ADP=120^\circ ，又平面ABCD\perp 平面PAD，平面ABCD\cap 平面PAD=AD，AB\perp AD，所以AB\perp 平面PAD，以A为坐标原点，AD，AB所在直线分别为y，z轴建立空间直角坐标系，则B(0，0，1)，D(0，2，0)，C(0，1，1)，P(\sqrt{3}，3，0)，所以\overrightarrow{PC}=(-\sqrt{3}，-2，1)，\overrightarrow{PB}=(-\sqrt{3}，-3，1).设平面PBC的法向量为n=(x,y,z).由\left\{\begin{matrix}\boldsymbol{n}\cdot \overrightarrow{PC}=0,\\ \boldsymbol{n}\cdot \overrightarrow{PB}=0\end{matrix}\right.得\left\{\begin{matrix}-\sqrt{3}x-2y+z=0,\\ -\sqrt{3}x-3y+z=0,\end{matrix}\right.令x=1，得平面PBC的一个法向量为n=(1,0,\sqrt{3})，同理可得平面PCD的一个法向量为m=(-1，\sqrt{3}，\sqrt{3}).则\\cos  \langle m,n\rangle =\frac{\boldsymbol{m}\cdot \boldsymbol{n}}{｜\boldsymbol{m}｜｜\boldsymbol{n}｜}=\frac{\sqrt{7}}{7}，设二面角B-PC-D的平面角为\theta ，则\\sin   \theta =\sqrt{1-\left(\frac{\sqrt{7}}{7}\right)^2}=\frac{\sqrt{42}}{7}.故二面角B-PC-D的正弦值为\frac{\sqrt{42}}{7}.$

Final answer: $\frac{\sqrt{42}}{7}$
Subfield: Solid Geometry


Question: $如图,以矩形ABCD的CD边为直径作半圆O,点E为半圆上一点，满足\angle EDC=60^\circ ,BC=1.现将半圆沿CD折起,使得半圆面和平面ABCD垂直.$<img_1854>
$(1)求证:平面ADE\perp 平面BCE；$

Solution:
$证明：\because 四边形ABCD为矩形,\therefore BC\perp CD，\because 平面CDE\perp 平面ABCD,且平面CDE\cap 平面ABCD=CD,\therefore BC\perp 平面CDE,而DE\subset 平面CDE,\therefore DE\perp BC，\because 点E在半圆O上,CD为直径，\therefore DE\perp EC,又BC\cap EC=C，\therefore DE\perp 平面BCE，而DE\subset 平面ADE,\therefore 平面ADE\perp 平面BCE.$
Subfield: Solid Geometry


Question: $如图,以矩形ABCD的CD边为直径作半圆O,点E为半圆上一点，满足\angle EDC=60^\circ ,BC=1.现将半圆沿CD折起,使得半圆面和平面ABCD垂直.$![Image](https://cdntk.53zaixian.com/tq/1715283960645369858/img/compression/M_358755d4c6ec0cafc931dc6042af6e2f.jpg)
$(1)求证:平面ADE\perp 平面BCE；$

Solution:
$证明：\because 四边形ABCD为矩形,\therefore BC\perp CD，\because 平面CDE\perp 平面ABCD,且平面CDE\cap 平面ABCD=CD,\therefore BC\perp 平面CDE,而DE\subset 平面CDE,\therefore DE\perp BC，\because 点E在半圆O上,CD为直径，\therefore DE\perp EC,又BC\cap EC=C，\therefore DE\perp 平面BCE，而DE\subset 平面ADE,\therefore 平面ADE\perp 平面BCE.$

Final answer: 证明题
Subfield: Solid Geometry


Question: $如图,以矩形ABCD的CD边为直径作半圆O,点E为半圆上一点，满足\angle EDC=60^\circ ,BC=1.现将半圆沿CD折起,使得半圆面和平面ABCD垂直.$<img_1854>
$(2)若P是半圆弧CD上的一点(不包含C、D两个端点),且异面直线AE与BC所成角的余弦值为\frac{\sqrt{5}}{5},是否存在一点P,使得二面角A-PC-B的余弦值为\frac{\sqrt{13}}{13}?若存在,求出线段PC的长度,若不存在,请说明理由.$

Solution:
$\because 四边形ABCD为矩形,\therefore AD\parallel BC,\therefore \angle EAD即为异面直线AE与BC所成角，即\\cos  \angle EAD=\frac{\sqrt{5}}{5}.由(1)易得DE\perp AD,设CD=b,则DE=\frac{1}{2}b,在Rt\triangle ADE中，\\cos  \angle EAD=\frac{AD}{AE}=\frac{1}{\sqrt{1^2+\frac{1}{4}b^2}}=\frac{\sqrt{5}}{5}，解得b=4,即CD=4,\therefore OP=2,以点D为原点,DA,DC所在直线为x轴,y轴，经过点D且垂直于平面ABCD的直线为z轴建立空间直角坐标系如图,设\angle POC=\theta ,$<img_3051>$则A(1,0,0),B(1,4,0),C(0,4,0),P(0,2+2\\cos   \theta , 2\\sin   \theta ).\therefore \overrightarrow{PC}=(0,2-2\\cos   \theta ,-2\\sin   \theta ),\overrightarrow{CA}=(1,-4,0),\overrightarrow{CB}=(1,0,0)，设平面PAC的一个法向量为m=(x_{1},y_{1},z_{1})，则\left\{\begin{matrix}\boldsymbol{m}\cdot \overrightarrow{PC}=(2-2\\\cos    \theta )y_1-2\\\sin    \theta \cdot z_1=0,\\ \boldsymbol{m}\cdot \overrightarrow{CA}=x_1-4y_1=0,\end{matrix}\right.令y_{1}=1,则x_{1}=4,z_{1}=\frac{1-\\\cos    \theta }{\\\sin    \theta }.\therefore m=\left(4,1,\frac{1-\\\cos    \theta }{\\\sin    \theta }\right).设平面PBC的一个法向量为n=(x_{2},y_{2}，z_{2})，则\left\{\begin{matrix}\boldsymbol{n}\cdot \overrightarrow{PC}=(2-2\\\cos    \theta )y_2-2\\\sin    \theta \cdot z_2=0,\\ \boldsymbol{n}\cdot \overrightarrow{CB}=x_2=0,\end{matrix}\right.令y_{2}=1,则z_{2}=\frac{1-\\\cos    \theta }{\\\sin    \theta },x_{2}=0,\therefore n=\left(0,1,\frac{1-\\\cos    \theta }{\\\sin    \theta }\right),\\cos  \langle m,n\rangle =\frac{1+\left(\frac{1-\\\cos     \theta }{\\\sin     \theta }\right)^2}{\sqrt{17+\left(\frac{1-\\\cos     \theta }{\\\sin     \theta }\right)^2}\sqrt{1+\left(\frac{1-\\\cos     \theta }{\\\sin     \theta }\right)^2}}=\frac{\sqrt{13}}{13}.解得\\cos   \theta =\frac{1}{2}，\therefore \theta =60^\circ ,\therefore \triangle OPC为等边三角形，\therefore PC=2.故存在点P，使得二面角A-PC-B的余弦值为\frac{\sqrt{13}}{13}，此时PC=2.$

Final answer: $2$
Subfield: Solid Geometry


Question: $如图,以矩形ABCD的CD边为直径作半圆O,点E为半圆上一点，满足\angle EDC=60^\circ ,BC=1.现将半圆沿CD折起,使得半圆面和平面ABCD垂直.$![Image](https://cdntk.53zaixian.com/tq/1715283960645369858/img/compression/M_358755d4c6ec0cafc931dc6042af6e2f.jpg)
$(2)若P是半圆弧CD上的一点(不包含C、D两个端点),且异面直线AE与BC所成角的余弦值为\frac{\sqrt{5}}{5},是否存在一点P,使得二面角A-PC-B的余弦值为\frac{\sqrt{13}}{13}?若存在,求出线段PC的长度,若不存在,请说明理由.$

Solution:
$\because 四边形ABCD为矩形,\therefore AD\parallel BC,\therefore \angle EAD即为异面直线AE与BC所成角，即\\cos  \angle EAD=\frac{\sqrt{5}}{5}.由(1)易得DE\perp AD,设CD=b,则DE=\frac{1}{2}b,在Rt\triangle ADE中，\\cos  \angle EAD=\frac{AD}{AE}=\frac{1}{\sqrt{1^2+\frac{1}{4}b^2}}=\frac{\sqrt{5}}{5}，解得b=4,即CD=4,\therefore OP=2,以点D为原点,DA,DC所在直线为x轴,y轴，经过点D且垂直于平面ABCD的直线为z轴建立空间直角坐标系如图,设\angle POC=\theta ,$![Image](https://cdntk.53zaixian.com/tq/1715283961018662913/img/compression/M_e32e3a5b583841e3f496f6fa60833c0a.jpg)$则A(1,0,0),B(1,4,0),C(0,4,0),P(0,2+2\\cos   \theta , 2\\sin   \theta ).\therefore \overrightarrow{PC}=(0,2-2\\cos   \theta ,-2\\sin   \theta ),\overrightarrow{CA}=(1,-4,0),\overrightarrow{CB}=(1,0,0)，设平面PAC的一个法向量为m=(x_{1},y_{1},z_{1})，则\left\{\begin{matrix}\boldsymbol{m}\cdot \overrightarrow{PC}=(2-2\\\cos    \theta )y_1-2\\\sin    \theta \cdot z_1=0,\\ \boldsymbol{m}\cdot \overrightarrow{CA}=x_1-4y_1=0,\end{matrix}\right.令y_{1}=1,则x_{1}=4,z_{1}=\frac{1-\\\cos    \theta }{\\\sin    \theta }.\therefore m=\left(4,1,\frac{1-\\\cos    \theta }{\\\sin    \theta }\right).设平面PBC的一个法向量为n=(x_{2},y_{2}，z_{2})，则\left\{\begin{matrix}\boldsymbol{n}\cdot \overrightarrow{PC}=(2-2\\\cos    \theta )y_2-2\\\sin    \theta \cdot z_2=0,\\ \boldsymbol{n}\cdot \overrightarrow{CB}=x_2=0,\end{matrix}\right.令y_{2}=1,则z_{2}=\frac{1-\\\cos    \theta }{\\\sin    \theta },x_{2}=0,\therefore n=\left(0,1,\frac{1-\\\cos    \theta }{\\\sin    \theta }\right),\\cos  \langle m,n\rangle =\frac{1+\left(\frac{1-\\\cos     \theta }{\\\sin     \theta }\right)^2}{\sqrt{17+\left(\frac{1-\\\cos     \theta }{\\\sin     \theta }\right)^2}\sqrt{1+\left(\frac{1-\\\cos     \theta }{\\\sin     \theta }\right)^2}}=\frac{\sqrt{13}}{13}.解得\\cos   \theta =\frac{1}{2}，\therefore \theta =60^\circ ,\therefore \triangle OPC为等边三角形，\therefore PC=2.故存在点P，使得二面角A-PC-B的余弦值为\frac{\sqrt{13}}{13}，此时PC=2.$

Final answer: $2$
Subfield: Solid Geometry


Question: $如图，在三棱台ABC-DEF中，平面ACFD\perp 平面ABC，\angle ACB=\angle ACD=45^\circ ，DC=2BC.$<img_3043>
$(1)证明：EF\perp DB；$

Solution:
$证明：如图，过点D作DO\perp AC，交直线AC于点O，连接OB.$<img_2338>$由\angle ACD=45^\circ ，DO\perp AC得CD=\sqrt{2}CO，由平面ACFD\perp 平面ABC且交线为AC得DO\perp 平面ABC，所以DO\perp BC.由\angle ACB=45^\circ ，BC=\frac{1}{2}CD=\frac{\sqrt{2}}{2}CO得BO\perp BC.又BO\cap DO=O，所以BC\perp 平面BDO，因为DB\subset 平面BDO，故BC\perp DB.由三棱台ABC-DEF得BC\parallel EF，所以EF\perp DB.$
Subfield: Solid Geometry


Question: $如图，在三棱台ABC-DEF中，平面ACFD\perp 平面ABC，\angle ACB=\angle ACD=45^\circ ，DC=2BC.$![Image](https://cdntk.53zaixian.com/tq/1715283937438285825/img/compression/M_aceee3984001f83cc56c3b7cee3a8581.jpg)
$(1)证明：EF\perp DB；$

Solution:
$证明：如图，过点D作DO\perp AC，交直线AC于点O，连接OB.$![Image](https://cdntk.53zaixian.com/tq/1715283937786413058/img/compression/M_536d25f0c407f8564ba40a24db2882c5.jpg)$由\angle ACD=45^\circ ，DO\perp AC得CD=\sqrt{2}CO，由平面ACFD\perp 平面ABC且交线为AC得DO\perp 平面ABC，所以DO\perp BC.由\angle ACB=45^\circ ，BC=\frac{1}{2}CD=\frac{\sqrt{2}}{2}CO得BO\perp BC.又BO\cap DO=O，所以BC\perp 平面BDO，因为DB\subset 平面BDO，故BC\perp DB.由三棱台ABC-DEF得BC\parallel EF，所以EF\perp DB.$

Final answer: 证明题
Subfield: Solid Geometry


Question: $如图，在三棱台ABC-DEF中，平面ACFD\perp 平面ABC，\angle ACB=\angle ACD=45^\circ ，DC=2BC.$<img_3043>
$(2)求直线DF与平面DBC所成角的正弦值.$

Solution:
$过点O作OH\perp BD，交直线BD于点H，连接CH.由三棱台ABC-DEF得DF\parallel CO，所以直线DF与平面DBC所成角等于直线CO与平面DBC所成角.由BC\perp 平面BDO且OH\subset 平面BDO得OH\perp BC，又BC\cap BD=B，故OH\perp 平面BCD，所以\angle OCH为直线CO与平面DBC所成角.设CD=2\sqrt{2}，由DO=OC=2，BO=BC=\sqrt{2}，得BD=\sqrt{6}，OH=\frac{2\sqrt{3}}{3}，所以\\sin  \angle OCH=\frac{OH}{OC}=\frac{\sqrt{3}}{3}，因此，直线DF与平面DBC所成角的正弦值为\frac{\sqrt{3}}{3}.$

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


Question: $如图，在三棱台ABC-DEF中，平面ACFD\perp 平面ABC，\angle ACB=\angle ACD=45^\circ ，DC=2BC.$![Image](https://cdntk.53zaixian.com/tq/1715283937438285825/img/compression/M_aceee3984001f83cc56c3b7cee3a8581.jpg)
$(2)求直线DF与平面DBC所成角的正弦值.$

Solution:
$过点O作OH\perp BD，交直线BD于点H，连接CH.由三棱台ABC-DEF得DF\parallel CO，所以直线DF与平面DBC所成角等于直线CO与平面DBC所成角.由BC\perp 平面BDO且OH\subset 平面BDO得OH\perp BC，又BC\cap BD=B，故OH\perp 平面BCD，所以\angle OCH为直线CO与平面DBC所成角.设CD=2\sqrt{2}，由DO=OC=2，BO=BC=\sqrt{2}，得BD=\sqrt{6}，OH=\frac{2\sqrt{3}}{3}，所以\\sin  \angle OCH=\frac{OH}{OC}=\frac{\sqrt{3}}{3}，因此，直线DF与平面DBC所成角的正弦值为\frac{\sqrt{3}}{3}.$

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


Question: $如图，长方体ABCD-A_{1}B_{1}C_{1}D_{1}的底面ABCD是正方形，点E在棱AA_{1}上，BE\perp EC_{1}.$<img_1813>
$(1)证明：BE\perp 平面EB_{1}C_{1};$

Solution:
$证明：由已知得，B_{1}C_{1}\perp 平面ABB_{1}A_{1},因为BE\subset 平面ABB_{1}A_{1}，故B_{1}C_{1}\perp BE.又BE\perp EC_{1}，B_{1}C_{1}\cap EC_{1}=C_{1}，所以BE\perp 平面EB_{1}C_{1}.$
Subfield: Solid Geometry


Question: $如图，长方体ABCD-A_{1}B_{1}C_{1}D_{1}的底面ABCD是正方形，点E在棱AA_{1}上，BE\perp EC_{1}.$![Image](https://cdntk.53zaixian.com/tq/1715283938130345987/img/compression/M_3a139046bba83d337a934f8ed1c86fb9.jpg)
$(1)证明：BE\perp 平面EB_{1}C_{1};$

Solution:
$证明：由已知得，B_{1}C_{1}\perp 平面ABB_{1}A_{1},因为BE\subset 平面ABB_{1}A_{1}，故B_{1}C_{1}\perp BE.又BE\perp EC_{1}，B_{1}C_{1}\cap EC_{1}=C_{1}，所以BE\perp 平面EB_{1}C_{1}.$

Final answer: 证明题
Subfield: Solid Geometry


Question: $如图，长方体ABCD-A_{1}B_{1}C_{1}D_{1}的底面ABCD是正方形，点E在棱AA_{1}上，BE\perp EC_{1}.$![Image](https://cdntk.53zaixian.com/tq/1715283938130345987/img/compression/M_3a139046bba83d337a934f8ed1c86fb9.jpg)
$(2)若AE=A_{1}E,求二面角B-EC-C_{1}的正弦值.$

Solution:
$由(1)知\angle BEB_{1}=90^\circ .由题设知Rt\triangle ABE\cong Rt\triangle A_{1}B_{1}E,所以\angle AEB=45^\circ ，故AE=AB，AA_{1}=2AB.以D为坐标原点，\overrightarrow{DA}的方向为x轴正方向，|\overrightarrow{DA}|为单位长，建立如图所示的空间直角坐标系D-xyz，则C(0，1，0)，B(1，1，0)，C_{1}(0，1，2)，E(1，0，1)，则\overrightarrow{CB}=(1,0,0),\overrightarrow{CE}=(1,-1,1),\overrightarrow{CC_1}=(0,0,2).设平面EBC的法向量为n=(x,y,z),则\left\{\begin{matrix}\overrightarrow{CB}\cdot \boldsymbol{n}=0,\\ \overrightarrow{CE}\cdot \boldsymbol{n}=0,\end{matrix}\right.即\left\{\begin{matrix}x=0,\\ x-y+z=0,\end{matrix}\right.所以可取n=(0,-1,-1).设平面ECC_{1}的法向量为m=(x',y',z'),则\left\{\begin{matrix}\overrightarrow{CC_1}\cdot \boldsymbol{m}=0,\\ \overrightarrow{CE}\cdot \boldsymbol{m}=0,\end{matrix}\right.即\left\{\begin{matrix}2z'=0,\\ x'-y'+z'=0,\end{matrix}\right.所以可取m=(1,1,0).于是\\cos  \langle n,m\rangle =\frac{\boldsymbol{n}\cdot \boldsymbol{m}}{|\boldsymbol{n}||\boldsymbol{m}|}=-\frac{1}{2}.所以，二面角B－EC－C_{1}的正弦值为\frac{\sqrt{3}}{2}.$![Image](https://cdntk.53zaixian.com/tq/1715283938474278913/img/compression/M_b7a6d812e14b9bd568b3eb111260e868.jpg)

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


Question: $如图，长方体ABCD-A_{1}B_{1}C_{1}D_{1}的底面ABCD是正方形，点E在棱AA_{1}上，BE\perp EC_{1}.$<img_1813>
$(2)若AE=A_{1}E,求二面角B-EC-C_{1}的正弦值.$

Solution:
$由(1)知\angle BEB_{1}=90^\circ .由题设知Rt\triangle ABE\cong Rt\triangle A_{1}B_{1}E,所以\angle AEB=45^\circ ，故AE=AB，AA_{1}=2AB.以D为坐标原点，\overrightarrow{DA}的方向为x轴正方向，|\overrightarrow{DA}|为单位长，建立如图所示的空间直角坐标系D-xyz，则C(0，1，0)，B(1，1，0)，C_{1}(0，1，2)，E(1，0，1)，则\overrightarrow{CB}=(1,0,0),\overrightarrow{CE}=(1,-1,1),\overrightarrow{CC_1}=(0,0,2).设平面EBC的法向量为n=(x,y,z),则\left\{\begin{matrix}\overrightarrow{CB}\cdot \boldsymbol{n}=0,\\ \overrightarrow{CE}\cdot \boldsymbol{n}=0,\end{matrix}\right.即\left\{\begin{matrix}x=0,\\ x-y+z=0,\end{matrix}\right.所以可取n=(0,-1,-1).设平面ECC_{1}的法向量为m=(x',y',z'),则\left\{\begin{matrix}\overrightarrow{CC_1}\cdot \boldsymbol{m}=0,\\ \overrightarrow{CE}\cdot \boldsymbol{m}=0,\end{matrix}\right.即\left\{\begin{matrix}2z'=0,\\ x'-y'+z'=0,\end{matrix}\right.所以可取m=(1,1,0).于是\\cos  \langle n,m\rangle =\frac{\boldsymbol{n}\cdot \boldsymbol{m}}{|\boldsymbol{n}||\boldsymbol{m}|}=-\frac{1}{2}.所以，二面角B－EC－C_{1}的正弦值为\frac{\sqrt{3}}{2}.$<img_1887>

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


Question: $如图，在三棱柱ABC-A_{1}B_{1}C_{1}中，A_{1}C\perp 平面ABC，\angle ACB=90^\circ ，AA_{1}=2，A_{1}到平面BCC_{1}B_{1}的距离为1.$<img_2247>
$(1)证明：A_{1}C=AC；$

Solution:
$证明：\because A_{1}C\perp 平面ABC,BC\subset 平面ABC，\therefore A_{1}C\perp BC，又\because BC\perp AC,AC\cap A_{1}C=C,\therefore BC\perp 平面ACC_{1}A_{1},又\because BC\subset 平面BCC_{1}B_{1}，\therefore 平面ACC_{1}A_{1}\perp 平面BCC_{1}B_{1}.过A_{1}作A_{1}H\perp CC_{1}，则A_{1}H\perp 平面BCC_{1}B_{1}，(面面垂直的性质)即A_{1}H的长为点A_{1}到平面BCC_{1}B_{1}的距离，\therefore A_{1}H=1.设AC=x，在Rt\triangle A_{1}CA中，A_{1}C=\sqrt{4-x^2},在Rt\triangle CA_{1}C_{1}中，A_{1}C_{1}\cdot A_{1}C=CC_{1}\cdot A_{1}H，即x\cdot \sqrt{4-x^2}=2\times 1,解得x=\sqrt{2},故A_{1}C=AC=\sqrt{2},\therefore A_{1}C=AC.$
Subfield: Solid Geometry


Question: $如图，在三棱柱ABC-A_{1}B_{1}C_{1}中，A_{1}C\perp 平面ABC，\angle ACB=90^\circ ，AA_{1}=2，A_{1}到平面BCC_{1}B_{1}的距离为1.$![Image](https://cdntk.53zaixian.com/tq/1715283939292168193/img/compression/M_7194d7a8e16bbc7410d09cb619c480a8.jpg)
$(1)证明：A_{1}C=AC；$

Solution:
$证明：\because A_{1}C\perp 平面ABC,BC\subset 平面ABC，\therefore A_{1}C\perp BC，又\because BC\perp AC,AC\cap A_{1}C=C,\therefore BC\perp 平面ACC_{1}A_{1},又\because BC\subset 平面BCC_{1}B_{1}，\therefore 平面ACC_{1}A_{1}\perp 平面BCC_{1}B_{1}.过A_{1}作A_{1}H\perp CC_{1}，则A_{1}H\perp 平面BCC_{1}B_{1}，(面面垂直的性质)即A_{1}H的长为点A_{1}到平面BCC_{1}B_{1}的距离，\therefore A_{1}H=1.设AC=x，在Rt\triangle A_{1}CA中，A_{1}C=\sqrt{4-x^2},在Rt\triangle CA_{1}C_{1}中，A_{1}C_{1}\cdot A_{1}C=CC_{1}\cdot A_{1}H，即x\cdot \sqrt{4-x^2}=2\times 1,解得x=\sqrt{2},故A_{1}C=AC=\sqrt{2},\therefore A_{1}C=AC.$

Final answer: 证明题
Subfield: Solid Geometry


Question: $如图，在三棱柱ABC-A_{1}B_{1}C_{1}中，A_{1}C\perp 平面ABC，\angle ACB=90^\circ ，AA_{1}=2，A_{1}到平面BCC_{1}B_{1}的距离为1.$<img_2247>
$(2)已知AA_{1}与BB_{1}的距离为2，求AB_{1}与平面BCC_{1}B_{1}所成角的正弦值.$

Solution:
$连接B_{1}C.过C作CQ\perp AA_{1}，垂足为Q,连接BQ.由(1)知BC\perp 平面ACC_{1}A_{1}，又AA_{1}\subset 平面ACC_{1}A_{1},\therefore BC\perp AA_{1}.又CQ\cap BC=C，\therefore AA_{1}\perp 平面BCQ，\because BQ\subset 平面BCQ,\therefore AA_{1}\perp BQ，又\because AA_{1}\parallel BB_{1}，\therefore BB_{1}\perp BQ，\therefore BQ的长为直线AA_{1}与BB_{1}之间的距离，即BQ=2.在Rt\triangle BCQ中，BQ=2，CQ=1，\therefore BC=\sqrt{3}.由(1)知CA、CB、CA_{1}两两垂直，\therefore 以直线CA，CB，CA_{1}分别为x轴，y轴，z轴建立空间直角坐标系，如图所示，$<img_1489>$则C(0,0,0),A(\sqrt{2},0,0),A_{1}(0,0,\sqrt{2}),B(0,\sqrt{3},0),\therefore \overrightarrow{CB}=(0,\sqrt{3},0),\overrightarrow{AA_1}=(-\sqrt{2},0,\sqrt{2}),又知\overrightarrow{BB_1}=\overrightarrow{AA_1}=(-\sqrt{2},0,\sqrt{2}),\therefore B_{1}(-\sqrt{2},\sqrt{3},\sqrt{2}).\therefore \overrightarrow{CB_1}=(-\sqrt{2},\sqrt{3},\sqrt{2}),\overrightarrow{AB_1}=(-2\sqrt{2},\sqrt{3},\sqrt{2}).设平面BCC_{1}B_{1}的法向量为n=(x,y,z),则\left\{\begin{matrix}\boldsymbol{n}\cdot \overrightarrow{CB}=0,\\ \boldsymbol{n}\cdot \overrightarrow{CB_1}=0,\end{matrix}\right.即\left\{\begin{matrix}\sqrt{3}y=0,\\ -\sqrt{2}x+\sqrt{3}y+\sqrt{2}z=0.\end{matrix}\right.令x=1,则z=1,则n=(1,0,1),设直线AB_{1}与平面BCC_{1}B_{1}所成角为\theta ，则\\sin   \theta =|\\cos  \langle \overrightarrow{AB_1},n\rangle |=\frac{|\overrightarrow{AB_1}\cdot \boldsymbol{n}|}{|\overrightarrow{AB_1}||\boldsymbol{n}|}=\frac{|-2\sqrt{2}\times 1+\sqrt{2}\times 1|}{\sqrt{8+3+2}\times \sqrt{1+1}}=\frac{\sqrt{2}}{\sqrt{13}\times \sqrt{2}}=\frac{\sqrt{13}}{13},即AB_{1}与平面BCC_{1}B_{1}所成角的正弦值为\frac{\sqrt{13}}{13}.$

Final answer: $\frac{\sqrt{13}}{13}$
Subfield: Solid Geometry


Question: $如图，在三棱柱ABC-A_{1}B_{1}C_{1}中，A_{1}C\perp 平面ABC，\angle ACB=90^\circ ，AA_{1}=2，A_{1}到平面BCC_{1}B_{1}的距离为1.$![Image](https://cdntk.53zaixian.com/tq/1715283939292168193/img/compression/M_7194d7a8e16bbc7410d09cb619c480a8.jpg)
$(2)已知AA_{1}与BB_{1}的距离为2，求AB_{1}与平面BCC_{1}B_{1}所成角的正弦值.$

Solution:
$连接B_{1}C.过C作CQ\perp AA_{1}，垂足为Q,连接BQ.由(1)知BC\perp 平面ACC_{1}A_{1}，又AA_{1}\subset 平面ACC_{1}A_{1},\therefore BC\perp AA_{1}.又CQ\cap BC=C，\therefore AA_{1}\perp 平面BCQ，\because BQ\subset 平面BCQ,\therefore AA_{1}\perp BQ，又\because AA_{1}\parallel BB_{1}，\therefore BB_{1}\perp BQ，\therefore BQ的长为直线AA_{1}与BB_{1}之间的距离，即BQ=2.在Rt\triangle BCQ中，BQ=2，CQ=1，\therefore BC=\sqrt{3}.由(1)知CA、CB、CA_{1}两两垂直，\therefore 以直线CA，CB，CA_{1}分别为x轴，y轴，z轴建立空间直角坐标系，如图所示，$![Image](https://cdntk.53zaixian.com/tq/1715283939644489730/img/compression/M_0a5e617287e4a0a719eafd3092288406.jpg)$则C(0,0,0),A(\sqrt{2},0,0),A_{1}(0,0,\sqrt{2}),B(0,\sqrt{3},0),\therefore \overrightarrow{CB}=(0,\sqrt{3},0),\overrightarrow{AA_1}=(-\sqrt{2},0,\sqrt{2}),又知\overrightarrow{BB_1}=\overrightarrow{AA_1}=(-\sqrt{2},0,\sqrt{2}),\therefore B_{1}(-\sqrt{2},\sqrt{3},\sqrt{2}).\therefore \overrightarrow{CB_1}=(-\sqrt{2},\sqrt{3},\sqrt{2}),\overrightarrow{AB_1}=(-2\sqrt{2},\sqrt{3},\sqrt{2}).设平面BCC_{1}B_{1}的法向量为n=(x,y,z),则\left\{\begin{matrix}\boldsymbol{n}\cdot \overrightarrow{CB}=0,\\ \boldsymbol{n}\cdot \overrightarrow{CB_1}=0,\end{matrix}\right.即\left\{\begin{matrix}\sqrt{3}y=0,\\ -\sqrt{2}x+\sqrt{3}y+\sqrt{2}z=0.\end{matrix}\right.令x=1,则z=1,则n=(1,0,1),设直线AB_{1}与平面BCC_{1}B_{1}所成角为\theta ，则\\sin   \theta =|\\cos  \langle \overrightarrow{AB_1},n\rangle |=\frac{|\overrightarrow{AB_1}\cdot \boldsymbol{n}|}{|\overrightarrow{AB_1}||\boldsymbol{n}|}=\frac{|-2\sqrt{2}\times 1+\sqrt{2}\times 1|}{\sqrt{8+3+2}\times \sqrt{1+1}}=\frac{\sqrt{2}}{\sqrt{13}\times \sqrt{2}}=\frac{\sqrt{13}}{13},即AB_{1}与平面BCC_{1}B_{1}所成角的正弦值为\frac{\sqrt{13}}{13}.$

Final answer: $\frac{\sqrt{13}}{13}$
Subfield: Solid Geometry


Question: $如图，三棱锥A-BCD中，DA=DB=DC,BD\perp CD，\angle ADB=\angle ADC=60^\circ ，E为BC的中点.$![Image](https://cdntk.53zaixian.com/tq/1715283941481594883/img/compression/M_de29da90feca03c43a87ef0366594e17.jpg)
$(1)证明:BC\perp DA；$

Solution:
$证明：连接AE，DE，\because DB=DC,E为BC的中点，\therefore DE\perp BC.又\because DA=DB=DC,\angle ADB=\angle ADC=60^\circ ,\therefore \triangle ACD与\triangle ABD均为等边三角形,\therefore AC=AB,\therefore AE\perp BC.又\because AE\cap DE=E,AE\subset 平面ADE,DE\subset 平面ADE,\therefore BC\perp 平面ADE,又\because DA\subset 平面ADE,\therefore BC\perp DA.$

Final answer: 证明题
Subfield: Solid Geometry


Question: $如图，三棱锥A-BCD中，DA=DB=DC,BD\perp CD，\angle ADB=\angle ADC=60^\circ ，E为BC的中点.$<img_2719>
$(1)证明:BC\perp DA；$

Solution:
$证明：连接AE，DE，\because DB=DC,E为BC的中点，\therefore DE\perp BC.又\because DA=DB=DC,\angle ADB=\angle ADC=60^\circ ,\therefore \triangle ACD与\triangle ABD均为等边三角形,\therefore AC=AB,\therefore AE\perp BC.又\because AE\cap DE=E,AE\subset 平面ADE,DE\subset 平面ADE,\therefore BC\perp 平面ADE,又\because DA\subset 平面ADE,\therefore BC\perp DA.$