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


Question: $如图，在直三棱柱ABC-A_{1}B_{1}C_{1}中，\triangle ABC是以BC为斜边的等腰直角三角形，AA_{1}=AB=3，D，E分别为BC，B_{1}C_{1}上的点，且\frac{BD}{BC}=\frac{C_1E}{C_1B_1}=t(0$![Image](https://cdntk.53zaixian.com/tq/1701760639702224898/img/compression/M_cf0de9bcf8bc6c067cdf2b50300598f8.jpg)
$(3)若平面AC_{1}D与平面ACD的夹角为\frac{\pi }{3}，求实数t的值.$

Solution:
$\overrightarrow{BD}=t\overrightarrow{BC}=(-3t,3t,0)(0(10分)
\overrightarrow{AC_1}=(0,3,3)，\overrightarrow{AD}=(3-3t,3t,0).设平面AC_{1}D的法向量为m=(x_{2},y_{2},z_{2})，则\left\{\begin{matrix}\boldsymbol{m}\cdot \overrightarrow{AC_1}=0,\\ \boldsymbol{m}\cdot \overrightarrow{AD}=0,\end{matrix}\right.即\left\{\begin{matrix}3y_2+3z_2=0,\\ (3-3t)x_2+3ty_2=0,\end{matrix}\right.取x_{2}=t，得m=(t,t-1,1-t).
(12分)
易得平面ACD的一个法向量为\overrightarrow{AA_1}=(0,0,3).\because 平面AC_{1}D与平面ACD的夹角为\frac{\pi }{3}，\therefore \\cos  \frac{\pi }{3}=|\\cos  \langle m,\overrightarrow{AA_1}\rangle |=\frac{|\boldsymbol{m}\cdot \overrightarrow{AA_1}|}{|\boldsymbol{m}||\overrightarrow{AA_1}|}=\frac{3(1-t)}{\sqrt{3t^2-4t+2}\times 3}=\frac{1}{2},
(14分)
即t^{2}-4t+2=0,解得t=2+\sqrt{2}或t=2-\sqrt{2}，又0<t<1，\therefore t=2-\sqrt{2}.
(15分)$

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


Question: $如图，AD\parallel BC且AD=2BC，AD\perp CD，EG\parallel AD且EG=AD，CD\parallel FG且CD=2FG，DG\perp 平面ABCD，DA=DC=DG=2.$![Image](https://cdntk.53zaixian.com/tq/1701760640616583170/img/compression/M_8e90692a94f7f5bb30f5fd4709f2e680.jpg)
$(1)若M为CF的中点，N为EG的中点，求证:MN\parallel 平面CDE;$

Solution:
$以D为原点，\overrightarrow{DA}，\overrightarrow{DC},\overrightarrow{DG}的方向分别为x轴,y轴,z轴的正方向建立空间直角坐标系(如图)，$![Image](https://cdntk.53zaixian.com/tq/1701760639702224898/img/compression/M_e16cdbcc11ff51117891f1c800f0b2af.jpg)$可得D(0，0,0)，A(2,0,0)，B(1,2,0)，C(0,2,0),E(2,0,2)，F(0,1,2)，G(0,0,2).证明:由题得M\left(\right.0，\frac{3}{2}，1\left)\right.，N(1，0，2)，\overrightarrow{DC}=(0,2,0),\overrightarrow{DE}=(2,0,2),设n_{0}=(x_{0},y_{0},z_{0})为平面CDE的法向量，则\left\{\begin{matrix}\boldsymbol{n}_0\cdot \overrightarrow{DC}=0,\\ \boldsymbol{n}_0\cdot \overrightarrow{DE}=0,\end{matrix}\right.即\left\{\begin{matrix}2y_0=0,\\ 2x_0+2z_0=0,\end{matrix}\right.不妨令z_{0}=-1,得n_{0}=(1，0,-1).又\overrightarrow{MN}=\left(1,-\frac{3}{2},1\right)，所以\overrightarrow{MN}\cdot n_{0}=0,又因为直线MN\not\subset 平面CDE,所以MN\parallel 平面CDE.$

Final answer: 证明题
Subfield: Solid Geometry


Question: $如图，AD\parallel BC且AD=2BC，AD\perp CD，EG\parallel AD且EG=AD，CD\parallel FG且CD=2FG，DG\perp 平面ABCD，DA=DC=DG=2.$<img_2546>
$(1)若M为CF的中点，N为EG的中点，求证:MN\parallel 平面CDE;$

Solution:
$以D为原点，\overrightarrow{DA}，\overrightarrow{DC},\overrightarrow{DG}的方向分别为x轴,y轴,z轴的正方向建立空间直角坐标系(如图)，$<img_3020>$可得D(0，0,0)，A(2,0,0)，B(1,2,0)，C(0,2,0),E(2,0,2)，F(0,1,2)，G(0,0,2).证明:由题得M\left(\right.0，\frac{3}{2}，1\left)\right.，N(1，0，2)，\overrightarrow{DC}=(0,2,0),\overrightarrow{DE}=(2,0,2),设n_{0}=(x_{0},y_{0},z_{0})为平面CDE的法向量，则\left\{\begin{matrix}\boldsymbol{n}_0\cdot \overrightarrow{DC}=0,\\ \boldsymbol{n}_0\cdot \overrightarrow{DE}=0,\end{matrix}\right.即\left\{\begin{matrix}2y_0=0,\\ 2x_0+2z_0=0,\end{matrix}\right.不妨令z_{0}=-1,得n_{0}=(1，0,-1).又\overrightarrow{MN}=\left(1,-\frac{3}{2},1\right)，所以\overrightarrow{MN}\cdot n_{0}=0,又因为直线MN\not\subset 平面CDE,所以MN\parallel 平面CDE.$
Subfield: Solid Geometry


Question: $如图，AD\parallel BC且AD=2BC，AD\perp CD，EG\parallel AD且EG=AD，CD\parallel FG且CD=2FG，DG\perp 平面ABCD，DA=DC=DG=2.$![Image](https://cdntk.53zaixian.com/tq/1701760640616583170/img/compression/M_8e90692a94f7f5bb30f5fd4709f2e680.jpg)
$(2)求平面EBC与平面BCF的夹角的正弦值;$

Solution:
$依题意，可得\overrightarrow{BC}=(-1，0，0)，\overrightarrow{BE}=(1，-2，2)，\overrightarrow{CF}=(0，-1，2).设n=(x_{1},y_{1},z_{1})为平面BCE的法向量，则\left\{\begin{matrix}\boldsymbol{n}\cdot \overrightarrow{BC}=0,\\ \boldsymbol{n}\cdot \overrightarrow{BE}=0,\end{matrix}\right.即\left\{\begin{matrix}-x_1=0,\\ x_1-2y_1+2z_1=0,\end{matrix}\right.不妨令z_{1}=1,得n=(0，1，1).设m=(x_{2},y_{2},z_{2})为平面BCF的法向量,则\left\{\begin{matrix}\boldsymbol{m}\cdot \overrightarrow{BC}=0,\\ \boldsymbol{m}\cdot \overrightarrow{CF}=0,\end{matrix}\right.即\left\{\begin{matrix}-x_2=0,\\ -y_2+2z_2=0,\end{matrix}\right.不妨令z_{2}=1,得m=(0，2，1).所以\\cos  \langle m,n\rangle =\frac{\boldsymbol{m}\cdot \boldsymbol{n}}{|\boldsymbol{m}||\boldsymbol{n}|}=\frac{3\sqrt{10}}{10},所以\\sin  \langle m,n\rangle =\frac{\sqrt{10}}{10}.所以平面EBC与平面BCF的夹角的正弦值为\frac{\sqrt{10}}{10}.$

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


Question: $如图，AD\parallel BC且AD=2BC，AD\perp CD，EG\parallel AD且EG=AD，CD\parallel FG且CD=2FG，DG\perp 平面ABCD，DA=DC=DG=2.$<img_2546>
$(2)求平面EBC与平面BCF的夹角的正弦值;$

Solution:
$依题意，可得\overrightarrow{BC}=(-1，0，0)，\overrightarrow{BE}=(1，-2，2)，\overrightarrow{CF}=(0，-1，2).设n=(x_{1},y_{1},z_{1})为平面BCE的法向量，则\left\{\begin{matrix}\boldsymbol{n}\cdot \overrightarrow{BC}=0,\\ \boldsymbol{n}\cdot \overrightarrow{BE}=0,\end{matrix}\right.即\left\{\begin{matrix}-x_1=0,\\ x_1-2y_1+2z_1=0,\end{matrix}\right.不妨令z_{1}=1,得n=(0，1，1).设m=(x_{2},y_{2},z_{2})为平面BCF的法向量,则\left\{\begin{matrix}\boldsymbol{m}\cdot \overrightarrow{BC}=0,\\ \boldsymbol{m}\cdot \overrightarrow{CF}=0,\end{matrix}\right.即\left\{\begin{matrix}-x_2=0,\\ -y_2+2z_2=0,\end{matrix}\right.不妨令z_{2}=1,得m=(0，2，1).所以\\cos  \langle m,n\rangle =\frac{\boldsymbol{m}\cdot \boldsymbol{n}}{|\boldsymbol{m}||\boldsymbol{n}|}=\frac{3\sqrt{10}}{10},所以\\sin  \langle m,n\rangle =\frac{\sqrt{10}}{10}.所以平面EBC与平面BCF的夹角的正弦值为\frac{\sqrt{10}}{10}.$

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


Question: $如图，AD\parallel BC且AD=2BC，AD\perp CD，EG\parallel AD且EG=AD，CD\parallel FG且CD=2FG，DG\perp 平面ABCD，DA=DC=DG=2.$![Image](https://cdntk.53zaixian.com/tq/1701760640616583170/img/compression/M_8e90692a94f7f5bb30f5fd4709f2e680.jpg)
$(3)若点P在线段DG上，且直线BP与平面ADGE所成的角为60^\circ ，求线段DP的长.$

Solution:
$设线段DP的长为h(h\in [0,2])，则点P的坐标为(0,0,h)，可得\overrightarrow{BP}=(-1,-2,h).易知\overrightarrow{DC}=(0,2,0)为平面ADGE的一个法向量，故|\\cos  \langle \overrightarrow{BP}，\overrightarrow{DC}\rangle |=\frac{|\overrightarrow{BP}\cdot \overrightarrow{DC}|}{|\overrightarrow{BP}||\overrightarrow{DC}|}=\frac{2}{\sqrt{h^2+5}}，由题意得\frac{2}{\sqrt{h^2+5}}=\\sin   60^\circ =\frac{\sqrt{3}}{2}，所以h=\frac{\sqrt{3}}{3}(负值舍去).所以线段DP的长为\frac{\sqrt{3}}{3}.$

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


Question: $如图，AD\parallel BC且AD=2BC，AD\perp CD，EG\parallel AD且EG=AD，CD\parallel FG且CD=2FG，DG\perp 平面ABCD，DA=DC=DG=2.$<img_2546>
$(3)若点P在线段DG上，且直线BP与平面ADGE所成的角为60^\circ ，求线段DP的长.$

Solution:
$设线段DP的长为h(h\in [0,2])，则点P的坐标为(0,0,h)，可得\overrightarrow{BP}=(-1,-2,h).易知\overrightarrow{DC}=(0,2,0)为平面ADGE的一个法向量，故|\\cos  \langle \overrightarrow{BP}，\overrightarrow{DC}\rangle |=\frac{|\overrightarrow{BP}\cdot \overrightarrow{DC}|}{|\overrightarrow{BP}||\overrightarrow{DC}|}=\frac{2}{\sqrt{h^2+5}}，由题意得\frac{2}{\sqrt{h^2+5}}=\\sin   60^\circ =\frac{\sqrt{3}}{2}，所以h=\frac{\sqrt{3}}{3}(负值舍去).所以线段DP的长为\frac{\sqrt{3}}{3}.$

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


Question: $如图，在四棱锥P-ABCD中，底面ABCD是矩形，AB=2AD=2，PA\perp 平面ABCD，E为PD的中点.$<img_2905>
$(1)若PA=1.(i)求证:AE\perp 平面PCD；(ii)求直线BE与平面PCD所成角的正弦值.$

Solution:
$以A为原点，AB，AD，AP所在直线分别为x轴，y轴，z轴建立空间直角坐标系，
(1分)
则A(0，0，0)，B(2，0，0)，C(2，1，0)，D(0，1，0).
(2分)
由题得P(0，0，1)，E\left(0,\frac{1}{2},\frac{1}{2}\right).(i)证明：设平面PCD的法向量为t=(a,b,c),\because \overrightarrow{DC}=(2,0,0)，\overrightarrow{PD}=(0,1,-1)，则\left\{\begin{matrix}\boldsymbol{t}\cdot \overrightarrow{DC}=0,\\ \boldsymbol{t}\cdot \overrightarrow{PD}=0,\end{matrix}\right.即\left\{\begin{matrix}2a=0,\\ b-c=0,\end{matrix}\right.令b=1，则c=1，\therefore t=(0,1,1)，
(4分)
\because \overrightarrow{AE}=\frac{1}{2}t，\therefore \overrightarrow{AE}\parallel t，\therefore AE\perp 平面PCD.
(5分)
(ii)由(i)得AE\perp 平面PCD,\therefore \overrightarrow{AE}=\left(0,\frac{1}{2},\frac{1}{2}\right)为平面PCD的一个法向量，
(6分)
设直线BE与平面PCD所成角为\beta ，\overrightarrow{BE}=\left(-2,\frac{1}{2},\frac{1}{2}\right)，则\\sin   \beta =|\\cos  \langle \overrightarrow{BE}，\overrightarrow{AE}\rangle |=\frac{|\overrightarrow{BE}\cdot \overrightarrow{AE}|}{|\overrightarrow{BE}||\overrightarrow{AE}|}=\frac{\left|\frac{1}{4}+\frac{1}{4}\right|}{\sqrt{\frac{1}{4}+\frac{1}{4}}\cdot \sqrt{4+\frac{1}{4}+\frac{1}{4}}}=\frac{1}{3}，\therefore 直线BE与平面PCD所成角的正弦值为\frac{1}{3}.
(8分)$
Subfield: Solid Geometry


Question: $如图，在四棱锥P-ABCD中，底面ABCD是矩形，AB=2AD=2，PA\perp 平面ABCD，E为PD的中点.$![Image](https://cdntk.53zaixian.com/tq/1701760641002459139/img/compression/M_5cf17422b8c15f5cfd7b4c4ad5f751a7.jpg)
$(1)若PA=1.(i)求证:AE\perp 平面PCD；(ii)求直线BE与平面PCD所成角的正弦值.$

Solution:
$以A为原点，AB，AD，AP所在直线分别为x轴，y轴，z轴建立空间直角坐标系，
(1分)
则A(0，0，0)，B(2，0，0)，C(2，1，0)，D(0，1，0).
(2分)
由题得P(0，0，1)，E\left(0,\frac{1}{2},\frac{1}{2}\right).(i)证明：设平面PCD的法向量为t=(a,b,c),\because \overrightarrow{DC}=(2,0,0)，\overrightarrow{PD}=(0,1,-1)，则\left\{\begin{matrix}\boldsymbol{t}\cdot \overrightarrow{DC}=0,\\ \boldsymbol{t}\cdot \overrightarrow{PD}=0,\end{matrix}\right.即\left\{\begin{matrix}2a=0,\\ b-c=0,\end{matrix}\right.令b=1，则c=1，\therefore t=(0,1,1)，
(4分)
\because \overrightarrow{AE}=\frac{1}{2}t，\therefore \overrightarrow{AE}\parallel t，\therefore AE\perp 平面PCD.
(5分)
(ii)由(i)得AE\perp 平面PCD,\therefore \overrightarrow{AE}=\left(0,\frac{1}{2},\frac{1}{2}\right)为平面PCD的一个法向量，
(6分)
设直线BE与平面PCD所成角为\beta ，\overrightarrow{BE}=\left(-2,\frac{1}{2},\frac{1}{2}\right)，则\\sin   \beta =|\\cos  \langle \overrightarrow{BE}，\overrightarrow{AE}\rangle |=\frac{|\overrightarrow{BE}\cdot \overrightarrow{AE}|}{|\overrightarrow{BE}||\overrightarrow{AE}|}=\frac{\left|\frac{1}{4}+\frac{1}{4}\right|}{\sqrt{\frac{1}{4}+\frac{1}{4}}\cdot \sqrt{4+\frac{1}{4}+\frac{1}{4}}}=\frac{1}{3}，\therefore 直线BE与平面PCD所成角的正弦值为\frac{1}{3}.
(8分)$

Final answer: 证明题
Subfield: Solid Geometry


Question: $如图，在四棱锥P-ABCD中，底面ABCD是矩形，AB=2AD=2，PA\perp 平面ABCD，E为PD的中点.$![Image](https://cdntk.53zaixian.com/tq/1701760641002459139/img/compression/M_5cf17422b8c15f5cfd7b4c4ad5f751a7.jpg)
$(2)若平面BCE与平面CED夹角的正弦值为\frac{\sqrt{21}}{5}，求PA的长.$

Solution:
$\overrightarrow{BC}=(0，1，0),设PA=a(a>0)，则E\left(0,\frac{1}{2},\frac{a}{2}\right)，\therefore \overrightarrow{AE}=\left(0,\frac{1}{2},\frac{a}{2}\right),\overrightarrow{BE}=\left(-2,\frac{1}{2},\frac{a}{2}\right)，设平面BCE的法向量为n=(x,y,z),则\left\{\begin{matrix}\boldsymbol{n}\cdot \overrightarrow{BC}=0,\\ \boldsymbol{n}\cdot \overrightarrow{BE}=0,\end{matrix}\right.即\left\{\begin{matrix}y=0,\\ -2x+\frac{1}{2}y+\frac{a}{2}z=0,\end{matrix}\right.令z=2，解得\left\{\begin{matrix}x=\frac{a}{2},\\ y=0,\end{matrix}\right.\therefore n=\left(\frac{a}{2},0,2\right),
(10分)
设平面CED的法向量为m=(x_{0},y_{0},z_{0})，又\overrightarrow{CP}=(-2,-1,a)，\overrightarrow{CD}=\overrightarrow{BA}=(-2,0,0),\therefore \left\{\begin{matrix}\boldsymbol{m}\cdot \overrightarrow{CP}=0,\\ \boldsymbol{m}\cdot \overrightarrow{CD}=0,\end{matrix}\right.即\left\{\begin{matrix}-2x_0-y_0+az_0=0,\\ -2x_0=0,\end{matrix}\right.令z_{0}=1,解得\left\{\begin{matrix}x_0=0,\\ y_0=a,\end{matrix}\right.\therefore m=(0,a,1)，
(12分)
设平面BCE与平面CED的夹角为\theta ，则\\cos   \theta =|\\cos  \langle n,m\rangle |=\left|\frac{\boldsymbol{n}\cdot \boldsymbol{m}}{|\boldsymbol{n}||\boldsymbol{m}|}\right|=\frac{2}{\sqrt{\frac{a^2}{4}+4}\cdot \sqrt{a^2+1}}=\frac{4}{\sqrt{(a^2+16)(a^2+1)}},因为\\sin   \theta =\frac{\sqrt{21}}{5}，所以\\cos   \theta =\frac{2}{5}，
(13分)
则\frac{4}{\sqrt{(a^2+16)(a^2+1)}}=\frac{2}{5},整理得(a^{2}-4)(a^{2}+21)=0,\therefore a=2,即PA=2.
(15分)$

Final answer: $2$
Subfield: Solid Geometry


Question: $如图，在四棱锥P-ABCD中，底面ABCD是矩形，AB=2AD=2，PA\perp 平面ABCD，E为PD的中点.$<img_2905>
$(2)若平面BCE与平面CED夹角的正弦值为\frac{\sqrt{21}}{5}，求PA的长.$

Solution:
$\overrightarrow{BC}=(0，1，0),设PA=a(a>0)，则E\left(0,\frac{1}{2},\frac{a}{2}\right)，\therefore \overrightarrow{AE}=\left(0,\frac{1}{2},\frac{a}{2}\right),\overrightarrow{BE}=\left(-2,\frac{1}{2},\frac{a}{2}\right)，设平面BCE的法向量为n=(x,y,z),则\left\{\begin{matrix}\boldsymbol{n}\cdot \overrightarrow{BC}=0,\\ \boldsymbol{n}\cdot \overrightarrow{BE}=0,\end{matrix}\right.即\left\{\begin{matrix}y=0,\\ -2x+\frac{1}{2}y+\frac{a}{2}z=0,\end{matrix}\right.令z=2，解得\left\{\begin{matrix}x=\frac{a}{2},\\ y=0,\end{matrix}\right.\therefore n=\left(\frac{a}{2},0,2\right),
(10分)
设平面CED的法向量为m=(x_{0},y_{0},z_{0})，又\overrightarrow{CP}=(-2,-1,a)，\overrightarrow{CD}=\overrightarrow{BA}=(-2,0,0),\therefore \left\{\begin{matrix}\boldsymbol{m}\cdot \overrightarrow{CP}=0,\\ \boldsymbol{m}\cdot \overrightarrow{CD}=0,\end{matrix}\right.即\left\{\begin{matrix}-2x_0-y_0+az_0=0,\\ -2x_0=0,\end{matrix}\right.令z_{0}=1,解得\left\{\begin{matrix}x_0=0,\\ y_0=a,\end{matrix}\right.\therefore m=(0,a,1)，
(12分)
设平面BCE与平面CED的夹角为\theta ，则\\cos   \theta =|\\cos  \langle n,m\rangle |=\left|\frac{\boldsymbol{n}\cdot \boldsymbol{m}}{|\boldsymbol{n}||\boldsymbol{m}|}\right|=\frac{2}{\sqrt{\frac{a^2}{4}+4}\cdot \sqrt{a^2+1}}=\frac{4}{\sqrt{(a^2+16)(a^2+1)}},因为\\sin   \theta =\frac{\sqrt{21}}{5}，所以\\cos   \theta =\frac{2}{5}，
(13分)
则\frac{4}{\sqrt{(a^2+16)(a^2+1)}}=\frac{2}{5},整理得(a^{2}-4)(a^{2}+21)=0,\therefore a=2,即PA=2.
(15分)$

Final answer: $2$
Subfield: Solid Geometry


Question: $如图,在多面体ABCDEF中,四边形ABCD为正方形，DE\perp 平面ABCD，DE\parallel BF，AD=DE=2，BF=\frac{1}{2}.$![Image](https://cdntk.53zaixian.com/tq/1701760641358974978/img/compression/M_5d9ce43501867cf077b8df37503d1913.jpg)
$(1)求证:AC\perp EF;$

Solution:
$依题意，以D为原点，\overrightarrow{DA}，\overrightarrow{DC}，\overrightarrow{DE}的方向分别为x轴，y轴，z轴的正方向建立空间直角坐标系，可得D(0,0,0)，A(2,0,0)，B(2,2,0)，C(0，2，0)，E(0，0，2)，F\left(2,2,\frac{1}{2}\right).
(1分)
证明：\overrightarrow{AC}=(-2,2,0)，\overrightarrow{EF}=\left(2,2,-\frac{3}{2}\right)，
(2分)
所以\overrightarrow{AC}\cdot \overrightarrow{EF}=(-2)\times 2+2\times 2+0=0,
(3分)
所以\overrightarrow{AC}\perp \overrightarrow{EF}，即AC\perp EF.
(4分)$

Final answer: 证明题
Subfield: Solid Geometry


Question: $如图,在多面体ABCDEF中,四边形ABCD为正方形，DE\perp 平面ABCD，DE\parallel BF，AD=DE=2，BF=\frac{1}{2}.$<img_1905>
$(1)求证:AC\perp EF;$

Solution:
$依题意，以D为原点，\overrightarrow{DA}，\overrightarrow{DC}，\overrightarrow{DE}的方向分别为x轴，y轴，z轴的正方向建立空间直角坐标系，可得D(0,0,0)，A(2,0,0)，B(2,2,0)，C(0，2，0)，E(0，0，2)，F\left(2,2,\frac{1}{2}\right).
(1分)
证明：\overrightarrow{AC}=(-2,2,0)，\overrightarrow{EF}=\left(2,2,-\frac{3}{2}\right)，
(2分)
所以\overrightarrow{AC}\cdot \overrightarrow{EF}=(-2)\times 2+2\times 2+0=0,
(3分)
所以\overrightarrow{AC}\perp \overrightarrow{EF}，即AC\perp EF.
(4分)$
Subfield: Solid Geometry


Question: $如图,在多面体ABCDEF中,四边形ABCD为正方形，DE\perp 平面ABCD，DE\parallel BF，AD=DE=2，BF=\frac{1}{2}.$![Image](https://cdntk.53zaixian.com/tq/1701760641358974978/img/compression/M_5d9ce43501867cf077b8df37503d1913.jpg)
$(2)求直线EC与平面ACF所成角的正弦值;$

Solution:
$\overrightarrow{AC}=(-2,2,0)，\overrightarrow{AF}=\left(0,2,\frac{1}{2}\right)，\overrightarrow{EC}=(0,2,-2),
(5分)
设n=(x,y,z)为平面ACF的法向量, 则\left\{\begin{matrix}\boldsymbol{n}\cdot \overrightarrow{AC}=0,\\ \boldsymbol{n}\cdot \overrightarrow{AF}=0,\end{matrix}\right.即\left\{\begin{matrix}-2x+2y=0,\\ 2y+\frac{1}{2}z=0,\end{matrix}\right.不妨令x=1，可得n=(1,1,-4).
(7分)
设直线EC与平面ACF所成角为\theta ,则\\sin   \theta =|\\cos  \langle \overrightarrow{EC},n\rangle |=\left|\frac{10}{2\sqrt{2}\times 3\sqrt{2}}\right|=\frac{5}{6}，所以直线EC与平面ACF所成角的正弦值为\frac{5}{6}.
(9分)$

Final answer: $\frac{5}{6}$
Subfield: Solid Geometry


Question: $如图,在多面体ABCDEF中,四边形ABCD为正方形，DE\perp 平面ABCD，DE\parallel BF，AD=DE=2，BF=\frac{1}{2}.$<img_1905>
$(2)求直线EC与平面ACF所成角的正弦值;$

Solution:
$\overrightarrow{AC}=(-2,2,0)，\overrightarrow{AF}=\left(0,2,\frac{1}{2}\right)，\overrightarrow{EC}=(0,2,-2),
(5分)
设n=(x,y,z)为平面ACF的法向量, 则\left\{\begin{matrix}\boldsymbol{n}\cdot \overrightarrow{AC}=0,\\ \boldsymbol{n}\cdot \overrightarrow{AF}=0,\end{matrix}\right.即\left\{\begin{matrix}-2x+2y=0,\\ 2y+\frac{1}{2}z=0,\end{matrix}\right.不妨令x=1，可得n=(1,1,-4).
(7分)
设直线EC与平面ACF所成角为\theta ,则\\sin   \theta =|\\cos  \langle \overrightarrow{EC},n\rangle |=\left|\frac{10}{2\sqrt{2}\times 3\sqrt{2}}\right|=\frac{5}{6}，所以直线EC与平面ACF所成角的正弦值为\frac{5}{6}.
(9分)$

Final answer: $\frac{5}{6}$
Subfield: Solid Geometry


Question: $如图,在多面体ABCDEF中,四边形ABCD为正方形，DE\perp 平面ABCD，DE\parallel BF，AD=DE=2，BF=\frac{1}{2}.$<img_1905>
$(3)在线段DE上是否存在点G，使得直线BG与AD所成角的余弦值为\frac{2}{3}？若存在，求出点G到平面ACF的距离；若不存在，请说明理由.$

Solution:
$假设线段DE上存在一点G(0,0,h)(0\leq h\leq 2),使得直线BG与AD所成角的余弦值为\frac{2}{3}，则\overrightarrow{BG}=(-2,-2,h).又\overrightarrow{AD}=(-2,0,0)，所以|\\cos  \langle \overrightarrow{BG},\overrightarrow{AD}\rangle |=\left|\frac{4}{\sqrt{8+h^2}\times 2}\right|=\frac{2}{3}，所以h=1.
(12分)
所以存在点G(0,0,1)满足条件,所以\overrightarrow{AG}=(-2,0,1),
(13分)
由(2)可知平面ACF的一个法向量为n=(1,1,-4),所以点G到平面ACF的距离为\frac{|\overrightarrow{AG}\cdot \boldsymbol{n}|}{|\boldsymbol{n}|}=\frac{|(-2,0,1)\cdot (1,1,-4)|}{3\sqrt{2}}=\sqrt{2}.
(15分)$

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


Question: $如图,在多面体ABCDEF中,四边形ABCD为正方形，DE\perp 平面ABCD，DE\parallel BF，AD=DE=2，BF=\frac{1}{2}.$![Image](https://cdntk.53zaixian.com/tq/1701760641358974978/img/compression/M_5d9ce43501867cf077b8df37503d1913.jpg)
$(3)在线段DE上是否存在点G，使得直线BG与AD所成角的余弦值为\frac{2}{3}？若存在，求出点G到平面ACF的距离；若不存在，请说明理由.$

Solution:
$假设线段DE上存在一点G(0,0,h)(0\leq h\leq 2),使得直线BG与AD所成角的余弦值为\frac{2}{3}，则\overrightarrow{BG}=(-2,-2,h).又\overrightarrow{AD}=(-2,0,0)，所以|\\cos  \langle \overrightarrow{BG},\overrightarrow{AD}\rangle |=\left|\frac{4}{\sqrt{8+h^2}\times 2}\right|=\frac{2}{3}，所以h=1.
(12分)
所以存在点G(0,0,1)满足条件,所以\overrightarrow{AG}=(-2,0,1),
(13分)
由(2)可知平面ACF的一个法向量为n=(1,1,-4),所以点G到平面ACF的距离为\frac{|\overrightarrow{AG}\cdot \boldsymbol{n}|}{|\boldsymbol{n}|}=\frac{|(-2,0,1)\cdot (1,1,-4)|}{3\sqrt{2}}=\sqrt{2}.
(15分)$

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


Question: $如图，四棱锥P-ABCD中，平面PAB\perp 平面ABCD，AB\parallel CD,AB\perp AD,AB=3,AD=\sqrt{3}，AP=CD=2，\angle PAB=60^\circ ，M是CD的中点，N是PB上一点.$<img_1994>
$(1)当PN=\frac{1}{3}PB时.(i)证明:MN\parallel 平面PAD;(ii)求直线PM与平面PAD所成角的正弦值.$

Solution:
$如图，以A为坐标原点，AB所在直线为x轴，AD所在直线为y轴，过点A与平面ABCD垂直的直线为z轴建立空间直角坐标系，则B(3,0,0)，D(0,\sqrt{3},0),P(1,0，\sqrt{3}),M(1,\sqrt{3},0).
(2分)
$<img_1739>$\overrightarrow{PB}=(2,0，-\sqrt{3})，\overrightarrow{PM}=(0，\sqrt{3}，-\sqrt{3})，由PN=\frac{1}{3}PB,可得\overrightarrow{PN}=\left(\frac{2}{3},0,-\frac{\sqrt{3}}{3}\right)，\overrightarrow{MN}=\overrightarrow{PN}-\overrightarrow{PM}=\left(\frac{2}{3},-\sqrt{3},\frac{2\sqrt{3}}{3}\right).
(3分)
(i)证明：设平面PAD的法向量为m=(x,y,z),\overrightarrow{AP}=(1，0，\sqrt{3})，\overrightarrow{AD}=(0，\sqrt{3}，0)，则\left\{\begin{matrix}\overrightarrow{AP}\cdot \boldsymbol{m}=x+\sqrt{3}z=0,\\ \overrightarrow{AD}\cdot \boldsymbol{m}=\sqrt{3}y=0,\end{matrix}\right.解得\left\{\begin{matrix}x=-\sqrt{3}z,\\ y=0,\end{matrix}\right.令z=1，得m=(-\sqrt{3},0,1).因为\overrightarrow{MN}\cdot m=-\frac{2\sqrt{3}}{3}+0+\frac{2\sqrt{3}}{3}=0,所以\overrightarrow{MN}\perp m.又MN\not\subset 平面PAD,所以MN\parallel 平面PAD.
(6分)
(ii)\\cos  \langle \overrightarrow{PM}，m\rangle =\frac{\overrightarrow{PM}\cdot \boldsymbol{m}}{|\overrightarrow{PM}|\cdot |\boldsymbol{m}|}=-\frac{\sqrt{2}}{4}，所以直线PM与平面PAD所成角的正弦值为\frac{\sqrt{2}}{4}.
(9分)$
Subfield: Solid Geometry


Question: $如图，四棱锥P-ABCD中，平面PAB\perp 平面ABCD，AB\parallel CD,AB\perp AD,AB=3,AD=\sqrt{3}，AP=CD=2，\angle PAB=60^\circ ，M是CD的中点，N是PB上一点.$![Image](https://cdntk.53zaixian.com/tq/1701760643087028225/img/compression/M_ee16745d043366aa77e41d6947f064c4.jpg)
$(1)当PN=\frac{1}{3}PB时.(i)证明:MN\parallel 平面PAD;(ii)求直线PM与平面PAD所成角的正弦值.$

Solution:
$如图，以A为坐标原点，AB所在直线为x轴，AD所在直线为y轴，过点A与平面ABCD垂直的直线为z轴建立空间直角坐标系，则B(3,0,0)，D(0,\sqrt{3},0),P(1,0，\sqrt{3}),M(1,\sqrt{3},0).
(2分)
$![Image](https://cdntk.53zaixian.com/tq/1701760643087028225/img/compression/M_86bfa51eb45f57ac6d51cdcac7ed8a60.jpg)$\overrightarrow{PB}=(2,0，-\sqrt{3})，\overrightarrow{PM}=(0，\sqrt{3}，-\sqrt{3})，由PN=\frac{1}{3}PB,可得\overrightarrow{PN}=\left(\frac{2}{3},0,-\frac{\sqrt{3}}{3}\right)，\overrightarrow{MN}=\overrightarrow{PN}-\overrightarrow{PM}=\left(\frac{2}{3},-\sqrt{3},\frac{2\sqrt{3}}{3}\right).
(3分)
(i)证明：设平面PAD的法向量为m=(x,y,z),\overrightarrow{AP}=(1，0，\sqrt{3})，\overrightarrow{AD}=(0，\sqrt{3}，0)，则\left\{\begin{matrix}\overrightarrow{AP}\cdot \boldsymbol{m}=x+\sqrt{3}z=0,\\ \overrightarrow{AD}\cdot \boldsymbol{m}=\sqrt{3}y=0,\end{matrix}\right.解得\left\{\begin{matrix}x=-\sqrt{3}z,\\ y=0,\end{matrix}\right.令z=1，得m=(-\sqrt{3},0,1).因为\overrightarrow{MN}\cdot m=-\frac{2\sqrt{3}}{3}+0+\frac{2\sqrt{3}}{3}=0,所以\overrightarrow{MN}\perp m.又MN\not\subset 平面PAD,所以MN\parallel 平面PAD.
(6分)
(ii)\\cos  \langle \overrightarrow{PM}，m\rangle =\frac{\overrightarrow{PM}\cdot \boldsymbol{m}}{|\overrightarrow{PM}|\cdot |\boldsymbol{m}|}=-\frac{\sqrt{2}}{4}，所以直线PM与平面PAD所成角的正弦值为\frac{\sqrt{2}}{4}.
(9分)$

Final answer: 证明题
Subfield: Solid Geometry


Question: $如图，四棱锥P-ABCD中，平面PAB\perp 平面ABCD，AB\parallel CD,AB\perp AD,AB=3,AD=\sqrt{3}，AP=CD=2，\angle PAB=60^\circ ，M是CD的中点，N是PB上一点.$<img_1994>
$(2)平面PAD与平面AMN夹角的余弦值为\frac{4}{5}，求\frac{PN}{PB}的值.$

Solution:
$\overrightarrow{AM}=(1，\sqrt{3}，0)，设\overrightarrow{PN}=t\overrightarrow{PB}=(2t,0, -\sqrt{3}t)，t\in [0，1],则\overrightarrow{AN}=\overrightarrow{AP}+\overrightarrow{PN}=(1+2t,0,\sqrt{3}-\sqrt{3}t)，设平面AMN的法向量为n=(x_{1}，y_{1},z_{1})，则\left\{\begin{matrix}\boldsymbol{n}\cdot \overrightarrow{AM}=x_1+\sqrt{3}y_1=0,\\ \boldsymbol{n}\cdot \overrightarrow{AN}=(1+2t)x_1+\sqrt{3}(1-t)z_1=0,\end{matrix}\right.取x_{1}=\sqrt{3}(t-1)，则n=(\sqrt{3}(t-1)，1-t,2t+1)，则|\\cos  \langle m,n\rangle |=\frac{|\boldsymbol{m}\cdot \boldsymbol{n}|}{|\boldsymbol{m}|\cdot |\boldsymbol{n}|}=\frac{|3-3t+2t+1|}{2\sqrt{3(t-1)^2+(1-t)^2+(2t+1)^2}}=\frac{4}{5}，化简得487t^{2}-56t-80=0，解得t=\frac{28+24\sqrt{69}}{487}或t=\frac{28-24\sqrt{69}}{487}(舍去).所以\frac{PN}{PB}=\frac{28+24\sqrt{69}}{487}.
(15分)$

Final answer: $\frac{28+24\sqrt{69}}{487}$
Subfield: Solid Geometry


Question: $如图，四棱锥P-ABCD中，平面PAB\perp 平面ABCD，AB\parallel CD,AB\perp AD,AB=3,AD=\sqrt{3}，AP=CD=2，\angle PAB=60^\circ ，M是CD的中点，N是PB上一点.$![Image](https://cdntk.53zaixian.com/tq/1701760643087028225/img/compression/M_ee16745d043366aa77e41d6947f064c4.jpg)
$(2)平面PAD与平面AMN夹角的余弦值为\frac{4}{5}，求\frac{PN}{PB}的值.$

Solution:
$\overrightarrow{AM}=(1，\sqrt{3}，0)，设\overrightarrow{PN}=t\overrightarrow{PB}=(2t,0, -\sqrt{3}t)，t\in [0，1],则\overrightarrow{AN}=\overrightarrow{AP}+\overrightarrow{PN}=(1+2t,0,\sqrt{3}-\sqrt{3}t)，设平面AMN的法向量为n=(x_{1}，y_{1},z_{1})，则\left\{\begin{matrix}\boldsymbol{n}\cdot \overrightarrow{AM}=x_1+\sqrt{3}y_1=0,\\ \boldsymbol{n}\cdot \overrightarrow{AN}=(1+2t)x_1+\sqrt{3}(1-t)z_1=0,\end{matrix}\right.取x_{1}=\sqrt{3}(t-1)，则n=(\sqrt{3}(t-1)，1-t,2t+1)，则|\\cos  \langle m,n\rangle |=\frac{|\boldsymbol{m}\cdot \boldsymbol{n}|}{|\boldsymbol{m}|\cdot |\boldsymbol{n}|}=\frac{|3-3t+2t+1|}{2\sqrt{3(t-1)^2+(1-t)^2+(2t+1)^2}}=\frac{4}{5}，化简得487t^{2}-56t-80=0，解得t=\frac{28+24\sqrt{69}}{487}或t=\frac{28-24\sqrt{69}}{487}(舍去).所以\frac{PN}{PB}=\frac{28+24\sqrt{69}}{487}.
(15分)$

Final answer: $\frac{28+24\sqrt{69}}{487}$
Subfield: Solid Geometry


Question: $如图，在四棱锥P-ABCD中，M是PA的中点，AB\parallel CD,AD\perp CD,PD\perp 平面ABCD，且CD=3，AD=PD=2,AB=1.$![Image](https://cdntk.53zaixian.com/tq/1701760645473587202/img/compression/M_2b942972cbbf5af2b16031688417e9e2.jpg)
$(1)求证：PA\perp CD；$

Solution:
$由题意知AD，CD,PD两两垂直，故以D为原点，分别以DA,DC,DP所在直线为x轴，y轴，z轴建立空间直角坐标系，$![Image](https://cdntk.53zaixian.com/tq/1701760645473587202/img/compression/M_9393656dd43d74fb2a6f8aabb299cfe7.jpg)$则D(0,0,0),A(2,0,0),C(0,3,0),P(0,0,2),M(1,0,1),B(2,1,0).证明：因为\overrightarrow{CD}=(0,-3,0),\overrightarrow{PA}=(2,0,-2)，所以\overrightarrow{PA}\cdot \overrightarrow{CD}=2\times 0+0\times (-3)+(-2)\times 0=0，所以\overrightarrow{PA}\perp \overrightarrow{CD}，所以PA\perp CD.$

Final answer: 证明题
Subfield: Solid Geometry


Question: $如图，在四棱锥P-ABCD中，M是PA的中点，AB\parallel CD,AD\perp CD,PD\perp 平面ABCD，且CD=3，AD=PD=2,AB=1.$<img_2177>
$(1)求证：PA\perp CD；$

Solution:
$由题意知AD，CD,PD两两垂直，故以D为原点，分别以DA,DC,DP所在直线为x轴，y轴，z轴建立空间直角坐标系，$<img_2816>$则D(0,0,0),A(2,0,0),C(0,3,0),P(0,0,2),M(1,0,1),B(2,1,0).证明：因为\overrightarrow{CD}=(0,-3,0),\overrightarrow{PA}=(2,0,-2)，所以\overrightarrow{PA}\cdot \overrightarrow{CD}=2\times 0+0\times (-3)+(-2)\times 0=0，所以\overrightarrow{PA}\perp \overrightarrow{CD}，所以PA\perp CD.$
Subfield: Solid Geometry


Question: $如图，在四棱锥P-ABCD中，M是PA的中点，AB\parallel CD,AD\perp CD,PD\perp 平面ABCD，且CD=3，AD=PD=2,AB=1.$<img_2177>
$(2)求直线PA与平面CMB所成角的正弦值；$

Solution:
$\overrightarrow{CB}=(2,-2,0),\overrightarrow{BM}=(-1,-1,1),\overrightarrow{PA}=(2,0,-2),|\overrightarrow{PA}|=2\sqrt{2}，设平面CMB的法向量为n=(x,y,z)，则\left\{\begin{matrix}\boldsymbol{n}\cdot \overrightarrow{CB}=0,\\ \boldsymbol{n}\cdot \overrightarrow{BM}=0,\end{matrix}\right.即\left\{\begin{matrix}2x-2y=0,\\ -x-y+z=0,\end{matrix}\right.令y=1，则x=1，z=2，\therefore n=(1,1,2),|n|=\sqrt{6},设直线PA与平面CMB所成角为\theta ，则\\sin   \theta =|\\cos  \langle \overrightarrow{PA},n\rangle |=\frac{|\overrightarrow{PA}\cdot \boldsymbol{n}|}{|\overrightarrow{PA}||\boldsymbol{n}|}=\frac{2}{2\sqrt{2}\times \sqrt{6}}=\frac{\sqrt{3}}{6},所以直线PA与平面CMB所成角的正弦值为\frac{\sqrt{3}}{6}.$

Final answer: $\frac{\sqrt{3}}{6}$