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


Question: $如图，在正三棱柱ABC-A_{1}B_{1}C_{1}中，D为棱AA_{1}上的点，E，F，G分别为AC，A_{1}C_{1}，BB_{1}的中点，AC=AA_{1}=2.$![Image](https://cdntk.53zaixian.com/tq/1701760614559633410/img/compression/M_f03a87e835f56775a1b1fbf4eed2e193.jpg)
$(3)若FG\parallel 平面BCD，求二面角B-CD-C_{1}的余弦值.$

Solution:
$由(2)得B(0,\sqrt{3},0)，C(-1,0,0),C_{1}(-1，0，2)，设D(1,0,a)(0\leq a\leq 2)，所以\overrightarrow{BC}=(-1，-\sqrt{3}，0)，\overrightarrow{BD}=(1，-\sqrt{3}，a).设平面BCD的法向量为n=(x，y，z)，则\left\{\begin{matrix}\overrightarrow{BC}\cdot \boldsymbol{n}=0,\\ \overrightarrow{BD}\cdot \boldsymbol{n}=0,\end{matrix}\right.即\left\{\begin{matrix}-x-\sqrt{3}y=0,\\ x-\sqrt{3}y+az=0,\end{matrix}\right.取z=6，可得n=(-3a，\sqrt{3}a，6)，因为FG\parallel 平面BCD，所以\overrightarrow{FG}\cdot n=0，即3a-6=0，解得a=2，所以n=(-6，2\sqrt{3}，6)，易知平面CDC_{1}的一个法向量为m=(0,1，0)，所以\\cos  \langle m,n\rangle =\frac{\boldsymbol{m}\cdot \boldsymbol{n}}{|\boldsymbol{m}||\boldsymbol{n}|}=\frac{2\sqrt{3}}{2\sqrt{21}}=\frac{\sqrt{7}}{7},由图可知二面角B-CD-C_{1}为钝二面角，所以二面角B-CD-C_{1}的余弦值为-\frac{\sqrt{7}}{7}.$

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


Question: $如图，在正三棱柱ABC-A_{1}B_{1}C_{1}中，D为棱AA_{1}上的点，E，F，G分别为AC，A_{1}C_{1}，BB_{1}的中点，AC=AA_{1}=2.$<img_2794>
$(3)若FG\parallel 平面BCD，求二面角B-CD-C_{1}的余弦值.$

Solution:
$由(2)得B(0,\sqrt{3},0)，C(-1,0,0),C_{1}(-1，0，2)，设D(1,0,a)(0\leq a\leq 2)，所以\overrightarrow{BC}=(-1，-\sqrt{3}，0)，\overrightarrow{BD}=(1，-\sqrt{3}，a).设平面BCD的法向量为n=(x，y，z)，则\left\{\begin{matrix}\overrightarrow{BC}\cdot \boldsymbol{n}=0,\\ \overrightarrow{BD}\cdot \boldsymbol{n}=0,\end{matrix}\right.即\left\{\begin{matrix}-x-\sqrt{3}y=0,\\ x-\sqrt{3}y+az=0,\end{matrix}\right.取z=6，可得n=(-3a，\sqrt{3}a，6)，因为FG\parallel 平面BCD，所以\overrightarrow{FG}\cdot n=0，即3a-6=0，解得a=2，所以n=(-6，2\sqrt{3}，6)，易知平面CDC_{1}的一个法向量为m=(0,1，0)，所以\\cos  \langle m,n\rangle =\frac{\boldsymbol{m}\cdot \boldsymbol{n}}{|\boldsymbol{m}||\boldsymbol{n}|}=\frac{2\sqrt{3}}{2\sqrt{21}}=\frac{\sqrt{7}}{7},由图可知二面角B-CD-C_{1}为钝二面角，所以二面角B-CD-C_{1}的余弦值为-\frac{\sqrt{7}}{7}.$

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


Question: $如图，四棱锥P-ABCD的底面是矩形，PD\perp 底面ABCD，M为BC的中点，且PB\perp AM.$![Image](https://cdntk.53zaixian.com/tq/1701760603251789826/img/compression/M_839dc4368fd375946cfd61960880d1d3.jpg)
$(1)证明：平面PAM\perp 平面PBD；$

Solution:
$证明：由于PD\perp 平面ABCD，AM\subset 平面ABCD，则PD\perp AM，又PB\perp AM，PB\cap PD=P，PB，PD\subset 平面PBD，所以AM\perp 平面PBD，因为AM\subset 平面PAM，所以平面PAM\perp 平面PBD.$

Final answer: 证明题
Subfield: Solid Geometry


Question: $如图，四棱锥P-ABCD的底面是矩形，PD\perp 底面ABCD，M为BC的中点，且PB\perp AM.$<img_1661>
$(1)证明：平面PAM\perp 平面PBD；$

Solution:
$证明：由于PD\perp 平面ABCD，AM\subset 平面ABCD，则PD\perp AM，又PB\perp AM，PB\cap PD=P，PB，PD\subset 平面PBD，所以AM\perp 平面PBD，因为AM\subset 平面PAM，所以平面PAM\perp 平面PBD.$
Subfield: Solid Geometry


Question: $如图，四棱锥P-ABCD的底面是矩形，PD\perp 底面ABCD，M为BC的中点，且PB\perp AM.$![Image](https://cdntk.53zaixian.com/tq/1701760603251789826/img/compression/M_839dc4368fd375946cfd61960880d1d3.jpg)
$(2)若PD=DC=1，求四棱锥P-ABCD的体积.$

Solution:
$由(1)知AM\perp 平面PBD，因为BD\subset 平面PBD，所以AM\perp BD，所以\angle MAB+\angle ABD=90^\circ ，因为四边形ABCD为矩形，所以\angle DAB=\angle ABM，所以\angle MAB+\angle AMB=90^\circ ，所以\angle ABD=\angle AMB，则\triangle DAB∽\triangle ABM,则\frac{DA}{AB}=\frac{AB}{BM}，又AB=DC=1，M为BC的中点,所以AD=\sqrt{2}，所以S_{矩形ABCD}=AB\cdot AD=\sqrt{2}，所以V_{四棱锥P-ABCD}=\frac{1}{3}S_{矩形ABCD}\cdot PD=\frac{1}{3}\times \sqrt{2}\times 1=\frac{\sqrt{2}}{3}.$

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


Question: $如图，四棱锥P-ABCD的底面是矩形，PD\perp 底面ABCD，M为BC的中点，且PB\perp AM.$<img_1661>
$(2)若PD=DC=1，求四棱锥P-ABCD的体积.$

Solution:
$由(1)知AM\perp 平面PBD，因为BD\subset 平面PBD，所以AM\perp BD，所以\angle MAB+\angle ABD=90^\circ ，因为四边形ABCD为矩形，所以\angle DAB=\angle ABM，所以\angle MAB+\angle AMB=90^\circ ，所以\angle ABD=\angle AMB，则\triangle DAB∽\triangle ABM,则\frac{DA}{AB}=\frac{AB}{BM}，又AB=DC=1，M为BC的中点,所以AD=\sqrt{2}，所以S_{矩形ABCD}=AB\cdot AD=\sqrt{2}，所以V_{四棱锥P-ABCD}=\frac{1}{3}S_{矩形ABCD}\cdot PD=\frac{1}{3}\times \sqrt{2}\times 1=\frac{\sqrt{2}}{3}.$

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


Question: $如图，在直三棱柱ABC-A_{1}B_{1}C_{1}中,AC\perp AB，点D，E，F分别为A_{1}B_{1},AA_{1},CD的中点，AB=AC=AA_{1}=2.$<img_2510>
$(1)求证:EF\parallel 平面ABC;$

Solution:
$证明：由题意知AC，AB，AA_{1}两两垂直.以A为原点，分别以直线AB，AA_{1}，AC为x,y,z轴建立如图所示的空间直角坐标系.则A(0，0，0)，B(2，0，0)，B_{1}(2，2，0)，A_{1}(0，2，0)，C(0，0，2)，C_{1}(0，2，2)，D(1，2，0)，E(0，1，0)，F\left(\frac{1}{2},1,1\right).\because AA_{1}\perp 平面ABC,\therefore 平面ABC的一个法向量为n_{1}=(0，1，0).\because \overrightarrow{EF}=\left(\frac{1}{2},0,1\right),\therefore \overrightarrow{EF}\cdot n_{1}=0.又\because EF\not\subset 平面ABC,\therefore EF\parallel 平面ABC.$<img_1785>
Subfield: Solid Geometry


Question: $如图，在直三棱柱ABC-A_{1}B_{1}C_{1}中,AC\perp AB，点D，E，F分别为A_{1}B_{1},AA_{1},CD的中点，AB=AC=AA_{1}=2.$![Image](https://cdntk.53zaixian.com/tq/1701760626725425154/img/compression/M_a90d8ea299b394e6bb080b143da7dd27.jpg)
$(1)求证:EF\parallel 平面ABC;$

Solution:
$证明：由题意知AC，AB，AA_{1}两两垂直.以A为原点，分别以直线AB，AA_{1}，AC为x,y,z轴建立如图所示的空间直角坐标系.则A(0，0，0)，B(2，0，0)，B_{1}(2，2，0)，A_{1}(0，2，0)，C(0，0，2)，C_{1}(0，2，2)，D(1，2，0)，E(0，1，0)，F\left(\frac{1}{2},1,1\right).\because AA_{1}\perp 平面ABC,\therefore 平面ABC的一个法向量为n_{1}=(0，1，0).\because \overrightarrow{EF}=\left(\frac{1}{2},0,1\right),\therefore \overrightarrow{EF}\cdot n_{1}=0.又\because EF\not\subset 平面ABC,\therefore EF\parallel 平面ABC.$![Image](https://cdntk.53zaixian.com/tq/1701760627056775169/img/compression/M_f357c296a7f31ab1296e74529d7f6614.jpg)

Final answer: 证明题
Subfield: Solid Geometry


Question: $如图，在直三棱柱ABC-A_{1}B_{1}C_{1}中,AC\perp AB，点D，E，F分别为A_{1}B_{1},AA_{1},CD的中点，AB=AC=AA_{1}=2.$![Image](https://cdntk.53zaixian.com/tq/1701760626725425154/img/compression/M_a90d8ea299b394e6bb080b143da7dd27.jpg)
$(2)求直线BE与平面CC_{1}D所成角的正弦值；$

Solution:
$设平面CC_{1}D的法向量为n_{2}=(x,y,z),\because \overrightarrow{CC_1}=(0,2,0),\overrightarrow{CD}=(1,2,-2),\therefore \left\{\begin{matrix}\boldsymbol{n}_2\cdot \overrightarrow{CC_1}=0,\\ \boldsymbol{n}_2\cdot \overrightarrow{CD}=0,\end{matrix}\right.即\left\{\begin{matrix}y=0,\\ x+2y-2z=0.\end{matrix}\right.令x=2,得z=1,\therefore n_{2}=(2,0,1).设直线BE与平面CC_{1}D所成的角为\theta ，\because \overrightarrow{BE}=(-2,1,0),\therefore \\sin   \theta =|\\cos  \langle \overrightarrow{BE}，n_{2}\rangle |=\frac{|\boldsymbol{n}_2\cdot \overrightarrow{BE}|}{|\boldsymbol{n}_2|\cdot |\overrightarrow{BE}|}=\frac{4}{\sqrt{5}\times \sqrt{5}}=\frac{4}{5},\therefore 直线BE与平面CC_{1}D所成角的正弦值为\frac{4}{5}.$

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


Question: $如图，在直三棱柱ABC-A_{1}B_{1}C_{1}中,AC\perp AB，点D，E，F分别为A_{1}B_{1},AA_{1},CD的中点，AB=AC=AA_{1}=2.$<img_2510>
$(2)求直线BE与平面CC_{1}D所成角的正弦值；$

Solution:
$设平面CC_{1}D的法向量为n_{2}=(x,y,z),\because \overrightarrow{CC_1}=(0,2,0),\overrightarrow{CD}=(1,2,-2),\therefore \left\{\begin{matrix}\boldsymbol{n}_2\cdot \overrightarrow{CC_1}=0,\\ \boldsymbol{n}_2\cdot \overrightarrow{CD}=0,\end{matrix}\right.即\left\{\begin{matrix}y=0,\\ x+2y-2z=0.\end{matrix}\right.令x=2,得z=1,\therefore n_{2}=(2,0,1).设直线BE与平面CC_{1}D所成的角为\theta ，\because \overrightarrow{BE}=(-2,1,0),\therefore \\sin   \theta =|\\cos  \langle \overrightarrow{BE}，n_{2}\rangle |=\frac{|\boldsymbol{n}_2\cdot \overrightarrow{BE}|}{|\boldsymbol{n}_2|\cdot |\overrightarrow{BE}|}=\frac{4}{\sqrt{5}\times \sqrt{5}}=\frac{4}{5},\therefore 直线BE与平面CC_{1}D所成角的正弦值为\frac{4}{5}.$

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


Question: $如图，在直三棱柱ABC-A_{1}B_{1}C_{1}中,AC\perp AB，点D，E，F分别为A_{1}B_{1},AA_{1},CD的中点，AB=AC=AA_{1}=2.$![Image](https://cdntk.53zaixian.com/tq/1701760626725425154/img/compression/M_a90d8ea299b394e6bb080b143da7dd27.jpg)
$(3)求平面A_{1}CD与平面CC_{1}D夹角的余弦值.$

Solution:
$设平面A_{1}CD的法向量为n_{3}=(x_{1},y_{1},z_{1}),\because \overrightarrow{A_1C}=(0,-2,2),\overrightarrow{CD}=(1,2,-2),\therefore \left\{\begin{matrix}\boldsymbol{n}_3\cdot \overrightarrow{A_1C}=0,\\ \boldsymbol{n}_3\cdot \overrightarrow{CD}=0,\end{matrix}\right.即\left\{\begin{matrix}-2y_1+2z_1=0,\\ x_1+2y_1-2z_1=0,\end{matrix}\right.\therefore x_{1}=0,令y_{1}=1,得z_{1}=1,\therefore n_{3}=(0,1,1).设平面A_{1}CD与平面CC_{1}D的夹角为\alpha ,\therefore \\cos   \alpha =|\\cos  \langle n_{2}，n_{3}\rangle |=\frac{|\boldsymbol{n}_2\cdot \boldsymbol{n}_3|}{|\boldsymbol{n}_2||\boldsymbol{n}_3|}=\frac{1}{\sqrt{5}\times \sqrt{2}}=\frac{\sqrt{10}}{10}.\therefore 平面A_{1}CD与平面CC_{1}D夹角的余弦值为\frac{\sqrt{10}}{10}.$

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


Question: $如图，在直三棱柱ABC-A_{1}B_{1}C_{1}中,AC\perp AB，点D，E，F分别为A_{1}B_{1},AA_{1},CD的中点，AB=AC=AA_{1}=2.$<img_2510>
$(3)求平面A_{1}CD与平面CC_{1}D夹角的余弦值.$

Solution:
$设平面A_{1}CD的法向量为n_{3}=(x_{1},y_{1},z_{1}),\because \overrightarrow{A_1C}=(0,-2,2),\overrightarrow{CD}=(1,2,-2),\therefore \left\{\begin{matrix}\boldsymbol{n}_3\cdot \overrightarrow{A_1C}=0,\\ \boldsymbol{n}_3\cdot \overrightarrow{CD}=0,\end{matrix}\right.即\left\{\begin{matrix}-2y_1+2z_1=0,\\ x_1+2y_1-2z_1=0,\end{matrix}\right.\therefore x_{1}=0,令y_{1}=1,得z_{1}=1,\therefore n_{3}=(0,1,1).设平面A_{1}CD与平面CC_{1}D的夹角为\alpha ,\therefore \\cos   \alpha =|\\cos  \langle n_{2}，n_{3}\rangle |=\frac{|\boldsymbol{n}_2\cdot \boldsymbol{n}_3|}{|\boldsymbol{n}_2||\boldsymbol{n}_3|}=\frac{1}{\sqrt{5}\times \sqrt{2}}=\frac{\sqrt{10}}{10}.\therefore 平面A_{1}CD与平面CC_{1}D夹角的余弦值为\frac{\sqrt{10}}{10}.$

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


Question: $如图所示，在直三棱柱ABC-A_{1}B_{1}C_{1}中，M为棱BB_{1}的中点，\angle ACB=90^\circ ，CC_{1}=AC=2BC=2.$<img_3019>
$(1)求证：C_{1}M\perp 平面AMC；$

Solution:
$如图，以C为坐标原点，CA，CB，CC_{1}所在直线分别为x轴，y轴,z轴建立空间直角坐标系C-xyz,$<img_1499>$则C(0,0,0),A(2,0,0),B(0,1,0),C_{1}(0,0,2),B_{1}(0,1,2),M(0,1,1),A_{1}(2,0,2).
(3分)
证明:易得\overrightarrow{CM}=(0,1,1),\overrightarrow{AM}=(-2,1,1),设n=(x_{1},y_{1},z_{1})为平面AMC的法向量，则\left\{\begin{matrix}\boldsymbol{n}\cdot \overrightarrow{CM}=y_1+z_1=0,\\ \boldsymbol{n}\cdot \overrightarrow{AM}=-2x_1+y_1+z_1=0,\end{matrix}\right.令y_{1}=1，则n=(0,1,-1),因为\overrightarrow{C_1M}=(0,1,-1)=n,所以C_{1}M\perp 平面AMC.
(7分)$
Subfield: Solid Geometry


Question: $如图所示，在直三棱柱ABC-A_{1}B_{1}C_{1}中，M为棱BB_{1}的中点，\angle ACB=90^\circ ，CC_{1}=AC=2BC=2.$![Image](https://cdntk.53zaixian.com/tq/1701760637043036161/img/compression/M_f03e661a7ce4fd185aebc4cae506ad40.jpg)
$(1)求证：C_{1}M\perp 平面AMC；$

Solution:
$如图，以C为坐标原点，CA，CB，CC_{1}所在直线分别为x轴，y轴,z轴建立空间直角坐标系C-xyz,$![Image](https://cdntk.53zaixian.com/tq/1701760637043036161/img/compression/M_164028349dee332777f51c8a19bcded2.jpg)$则C(0,0,0),A(2,0,0),B(0,1,0),C_{1}(0,0,2),B_{1}(0,1,2),M(0,1,1),A_{1}(2,0,2).
(3分)
证明:易得\overrightarrow{CM}=(0,1,1),\overrightarrow{AM}=(-2,1,1),设n=(x_{1},y_{1},z_{1})为平面AMC的法向量，则\left\{\begin{matrix}\boldsymbol{n}\cdot \overrightarrow{CM}=y_1+z_1=0,\\ \boldsymbol{n}\cdot \overrightarrow{AM}=-2x_1+y_1+z_1=0,\end{matrix}\right.令y_{1}=1，则n=(0,1,-1),因为\overrightarrow{C_1M}=(0,1,-1)=n,所以C_{1}M\perp 平面AMC.
(7分)$

Final answer: 证明题
Subfield: Solid Geometry


Question: $如图所示，在直三棱柱ABC-A_{1}B_{1}C_{1}中，M为棱BB_{1}的中点，\angle ACB=90^\circ ，CC_{1}=AC=2BC=2.$![Image](https://cdntk.53zaixian.com/tq/1701760637043036161/img/compression/M_f03e661a7ce4fd185aebc4cae506ad40.jpg)
$(2)求异面直线AM与BC_{1}所成角的余弦值；$

Solution:
$因为\overrightarrow{BC_1}=(0,-1,2),\overrightarrow{AM}=(-2,1,1)，所以|\\cos  \langle \overrightarrow{BC_1},\overrightarrow{AM}\rangle |=\left|\frac{\overrightarrow{BC_1}\cdot \overrightarrow{AM}}{|\overrightarrow{BC_1}||\overrightarrow{AM}|}\right|=\frac{1}{\sqrt{5}\times \sqrt{6}}=\frac{\sqrt{30}}{30}，所以异面直线AM与BC_{1}所成角的余弦值为\frac{\sqrt{30}}{30}.
(11分)$

Final answer: $\frac{\sqrt{30}}{30}$
Subfield: Solid Geometry


Question: $如图所示，在直三棱柱ABC-A_{1}B_{1}C_{1}中，M为棱BB_{1}的中点，\angle ACB=90^\circ ，CC_{1}=AC=2BC=2.$<img_3019>
$(2)求异面直线AM与BC_{1}所成角的余弦值；$

Solution:
$因为\overrightarrow{BC_1}=(0,-1,2),\overrightarrow{AM}=(-2,1,1)，所以|\\cos  \langle \overrightarrow{BC_1},\overrightarrow{AM}\rangle |=\left|\frac{\overrightarrow{BC_1}\cdot \overrightarrow{AM}}{|\overrightarrow{BC_1}||\overrightarrow{AM}|}\right|=\frac{1}{\sqrt{5}\times \sqrt{6}}=\frac{\sqrt{30}}{30}，所以异面直线AM与BC_{1}所成角的余弦值为\frac{\sqrt{30}}{30}.
(11分)$

Final answer: $\frac{\sqrt{30}}{30}$
Subfield: Solid Geometry


Question: $如图所示，在直三棱柱ABC-A_{1}B_{1}C_{1}中，M为棱BB_{1}的中点，\angle ACB=90^\circ ，CC_{1}=AC=2BC=2.$<img_3019>
$(3)求平面AMC与平面A_{1}BC_{1}的夹角的余弦值.$

Solution:
$易得\overrightarrow{A_1B}=(-2,1,-2),\overrightarrow{A_1C_1}=(-2,0,0)，设m=(x_{2},y_{2},z_{2})为平面A_{1}BC_{1}的法向量，则\left\{\begin{matrix}\boldsymbol{m}\cdot \overrightarrow{A_1B}=-2x_2+y_2-2z_2=0,\\ \boldsymbol{m}\cdot \overrightarrow{A_1C_1}=-2x_2=0,\end{matrix}\right.令y_{2}=2，则m=(0,2,1),由(1)知平面AMC的一个法向量为n=(0，1，-1)，所以|\\cos  \langle m,n\rangle |=\left|\frac{\boldsymbol{m}\cdot \boldsymbol{n}}{|\boldsymbol{m}||\boldsymbol{n}|}\right|=\frac{1}{\sqrt{5}\times \sqrt{2}}=\frac{\sqrt{10}}{10}，所以平面AMC与平面A_{1}BC_{1}的夹角的余弦值为\frac{\sqrt{10}}{10}.
(15分)$

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


Question: $如图所示，在直三棱柱ABC-A_{1}B_{1}C_{1}中，M为棱BB_{1}的中点，\angle ACB=90^\circ ，CC_{1}=AC=2BC=2.$![Image](https://cdntk.53zaixian.com/tq/1701760637043036161/img/compression/M_f03e661a7ce4fd185aebc4cae506ad40.jpg)
$(3)求平面AMC与平面A_{1}BC_{1}的夹角的余弦值.$

Solution:
$易得\overrightarrow{A_1B}=(-2,1,-2),\overrightarrow{A_1C_1}=(-2,0,0)，设m=(x_{2},y_{2},z_{2})为平面A_{1}BC_{1}的法向量，则\left\{\begin{matrix}\boldsymbol{m}\cdot \overrightarrow{A_1B}=-2x_2+y_2-2z_2=0,\\ \boldsymbol{m}\cdot \overrightarrow{A_1C_1}=-2x_2=0,\end{matrix}\right.令y_{2}=2，则m=(0,2,1),由(1)知平面AMC的一个法向量为n=(0，1，-1)，所以|\\cos  \langle m,n\rangle |=\left|\frac{\boldsymbol{m}\cdot \boldsymbol{n}}{|\boldsymbol{m}||\boldsymbol{n}|}\right|=\frac{1}{\sqrt{5}\times \sqrt{2}}=\frac{\sqrt{10}}{10}，所以平面AMC与平面A_{1}BC_{1}的夹角的余弦值为\frac{\sqrt{10}}{10}.
(15分)$

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


Question: $如图，在直三棱柱ABC-A_{1}B_{1}C_{1}中，AC\perp AB,AC=2AB=2AA_{1}=2,M为AC的中点，A_{1}N\perp B_{1}C_{1},垂足为N.$<img_2426>
$(1)求证:B_{1}C\parallel 平面A_{1}BM;$

Solution:
$如图，以A为坐标原点，直线AB，AA_{1}，AC分别为x轴，y轴，z轴建立空间直角坐标系，则B(1,0,0),C(0,0,2)，A_{1}(0，1,0)，B_{1}(1，1,0)，C_{1}(0，1,2)，M(0，0，1).
(2分)
$<img_2526>$证明：易得\overrightarrow{A_1B}=(1,-1,0)，\overrightarrow{BM}=(-1,0，1)，\overrightarrow{B_1C}=(-1,-1,2)，设平面A_{1}BM的法向量为m=(x_{1}，y_{1}，z_{1}),则\left\{\begin{matrix}\overrightarrow{A_1B}\cdot \boldsymbol{m}=x_1-y_1=0,\\ \overrightarrow{BM}\cdot \boldsymbol{m}=-x_1+z_1=0,\end{matrix}\right.令x_{1}=1,得m=(1，1，1).因为\overrightarrow{B_1C}\cdot m=-1-1+2=0，所以\overrightarrow{B_1C}\perp m.又B_{1}C\not\subset 平面A_{1}BM,所以B_{1}C\parallel 平面A_{1}BM.
(6分)$
Subfield: Solid Geometry


Question: $如图，在直三棱柱ABC-A_{1}B_{1}C_{1}中，AC\perp AB,AC=2AB=2AA_{1}=2,M为AC的中点，A_{1}N\perp B_{1}C_{1},垂足为N.$![Image](https://cdntk.53zaixian.com/tq/1701760638540402690/img/compression/M_022a2edfe1a32f3ec184c9b95df252ac.jpg)
$(1)求证:B_{1}C\parallel 平面A_{1}BM;$

Solution:
$如图，以A为坐标原点，直线AB，AA_{1}，AC分别为x轴，y轴，z轴建立空间直角坐标系，则B(1,0,0),C(0,0,2)，A_{1}(0，1,0)，B_{1}(1，1,0)，C_{1}(0，1,2)，M(0，0，1).
(2分)
$![Image](https://cdntk.53zaixian.com/tq/1701760638540402690/img/compression/M_9327986f9ad62fd13cf800fac5af3771.jpg)$证明：易得\overrightarrow{A_1B}=(1,-1,0)，\overrightarrow{BM}=(-1,0，1)，\overrightarrow{B_1C}=(-1,-1,2)，设平面A_{1}BM的法向量为m=(x_{1}，y_{1}，z_{1}),则\left\{\begin{matrix}\overrightarrow{A_1B}\cdot \boldsymbol{m}=x_1-y_1=0,\\ \overrightarrow{BM}\cdot \boldsymbol{m}=-x_1+z_1=0,\end{matrix}\right.令x_{1}=1,得m=(1，1，1).因为\overrightarrow{B_1C}\cdot m=-1-1+2=0，所以\overrightarrow{B_1C}\perp m.又B_{1}C\not\subset 平面A_{1}BM,所以B_{1}C\parallel 平面A_{1}BM.
(6分)$

Final answer: 证明题
Subfield: Solid Geometry


Question: $如图，在直三棱柱ABC-A_{1}B_{1}C_{1}中，AC\perp AB,AC=2AB=2AA_{1}=2,M为AC的中点，A_{1}N\perp B_{1}C_{1},垂足为N.$<img_2426>
$(2)求直线BN与平面A_{1}BM所成角的正弦值;$

Solution:
$易得\overrightarrow{B_1C_1}=(-1，0，2)，\overrightarrow{A_1B_1}=(1，0，0)，\overrightarrow{BB_1}=(0，1，0)，设\overrightarrow{B_1N}=\lambda \overrightarrow{B_1C_1}=(-\lambda ,0,2\lambda )，则\overrightarrow{A_1N}=\overrightarrow{A_1B_1}+\overrightarrow{B_1N}=(1-\lambda ,0,2\lambda ).因为A_{1}N\perp B_{1}C_{1}，所以\overrightarrow{A_1N}\cdot \overrightarrow{B_1C_1}=(1-\lambda ,0,2\lambda )\cdot (-1,0,2)=0，即5\lambda -1=0，解得\lambda =\frac{1}{5}，所以\overrightarrow{B_1N}=\left(-\frac{1}{5},0,\frac{2}{5}\right),所以\overrightarrow{BN}=\overrightarrow{BB_1}+\overrightarrow{B_1N}=\left(-\frac{1}{5},1,\frac{2}{5}\right).
(9分)
所以|\\cos  \langle \overrightarrow{BN},m\rangle |=\frac{|\overrightarrow{BN}\cdot \boldsymbol{m}|}{|\overrightarrow{BN}||\boldsymbol{m}|}=\frac{\sqrt{10}}{5}，所以直线BN与平面A_{1}BM所成角的正弦值为\frac{\sqrt{10}}{5}.
(11分)$

Final answer: $\frac{\sqrt{10}}{5}$
Subfield: Solid Geometry


Question: $如图，在直三棱柱ABC-A_{1}B_{1}C_{1}中，AC\perp AB,AC=2AB=2AA_{1}=2,M为AC的中点，A_{1}N\perp B_{1}C_{1},垂足为N.$![Image](https://cdntk.53zaixian.com/tq/1701760638540402690/img/compression/M_022a2edfe1a32f3ec184c9b95df252ac.jpg)
$(2)求直线BN与平面A_{1}BM所成角的正弦值;$

Solution:
$易得\overrightarrow{B_1C_1}=(-1，0，2)，\overrightarrow{A_1B_1}=(1，0，0)，\overrightarrow{BB_1}=(0，1，0)，设\overrightarrow{B_1N}=\lambda \overrightarrow{B_1C_1}=(-\lambda ,0,2\lambda )，则\overrightarrow{A_1N}=\overrightarrow{A_1B_1}+\overrightarrow{B_1N}=(1-\lambda ,0,2\lambda ).因为A_{1}N\perp B_{1}C_{1}，所以\overrightarrow{A_1N}\cdot \overrightarrow{B_1C_1}=(1-\lambda ,0,2\lambda )\cdot (-1,0,2)=0，即5\lambda -1=0，解得\lambda =\frac{1}{5}，所以\overrightarrow{B_1N}=\left(-\frac{1}{5},0,\frac{2}{5}\right),所以\overrightarrow{BN}=\overrightarrow{BB_1}+\overrightarrow{B_1N}=\left(-\frac{1}{5},1,\frac{2}{5}\right).
(9分)
所以|\\cos  \langle \overrightarrow{BN},m\rangle |=\frac{|\overrightarrow{BN}\cdot \boldsymbol{m}|}{|\overrightarrow{BN}||\boldsymbol{m}|}=\frac{\sqrt{10}}{5}，所以直线BN与平面A_{1}BM所成角的正弦值为\frac{\sqrt{10}}{5}.
(11分)$

Final answer: $\frac{\sqrt{10}}{5}$
Subfield: Solid Geometry


Question: $如图，在直三棱柱ABC-A_{1}B_{1}C_{1}中，AC\perp AB,AC=2AB=2AA_{1}=2,M为AC的中点，A_{1}N\perp B_{1}C_{1},垂足为N.$<img_2426>
$(3)求平面A_{1}BN与平面A_{1}BM的夹角.$

Solution:
$设平面A_{1}BN的法向量为n=(x_{2}，y_{2},z_{2}),则\left\{\begin{matrix}\overrightarrow{A_1B}\cdot \boldsymbol{n}=x_2-y_2=0,\\ \overrightarrow{BN}\cdot \boldsymbol{n}=-\frac{1}{5}x_2+y_2+\frac{2}{5}z_2=0,\end{matrix}\right.令x_{2}=1,得n=(1，1,-2).因为m\cdot n=1+1-2=0，所以m\perp n.所以平面A_{1}BN与平面A_{1}BM的夹角为90^\circ .
(15分)$

Final answer: $90^\circ $
Subfield: Solid Geometry


Question: $如图，在直三棱柱ABC-A_{1}B_{1}C_{1}中，AC\perp AB,AC=2AB=2AA_{1}=2,M为AC的中点，A_{1}N\perp B_{1}C_{1},垂足为N.$![Image](https://cdntk.53zaixian.com/tq/1701760638540402690/img/compression/M_022a2edfe1a32f3ec184c9b95df252ac.jpg)
$(3)求平面A_{1}BN与平面A_{1}BM的夹角.$

Solution:
$设平面A_{1}BN的法向量为n=(x_{2}，y_{2},z_{2}),则\left\{\begin{matrix}\overrightarrow{A_1B}\cdot \boldsymbol{n}=x_2-y_2=0,\\ \overrightarrow{BN}\cdot \boldsymbol{n}=-\frac{1}{5}x_2+y_2+\frac{2}{5}z_2=0,\end{matrix}\right.令x_{2}=1,得n=(1，1,-2).因为m\cdot n=1+1-2=0，所以m\perp n.所以平面A_{1}BN与平面A_{1}BM的夹角为90^\circ .
(15分)$

Final answer: $90^\circ $
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)
$(1)若t=\frac{1}{2}，求证:AD\parallel 平面A_{1}BE;$

Solution:
$以A为坐标原点，AB，AC，AA_{1}所在直线分别为x轴，y轴，z轴，建立空间直角坐标系，$![Image](https://cdntk.53zaixian.com/tq/1701760638540402690/img/compression/M_a00154ff47fac924a08050a6c2de09e6.jpg)$则A(0,0,0)，B(3,0,0)，C(0,3,0)，A_{1}(0,0,3)，B_{1}(3,0,3)，C_{1}(0,3,3).证明：\because t=\frac{1}{2}，\therefore D\left(\frac{3}{2},\frac{3}{2},0\right)，E\left(\frac{3}{2},\frac{3}{2},3\right).易得\overrightarrow{A_1B}=(3，0，-3)，\overrightarrow{A_1E}=\left(\frac{3}{2},\frac{3}{2},0\right).设平面A_{1}BE的法向量为n=(x_{1},y_{1},z_{1}),则\left\{\begin{matrix}\boldsymbol{n}\cdot \overrightarrow{A_1B}=0,\\ \boldsymbol{n}\cdot \overrightarrow{A_1E}=0,\end{matrix}\right.即\left\{\begin{matrix}3x_1-3z_1=0,\\ \frac{3}{2}x_1+\frac{3}{2}y_1=0.\end{matrix}\right.取x_{1}=1，得n=(1,-1,1).
(3分)
\because \overrightarrow{AD}=\left(\frac{3}{2},\frac{3}{2},0\right)，\overrightarrow{AD}\cdot n=0，\therefore \overrightarrow{AD}\perp n.
(4分)
又AD\not\subset 平面A_{1}BE,\therefore AD\parallel 平面A_{1}BE.
(5分)$

Final answer: 证明题
Subfield: Solid Geometry


Question: $如图，在直三棱柱ABC-A_{1}B_{1}C_{1}中，\triangle ABC是以BC为斜边的等腰直角三角形，AA_{1}=AB=3，D，E分别为BC，B_{1}C_{1}上的点，且\frac{BD}{BC}=\frac{C_1E}{C_1B_1}=t(0$<img_3125>
$(1)若t=\frac{1}{2}，求证:AD\parallel 平面A_{1}BE;$

Solution:
$以A为坐标原点，AB，AC，AA_{1}所在直线分别为x轴，y轴，z轴，建立空间直角坐标系，$<img_2920>$则A(0,0,0)，B(3,0,0)，C(0,3,0)，A_{1}(0,0,3)，B_{1}(3,0,3)，C_{1}(0,3,3).证明：\because t=\frac{1}{2}，\therefore D\left(\frac{3}{2},\frac{3}{2},0\right)，E\left(\frac{3}{2},\frac{3}{2},3\right).易得\overrightarrow{A_1B}=(3，0，-3)，\overrightarrow{A_1E}=\left(\frac{3}{2},\frac{3}{2},0\right).设平面A_{1}BE的法向量为n=(x_{1},y_{1},z_{1}),则\left\{\begin{matrix}\boldsymbol{n}\cdot \overrightarrow{A_1B}=0,\\ \boldsymbol{n}\cdot \overrightarrow{A_1E}=0,\end{matrix}\right.即\left\{\begin{matrix}3x_1-3z_1=0,\\ \frac{3}{2}x_1+\frac{3}{2}y_1=0.\end{matrix}\right.取x_{1}=1，得n=(1,-1,1).
(3分)
\because \overrightarrow{AD}=\left(\frac{3}{2},\frac{3}{2},0\right)，\overrightarrow{AD}\cdot n=0，\therefore \overrightarrow{AD}\perp n.
(4分)
又AD\not\subset 平面A_{1}BE,\therefore AD\parallel 平面A_{1}BE.
(5分)$
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)
$(2)若t=\frac{1}{2}，求直线A_{1}C与平面A_{1}BE所成角的正弦值;$

Solution:
$由(1)知，平面A_{1}BE的一个法向量为n=(1,-1,1).设直线A_{1}C与平面A_{1}BE所成的角为\alpha ，\because \overrightarrow{A_1C}=(0,3,-3).
(6分)
\therefore \\sin   \alpha =|\\cos  \langle \overrightarrow{A_1C},n\rangle |=\frac{|\overrightarrow{A_1C}\cdot \boldsymbol{n}|}{|\overrightarrow{A_1C}||\boldsymbol{n}|}=\frac{6}{3\sqrt{2}\times \sqrt{3}}=\frac{\sqrt{6}}{3}.\therefore 直线A_{1}C与平面A_{1}BE所成角的正弦值为\frac{\sqrt{6}}{3}.
(9分)$

Final answer: $\frac{\sqrt{6}}{3}$
Subfield: Solid Geometry


Question: $如图，在直三棱柱ABC-A_{1}B_{1}C_{1}中，\triangle ABC是以BC为斜边的等腰直角三角形，AA_{1}=AB=3，D，E分别为BC，B_{1}C_{1}上的点，且\frac{BD}{BC}=\frac{C_1E}{C_1B_1}=t(0$<img_3125>
$(2)若t=\frac{1}{2}，求直线A_{1}C与平面A_{1}BE所成角的正弦值;$

Solution:
$由(1)知，平面A_{1}BE的一个法向量为n=(1,-1,1).设直线A_{1}C与平面A_{1}BE所成的角为\alpha ，\because \overrightarrow{A_1C}=(0,3,-3).
(6分)
\therefore \\sin   \alpha =|\\cos  \langle \overrightarrow{A_1C},n\rangle |=\frac{|\overrightarrow{A_1C}\cdot \boldsymbol{n}|}{|\overrightarrow{A_1C}||\boldsymbol{n}|}=\frac{6}{3\sqrt{2}\times \sqrt{3}}=\frac{\sqrt{6}}{3}.\therefore 直线A_{1}C与平面A_{1}BE所成角的正弦值为\frac{\sqrt{6}}{3}.
(9分)$

Final answer: $\frac{\sqrt{6}}{3}$
Subfield: Solid Geometry


Question: $如图，在直三棱柱ABC-A_{1}B_{1}C_{1}中，\triangle ABC是以BC为斜边的等腰直角三角形，AA_{1}=AB=3，D，E分别为BC，B_{1}C_{1}上的点，且\frac{BD}{BC}=\frac{C_1E}{C_1B_1}=t(0$<img_3125>
$(3)若平面AC_{1}D与平面ACD的夹角为\frac{\pi }{3}，求实数t的值.$

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: $2-\sqrt{2}$