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


Question: $如图，在长方体ABCD-A_{1}B_{1}C_{1}D_{1}中，点E，F分别在棱DD_{1}，BB_{1}上，且2DE=ED_{1}，BF=2FB_{1}.$<img_2011>
$(2)若AB=2，AD=1，AA_{1}=3，求二面角A-EF-A_{1}的正弦值.$

Solution:
$建系同(1).由已知得A(2，1，3)，E(2，0，2)，F(0，1，1)，A_{1}(2，1，0)，\overrightarrow{AE}=(0，-1， -1)，\overrightarrow{AF}=(-2，0，-2)，\overrightarrow{A_1E}=(0，-1，2)，\overrightarrow{A_1F}=(-2，0，1).设n_{1}=(x,y,z)为平面AEF的法向量，则\left\{\begin{matrix}\boldsymbol{n}_1\cdot A\overrightarrow{E}=0,\\ \boldsymbol{n}_1\cdot A\overrightarrow{F}=0,\end{matrix}\right.即\left\{\begin{matrix}-y-z=0,\\ -2x-2z=0,\end{matrix}\right.可取n_{1}=(-1,-1,1).设n_{2}为平面A_{1}EF的法向量，则\left\{\begin{matrix}\boldsymbol{n}_2\cdot \overrightarrow{A_1E}=0,\\ \boldsymbol{n}_2\cdot \overrightarrow{A_1F}=0,\end{matrix}\right.同理可取n_{2}=\left(\frac{1}{2},2,1\right).因为\\cos  \langle n_{1},n_{2}\rangle =\frac{\boldsymbol{n}_1\cdot \boldsymbol{n}_2}{|\boldsymbol{n}_1|\cdot |\boldsymbol{n}_2|}=-\frac{\sqrt{7}}{7}，所以二面角A-EF-A_{1}的正弦值为\frac{\sqrt{42}}{7}.$

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


Question: $在正方体ABCD-A_{1}B_{1}C_{1}D_{1}中，点E为A_{1}D_{1}的中点，B_{1}C_{1}与平面CDE交于点F.$<img_2533>
$(1)求证：F为B_{1}C_{1}的中点；$

Solution:
$证明：在正方体ABCD-A_{1}B_{1}C_{1}D_{1}中，CD\parallel C_{1}D_{1}，因为CD\not\subset 平面A_{1}B_{1}C_{1}D_{1}，C_{1}D_{1}\subset 平面A_{1}B_{1}C_{1}D_{1}，所以CD\parallel 平面A_{1}B_{1}C_{1}D_{1}，又CD\subset 平面CDE，平面A_{1}B_{1}C_{1}D_{1}\cap 平面CDE=EF，所以CD\parallel EF，则EF\parallel C_{1}D_{1}.在正方体ABCD-A_{1}B_{1}C_{1}D_{1}中，A_{1}D_{1}\parallel B_{1}C_{1}，所以四边形EFC_{1}D_{1}是平行四边形，则FC_{1}=ED_{1}，又E是A_{1}D_{1}的中点，所以FC_{1}=ED_{1}=\frac{1}{2}A_{1}D_{1}=\frac{1}{2}B_{1}C_{1}，即点F是B_{1}C_{1}的中点.$
Subfield: Solid Geometry


Question: $在正方体ABCD-A_{1}B_{1}C_{1}D_{1}中，点E为A_{1}D_{1}的中点，B_{1}C_{1}与平面CDE交于点F.$![Image](https://cdntk.53zaixian.com/tq/1715283946867081218/img/compression/M_a21b01e9af5f1573350158920a4bc8ec.jpg)
$(1)求证：F为B_{1}C_{1}的中点；$

Solution:
$证明：在正方体ABCD-A_{1}B_{1}C_{1}D_{1}中，CD\parallel C_{1}D_{1}，因为CD\not\subset 平面A_{1}B_{1}C_{1}D_{1}，C_{1}D_{1}\subset 平面A_{1}B_{1}C_{1}D_{1}，所以CD\parallel 平面A_{1}B_{1}C_{1}D_{1}，又CD\subset 平面CDE，平面A_{1}B_{1}C_{1}D_{1}\cap 平面CDE=EF，所以CD\parallel EF，则EF\parallel C_{1}D_{1}.在正方体ABCD-A_{1}B_{1}C_{1}D_{1}中，A_{1}D_{1}\parallel B_{1}C_{1}，所以四边形EFC_{1}D_{1}是平行四边形，则FC_{1}=ED_{1}，又E是A_{1}D_{1}的中点，所以FC_{1}=ED_{1}=\frac{1}{2}A_{1}D_{1}=\frac{1}{2}B_{1}C_{1}，即点F是B_{1}C_{1}的中点.$

Final answer: 证明题
Subfield: Solid Geometry


Question: $在正方体ABCD-A_{1}B_{1}C_{1}D_{1}中，点E为A_{1}D_{1}的中点，B_{1}C_{1}与平面CDE交于点F.$![Image](https://cdntk.53zaixian.com/tq/1715283946867081218/img/compression/M_a21b01e9af5f1573350158920a4bc8ec.jpg)
$(2)若M是棱A_{1}B_{1}上一点，且二面角M-FC-E的余弦值为\frac{\sqrt{5}}{3}，求\frac{A_1M}{A_1B_1}的值.$

Solution:
$以D为原点，\overrightarrow{DA}，\overrightarrow{DC}，\overrightarrow{DD_1}的方向分别为x轴,y轴,z轴的正方向建立空间直角坐标系D-xyz(图略).设正方体ABCD-A_{1}B_{1}C_{1}D_{1}的棱长为2，\overrightarrow{A_1M}=\lambda \overrightarrow{A_1B_1}(0<\lambda <1)，则C(0,2,0)，A_{1}(2,0,2)，B_{1}(2,2,2)，C_{1}(0,2,2)，D_{1}(0,0,2)，F(1,2,2)，E(1,0,2)，所以\overrightarrow{CF}=(1,0,2)，\overrightarrow{EF}=(0,2,0)，\overrightarrow{CM}=\overrightarrow{CA_1}+\overrightarrow{A_1M}=\overrightarrow{CA_1}+\lambda \overrightarrow{A_1B_1}=(2,-2,2)+\lambda (0,2,0)=(2,-2+2\lambda ,2).设平面CEF的法向量为n_{1}=(x_{1},y_{1},z_{1})，则\left\{\begin{matrix}\boldsymbol{n}_1\cdot \overrightarrow{CF}=0,\\ \boldsymbol{n}_1\cdot \overrightarrow{EF}=0,\end{matrix}\right.即\left\{\begin{matrix}x_1+2z_1=0,\\ 2y_1=0,\end{matrix}\right.令z_{1}=1，得x_{1}=-2,所以平面CEF的一个法向量为n_{1}=(-2,0,1).设平面CFM的法向量为n_{2}=(x_{2},y_{2},z_{2})，则\left\{\begin{matrix}\boldsymbol{n}_2\cdot \overrightarrow{CF}=0,\\ \boldsymbol{n}_2\cdot \overrightarrow{CM}=0,\end{matrix}\right.即\left\{\begin{matrix}x_2+2z_2=0,\\ 2x_2+(-2+2\lambda )y_2+2z_2=0,\end{matrix}\right.令z_{2}=1，得x_{2}=-2,y_{2}=\frac{1}{\lambda -1}，所以平面CFM的一个法向量为n_{2}=\left(-2,\frac{1}{\lambda -1},1\right).因为二面角M-FC-E的余弦值为\frac{\sqrt{5}}{3}，所以\frac{|\boldsymbol{n}_1\cdot \boldsymbol{n}_2|}{|\boldsymbol{n}_1|\cdot |\boldsymbol{n}_2|}=\frac{\sqrt{5}}{3}，即\frac{|(-2)^2+1^2|}{\sqrt{(-2)^2+0^2+1^2}\cdot \sqrt{(-2)^2+\left(\frac{1}{\lambda -1}\right)^2+1^2}}=\frac{\sqrt{5}}{3}，又0<\lambda <1，解得\lambda =\frac{1}{2}，即\frac{A_1M}{A_1B_1}=\frac{1}{2}.$

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


Question: $在正方体ABCD-A_{1}B_{1}C_{1}D_{1}中，点E为A_{1}D_{1}的中点，B_{1}C_{1}与平面CDE交于点F.$<img_2533>
$(2)若M是棱A_{1}B_{1}上一点，且二面角M-FC-E的余弦值为\frac{\sqrt{5}}{3}，求\frac{A_1M}{A_1B_1}的值.$

Solution:
$以D为原点，\overrightarrow{DA}，\overrightarrow{DC}，\overrightarrow{DD_1}的方向分别为x轴,y轴,z轴的正方向建立空间直角坐标系D-xyz(图略).设正方体ABCD-A_{1}B_{1}C_{1}D_{1}的棱长为2，\overrightarrow{A_1M}=\lambda \overrightarrow{A_1B_1}(0<\lambda <1)，则C(0,2,0)，A_{1}(2,0,2)，B_{1}(2,2,2)，C_{1}(0,2,2)，D_{1}(0,0,2)，F(1,2,2)，E(1,0,2)，所以\overrightarrow{CF}=(1,0,2)，\overrightarrow{EF}=(0,2,0)，\overrightarrow{CM}=\overrightarrow{CA_1}+\overrightarrow{A_1M}=\overrightarrow{CA_1}+\lambda \overrightarrow{A_1B_1}=(2,-2,2)+\lambda (0,2,0)=(2,-2+2\lambda ,2).设平面CEF的法向量为n_{1}=(x_{1},y_{1},z_{1})，则\left\{\begin{matrix}\boldsymbol{n}_1\cdot \overrightarrow{CF}=0,\\ \boldsymbol{n}_1\cdot \overrightarrow{EF}=0,\end{matrix}\right.即\left\{\begin{matrix}x_1+2z_1=0,\\ 2y_1=0,\end{matrix}\right.令z_{1}=1，得x_{1}=-2,所以平面CEF的一个法向量为n_{1}=(-2,0,1).设平面CFM的法向量为n_{2}=(x_{2},y_{2},z_{2})，则\left\{\begin{matrix}\boldsymbol{n}_2\cdot \overrightarrow{CF}=0,\\ \boldsymbol{n}_2\cdot \overrightarrow{CM}=0,\end{matrix}\right.即\left\{\begin{matrix}x_2+2z_2=0,\\ 2x_2+(-2+2\lambda )y_2+2z_2=0,\end{matrix}\right.令z_{2}=1，得x_{2}=-2,y_{2}=\frac{1}{\lambda -1}，所以平面CFM的一个法向量为n_{2}=\left(-2,\frac{1}{\lambda -1},1\right).因为二面角M-FC-E的余弦值为\frac{\sqrt{5}}{3}，所以\frac{|\boldsymbol{n}_1\cdot \boldsymbol{n}_2|}{|\boldsymbol{n}_1|\cdot |\boldsymbol{n}_2|}=\frac{\sqrt{5}}{3}，即\frac{|(-2)^2+1^2|}{\sqrt{(-2)^2+0^2+1^2}\cdot \sqrt{(-2)^2+\left(\frac{1}{\lambda -1}\right)^2+1^2}}=\frac{\sqrt{5}}{3}，又0<\lambda <1，解得\lambda =\frac{1}{2}，即\frac{A_1M}{A_1B_1}=\frac{1}{2}.$

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


Question: $已知直三棱柱ABC-A_{1}B_{1}C_{1}中，侧面AA_{1}B_{1}B为正方形，AB=BC=2，E，F分别为AC和CC_{1}的中点，D为棱A_{1}B_{1}上的点，BF\perp A_{1}B_{1}.$![Image](https://cdntk.53zaixian.com/tq/1715283947596890114/img/compression/M_52ffd97de67cddc428db85e627ad6f35.jpg)
$(1)证明：BF\perp DE；$

Solution:
$第一步：证明线线垂直，为建系做铺垫.\because BF\perp A_{1}B_{1}，B_{1}B\perp A_{1}B_{1}，BF\cap B_{1}B=B，\therefore A_{1}B_{1}\perp 平面B_{1}C_{1}CB，\because AB\parallel A_{1}B_{1}，\therefore AB\perp 平面B_{1}C_{1}CB，又\because BC\subset 平面B_{1}C_{1}CB，\therefore AB\perp BC.第二步：建立空间直角坐标系，求出相关量的坐标.以B为坐标原点，BA，BC，BB_{1}所在直线分别为x轴，y轴，z轴，建立如图所示的空间直角坐标系，则B(0，0，0)，F(0，2，1)，E(1，1，0)，\therefore \overrightarrow{BF}=(0,2,1)，设B_{1}D=a(0\leq a\leq 2)，则D(a,0,2),则\overrightarrow{DE}=(1-a,1,-2).$![Image](https://cdntk.53zaixian.com/tq/1715283947194236929/img/compression/M_6f304767d47e0977ce65c3c89f58cff4.jpg)$第三步：利用a\cdot b=0判定线线垂直.证明：\because \overrightarrow{BF}\cdot \overrightarrow{DE}=(0，2，1)\cdot (1-a,1,-2)=0\times (1-a)+2\times 1+1\times (-2)=0,\therefore BF\perp DE.$

Final answer: 证明题
Subfield: Solid Geometry


Question: $已知直三棱柱ABC-A_{1}B_{1}C_{1}中，侧面AA_{1}B_{1}B为正方形，AB=BC=2，E，F分别为AC和CC_{1}的中点，D为棱A_{1}B_{1}上的点，BF\perp A_{1}B_{1}.$<img_1598>
$(1)证明：BF\perp DE；$

Solution:
$第一步：证明线线垂直，为建系做铺垫.\because BF\perp A_{1}B_{1}，B_{1}B\perp A_{1}B_{1}，BF\cap B_{1}B=B，\therefore A_{1}B_{1}\perp 平面B_{1}C_{1}CB，\because AB\parallel A_{1}B_{1}，\therefore AB\perp 平面B_{1}C_{1}CB，又\because BC\subset 平面B_{1}C_{1}CB，\therefore AB\perp BC.第二步：建立空间直角坐标系，求出相关量的坐标.以B为坐标原点，BA，BC，BB_{1}所在直线分别为x轴，y轴，z轴，建立如图所示的空间直角坐标系，则B(0，0，0)，F(0，2，1)，E(1，1，0)，\therefore \overrightarrow{BF}=(0,2,1)，设B_{1}D=a(0\leq a\leq 2)，则D(a,0,2),则\overrightarrow{DE}=(1-a,1,-2).$<img_1991>$第三步：利用a\cdot b=0判定线线垂直.证明：\because \overrightarrow{BF}\cdot \overrightarrow{DE}=(0，2，1)\cdot (1-a,1,-2)=0\times (1-a)+2\times 1+1\times (-2)=0,\therefore BF\perp DE.$
Subfield: Solid Geometry


Question: $已知直三棱柱ABC-A_{1}B_{1}C_{1}中，侧面AA_{1}B_{1}B为正方形，AB=BC=2，E，F分别为AC和CC_{1}的中点，D为棱A_{1}B_{1}上的点，BF\perp A_{1}B_{1}.$![Image](https://cdntk.53zaixian.com/tq/1715283947596890114/img/compression/M_52ffd97de67cddc428db85e627ad6f35.jpg)
$(2)当B_{1}D为何值时，面BB_{1}C_{1}C与面DFE所成的二面角的正弦值最小？$

Solution:
$第四步：求出相关平面的法向量.\overrightarrow{EF}=(-1，1，1)，\overrightarrow{FD}=(a，-2，1)，设平面DFE的法向量为n=(x,y,z)，则\left\{\begin{matrix}\overrightarrow{EF}\cdot \boldsymbol{n}=-x+y+z=0,\\ \overrightarrow{FD}\cdot \boldsymbol{n}=ax-2y+z=0,\end{matrix}\right.不妨设x=1，则y=\frac{a+1}{3},z=\frac{2-a}{3},\therefore n=\left(1,\frac{a+1}{3},\frac{2-a}{3}\right).易知m=(1,0,0)是平面BB_{1}C_{1}C的一个法向量.第五步：由向量夹角公式表示二面角大小，利用函数思想求其最值.设平面BB_{1}C_{1}C与平面DEF所成锐二面角的大小为\theta ，则\\cos   \theta =|\\cos  \langle m,n\rangle |=\frac{|\boldsymbol{m}\cdot \boldsymbol{n}|}{|\boldsymbol{m}|\cdot |\boldsymbol{n}|}=\frac{1}{\sqrt{1+\left(\frac{a+1}{3}\right)^2+\left(\frac{2-a}{3}\right)^2}}=\frac{3}{\sqrt{2\left(a-\frac{1}{2}\right)^2+\frac{27}{2}}}\leq \frac{3}{\sqrt{\frac{27}{2}}}=\frac{\sqrt{6}}{3}\left({\text{当}}a=\frac{1}{2}{\text{时取等号}}\right),\therefore \\sin   \theta =\sqrt{1-\\\cos   ^2\theta }\geq \frac{\sqrt{3}}{3}，故当a=\frac{1}{2},即B_{1}D=\frac{1}{2}时，平面BB_{1}C_{1}C与平面DFE所成的二面角的正弦值最小，最小值为\frac{\sqrt{3}}{3}.$

Final answer: $当$B_{1}D=\frac{1}{2}$时，平面BB_{1}C_{1}C与平面DFE所成的二面角的正弦值最小，最小值为$\frac{\sqrt{3}}{3}$
Subfield: Solid Geometry


Question: $已知直三棱柱ABC-A_{1}B_{1}C_{1}中，侧面AA_{1}B_{1}B为正方形，AB=BC=2，E，F分别为AC和CC_{1}的中点，D为棱A_{1}B_{1}上的点，BF\perp A_{1}B_{1}.$<img_1598>
$(2)当B_{1}D为何值时，面BB_{1}C_{1}C与面DFE所成的二面角的正弦值最小？$

Solution:
$第四步：求出相关平面的法向量.\overrightarrow{EF}=(-1，1，1)，\overrightarrow{FD}=(a，-2，1)，设平面DFE的法向量为n=(x,y,z)，则\left\{\begin{matrix}\overrightarrow{EF}\cdot \boldsymbol{n}=-x+y+z=0,\\ \overrightarrow{FD}\cdot \boldsymbol{n}=ax-2y+z=0,\end{matrix}\right.不妨设x=1，则y=\frac{a+1}{3},z=\frac{2-a}{3},\therefore n=\left(1,\frac{a+1}{3},\frac{2-a}{3}\right).易知m=(1,0,0)是平面BB_{1}C_{1}C的一个法向量.第五步：由向量夹角公式表示二面角大小，利用函数思想求其最值.设平面BB_{1}C_{1}C与平面DEF所成锐二面角的大小为\theta ，则\\cos   \theta =|\\cos  \langle m,n\rangle |=\frac{|\boldsymbol{m}\cdot \boldsymbol{n}|}{|\boldsymbol{m}|\cdot |\boldsymbol{n}|}=\frac{1}{\sqrt{1+\left(\frac{a+1}{3}\right)^2+\left(\frac{2-a}{3}\right)^2}}=\frac{3}{\sqrt{2\left(a-\frac{1}{2}\right)^2+\frac{27}{2}}}\leq \frac{3}{\sqrt{\frac{27}{2}}}=\frac{\sqrt{6}}{3}\left({\text{当}}a=\frac{1}{2}{\text{时取等号}}\right),\therefore \\sin   \theta =\sqrt{1-\\\cos   ^2\theta }\geq \frac{\sqrt{3}}{3}，故当a=\frac{1}{2},即B_{1}D=\frac{1}{2}时，平面BB_{1}C_{1}C与平面DFE所成的二面角的正弦值最小，最小值为\frac{\sqrt{3}}{3}.$





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


Question: $如图，四棱锥P-ABCD的底面为正方形，PD\perp 底面ABCD.设平面PAD与平面PBC的交线为l.$<img_2165>
$(1)证明：l\perp 平面PDC；$

Solution:
$证明：因为PD\perp 底面ABCD，所以PD\perp AD.又底面ABCD为正方形，所以AD\perp DC.因此AD\perp 平面PDC.因为AD\parallel BC，AD\not\subset 平面PBC，所以AD\parallel 平面PBC.由已知得l\parallel AD.因此l\perp 平面PDC.$
Subfield: Solid Geometry


Question: $如图，四棱锥P-ABCD的底面为正方形，PD\perp 底面ABCD.设平面PAD与平面PBC的交线为l.$![Image](https://cdntk.53zaixian.com/tq/1715283947961794561/img/compression/M_62a83a6b173937a55f672ab06eccda08.jpg)
$(1)证明：l\perp 平面PDC；$

Solution:
$证明：因为PD\perp 底面ABCD，所以PD\perp AD.又底面ABCD为正方形，所以AD\perp DC.因此AD\perp 平面PDC.因为AD\parallel BC，AD\not\subset 平面PBC，所以AD\parallel 平面PBC.由已知得l\parallel AD.因此l\perp 平面PDC.$

Final answer: 证明题
Subfield: Solid Geometry


Question: $如图，四棱锥P-ABCD的底面为正方形，PD\perp 底面ABCD.设平面PAD与平面PBC的交线为l.$<img_2165>
$(2)已知PD=AD=1，Q为l上的点，求PB与平面QCD所成角的正弦值的最大值.$

Solution:
$以D为坐标原点，\overrightarrow{DA}的方向为x轴正方向，建立如图所示的空间直角坐标系D-xyz.则D(0，0，0)，C(0，1，0)，B(1，1，0)，P(0，0，1)，\overrightarrow{DC}=(0，1，0)，\overrightarrow{PB}=(1，1，-1).$<img_2666>$由(1)可设Q(a,0，1)，则\overrightarrow{DQ}=(a,0,1).设n=(x,y,z)是平面QCD的法向量，则\left\{\begin{matrix}\boldsymbol{n}\cdot \overrightarrow{DQ}=0,\\ \boldsymbol{n}\cdot \overrightarrow{DC}=0,\end{matrix}\right.即\left\{\begin{matrix}ax+z=0,\\ y=0.\end{matrix}\right.可取n=(-1,0,a).所以\\cos  \langle n，\overrightarrow{PB}\rangle =\frac{\boldsymbol{n}\cdot \overrightarrow{PB}}{|\boldsymbol{n}|\cdot |\overrightarrow{PB}|}=\frac{-1-a}{\sqrt{3}\sqrt{1+a^2}}.设PB与平面QCD所成角为\theta ，则\\sin   \theta =\frac{\sqrt{3}}{3}\times \frac{|a+1|}{\sqrt{1+a^2}}=\frac{\sqrt{3}}{3}\sqrt{1+\frac{2a}{a^2+1}}(易错：余弦值转化为正弦值后应该为正).因为\frac{\sqrt{3}}{3}\sqrt{1+\frac{2a}{a^2+1}}\leq \frac{\sqrt{6}}{3}，当且仅当a=1时等号成立，所以PB与平面QCD所成角的正弦值的最大值为\frac{\sqrt{6}}{3}.$

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


Question: $如图，四棱锥P-ABCD的底面为正方形，PD\perp 底面ABCD.设平面PAD与平面PBC的交线为l.$![Image](https://cdntk.53zaixian.com/tq/1715283947961794561/img/compression/M_62a83a6b173937a55f672ab06eccda08.jpg)
$(2)已知PD=AD=1，Q为l上的点，求PB与平面QCD所成角的正弦值的最大值.$

Solution:
$以D为坐标原点，\overrightarrow{DA}的方向为x轴正方向，建立如图所示的空间直角坐标系D-xyz.则D(0，0，0)，C(0，1，0)，B(1，1，0)，P(0，0，1)，\overrightarrow{DC}=(0，1，0)，\overrightarrow{PB}=(1，1，-1).$![Image](https://cdntk.53zaixian.com/tq/1715283948314116098/img/compression/M_34a85c7371a708c8193d4b0da099f5d8.jpg)$由(1)可设Q(a,0，1)，则\overrightarrow{DQ}=(a,0,1).设n=(x,y,z)是平面QCD的法向量，则\left\{\begin{matrix}\boldsymbol{n}\cdot \overrightarrow{DQ}=0,\\ \boldsymbol{n}\cdot \overrightarrow{DC}=0,\end{matrix}\right.即\left\{\begin{matrix}ax+z=0,\\ y=0.\end{matrix}\right.可取n=(-1,0,a).所以\\cos  \langle n，\overrightarrow{PB}\rangle =\frac{\boldsymbol{n}\cdot \overrightarrow{PB}}{|\boldsymbol{n}|\cdot |\overrightarrow{PB}|}=\frac{-1-a}{\sqrt{3}\sqrt{1+a^2}}.设PB与平面QCD所成角为\theta ，则\\sin   \theta =\frac{\sqrt{3}}{3}\times \frac{|a+1|}{\sqrt{1+a^2}}=\frac{\sqrt{3}}{3}\sqrt{1+\frac{2a}{a^2+1}}(易错：余弦值转化为正弦值后应该为正).因为\frac{\sqrt{3}}{3}\sqrt{1+\frac{2a}{a^2+1}}\leq \frac{\sqrt{6}}{3}，当且仅当a=1时等号成立，所以PB与平面QCD所成角的正弦值的最大值为\frac{\sqrt{6}}{3}.$

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


Question: $如图，直三棱柱ABC-A_{1}B_{1}C_{1}的体积为4，\triangle A_{1}BC的面积为2\sqrt{2}.$<img_3098>
$(1)求A到平面A_{1}BC的距离；$

Solution:
$设A到平面A_{1}BC的距离为d.因为V_{A_{1}-ABC}=V_{A-A_{1}BC}=\frac{1}{3}S_{\triangle ABC}\cdot AA_{1}=\frac{1}{3}S_{\triangle A_{1}BC}\cdot d=\frac{1}{3}\cdot V_{ABC-A_{1}B_{1}C_{1}}=\frac{4}{3},S_{\triangle A_{1}BC}=2\sqrt{2},所以d=\frac{3\times \frac{4}{3}}{2\sqrt{2}}=\sqrt{2}.$

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


Question: $如图，直三棱柱ABC-A_{1}B_{1}C_{1}的体积为4，\triangle A_{1}BC的面积为2\sqrt{2}.$![Image](https://cdntk.53zaixian.com/tq/1715283948691603459/img/compression/M_efb4efe5c251c2ff4685e574363344a8.jpg)
$(1)求A到平面A_{1}BC的距离；$

Solution:
$设A到平面A_{1}BC的距离为d.因为V_{A_{1}-ABC}=V_{A-A_{1}BC}=\frac{1}{3}S_{\triangle ABC}\cdot AA_{1}=\frac{1}{3}S_{\triangle A_{1}BC}\cdot d=\frac{1}{3}\cdot V_{ABC-A_{1}B_{1}C_{1}}=\frac{4}{3},S_{\triangle A_{1}BC}=2\sqrt{2},所以d=\frac{3\times \frac{4}{3}}{2\sqrt{2}}=\sqrt{2}.$

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


Question: $如图，直三棱柱ABC-A_{1}B_{1}C_{1}的体积为4，\triangle A_{1}BC的面积为2\sqrt{2}.$<img_3098>
$(2)设D为A_{1}C的中点，AA_{1}=AB，平面A_{1}BC\perp 平面ABB_{1}A_{1}，求二面角A-BD-C的正弦值.$

Solution:
$如图，取A_{1}B的中点E，连接AE.因为AA_{1}=AB，所以AE\perp A_{1}B，又因为平面A_{1}BC\perp 平面ABB_{1}A_{1}，平面A_{1}BC\cap 平面ABB_{1}A_{1}=A_{1}B，AE\subset 平面ABB_{1}A_{1}，所以AE\perp 平面A_{1}BC.又BC\subset 平面A_{1}BC，所以AE\perp BC.由直三棱柱ABC-A_{1}B_{1}C_{1}得AA_{1}\perp 平面ABC，又BC\subset 平面ABC，所以AA_{1}\perp BC，又AA_{1}\cap AE=A，AA_{1},AE\subset 平面ABB_{1}A_{1}，所以BC\perp 平面ABB_{1}A_{1}，又AB\subset 平面ABB_{1}A_{1}，所以BC\perp AB.由(1)知AE=d=\sqrt{2}，所以AB=AA_{1}=2,A_{1}B=2\sqrt{2}，又因为\triangle A_{1}BC的面积为2\sqrt{2}，所以BC=2.以B为坐标原点，向量\overrightarrow{BC},\overrightarrow{BA},\overrightarrow{BB_1}的方向分别为x,y,z轴的正方向，建立空间直角坐标系B-xyz，则C(2,0,0),A(0,2,0),A_{1}(0,2,2)，D(1,1,1),B(0,0,0)，则\overrightarrow{BC}=(2,0,0),\overrightarrow{BA}=(0,2,0),\overrightarrow{BD}=(1,1,1).$<img_2489>$设平面ABD的法向量为n_{1}=(x_{1},y_{1},z_{1})，则\left\{\begin{matrix}\boldsymbol{n}_1\cdot \overrightarrow{BA}=0,\\ \boldsymbol{n}_1\cdot \overrightarrow{BD}=0,\end{matrix}\right.即\left\{\begin{matrix}2y_1=0,\\ x_1+y_1+z_1=0,\end{matrix}\right.令x_{1}=1，得z_{1}=-1，所以n_{1}=(1,0,-1).设平面BCD的法向量为n_{2}=(x_{2},y_{2},z_{2})，则\left\{\begin{matrix}\boldsymbol{n}_2\cdot \overrightarrow{BC}=0,\\ \boldsymbol{n}_2\cdot \overrightarrow{BD}=0,\end{matrix}\right.即\left\{\begin{matrix}2x_2=0,\\ x_2+y_2+z_2=0,\end{matrix}\right.令y_{2}=1，得z_{2}=-1，所以n_{2}=(0,1,-1).所以\\cos  \langle n_{1},n_{2}\rangle =\frac{\boldsymbol{n}_1\cdot \boldsymbol{n}_2}{|\boldsymbol{n}_1||\boldsymbol{n}_2|}=\frac{1}{2}，又\\sin  \langle n_{1},n_{2}\rangle >0,所以\\sin  \langle n_{1},n_{2}\rangle =\frac{\sqrt{3}}{2}.所以二面角A-BD-C的正弦值为\frac{\sqrt{3}}{2}.$

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


Question: $如图，直三棱柱ABC-A_{1}B_{1}C_{1}的体积为4，\triangle A_{1}BC的面积为2\sqrt{2}.$![Image](https://cdntk.53zaixian.com/tq/1715283948691603459/img/compression/M_efb4efe5c251c2ff4685e574363344a8.jpg)
$(2)设D为A_{1}C的中点，AA_{1}=AB，平面A_{1}BC\perp 平面ABB_{1}A_{1}，求二面角A-BD-C的正弦值.$

Solution:
$如图，取A_{1}B的中点E，连接AE.因为AA_{1}=AB，所以AE\perp A_{1}B，又因为平面A_{1}BC\perp 平面ABB_{1}A_{1}，平面A_{1}BC\cap 平面ABB_{1}A_{1}=A_{1}B，AE\subset 平面ABB_{1}A_{1}，所以AE\perp 平面A_{1}BC.又BC\subset 平面A_{1}BC，所以AE\perp BC.由直三棱柱ABC-A_{1}B_{1}C_{1}得AA_{1}\perp 平面ABC，又BC\subset 平面ABC，所以AA_{1}\perp BC，又AA_{1}\cap AE=A，AA_{1},AE\subset 平面ABB_{1}A_{1}，所以BC\perp 平面ABB_{1}A_{1}，又AB\subset 平面ABB_{1}A_{1}，所以BC\perp AB.由(1)知AE=d=\sqrt{2}，所以AB=AA_{1}=2,A_{1}B=2\sqrt{2}，又因为\triangle A_{1}BC的面积为2\sqrt{2}，所以BC=2.以B为坐标原点，向量\overrightarrow{BC},\overrightarrow{BA},\overrightarrow{BB_1}的方向分别为x,y,z轴的正方向，建立空间直角坐标系B-xyz，则C(2,0,0),A(0,2,0),A_{1}(0,2,2)，D(1,1,1),B(0,0,0)，则\overrightarrow{BC}=(2,0,0),\overrightarrow{BA}=(0,2,0),\overrightarrow{BD}=(1,1,1).$![Image](https://cdntk.53zaixian.com/tq/1715283949056507906/img/compression/M_ae0922fcd2a6b05cad232a43fe6c0a3a.jpg)$设平面ABD的法向量为n_{1}=(x_{1},y_{1},z_{1})，则\left\{\begin{matrix}\boldsymbol{n}_1\cdot \overrightarrow{BA}=0,\\ \boldsymbol{n}_1\cdot \overrightarrow{BD}=0,\end{matrix}\right.即\left\{\begin{matrix}2y_1=0,\\ x_1+y_1+z_1=0,\end{matrix}\right.令x_{1}=1，得z_{1}=-1，所以n_{1}=(1,0,-1).设平面BCD的法向量为n_{2}=(x_{2},y_{2},z_{2})，则\left\{\begin{matrix}\boldsymbol{n}_2\cdot \overrightarrow{BC}=0,\\ \boldsymbol{n}_2\cdot \overrightarrow{BD}=0,\end{matrix}\right.即\left\{\begin{matrix}2x_2=0,\\ x_2+y_2+z_2=0,\end{matrix}\right.令y_{2}=1，得z_{2}=-1，所以n_{2}=(0,1,-1).所以\\cos  \langle n_{1},n_{2}\rangle =\frac{\boldsymbol{n}_1\cdot \boldsymbol{n}_2}{|\boldsymbol{n}_1||\boldsymbol{n}_2|}=\frac{1}{2}，又\\sin  \langle n_{1},n_{2}\rangle >0,所以\\sin  \langle n_{1},n_{2}\rangle =\frac{\sqrt{3}}{2}.所以二面角A-BD-C的正弦值为\frac{\sqrt{3}}{2}.$

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


Question: $如图，已知三棱柱ABC-A_{1}B_{1}C_{1}的底面是正三角形，侧面BB_{1}C_{1}C是矩形，M，N分别为BC，B_{1}C_{1}的中点，P为AM上一点.过B_{1}C_{1}和P的平面交AB于E，交AC于F.$![Image](https://cdntk.53zaixian.com/tq/1715283949417218051/img/compression/M_aad79aa84346e3af8d174fd1de92cae9.jpg)
$(1)证明：AA_{1}\parallel MN，且平面A_{1}AMN\perp 平面EB_{1}C_{1}F；$

Solution:
$证明：因为M，N分别为BC，B_{1}C_{1}的中点，所以MN\parallel CC_{1}.又由已知得AA_{1}\parallel CC_{1}，故AA_{1}\parallel MN.因为\triangle A_{1}B_{1}C_{1}是正三角形，所以B_{1}C_{1}\perp A_{1}N.又B_{1}C_{1}\perp MN，故B_{1}C_{1}\perp 平面A_{1}AMN.又因为B_{1}C_{1}\subset 平面EB_{1}C_{1}F，所以平面A_{1}AMN\perp 平面EB_{1}C_{1}F.$

Final answer: 证明题
Subfield: Solid Geometry


Question: $如图，已知三棱柱ABC-A_{1}B_{1}C_{1}的底面是正三角形，侧面BB_{1}C_{1}C是矩形，M，N分别为BC，B_{1}C_{1}的中点，P为AM上一点.过B_{1}C_{1}和P的平面交AB于E，交AC于F.$<img_2871>
$(1)证明：AA_{1}\parallel MN，且平面A_{1}AMN\perp 平面EB_{1}C_{1}F；$

Solution:
$证明：因为M，N分别为BC，B_{1}C_{1}的中点，所以MN\parallel CC_{1}.又由已知得AA_{1}\parallel CC_{1}，故AA_{1}\parallel MN.因为\triangle A_{1}B_{1}C_{1}是正三角形，所以B_{1}C_{1}\perp A_{1}N.又B_{1}C_{1}\perp MN，故B_{1}C_{1}\perp 平面A_{1}AMN.又因为B_{1}C_{1}\subset 平面EB_{1}C_{1}F，所以平面A_{1}AMN\perp 平面EB_{1}C_{1}F.$
Subfield: Solid Geometry


Question: $如图，已知三棱柱ABC-A_{1}B_{1}C_{1}的底面是正三角形，侧面BB_{1}C_{1}C是矩形，M，N分别为BC，B_{1}C_{1}的中点，P为AM上一点.过B_{1}C_{1}和P的平面交AB于E，交AC于F.$<img_2871>
$(2)设O为\triangle A_{1}B_{1}C_{1}的中心.若AO\parallel 平面EB_{1}C_{1}F，且AO=AB，求直线B_{1}E与平面A_{1}AMN所成角的正弦值.$

Solution:
$由已知得AM\perp BC.以M为坐标原点，\overrightarrow{MA}的方向为x轴正方向，|\overrightarrow{MB}|为单位长，建立如图所示的空间直角坐标系M-xyz,则AB=2，AM=\sqrt{3}.$<img_1932>$连接NP，则四边形AONP为平行四边形，故PM=\frac{2\sqrt{3}}{3}，E\left(\frac{2\sqrt{3}}{3},\frac{1}{3},0\right).由(1)知平面A_{1}AMN\perp 平面ABC，作NQ\perp AM，垂足为Q，则NQ\perp 平面ABC.设Q(a,0,0)，则NQ=\sqrt{4-\left(\frac{2\sqrt{3}}{3}-a\right)^2},B_{1}\left(\right.a,1,\sqrt{4-\left(\frac{2\sqrt{3}}{3}-a\right)^2}\left)\right.,故\overrightarrow{B_1E}=\left(\right.\frac{2\sqrt{3}}{3}-a,-\frac{2}{3},-\sqrt{4-\left(\frac{2\sqrt{3}}{3}-a\right)^2}\left)\right.,|\overrightarrow{B_1E}|=\frac{2\sqrt{10}}{3}.又n=(0,-1,0)是平面A_{1}AMN的法向量，故\\sin  \left(\frac{\pi }{2}-\langle \boldsymbol{n},\overrightarrow{B_1E}\rangle \right)=\\cos  \langle n,\overrightarrow{B_1E}\rangle =\frac{\boldsymbol{n}\cdot \overrightarrow{B_1E}}{|\boldsymbol{n}|\cdot |\overrightarrow{B_1E}|}=\frac{\sqrt{10}}{10}.所以直线B_{1}E与平面A_{1}AMN所成角的正弦值为\frac{\sqrt{10}}{10}.$

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


Question: $如图，已知三棱柱ABC-A_{1}B_{1}C_{1}的底面是正三角形，侧面BB_{1}C_{1}C是矩形，M，N分别为BC，B_{1}C_{1}的中点，P为AM上一点.过B_{1}C_{1}和P的平面交AB于E，交AC于F.$![Image](https://cdntk.53zaixian.com/tq/1715283949417218051/img/compression/M_aad79aa84346e3af8d174fd1de92cae9.jpg)
$(2)设O为\triangle A_{1}B_{1}C_{1}的中心.若AO\parallel 平面EB_{1}C_{1}F，且AO=AB，求直线B_{1}E与平面A_{1}AMN所成角的正弦值.$

Solution:
$由已知得AM\perp BC.以M为坐标原点，\overrightarrow{MA}的方向为x轴正方向，|\overrightarrow{MB}|为单位长，建立如图所示的空间直角坐标系M-xyz,则AB=2，AM=\sqrt{3}.$![Image](https://cdntk.53zaixian.com/tq/1715283949777928194/img/compression/M_79cdec60e102eb1444c2e49a435f9146.jpg)$连接NP，则四边形AONP为平行四边形，故PM=\frac{2\sqrt{3}}{3}，E\left(\frac{2\sqrt{3}}{3},\frac{1}{3},0\right).由(1)知平面A_{1}AMN\perp 平面ABC，作NQ\perp AM，垂足为Q，则NQ\perp 平面ABC.设Q(a,0,0)，则NQ=\sqrt{4-\left(\frac{2\sqrt{3}}{3}-a\right)^2},B_{1}\left(\right.a,1,\sqrt{4-\left(\frac{2\sqrt{3}}{3}-a\right)^2}\left)\right.,故\overrightarrow{B_1E}=\left(\right.\frac{2\sqrt{3}}{3}-a,-\frac{2}{3},-\sqrt{4-\left(\frac{2\sqrt{3}}{3}-a\right)^2}\left)\right.,|\overrightarrow{B_1E}|=\frac{2\sqrt{10}}{3}.又n=(0,-1,0)是平面A_{1}AMN的法向量，故\\sin  \left(\frac{\pi }{2}-\langle \boldsymbol{n},\overrightarrow{B_1E}\rangle \right)=\\cos  \langle n,\overrightarrow{B_1E}\rangle =\frac{\boldsymbol{n}\cdot \overrightarrow{B_1E}}{|\boldsymbol{n}|\cdot |\overrightarrow{B_1E}|}=\frac{\sqrt{10}}{10}.所以直线B_{1}E与平面A_{1}AMN所成角的正弦值为\frac{\sqrt{10}}{10}.$

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


Question: $如图，在正方体ABCD-A_{1}B_{1}C_{1}D_{1}中，E为BB_{1}的中点.$![Image](https://cdntk.53zaixian.com/tq/1716267980298997762/img/compression/M_2a09cf9e5fedf1381160cc56c44fec02.jpg)
$(1)求证：BC_{1}\parallel 平面AD_{1}E；$

Solution:
$证明：\because ABCD-A_{1}B_{1}C_{1}D_{1}为正方体，\therefore D_{1}C_{1}\parallel A_{1}B_{1}，D_{1}C_{1}=A_{1}B_{1}.又AB\parallel A_{1}B_{1}，AB=A_{1}B_{1},\therefore D_{1}C_{1}\parallel AB,D_{1}C_{1}=AB,\therefore 四边形ABC_{1}D_{1}为平行四边形，\therefore AD_{1}\parallel BC_{1}，又AD_{1}\subset 平面AD_{1}E，BC_{1}\not\subset 平面AD_{1}E，\therefore BC_{1}\parallel 平面AD_{1}E.$

Final answer: 证明题
Subfield: Solid Geometry


Question: $如图，在正方体ABCD-A_{1}B_{1}C_{1}D_{1}中，E为BB_{1}的中点.$<img_2473>
$(1)求证：BC_{1}\parallel 平面AD_{1}E；$

Solution:
$证明：\because ABCD-A_{1}B_{1}C_{1}D_{1}为正方体，\therefore D_{1}C_{1}\parallel A_{1}B_{1}，D_{1}C_{1}=A_{1}B_{1}.又AB\parallel A_{1}B_{1}，AB=A_{1}B_{1},\therefore D_{1}C_{1}\parallel AB,D_{1}C_{1}=AB,\therefore 四边形ABC_{1}D_{1}为平行四边形，\therefore AD_{1}\parallel BC_{1}，又AD_{1}\subset 平面AD_{1}E，BC_{1}\not\subset 平面AD_{1}E，\therefore BC_{1}\parallel 平面AD_{1}E.$
Subfield: Solid Geometry


Question: $如图，在正方体ABCD-A_{1}B_{1}C_{1}D_{1}中，E为BB_{1}的中点.$![Image](https://cdntk.53zaixian.com/tq/1716267980298997762/img/compression/M_2a09cf9e5fedf1381160cc56c44fec02.jpg)
$(2)求直线AA_{1}与平面AD_{1}E所成角的正弦值.$

Solution:
$不妨设正方体的棱长为2，如图，以{\overrightarrow{AD}，\overrightarrow{AB}，\overrightarrow{AA_1}}为正交基底建立空间直角坐标系A-xyz,$![Image](https://cdntk.53zaixian.com/tq/1716267980936531969/img/compression/M_d0bf4a2e032fc2072dda35b533870ecd.jpg)$则A(0，0，0)，A_{1}(0，0，2)，D_{1}(2，0，2)，E(0，2，1)，\therefore \overrightarrow{AA_1}=(0，0，2)，\overrightarrow{AD_1}=(2，0，2)，\overrightarrow{AE}=(0，2，1)，设平面AD_{1}E的法向量为n=(x,y,z)，直线AA_{1}与平面AD_{1}E所成的角为\theta ，则\left\{\begin{matrix}\boldsymbol{n}\cdot \overrightarrow{AD_1}=0,\\ \boldsymbol{n}\cdot \overrightarrow{AE}=0,\end{matrix}\right.即\left\{\begin{matrix}2x+2z=0,\\ 2y+z=0,\end{matrix}\right.令z=-2，则x=2,y=1,此时n=(2，1，-2)，\therefore \\sin   \theta =|\\cos  \langle n,\overrightarrow{AA_1}\rangle |=\frac{|\boldsymbol{n}\cdot \overrightarrow{AA_1}|}{|\boldsymbol{n}||\overrightarrow{AA_1}|}=\frac{|-4|}{\sqrt{4+1+4}\times 2}=\frac{2}{3}，\therefore 直线AA_{1}与平面AD_{1}E所成角的正弦值为\frac{2}{3}.$

Final answer: $\frac{2}{3}$