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


Question: $如图,直三棱柱ABC-A_{1}B_{1}C_{1}中,\angle BAC=90^\circ ,AB=AC,D,E分别为AA_{1},B_{1}C的中点.$<img_1917>
$(2)已知B_{1}C与平面BCD所成角的大小为30^\circ ,求二面角D-BC-B_{1}的余弦值.$

Solution:
$解法一(空间向量法):由(1)知\overrightarrow{BC}=(-1,1,0),\overrightarrow{BD}=(-1,0,a),\overrightarrow{B_1C}=(-1,1,-2a),设平面BCD的法向量为n=(x,y,z),则\left\{\begin{matrix}\boldsymbol{n}\cdot \overrightarrow{BC}=0,\\ \boldsymbol{n}\cdot \overrightarrow{BD}=0,\end{matrix}\right.即\left\{\begin{matrix}-x+y=0,\\ -x+az=0,\end{matrix}\right.令x=1,得n=\left(1,1,\frac{1}{a}\right).因为B_{1}C与平面BCD所成角的大小为30^\circ ,所以|\\cos  \langle n,\overrightarrow{B_1C}\rangle |=\\sin   30^\circ =\frac{1}{2},即\frac{|\boldsymbol{n}\cdot \overrightarrow{B_1C}|}{|\boldsymbol{n}|\cdot |\overrightarrow{B_1C}|}=\frac{2}{\sqrt{2+\frac{1}{a^2}}\cdot \sqrt{2+4a^2}}=\frac{1}{2},解得a=\frac{\sqrt{2}}{2}(负值舍去),所以n=(1,1,\sqrt{2}).由(1)知平面BCB_{1}的一个法向量为\overrightarrow{DE}=\left(\frac{1}{2},\frac{1}{2},0\right),所以\\cos  \langle n,\overrightarrow{DE}\rangle =\frac{\boldsymbol{n}\cdot \overrightarrow{DE}}{|\boldsymbol{n}|\cdot |\overrightarrow{DE}|}=\frac{\frac{1}{2}+\frac{1}{2}}{\sqrt{1^2+1^2+{(\sqrt{2})^2}}\times \sqrt{\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^2}}=\frac{\sqrt{2}}{2},所以二面角D-BC-B_{1}的余弦值为\frac{\sqrt{2}}{2}.
解法二(几何法):设AB=AC=1,AA_{1}=2a(a>0),则AD=a,BB_{1}=2a,BC=\sqrt{2},AF=\frac{\sqrt{2}}{2},BD=DC=\sqrt{1+a^2},所以DF=\sqrt{AD^2+AF^2}=\sqrt{a^2+\frac{1}{2}},所以S_{\triangle BDC}=\frac{1}{2}BC\cdot DF=\frac{\sqrt{2a^2+1}}{2},S_{\triangle BCB1}=\frac{1}{2}BB_{1}\cdot BC=\sqrt{2}a.由(1)知D到平面BCB_{1}的距离为DE=\frac{\sqrt{2}}{2},设B_{1}到平面BCD的距离为d,由V_{D-BCB1}=V_{B1-BDC},得\frac{1}{3}S_{\triangle BCB1}\cdot DE=\frac{1}{3}S_{\triangle BDC}\cdot d,即\frac{1}{3}\times \sqrt{2}a\times \frac{\sqrt{2}}{2}=\frac{1}{3}\times \frac{\sqrt{2a^2+1}}{2}\times d,解得d=\frac{2a}{\sqrt{2a^2+1}}.因为B_{1}C与平面BCD所成角的大小为30^\circ ,所以B_{1}C=\frac{d}{\\\sin    30^{\circ}}=2d=\frac{4a}{\sqrt{2a^2+1}}.易知在直角三角形B_{1}BC中,B_{1}C=\sqrt{BB^2_1+BC^2}=\sqrt{4a^2+2},所以\sqrt{4a^2+2}=\frac{4a}{\sqrt{2a^2+1}},解得a=\frac{\sqrt{2}}{2}(负值舍去).由(1)知,AF\perp BC,EF\perp BC.又因为AF\cap EF=F,AF、EF\subset 平面DEFA，所以BC\perp 平面DEFA,又DF\subset 平面DEFA，所以DF\perp BC.又DF\subset 平面DBC,EF\subset 平面B_{1}BC,平面DBC\cap 平面B_{1}BC=BC,所以\angle EFD为二面角D-BC-B_{1}的平面角.因为DA=AF=\frac{\sqrt{2}}{2},DA\perp AF，所以平行四边形DAFE是正方形,所以\angle EFD=45^\circ ,所以二面角D-BC-B_{1}的余弦值为\frac{\sqrt{2}}{2}.$

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


Question: $如图，在四棱锥P-ABCD中，底面ABCD是边长为2的菱形，\angle ABC=\frac{\pi }{3}，平面PAB\perp 平面ABCD，PA=3，PB=\sqrt{13}，PF=2FA，E为CD的中点.$<img_3061>
$(1)求四棱锥P-ABCD的体积；$

Solution:
$因为PA=3，PB=\sqrt{13}，AB=2，所以PA^{2}+AB^{2}=PB^{2}，所以PA\perp AB.又平面PAB\perp 平面ABCD，平面PAB\cap 平面ABCD=AB，PA\subset 平面PAB，所以PA\perp 平面ABCD，则PA为四棱锥P-ABCD的高，因为底面是边长为2的菱形，且\angle ABC=\frac{\pi }{3}，所以S_{菱形ABCD}=2\times S_{\triangle ABC}=2\times \frac{1}{2}\times 2\times 2\times \\sin  \frac{\pi }{3}=2\sqrt{3}，(3分)故四棱锥P-ABCD的体积V=\frac{1}{3}S_{菱形ABCD}\cdot PA=\frac{1}{3}\times 2\sqrt{3}\times 3=2\sqrt{3}.(4分)$

Final answer: $2\sqrt 3$
Subfield: Solid Geometry


Question: $如图，在四棱锥P-ABCD中，底面ABCD是边长为2的菱形，\angle ABC=\frac{\pi }{3}，平面PAB\perp 平面ABCD，PA=3，PB=\sqrt{13}，PF=2FA，E为CD的中点.$![Image](https://cdntk.53zaixian.com/tq/1696694366328123393/img/compression/M_93929d8ed3c336f96d993323060ba32c.jpg)
$(1)求四棱锥P-ABCD的体积；$

Solution:
$因为PA=3，PB=\sqrt{13}，AB=2，所以PA^{2}+AB^{2}=PB^{2}，所以PA\perp AB.又平面PAB\perp 平面ABCD，平面PAB\cap 平面ABCD=AB，PA\subset 平面PAB，所以PA\perp 平面ABCD，则PA为四棱锥P-ABCD的高，因为底面是边长为2的菱形，且\angle ABC=\frac{\pi }{3}，所以S_{菱形ABCD}=2\times S_{\triangle ABC}=2\times \frac{1}{2}\times 2\times 2\times \\sin  \frac{\pi }{3}=2\sqrt{3}，(3分)故四棱锥P-ABCD的体积V=\frac{1}{3}S_{菱形ABCD}\cdot PA=\frac{1}{3}\times 2\sqrt{3}\times 3=2\sqrt{3}.(4分)$

Final answer: $2\sqrt 3$
Subfield: Solid Geometry


Question: $如图，在四棱锥P-ABCD中，底面ABCD是边长为2的菱形，\angle ABC=\frac{\pi }{3}，平面PAB\perp 平面ABCD，PA=3，PB=\sqrt{13}，PF=2FA，E为CD的中点.$![Image](https://cdntk.53zaixian.com/tq/1696694366328123393/img/compression/M_93929d8ed3c336f96d993323060ba32c.jpg)
$(2)求异面直线AB与DF所成角的余弦值；$

Solution:
$连接AC，设AC与BD交于点O，因为底面ABCD为菱形，所以BD\perp AC，由(1)可知，PA\perp 平面ABCD，故以点O为坐标原点建立空间直角坐标系，如图所示，$![Image](https://cdntk.53zaixian.com/tq/1696694366701416450/img/compression/M_30520dc7950a76c446a63b9b400976f2.jpg)$则A(0,-1,0)，B(\sqrt{3},0,0)，C(0,1,0)，D(-\sqrt{3},0,0)，E\left(-\frac{\sqrt{3}}{2},\frac{1}{2},0\right)，P(0,-1,3)，F(0,-1,1)，(6分)所以\overrightarrow{AB}=(\sqrt{3},1,0)，\overrightarrow{DF}=(\sqrt{3},-1,1)，则|\\cos  \langle \overrightarrow{AB},\overrightarrow{DF}\rangle |=\frac{|\overrightarrow{AB}\cdot \overrightarrow{DF}|}{|\overrightarrow{AB}||\overrightarrow{DF}|}=\frac{2}{\sqrt{3+1}\times \sqrt{3+1+1}}=\frac{\sqrt{5}}{5}，故异面直线AB与DF所成角的余弦值为\frac{\sqrt{5}}{5}.(8分)$

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


Question: $如图，在四棱锥P-ABCD中，底面ABCD是边长为2的菱形，\angle ABC=\frac{\pi }{3}，平面PAB\perp 平面ABCD，PA=3，PB=\sqrt{13}，PF=2FA，E为CD的中点.$<img_3061>
$(2)求异面直线AB与DF所成角的余弦值；$

Solution:
$连接AC，设AC与BD交于点O，因为底面ABCD为菱形，所以BD\perp AC，由(1)可知，PA\perp 平面ABCD，故以点O为坐标原点建立空间直角坐标系，如图所示，$<img_2321>$则A(0,-1,0)，B(\sqrt{3},0,0)，C(0,1,0)，D(-\sqrt{3},0,0)，E\left(-\frac{\sqrt{3}}{2},\frac{1}{2},0\right)，P(0,-1,3)，F(0,-1,1)，(6分)所以\overrightarrow{AB}=(\sqrt{3},1,0)，\overrightarrow{DF}=(\sqrt{3},-1,1)，则|\\cos  \langle \overrightarrow{AB},\overrightarrow{DF}\rangle |=\frac{|\overrightarrow{AB}\cdot \overrightarrow{DF}|}{|\overrightarrow{AB}||\overrightarrow{DF}|}=\frac{2}{\sqrt{3+1}\times \sqrt{3+1+1}}=\frac{\sqrt{5}}{5}，故异面直线AB与DF所成角的余弦值为\frac{\sqrt{5}}{5}.(8分)$

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


Question: $如图，在四棱锥P-ABCD中，底面ABCD是边长为2的菱形，\angle ABC=\frac{\pi }{3}，平面PAB\perp 平面ABCD，PA=3，PB=\sqrt{13}，PF=2FA，E为CD的中点.$![Image](https://cdntk.53zaixian.com/tq/1696694366328123393/img/compression/M_93929d8ed3c336f96d993323060ba32c.jpg)
$(3)判断直线EF与平面PBC的位置关系，请说明理由.$

Solution:
$直线EF与平面PBC相交，理由如下：由(2)可知，\overrightarrow{EF}=\left(\frac{\sqrt{3}}{2},-\frac{3}{2},1\right)，\overrightarrow{BC}=(-\sqrt{3},1,0)，\overrightarrow{BP}=(-\sqrt{3},-1,3)，(9分)设平面PBC的法向量为n=(x,y,z)，则\left\{\begin{matrix}\boldsymbol{n}\cdot \overrightarrow{BC}=0,\\ \boldsymbol{n}\cdot \overrightarrow{BP}=0,\end{matrix}\right.即\left\{\begin{matrix}-\sqrt{3}x+y=0,\\ -\sqrt{3}x-y+3z=0,\end{matrix}\right.令x=\sqrt{3}，则y=3，z=2，故n=(\sqrt{3},3,2)，(11分)所以\overrightarrow{EF}\cdot n=-1\neq 0，故直线EF与平面PBC相交.(12分)$

Final answer: $直线EF与平面PBC相交$
Subfield: Solid Geometry


Question: $如图，四棱锥P-ABCD中，底面ABCD是边长为2的菱形，PA=PC=2，PB=PD，且\angle ABC=60^\circ ，E为PD的中点.$![Image](https://cdntk.53zaixian.com/tq/1696694367297007618/img/compression/M_f1e8395ce41f3d5791516dee016be6d3.jpg)
$(1)求证：AC\perp PD；$

Solution:
$证明：连接BD，设BD与AC交于点O，连接PO.$![Image](https://cdntk.53zaixian.com/tq/1696694367695466498/img/compression/M_ff83a681ae40f6a446b6f31df2cdf2bc.jpg)$因为PA=PC，所以AC\perp PO.(2分)四棱锥P-ABCD中，底面ABCD是边长为2的菱形，则AC\perp BD.易知PO\cap BD=O，PO，BD\subset 平面PBD，所以AC\perp 平面PBD，又因为PD\subset 平面PBD，所以AC\perp PD.(4分)$

Final answer: 证明题
Subfield: Solid Geometry


Question: $如图，四棱锥P-ABCD中，底面ABCD是边长为2的菱形，PA=PC=2，PB=PD，且\angle ABC=60^\circ ，E为PD的中点.$<img_1798>
$(1)求证：AC\perp PD；$

Solution:
$证明：连接BD，设BD与AC交于点O，连接PO.$<img_2069>$因为PA=PC，所以AC\perp PO.(2分)四棱锥P-ABCD中，底面ABCD是边长为2的菱形，则AC\perp BD.易知PO\cap BD=O，PO，BD\subset 平面PBD，所以AC\perp 平面PBD，又因为PD\subset 平面PBD，所以AC\perp PD.(4分)$
Subfield: Solid Geometry


Question: $如图，四棱锥P-ABCD中，底面ABCD是边长为2的菱形，PA=PC=2，PB=PD，且\angle ABC=60^\circ ，E为PD的中点.$![Image](https://cdntk.53zaixian.com/tq/1696694367297007618/img/compression/M_f1e8395ce41f3d5791516dee016be6d3.jpg)
$(2)求二面角E-AC-D的大小；$

Solution:
$因为PB=PD，且由(1)易知O为BD的中点，所以BD\perp PO，所以由(1)知PO\perp 平面ABCD，以O为原点，\overrightarrow{OB}，\overrightarrow{OC}，\overrightarrow{OP}的方向分别为x轴，y轴，z轴的正方向，建立空间直角坐标系，$![Image](https://cdntk.53zaixian.com/tq/1696694368014233602/img/compression/M_9aea6fa475b8a26fab36d0b294a855b7.jpg)$则A(0,-1,0)，B(\sqrt{3},0,0)，C(0,1,0)，D(-\sqrt{3},0,0)，P(0,0,\sqrt{3})，E\left(-\frac{\sqrt{3}}{2},0,\frac{\sqrt{3}}{2}\right)，(5分)所以\overrightarrow{AC}=(0,2,0)，\overrightarrow{DC}=(\sqrt{3},1,0)，\overrightarrow{AE}=\left(-\frac{\sqrt{3}}{2},1,\frac{\sqrt{3}}{2}\right)，设平面AEC的法向量为n=(x,y,z)，则\left\{\begin{matrix}\boldsymbol{n}\cdot \overrightarrow{AE}=0,\\ \boldsymbol{n}\cdot \overrightarrow{AC}=0,\end{matrix}\right.即\left\{\begin{matrix}-\frac{\sqrt{3}}{2}x+y+\frac{\sqrt{3}}{2}z=0,\\ 2y=0,\end{matrix}\right.令x=1，则n=(1,0,1).(7分)易知平面ACD的一个法向量为\overrightarrow{OP}=(0,0,\sqrt{3})，\\cos  \langle n,\overrightarrow{OP}\rangle =\frac{\boldsymbol{n}\cdot \overrightarrow{OP}}{|\boldsymbol{n}||\overrightarrow{OP}|}=\frac{\sqrt{3}}{\sqrt{2}\times \sqrt{3}}=\frac{\sqrt{2}}{2}，所以二面角E-AC-D的大小为\frac{\pi }{4}.(8分)$

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


Question: $如图，四棱锥P-ABCD中，底面ABCD是边长为2的菱形，PA=PC=2，PB=PD，且\angle ABC=60^\circ ，E为PD的中点.$<img_1798>
$(2)求二面角E-AC-D的大小；$

Solution:
$因为PB=PD，且由(1)易知O为BD的中点，所以BD\perp PO，所以由(1)知PO\perp 平面ABCD，以O为原点，\overrightarrow{OB}，\overrightarrow{OC}，\overrightarrow{OP}的方向分别为x轴，y轴，z轴的正方向，建立空间直角坐标系，$<img_1918>$则A(0,-1,0)，B(\sqrt{3},0,0)，C(0,1,0)，D(-\sqrt{3},0,0)，P(0,0,\sqrt{3})，E\left(-\frac{\sqrt{3}}{2},0,\frac{\sqrt{3}}{2}\right)，(5分)所以\overrightarrow{AC}=(0,2,0)，\overrightarrow{DC}=(\sqrt{3},1,0)，\overrightarrow{AE}=\left(-\frac{\sqrt{3}}{2},1,\frac{\sqrt{3}}{2}\right)，设平面AEC的法向量为n=(x,y,z)，则\left\{\begin{matrix}\boldsymbol{n}\cdot \overrightarrow{AE}=0,\\ \boldsymbol{n}\cdot \overrightarrow{AC}=0,\end{matrix}\right.即\left\{\begin{matrix}-\frac{\sqrt{3}}{2}x+y+\frac{\sqrt{3}}{2}z=0,\\ 2y=0,\end{matrix}\right.令x=1，则n=(1,0,1).(7分)易知平面ACD的一个法向量为\overrightarrow{OP}=(0,0,\sqrt{3})，\\cos  \langle n,\overrightarrow{OP}\rangle =\frac{\boldsymbol{n}\cdot \overrightarrow{OP}}{|\boldsymbol{n}||\overrightarrow{OP}|}=\frac{\sqrt{3}}{\sqrt{2}\times \sqrt{3}}=\frac{\sqrt{2}}{2}，所以二面角E-AC-D的大小为\frac{\pi }{4}.(8分)$

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


Question: $如图，四棱锥P-ABCD中，底面ABCD是边长为2的菱形，PA=PC=2，PB=PD，且\angle ABC=60^\circ ，E为PD的中点.$<img_1798>
$(3)在侧棱PC上是否存在点F，使得点F到平面AEC的距离为\frac{\sqrt{6}}{6}？若存在，求出\frac{PF}{FC}的值；若不存在，请说明理由.$

Solution:
$存在点F到平面AEC的距离为\frac{\sqrt{6}}{6}，且\frac{PF}{FC}=2，理由如下：由(2)得\overrightarrow{CP}=(0,-1,\sqrt{3})，\overrightarrow{BC}=(-\sqrt{3},1,0)，设\overrightarrow{CF}=\lambda \overrightarrow{CP}=(0,-\lambda ,\sqrt{3}\lambda ),\lambda \in [0,1]，则\overrightarrow{OF}=\overrightarrow{OC}+\overrightarrow{CF}=(0,1-\lambda ,\sqrt{3}\lambda )，(10分)所以点F到平面AEC的距离d=\frac{|\overrightarrow{OF}\cdot \boldsymbol{n}|}{|\boldsymbol{n}|}=\frac{\sqrt{3}\lambda }{\sqrt{2}}=\frac{\sqrt{6}}{6}，解得\lambda =\frac{1}{3}，所以\overrightarrow{CF}=\frac{1}{3}\overrightarrow{CP}，所以\frac{PF}{FC}=2.(12分)$

Final answer: $\frac{\sqrt{6}}{6},2$
Subfield: Solid Geometry


Question: $如图，四棱锥P-ABCD中，底面ABCD是边长为2的菱形，PA=PC=2，PB=PD，且\angle ABC=60^\circ ，E为PD的中点.$![Image](https://cdntk.53zaixian.com/tq/1696694367297007618/img/compression/M_f1e8395ce41f3d5791516dee016be6d3.jpg)
$(3)在侧棱PC上是否存在点F，使得点F到平面AEC的距离为\frac{\sqrt{6}}{6}？若存在，求出\frac{PF}{FC}的值；若不存在，请说明理由.$

Solution:
$存在点F到平面AEC的距离为\frac{\sqrt{6}}{6}，且\frac{PF}{FC}=2，理由如下：由(2)得\overrightarrow{CP}=(0,-1,\sqrt{3})，\overrightarrow{BC}=(-\sqrt{3},1,0)，设\overrightarrow{CF}=\lambda \overrightarrow{CP}=(0,-\lambda ,\sqrt{3}\lambda ),\lambda \in [0,1]，则\overrightarrow{OF}=\overrightarrow{OC}+\overrightarrow{CF}=(0,1-\lambda ,\sqrt{3}\lambda )，(10分)所以点F到平面AEC的距离d=\frac{|\overrightarrow{OF}\cdot \boldsymbol{n}|}{|\boldsymbol{n}|}=\frac{\sqrt{3}\lambda }{\sqrt{2}}=\frac{\sqrt{6}}{6}，解得\lambda =\frac{1}{3}，所以\overrightarrow{CF}=\frac{1}{3}\overrightarrow{CP}，所以\frac{PF}{FC}=2.(12分)$

Final answer: $\frac{\sqrt{6}}{6},2$
Subfield: Solid Geometry


Question: $如图，平行六面体ABCD-A_{1}B_{1}C_{1}D_{1}中，底面ABCD是边长为1的正方形，AA_{1}=\sqrt{2}，设\overrightarrow{AB}=a，\overrightarrow{AD}=b，\overrightarrow{AA_1}=c.$<img_1594>
$(1)试用a，b，c表示向量\overrightarrow{AC}，\overrightarrow{BD_1}；$

Solution:
$由向量的加减运算法则知，在正方形ABCD中，\overrightarrow{AC}=\overrightarrow{AB}+\overrightarrow{AD}=a+b.(2分)连接AD_{1}，在平行六面体ABCD-A_{1}B_{1}C_{1}D_{1}中，\overrightarrow{BD_1}=\overrightarrow{AD_1}-\overrightarrow{AB}=\overrightarrow{AD}+\overrightarrow{DD_1}-\overrightarrow{AB}=\overrightarrow{AD}+\overrightarrow{AA_1}-\overrightarrow{AB}=b+c-a.(4分)$

Final answer: $a+b, b+c-a$
Subfield: Solid Geometry


Question: $如图，平行六面体ABCD-A_{1}B_{1}C_{1}D_{1}中，底面ABCD是边长为1的正方形，AA_{1}=\sqrt{2}，设\overrightarrow{AB}=a，\overrightarrow{AD}=b，\overrightarrow{AA_1}=c.$![Image](https://cdntk.53zaixian.com/tq/1646400097922793476/img/compression/M_ca76ab62bd2b3cd068e81b155b78dcd3.jpg)
$(1)试用a，b，c表示向量\overrightarrow{AC}，\overrightarrow{BD_1}；$

Solution:
$由向量的加减运算法则知，在正方形ABCD中，\overrightarrow{AC}=\overrightarrow{AB}+\overrightarrow{AD}=a+b.(2分)连接AD_{1}，在平行六面体ABCD-A_{1}B_{1}C_{1}D_{1}中，\overrightarrow{BD_1}=\overrightarrow{AD_1}-\overrightarrow{AB}=\overrightarrow{AD}+\overrightarrow{DD_1}-\overrightarrow{AB}=\overrightarrow{AD}+\overrightarrow{AA_1}-\overrightarrow{AB}=b+c-a.(4分)$

Final answer: $a+b, b+c-a$
Subfield: Solid Geometry


Question: $如图，平行六面体ABCD-A_{1}B_{1}C_{1}D_{1}中，底面ABCD是边长为1的正方形，AA_{1}=\sqrt{2}，设\overrightarrow{AB}=a，\overrightarrow{AD}=b，\overrightarrow{AA_1}=c.$<img_1594>
$(2)若\angle A_{1}AD=\angle A_{1}AB=120^\circ ，求直线AC与BD_{1}所成的角.$

Solution:
$由题意知|a|=|b|=1，|c|=\sqrt{2}，\langle a,b\rangle =90^\circ ，\langle a,c\rangle =120^\circ ，\langle b,c\rangle =120^\circ ，\overrightarrow{AC}\cdot \overrightarrow{BD_1}=(a+b)\cdot (b+c-a)=a\cdot c-a^{2}+b^{2}+b\cdot c=1\times \sqrt{2}\times \\cos   120^\circ -1+1+1\times \sqrt{2}\times \\cos   120^\circ =-\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}=-\sqrt{2}.(6分)又|\overrightarrow{AC}|=\sqrt{2}，|\overrightarrow{BD_1}|=\sqrt{{(\boldsymbol{b}+\boldsymbol{c}-\boldsymbol{a})}^2}=\sqrt{{\boldsymbol{b}}^2+{\boldsymbol{c}}^2+{\boldsymbol{a}}^2+2(\boldsymbol{b}\cdot \boldsymbol{c}-\boldsymbol{a}\cdot \boldsymbol{b}-\boldsymbol{a}\cdot \boldsymbol{c})}=\sqrt{1+2+1+2\times \left[\right.-\frac{\sqrt{2}}{2}-\left(\right.-\frac{\sqrt{2}}{2}\left)\right.\left]\right.}=\sqrt{4}=2，所以\\cos  \langle \overrightarrow{AC},\overrightarrow{BD_1}\rangle =\frac{\overrightarrow{AC}\cdot \overrightarrow{BD_1}}{|\overrightarrow{AC}|\cdot |\overrightarrow{BD_1}|}=\frac{-\sqrt{2}}{\sqrt{2}\times 2}=-\frac{1}{2}.因为0^\circ \leq \langle \overrightarrow{AC},\overrightarrow{BD_1}\rangle \leq 180^\circ ，所以\langle \overrightarrow{AC},\overrightarrow{BD_1}\rangle =120^\circ .(9分)所以AC与BD_{1}所成角的大小为60^\circ .(10分)$

Final answer: $60^\circ $
Subfield: Solid Geometry


Question: $如图，平行六面体ABCD-A_{1}B_{1}C_{1}D_{1}中，底面ABCD是边长为1的正方形，AA_{1}=\sqrt{2}，设\overrightarrow{AB}=a，\overrightarrow{AD}=b，\overrightarrow{AA_1}=c.$![Image](https://cdntk.53zaixian.com/tq/1646400097922793476/img/compression/M_ca76ab62bd2b3cd068e81b155b78dcd3.jpg)
$(2)若\angle A_{1}AD=\angle A_{1}AB=120^\circ ，求直线AC与BD_{1}所成的角.$

Solution:
$由题意知|a|=|b|=1，|c|=\sqrt{2}，\langle a,b\rangle =90^\circ ，\langle a,c\rangle =120^\circ ，\langle b,c\rangle =120^\circ ，\overrightarrow{AC}\cdot \overrightarrow{BD_1}=(a+b)\cdot (b+c-a)=a\cdot c-a^{2}+b^{2}+b\cdot c=1\times \sqrt{2}\times \\cos   120^\circ -1+1+1\times \sqrt{2}\times \\cos   120^\circ =-\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}=-\sqrt{2}.(6分)又|\overrightarrow{AC}|=\sqrt{2}，|\overrightarrow{BD_1}|=\sqrt{{(\boldsymbol{b}+\boldsymbol{c}-\boldsymbol{a})}^2}=\sqrt{{\boldsymbol{b}}^2+{\boldsymbol{c}}^2+{\boldsymbol{a}}^2+2(\boldsymbol{b}\cdot \boldsymbol{c}-\boldsymbol{a}\cdot \boldsymbol{b}-\boldsymbol{a}\cdot \boldsymbol{c})}=\sqrt{1+2+1+2\times \left[\right.-\frac{\sqrt{2}}{2}-\left(\right.-\frac{\sqrt{2}}{2}\left)\right.\left]\right.}=\sqrt{4}=2，所以\\cos  \langle \overrightarrow{AC},\overrightarrow{BD_1}\rangle =\frac{\overrightarrow{AC}\cdot \overrightarrow{BD_1}}{|\overrightarrow{AC}|\cdot |\overrightarrow{BD_1}|}=\frac{-\sqrt{2}}{\sqrt{2}\times 2}=-\frac{1}{2}.因为0^\circ \leq \langle \overrightarrow{AC},\overrightarrow{BD_1}\rangle \leq 180^\circ ，所以\langle \overrightarrow{AC},\overrightarrow{BD_1}\rangle =120^\circ .(9分)所以AC与BD_{1}所成角的大小为60^\circ .(10分)$

Final answer: $60^\circ $
Subfield: Solid Geometry


Question: $如图，已知在斜三棱柱ABC-A_{1}B_{1}C_{1}中，\angle BCA=90^\circ ，AC=BC=2，点A_{1}在底面ABC上的射影恰为AC的中点D，且A_{1}D=\sqrt{3}.$<img_2015>
$(1)求证：A_{1}B\perp AC_{1}；$

Solution:
$证明：过点D作DE\perp AC交AB于点E，分别以DE，DC，DA_{1}所在直线为x轴，y轴，z轴建立空间直角坐标系，如图，
$<img_2288>$则A(0，-1，0)，C(0，1，0)，B(2，1，0)，A_{1}(0，0，\sqrt{3})，C_{1}(0，2，\sqrt{3})，所以\overrightarrow{A_1B}=(2，1，-\sqrt{3})，\overrightarrow{AC_1}=(0,3,\sqrt{3})，所以\overrightarrow{A_1B}\cdot \overrightarrow{AC_1}=0+3-3=0，所以A_{1}B\perp AC_{1}.
(5分)
$
Subfield: Solid Geometry


Question: $如图，已知在斜三棱柱ABC-A_{1}B_{1}C_{1}中，\angle BCA=90^\circ ，AC=BC=2，点A_{1}在底面ABC上的射影恰为AC的中点D，且A_{1}D=\sqrt{3}.$![Image](https://cdntk.53zaixian.com/tq/1646400110656446466/img/compression/M_1e041397a5dcd6a43bba73d32ff0eb1f.jpg)
$(1)求证：A_{1}B\perp AC_{1}；$

Solution:
$证明：过点D作DE\perp AC交AB于点E，分别以DE，DC，DA_{1}所在直线为x轴，y轴，z轴建立空间直角坐标系，如图，
$![Image](https://cdntk.53zaixian.com/tq/1646400110656446466/img/compression/M_da11567530eab39c30fb8c9d7a0d361c.jpg)$则A(0，-1，0)，C(0，1，0)，B(2，1，0)，A_{1}(0，0，\sqrt{3})，C_{1}(0，2，\sqrt{3})，所以\overrightarrow{A_1B}=(2，1，-\sqrt{3})，\overrightarrow{AC_1}=(0,3,\sqrt{3})，所以\overrightarrow{A_1B}\cdot \overrightarrow{AC_1}=0+3-3=0，所以A_{1}B\perp AC_{1}.
(5分)
$

Final answer: 证明题
Subfield: Solid Geometry


Question: $如图，已知在斜三棱柱ABC-A_{1}B_{1}C_{1}中，\angle BCA=90^\circ ，AC=BC=2，点A_{1}在底面ABC上的射影恰为AC的中点D，且A_{1}D=\sqrt{3}.$![Image](https://cdntk.53zaixian.com/tq/1646400110656446466/img/compression/M_1e041397a5dcd6a43bba73d32ff0eb1f.jpg)
$(2)在直线C_{1}C上找一点M，使得二面角M-A_{1}B_{1}-C_{1}的平面角为90^\circ .$

Solution:
$
设\overrightarrow{CM}=\lambda \overrightarrow{CC_1}=(0,\lambda ,\sqrt{3}\lambda )(\lambda \neq 1)，易知\overrightarrow{A_1B_1}=\overrightarrow{AB}=(2,2,0),\overrightarrow{A_1M}=\overrightarrow{A_1C}+\overrightarrow{CM}=(0,\lambda +1,\sqrt{3}\lambda -\sqrt{3}),设平面MA_{1}B_{1}的法向量为n=(x,y,z)，则\left\{\begin{matrix}\overrightarrow{A_1B_1}\cdot \boldsymbol{n}=0，\\ \overrightarrow{A_1M}\cdot \boldsymbol{n}=0，\end{matrix}\right.即\left\{\begin{matrix}2x+2y=0,\\ (\lambda +1)y+(\sqrt{3}\lambda -\sqrt{3})z=0，\end{matrix}\right.解得\left\{\begin{matrix}x=-y,\\ z=-\frac{(\lambda +1)y}{\sqrt{3}\lambda -\sqrt{3}}，\end{matrix}\right.令y=-1，得x=1,z=\frac{\lambda +1}{\sqrt{3}\lambda -\sqrt{3}}，所以n=\left(1，-1，\frac{\lambda +1}{\sqrt{3}\lambda -\sqrt{3}}\right).
(8分)
易知平面A_{1}B_{1}C_{1}的一个法向量为m=(0，0，1)，所以m\cdot n=\frac{ \lambda +1}{\sqrt{3}\lambda -\sqrt{3}}=0，解得\lambda =-1,所以点M在C_{1}C的延长线上且|CM|=|CC_{1}|.
(10分)$

Final answer: $点M在C_{1}C的延长线上且|CM|=|CC_{1}|.
点M在C_{1}C的延长线上, |CM|=|CC_{1}.$
Subfield: Solid Geometry


Question: $如图，在直三棱柱ABC-A_{1}B_{1}C_{1}中，\triangle ABC是边长为6的等边三角形，D，E分别为AA_{1},BC的中点.$![Image](https://cdntk.53zaixian.com/tq/1646400132769071105/img/compression/M_fc13fa8489cd7d97e2ac6c36634b6caf.jpg)
$(1)证明：AE\parallel 平面BDC_{1}；$

Solution:
$证明：取BC_{1}的中点F，连接DF，EF，如图1所示，\because E为BC的中点，\therefore EF\parallel CC_{1}，EF=\frac{1}{2}CC_{1}.\because D为AA_{1}的中点，\therefore DA\parallel CC_{1}，DA=\frac{1}{2}CC_{1}，\therefore EF\parallel DA，且EF=DA，\therefore 四边形ADFE为平行四边形，\therefore AE\parallel DF.(3分)又\because AE不在平面BDC_{1}内，DF在平面BDC_{1}内，\therefore AE\parallel 平面BDC_{1}.(4分)$![Image](https://cdntk.53zaixian.com/tq/1646400133075255297/img/compression/M_76795b09a43378891d78ec7659518d33.jpg)$图1$

Final answer: 证明题
Subfield: Solid Geometry


Question: $如图，在直三棱柱ABC-A_{1}B_{1}C_{1}中，\triangle ABC是边长为6的等边三角形，D，E分别为AA_{1},BC的中点.$<img_2598>
$(1)证明：AE\parallel 平面BDC_{1}；$

Solution:
$证明：取BC_{1}的中点F，连接DF，EF，如图1所示，\because E为BC的中点，\therefore EF\parallel CC_{1}，EF=\frac{1}{2}CC_{1}.\because D为AA_{1}的中点，\therefore DA\parallel CC_{1}，DA=\frac{1}{2}CC_{1}，\therefore EF\parallel DA，且EF=DA，\therefore 四边形ADFE为平行四边形，\therefore AE\parallel DF.(3分)又\because AE不在平面BDC_{1}内，DF在平面BDC_{1}内，\therefore AE\parallel 平面BDC_{1}.(4分)$<img_2810>$图1$
Subfield: Solid Geometry


Question: $如图，在直三棱柱ABC-A_{1}B_{1}C_{1}中，\triangle ABC是边长为6的等边三角形，D，E分别为AA_{1},BC的中点.$![Image](https://cdntk.53zaixian.com/tq/1646400132769071105/img/compression/M_fc13fa8489cd7d97e2ac6c36634b6caf.jpg)
$(2)若异面直线BC_{1}与AC所成角的余弦值为\frac{\sqrt{3}}{4}，求DE与平面BDC_{1}所成角的正弦值.$

Solution:
$由(1)及题设可知，BC，EA，EF两两垂直，则以点E为坐标原点，EC，EA，EF所在直线分别为x轴，y轴，z轴建立如图2所示的空间直角坐标系，$![Image](https://cdntk.53zaixian.com/tq/1646400133331107842/img/compression/M_8ba3fca64b9ab4a9ac30efa25a81cdbc.jpg)$图2设AA_{1}=2t(t＞0)，则B(-3，0，0)，C_{1}(3，0，2t)，A(0，3\sqrt{3}，0)，C(3，0，0)，D(0，3\sqrt{3}，t)，(6分)\therefore \overrightarrow{BD}=(3,3\sqrt{3},t),\overrightarrow{BC_1}=(6,0,2t),\overrightarrow{AC}=(3,-3\sqrt{3},0)，\therefore |\\cos  \langle \overrightarrow{BC_1},\overrightarrow{AC}\rangle |=\frac{|\overrightarrow{BC_1}\cdot \overrightarrow{AC}|}{|\overrightarrow{BC_1}||\overrightarrow{AC}|}=\frac{6\times 3}{\sqrt{6^2+{(2t)}^2}\cdot \sqrt{3^2+{(-3\sqrt{3})}^2+0^2}}=\frac{\sqrt{3}}{4},解得t=\sqrt{3}，(9分)设平面BDC_{1}的法向量m=(x,y,z)，则\left\{\begin{matrix}\boldsymbol{m}\cdot \overrightarrow{BD}=3x+3\sqrt{3}y+\sqrt{3}z=0，\\ \boldsymbol{m}\cdot \overrightarrow{BC_1}=6x+2\sqrt{3}z=0，\end{matrix}\right.故可取x=1,则m=(1，0，-\sqrt{3})，易知D(0，3\sqrt{3}，\sqrt{3})，则\overrightarrow{ED}=(0,3\sqrt{3},\sqrt{3})，\therefore \\cos  \langle \overrightarrow{ED},m\rangle =\frac{\overrightarrow{ED}\cdot \boldsymbol{m}}{|\overrightarrow{ED}||\boldsymbol{m}|}=\frac{-3}{\sqrt{0^2+(3\sqrt{3})^2+(\sqrt{3})^2}\times \sqrt{1^2+0^2+(-\sqrt{3})^2}}=-\frac{\sqrt{30}}{20}，\therefore DE与平面BDC_{1}所成角的正弦值为\frac{\sqrt{30}}{20}.(12分)$

Final answer: $\frac{\sqrt{30}}{20}$
Subfield: Solid Geometry


Question: $如图，在直三棱柱ABC-A_{1}B_{1}C_{1}中，\triangle ABC是边长为6的等边三角形，D，E分别为AA_{1},BC的中点.$<img_2598>
$(2)若异面直线BC_{1}与AC所成角的余弦值为\frac{\sqrt{3}}{4}，求DE与平面BDC_{1}所成角的正弦值.$

Solution:
$由(1)及题设可知，BC，EA，EF两两垂直，则以点E为坐标原点，EC，EA，EF所在直线分别为x轴，y轴，z轴建立如图2所示的空间直角坐标系，$<img_2980>$图2设AA_{1}=2t(t＞0)，则B(-3，0，0)，C_{1}(3，0，2t)，A(0，3\sqrt{3}，0)，C(3，0，0)，D(0，3\sqrt{3}，t)，(6分)\therefore \overrightarrow{BD}=(3,3\sqrt{3},t),\overrightarrow{BC_1}=(6,0,2t),\overrightarrow{AC}=(3,-3\sqrt{3},0)，\therefore |\\cos  \langle \overrightarrow{BC_1},\overrightarrow{AC}\rangle |=\frac{|\overrightarrow{BC_1}\cdot \overrightarrow{AC}|}{|\overrightarrow{BC_1}||\overrightarrow{AC}|}=\frac{6\times 3}{\sqrt{6^2+{(2t)}^2}\cdot \sqrt{3^2+{(-3\sqrt{3})}^2+0^2}}=\frac{\sqrt{3}}{4},解得t=\sqrt{3}，(9分)设平面BDC_{1}的法向量m=(x,y,z)，则\left\{\begin{matrix}\boldsymbol{m}\cdot \overrightarrow{BD}=3x+3\sqrt{3}y+\sqrt{3}z=0，\\ \boldsymbol{m}\cdot \overrightarrow{BC_1}=6x+2\sqrt{3}z=0，\end{matrix}\right.故可取x=1,则m=(1，0，-\sqrt{3})，易知D(0，3\sqrt{3}，\sqrt{3})，则\overrightarrow{ED}=(0,3\sqrt{3},\sqrt{3})，\therefore \\cos  \langle \overrightarrow{ED},m\rangle =\frac{\overrightarrow{ED}\cdot \boldsymbol{m}}{|\overrightarrow{ED}||\boldsymbol{m}|}=\frac{-3}{\sqrt{0^2+(3\sqrt{3})^2+(\sqrt{3})^2}\times \sqrt{1^2+0^2+(-\sqrt{3})^2}}=-\frac{\sqrt{30}}{20}，\therefore DE与平面BDC_{1}所成角的正弦值为\frac{\sqrt{30}}{20}.(12分)$

Final answer: $\frac{\sqrt{30}}{20}$
Subfield: Solid Geometry


Question: $在如图所示的多面体中，AD\parallel BC且AD=2BC，DA\perp DC，EG\parallel AD且EG=AD，CD\parallel FG且CD=2FG，DG\perp 平面ABCD，DA=DC=DG=2.$![Image](https://cdntk.53zaixian.com/tq/1646400133821841410/img/compression/M_ef10168583c28dc0981fc4578b3ab557.jpg)
$(1)求点F到直线EC的距离；$

Solution:
$因为DG\perp 平面ABCD，DA\subset 平面ABCD，DC\subset 平面ABCD，所以DG\perp DA，DG\perp DC，(2分)又DA\perp DC，所以DG，DA，DC两两垂直，如图，以D为坐标原点建立空间直角坐标系，则D(0,0,0)，A(2,0,0)，C(0,2,0)，G(0,0,2)，E(2,0,2)，F(0,1,2)，B(1,2,0)，(4分)所以\overrightarrow{CE}=(2,-2,2)，\overrightarrow{EF}=(-2,1,0)，$![Image](https://cdntk.53zaixian.com/tq/1646400134119636994/img/compression/M_1d8613af9284b49cb605f821a29a515e.jpg)$所以点F到直线EC的距离为\sqrt{|\overrightarrow{EF}|^2-\left(\frac{|\overrightarrow{CE}\cdot \overrightarrow{EF}|}{|\overrightarrow{CE}|}\right)^2}=\sqrt{4+1+0-\left(\frac{-4-2+0}{\sqrt{4+4+4}}\right)^2}=\sqrt{2}.(6分)$

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


Question: $在如图所示的多面体中，AD\parallel BC且AD=2BC，DA\perp DC，EG\parallel AD且EG=AD，CD\parallel FG且CD=2FG，DG\perp 平面ABCD，DA=DC=DG=2.$<img_1824>
$(1)求点F到直线EC的距离；$

Solution:
$因为DG\perp 平面ABCD，DA\subset 平面ABCD，DC\subset 平面ABCD，所以DG\perp DA，DG\perp DC，(2分)又DA\perp DC，所以DG，DA，DC两两垂直，如图，以D为坐标原点建立空间直角坐标系，则D(0,0,0)，A(2,0,0)，C(0,2,0)，G(0,0,2)，E(2,0,2)，F(0,1,2)，B(1,2,0)，(4分)所以\overrightarrow{CE}=(2,-2,2)，\overrightarrow{EF}=(-2,1,0)，$<img_2397>$所以点F到直线EC的距离为\sqrt{|\overrightarrow{EF}|^2-\left(\frac{|\overrightarrow{CE}\cdot \overrightarrow{EF}|}{|\overrightarrow{CE}|}\right)^2}=\sqrt{4+1+0-\left(\frac{-4-2+0}{\sqrt{4+4+4}}\right)^2}=\sqrt{2}.(6分)$

Final answer: $\sqrt{2}$