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


Question: $某工厂拟建造如图所示的容器，容器的上部为半球形，其半径为r m,下部为圆柱形，其高为l m,该容器的容积为\frac{160\pi }{3} m^{3}，且l\geq 6r.假设该容器的建造费用仅与其表面积有关，材料厚度不计.已知圆柱形部分侧面的建造费用为每平方米2.25千元，半球形部分以及圆柱下底面的建造费用为每平方米m(m>2.25)千元.设该容器的建造费用为y千元.$<img_2116>
$(2)求该容器的建造费用最小时r的值.$

Solution:
$由(1)得y'=6\pi (m-1)r-\frac{240\pi }{r^2}=\frac{6\pi (m-1)}{r^2}\cdot \left(r^3-\frac{40}{m-1}\right)，0由于m>\frac{9}{4}，所以m-1>0.令r^{3}-\frac{40}{m-1}=0，得r=\sqrt[3]{\frac{40}{m-1}}.若\sqrt[3]{\frac{40}{m-1}}<2，即m>6，则当r\in \left(0,\sqrt[3]{\frac{40}{m-1}}\right)时，y'<0，原函数单调递减，当r\in \left(\sqrt[3]{\frac{40}{m-1}},2\right)时，y'>0，原函数单调递增，故原函数在r=\sqrt[3]{\frac{40}{m-1}}处取得极小值，也是最小值.若\sqrt[3]{\frac{40}{m-1}}\geq 2，即\frac{9}{4}综上所述，若\frac{9}{4}6，则建造费用最小时r=\sqrt[3]{\frac{40}{m-1}}.$

Final answer: $\sqrt[3]{\frac{40}{m-1}}.$
Subfield: Solid Geometry


Question: $某工厂拟建造如图所示的容器，容器的上部为半球形，其半径为r m,下部为圆柱形，其高为l m,该容器的容积为\frac{160\pi }{3} m^{3}，且l\geq 6r.假设该容器的建造费用仅与其表面积有关，材料厚度不计.已知圆柱形部分侧面的建造费用为每平方米2.25千元，半球形部分以及圆柱下底面的建造费用为每平方米m(m>2.25)千元.设该容器的建造费用为y千元.$![Image](https://cdntk.53zaixian.com/tq/1693820373075296258/img/compression/M_afdd8f61648bfc3b3dd95fbf5732ced7.jpg)
$(2)求该容器的建造费用最小时r的值.$

Solution:
$由(1)得y'=6\pi (m-1)r-\frac{240\pi }{r^2}=\frac{6\pi (m-1)}{r^2}\cdot \left(r^3-\frac{40}{m-1}\right)，0由于m>\frac{9}{4}，所以m-1>0.令r^{3}-\frac{40}{m-1}=0，得r=\sqrt[3]{\frac{40}{m-1}}.若\sqrt[3]{\frac{40}{m-1}}<2，即m>6，则当r\in \left(0,\sqrt[3]{\frac{40}{m-1}}\right)时，y'<0，原函数单调递减，当r\in \left(\sqrt[3]{\frac{40}{m-1}},2\right)时，y'>0，原函数单调递增，故原函数在r=\sqrt[3]{\frac{40}{m-1}}处取得极小值，也是最小值.若\sqrt[3]{\frac{40}{m-1}}\geq 2，即\frac{9}{4}综上所述，若\frac{9}{4}6，则建造费用最小时r=\sqrt[3]{\frac{40}{m-1}}.$

Final answer: $\sqrt[3]{\frac{40}{m-1}}.$
Subfield: Solid Geometry


Question: $在长方体ABCD-A_{1}B_{1}C_{1}D_{1}中,|AD|=2,|DC|=4,|DD_{1}|=3,以D为原点,建立如图所示的空间直角坐标系D-xyz.$<img_1908>
$(1)求点B_{1},C_{1},D_{1},A的坐标;$

Solution:
$设B_{1}(x_{1},y_{1},z_{1}).过点B_{1}向三个坐标平面yDz,xDz,xDy作垂线,分别交平面yDz,xDz,xDy于点C_{1},A_{1},B,故|x_{1}|=|B_{1}C_{1}|=2,|y_{1}|=|B_{1}A_{1}|=4,|z_{1}|=|B_{1}B|=3,由图可知x_{1}，y_{1}，z_{1}均为正数,故点B_{1}的坐标是(2,4,3).同理可求得C_{1}(0,4,3),D_{1}(0,0,3),A(2,0,0).$

Final answer: $(2,4,3),(0,4,3),(0,0,3),(2,0,0)$
Subfield: Solid Geometry


Question: $在长方体ABCD-A_{1}B_{1}C_{1}D_{1}中,|AD|=2,|DC|=4,|DD_{1}|=3,以D为原点,建立如图所示的空间直角坐标系D-xyz.$![Image](https://cdntk.53zaixian.com/tq/1693861339183194113/img/compression/M_50f050712bd305837a63113aedec95c8.jpg)
$(1)求点B_{1},C_{1},D_{1},A的坐标;$

Solution:
$设B_{1}(x_{1},y_{1},z_{1}).过点B_{1}向三个坐标平面yDz,xDz,xDy作垂线,分别交平面yDz,xDz,xDy于点C_{1},A_{1},B,故|x_{1}|=|B_{1}C_{1}|=2,|y_{1}|=|B_{1}A_{1}|=4,|z_{1}|=|B_{1}B|=3,由图可知x_{1}，y_{1}，z_{1}均为正数,故点B_{1}的坐标是(2,4,3).同理可求得C_{1}(0,4,3),D_{1}(0,0,3),A(2,0,0).$

Final answer: $(2,4,3),(0,4,3),(0,0,3),(2,0,0)$
Subfield: Solid Geometry


Question: $在长方体ABCD-A_{1}B_{1}C_{1}D_{1}中,|AD|=2,|DC|=4,|DD_{1}|=3,以D为原点,建立如图所示的空间直角坐标系D-xyz.$<img_1908>
$(2)若E为C_{1}D_{1}的中点,试求点E的坐标.$

Solution:
$因为E是C_{1}D_{1}的中点,C_{1}(0,4,3),D_{1}(0,0,3),所以由中点坐标公式得点E的坐标为(0,2,3).$

Final answer: $(0,2,3)$
Subfield: Solid Geometry


Question: $在长方体ABCD-A_{1}B_{1}C_{1}D_{1}中,|AD|=2,|DC|=4,|DD_{1}|=3,以D为原点,建立如图所示的空间直角坐标系D-xyz.$![Image](https://cdntk.53zaixian.com/tq/1693861339183194113/img/compression/M_50f050712bd305837a63113aedec95c8.jpg)
$(2)若E为C_{1}D_{1}的中点,试求点E的坐标.$

Solution:
$因为E是C_{1}D_{1}的中点,C_{1}(0,4,3),D_{1}(0,0,3),所以由中点坐标公式得点E的坐标为(0,2,3).$

Final answer: $(0,2,3)$
Subfield: Solid Geometry


Question: $如图，在平行六面体ABCD-A_{1}B_{1}C_{1}D_{1}中，底面ABCD是正方形，AD=AB=1，AA_{1}=2，\angle A_{1}AB=\angle DAA_{1}=60^\circ ，\overrightarrow{A_1C_1}=3\overrightarrow{NC_1}，\overrightarrow{D_1B}=2\overrightarrow{MB}，设\overrightarrow{AB}=a，\overrightarrow{AD}=b，\overrightarrow{AA_1}=c.$![Image](https://cdntk.53zaixian.com/tq/1693861363714813955/img/compression/M_8c8de06c406780cd00e382f63da14900.jpg)
$(1)试用a，b，c表示\overrightarrow{AN}；$

Solution:
$由题意得\overrightarrow{AN}=\overrightarrow{AA_1}+\overrightarrow{A_1N}=\overrightarrow{AA_1}+\frac{2}{3}\overrightarrow{A_1C_1}=\overrightarrow{AA_1}+\frac{2}{3}(\overrightarrow{A_1B_1}+\overrightarrow{A_1D_1})=\overrightarrow{AA_1}+\frac{2}{3}(\overrightarrow{AB}+\overrightarrow{AD})=\frac{2}{3}a+\frac{2}{3}b+c.$

Final answer: $\overrightarrow{AN}=\frac{2}{3}a+\frac{2}{3}b+c$
Subfield: Solid Geometry


Question: $如图，在平行六面体ABCD-A_{1}B_{1}C_{1}D_{1}中，底面ABCD是正方形，AD=AB=1，AA_{1}=2，\angle A_{1}AB=\angle DAA_{1}=60^\circ ，\overrightarrow{A_1C_1}=3\overrightarrow{NC_1}，\overrightarrow{D_1B}=2\overrightarrow{MB}，设\overrightarrow{AB}=a，\overrightarrow{AD}=b，\overrightarrow{AA_1}=c.$<img_1487>
$(1)试用a，b，c表示\overrightarrow{AN}；$

Solution:
$由题意得\overrightarrow{AN}=\overrightarrow{AA_1}+\overrightarrow{A_1N}=\overrightarrow{AA_1}+\frac{2}{3}\overrightarrow{A_1C_1}=\overrightarrow{AA_1}+\frac{2}{3}(\overrightarrow{A_1B_1}+\overrightarrow{A_1D_1})=\overrightarrow{AA_1}+\frac{2}{3}(\overrightarrow{AB}+\overrightarrow{AD})=\frac{2}{3}a+\frac{2}{3}b+c.$



Final answer: $\overrightarrow{AN}=\frac{2}{3}a+\frac{2}{3}b+c$
Subfield: Solid Geometry


Question: $如图，在平行六面体ABCD-A_{1}B_{1}C_{1}D_{1}中，底面ABCD是正方形，AD=AB=1，AA_{1}=2，\angle A_{1}AB=\angle DAA_{1}=60^\circ ，\overrightarrow{A_1C_1}=3\overrightarrow{NC_1}，\overrightarrow{D_1B}=2\overrightarrow{MB}，设\overrightarrow{AB}=a，\overrightarrow{AD}=b，\overrightarrow{AA_1}=c.$![Image](https://cdntk.53zaixian.com/tq/1693861363714813955/img/compression/M_8c8de06c406780cd00e382f63da14900.jpg)
$(2)求MN的长度.$

Solution:
$由题意得\overrightarrow{MN}=\overrightarrow{MD_1}+\overrightarrow{D_1C_1}+\overrightarrow{C_1N}=\frac{1}{2}\overrightarrow{BD_1}+\overrightarrow{AB}-\frac{1}{3}\overrightarrow{A_1C_1}=\frac{1}{2}(\overrightarrow{BA}+\overrightarrow{AA_1}+\overrightarrow{A_1D_1})+\overrightarrow{AB}-\frac{1}{3}(\overrightarrow{A_1B_1}+\overrightarrow{A_1D_1})=-\frac{1}{2}\overrightarrow{AB}+\frac{1}{2}\overrightarrow{AA_1}+\frac{1}{2}\overrightarrow{AD}+\overrightarrow{AB}-\frac{1}{3}\overrightarrow{AB}-\frac{1}{3}\overrightarrow{AD}=\frac{1}{6}\overrightarrow{AB}+\frac{1}{6}\overrightarrow{AD}+\frac{1}{2}\overrightarrow{AA_1}=\frac{1}{6}a+\frac{1}{6}b+\frac{1}{2}c.因为\angle A_{1}AB=\angle DAA_{1}=60^\circ ，且底面ABCD是正方形，AD=AB=1，AA_{1}=2，所以\overrightarrow{AB}\cdot \overrightarrow{AD}=a\cdot b=0，\overrightarrow{AD}\cdot \overrightarrow{AA_1}=b\cdot c=|b||c|\cdot \\cos   60^\circ =1，\overrightarrow{AB}\cdot \overrightarrow{AA_1}=a\cdot c=|a||c|\\cos   60^\circ =1.因此，|\overrightarrow{MN}|=\frac{1}{6}\sqrt{{(\boldsymbol{a}+\boldsymbol{b}+3\boldsymbol{c})}^2}=\frac{1}{6}\times \sqrt{{\boldsymbol{a}}^2+{\boldsymbol{b}}^2+9\boldsymbol{c}^2+2\boldsymbol{a}\cdot \boldsymbol{b}+6\boldsymbol{b}\cdot \boldsymbol{c}+6\boldsymbol{a}\cdot \boldsymbol{c}}=\frac{5\sqrt{2}}{6}，所以MN的长度是\frac{5\sqrt{2}}{6}.$

Final answer: $\frac{5\sqrt 2}{6}。$
Subfield: Solid Geometry


Question: $如图，在平行六面体ABCD-A_{1}B_{1}C_{1}D_{1}中，底面ABCD是正方形，AD=AB=1，AA_{1}=2，\angle A_{1}AB=\angle DAA_{1}=60^\circ ，\overrightarrow{A_1C_1}=3\overrightarrow{NC_1}，\overrightarrow{D_1B}=2\overrightarrow{MB}，设\overrightarrow{AB}=a，\overrightarrow{AD}=b，\overrightarrow{AA_1}=c.$<img_1487>
$(2)求MN的长度.$

Solution:
$由题意得\overrightarrow{MN}=\overrightarrow{MD_1}+\overrightarrow{D_1C_1}+\overrightarrow{C_1N}=\frac{1}{2}\overrightarrow{BD_1}+\overrightarrow{AB}-\frac{1}{3}\overrightarrow{A_1C_1}=\frac{1}{2}(\overrightarrow{BA}+\overrightarrow{AA_1}+\overrightarrow{A_1D_1})+\overrightarrow{AB}-\frac{1}{3}(\overrightarrow{A_1B_1}+\overrightarrow{A_1D_1})=-\frac{1}{2}\overrightarrow{AB}+\frac{1}{2}\overrightarrow{AA_1}+\frac{1}{2}\overrightarrow{AD}+\overrightarrow{AB}-\frac{1}{3}\overrightarrow{AB}-\frac{1}{3}\overrightarrow{AD}=\frac{1}{6}\overrightarrow{AB}+\frac{1}{6}\overrightarrow{AD}+\frac{1}{2}\overrightarrow{AA_1}=\frac{1}{6}a+\frac{1}{6}b+\frac{1}{2}c.因为\angle A_{1}AB=\angle DAA_{1}=60^\circ ，且底面ABCD是正方形，AD=AB=1，AA_{1}=2，所以\overrightarrow{AB}\cdot \overrightarrow{AD}=a\cdot b=0，\overrightarrow{AD}\cdot \overrightarrow{AA_1}=b\cdot c=|b||c|\cdot \\cos   60^\circ =1，\overrightarrow{AB}\cdot \overrightarrow{AA_1}=a\cdot c=|a||c|\\cos   60^\circ =1.因此，|\overrightarrow{MN}|=\frac{1}{6}\sqrt{{(\boldsymbol{a}+\boldsymbol{b}+3\boldsymbol{c})}^2}=\frac{1}{6}\times \sqrt{{\boldsymbol{a}}^2+{\boldsymbol{b}}^2+9\boldsymbol{c}^2+2\boldsymbol{a}\cdot \boldsymbol{b}+6\boldsymbol{b}\cdot \boldsymbol{c}+6\boldsymbol{a}\cdot \boldsymbol{c}}=\frac{5\sqrt{2}}{6}，所以MN的长度是\frac{5\sqrt{2}}{6}.$

Final answer: $\frac{5\sqrt 2}{6}。$
Subfield: Solid Geometry


Question: $如图,在空间直角坐标系O-xyz中有一长方体OABC-O'A'B'C',且OA=6,OC=8,OO'=5.$![Image](https://cdntk.53zaixian.com/tq/1693861364784361475/img/compression/M_e794bbf13918766e07dc559093eb5c4f.jpg)
$(1)写出点B'的坐标,并将\overrightarrow{OB'}用标准正交基{i,j,k}表示;$

Solution:
$因为OA=6,OC=8,OO'=5,所以点B'的坐标为(6，8，5)，从而\overrightarrow{OB'}=(6，8，5)=6i+8j+5k.$

Final answer: $点B'的坐标为(6，8，5)， \overrightarrow{OB'}=(6，8，5)=6i+8j+5k.$
Subfield: Solid Geometry


Question: $如图,在空间直角坐标系O-xyz中有一长方体OABC-O'A'B'C',且OA=6,OC=8,OO'=5.$<img_2240>
$(1)写出点B'的坐标$

Solution:
$因为OA=6,OC=8,OO'=5,所以点B'的坐标为(6，8，5)，从而\overrightarrow{OB'}=(6，8，5)=6i+8j+5k.$



Final answer: (6,8,5)
Subfield: Solid Geometry


Question: $如图,在空间直角坐标系O-xyz中有一长方体OABC-O'A'B'C',且OA=6,OC=8,OO'=5.$<img_2240>
$(2)求\overrightarrow{OC'}的坐标.$

Solution:
$易得点C'的坐标为(0，8，5)，所以\overrightarrow{OC'}=(0，8，5).$

Final answer: $(0，8，5)$
Subfield: Solid Geometry


Question: $如图,在空间直角坐标系O-xyz中有一长方体OABC-O'A'B'C',且OA=6,OC=8,OO'=5.$![Image](https://cdntk.53zaixian.com/tq/1693861364784361475/img/compression/M_e794bbf13918766e07dc559093eb5c4f.jpg)
$(2)求\overrightarrow{OC'}的坐标.$

Solution:
$易得点C'的坐标为(0，8，5)，所以\overrightarrow{OC'}=(0，8，5).$

Final answer: $(0，8，5)$
Subfield: Solid Geometry


Question: $如图，在直三棱柱ABC-A_{1}B_{1}C_{1}中，AB=AC=AA_{1}=1，AB\perp AC，D是棱BC的中点.$<img_3032>
$(1)求异面直线AB_{1}与DC_{1}所成角的余弦值；$

Solution:
$由题意知，AB，AC，AA_{1}两两垂直.以A为原点，\overrightarrow{AB},\overrightarrow{AC},\overrightarrow{AA_1}的方向分别为x,y,z轴的正方向，1为单位长度，建立如图所示的空间直角坐标系A-xyz,则A(0,0,0),B_{1}(1,0,1),D\left(\frac{1}{2},\frac{1}{2},0\right)，C_{1}(0,1,1),则\overrightarrow{AB_1}=(1,0,1),\overrightarrow{DC_1}=\left(-\frac{1}{2},\frac{1}{2},1\right).设异面直线AB_{1}与DC_{1}所成的角为\theta ，则\\cos   \theta =|\\cos  \langle \overrightarrow{AB_1},\overrightarrow{DC_1}\rangle |=\frac{|\overrightarrow{AB_1}\cdot \overrightarrow{DC_1}|}{|\overrightarrow{AB_1}|\cdot |\overrightarrow{DC_1}|}=\frac{\sqrt{3}}{6}，所以异面直线AB_{1}与DC_{1}所成角的余弦值为\frac{\sqrt{3}}{6}.$<img_1875>

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


Question: $如图，在直三棱柱ABC-A_{1}B_{1}C_{1}中，AB=AC=AA_{1}=1，AB\perp AC，D是棱BC的中点.$![Image](https://cdntk.53zaixian.com/tq/1693861402818183170/img/compression/M_7a9b2ffa50bbfe6fce291d49a9499823.jpg)
$(1)求异面直线AB_{1}与DC_{1}所成角的余弦值；$

Solution:
$由题意知，AB，AC，AA_{1}两两垂直.以A为原点，\overrightarrow{AB},\overrightarrow{AC},\overrightarrow{AA_1}的方向分别为x,y,z轴的正方向，1为单位长度，建立如图所示的空间直角坐标系A-xyz,则A(0,0,0),B_{1}(1,0,1),D\left(\frac{1}{2},\frac{1}{2},0\right)，C_{1}(0,1,1),则\overrightarrow{AB_1}=(1,0,1),\overrightarrow{DC_1}=\left(-\frac{1}{2},\frac{1}{2},1\right).设异面直线AB_{1}与DC_{1}所成的角为\theta ，则\\cos   \theta =|\\cos  \langle \overrightarrow{AB_1},\overrightarrow{DC_1}\rangle |=\frac{|\overrightarrow{AB_1}\cdot \overrightarrow{DC_1}|}{|\overrightarrow{AB_1}|\cdot |\overrightarrow{DC_1}|}=\frac{\sqrt{3}}{6}，所以异面直线AB_{1}与DC_{1}所成角的余弦值为\frac{\sqrt{3}}{6}.$![Image](https://cdntk.53zaixian.com/tq/1693861403162116098/img/compression/M_8531632f8a0535599e1fe5b60c854e3a.jpg)

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


Question: $如图，在直三棱柱ABC-A_{1}B_{1}C_{1}中，AB=AC=AA_{1}=1，AB\perp AC，D是棱BC的中点.$![Image](https://cdntk.53zaixian.com/tq/1693861402818183170/img/compression/M_7a9b2ffa50bbfe6fce291d49a9499823.jpg)
$(2)求平面B_{1}AD与平面AC_{1}D所成的角的余弦值.$

Solution:
$由(1)中所建坐标系可得\overrightarrow{AD}=\left(\frac{1}{2},\frac{1}{2},0\right),\overrightarrow{AB_1}=(1,0,1),\overrightarrow{AC_1}=(0，1，1).设m=(x,y,z)为平面B_{1}AD的法向量，则\left\{\begin{matrix}{\boldsymbol{m}}\cdot \overrightarrow{AD}=\frac{1}{2}x+\frac{1}{2}y=0,\\ {\boldsymbol{m}}\cdot \overrightarrow{AB_1}=x+z=0,\end{matrix}\right.所以\left\{\begin{matrix}x+y=0,\\ x+z=0,\end{matrix}\right.令x=1,则y=-1,z=-1，则m=(1,-1,-1)为平面B_{1}AD的一个法向量，设n=(x',y',z')为平面AC_{1}D的法向量，则\left\{\begin{matrix}\boldsymbol{n}\cdot \overrightarrow{AD}=\frac{1}{2}x'+\frac{1}{2}y'=0,\\ \boldsymbol{n}\cdot \overrightarrow{AC_1}=y'+z'=0,\end{matrix}\right.所以\left\{\begin{matrix}x'+y'=0,\\ y'+z'=0,\end{matrix}\right.令x'=1,则y'=-1,z'=1，则n=(1,-1,1)为平面AC_{1}D的一个法向量，则\\cos  \langle m,n\rangle =\frac{\boldsymbol{m}\cdot \boldsymbol{n}}{|\boldsymbol{m}|\cdot |\boldsymbol{n}|}=\frac{1}{\sqrt{3}\times \sqrt{3}}=\frac{1}{3},故平面B_{1}AD与平面AC_{1}D所成的角的余弦值为\frac{1}{3} .$

Final answer: $\frac{1}{3}$
Subfield: Solid Geometry


Question: $如图，在直三棱柱ABC-A_{1}B_{1}C_{1}中，AB=AC=AA_{1}=1，AB\perp AC，D是棱BC的中点.$<img_3032>
$(2)求平面B_{1}AD与平面AC_{1}D所成的角的余弦值.$

Solution:
$由(1)中所建坐标系可得\overrightarrow{AD}=\left(\frac{1}{2},\frac{1}{2},0\right),\overrightarrow{AB_1}=(1,0,1),\overrightarrow{AC_1}=(0，1，1).设m=(x,y,z)为平面B_{1}AD的法向量，则\left\{\begin{matrix}{\boldsymbol{m}}\cdot \overrightarrow{AD}=\frac{1}{2}x+\frac{1}{2}y=0,\\ {\boldsymbol{m}}\cdot \overrightarrow{AB_1}=x+z=0,\end{matrix}\right.所以\left\{\begin{matrix}x+y=0,\\ x+z=0,\end{matrix}\right.令x=1,则y=-1,z=-1，则m=(1,-1,-1)为平面B_{1}AD的一个法向量，设n=(x',y',z')为平面AC_{1}D的法向量，则\left\{\begin{matrix}\boldsymbol{n}\cdot \overrightarrow{AD}=\frac{1}{2}x'+\frac{1}{2}y'=0,\\ \boldsymbol{n}\cdot \overrightarrow{AC_1}=y'+z'=0,\end{matrix}\right.所以\left\{\begin{matrix}x'+y'=0,\\ y'+z'=0,\end{matrix}\right.令x'=1,则y'=-1,z'=1，则n=(1,-1,1)为平面AC_{1}D的一个法向量，则\\cos  \langle m,n\rangle =\frac{\boldsymbol{m}\cdot \boldsymbol{n}}{|\boldsymbol{m}|\cdot |\boldsymbol{n}|}=\frac{1}{\sqrt{3}\times \sqrt{3}}=\frac{1}{3},故平面B_{1}AD与平面AC_{1}D所成的角的余弦值为\frac{1}{3} .$

Final answer: $\frac{1}{3}$
Subfield: Solid Geometry


Question: $如图①，在\triangle ABC中，AB\perp BC，且AB=2BC，将\triangle ABC沿中位线EF折起，使得AE\perp BE，连接AB，AC，M为AC的中点，连接ME，MF(如图②).$<img_1763>
$(1)证明：MF\perp 平面ABC；$

Solution:
$证明：连接EC，设BC=2，则EF=1,AE=BE=2,AF=FC=\sqrt{5}，EC=2\sqrt{2}.\because AE\perp EF,AE\perp BE，EF\cap BE=E，\therefore AE\perp 平面BCFE，\because EC\subset 平面BCFE，\therefore AE\perp EC，连接BM，BF，\because AC=\sqrt{AE^2+EC^2}=\sqrt{4+8}=2\sqrt{3}，BC=2,AB=\sqrt{AE^2+BE^2}=2\sqrt{2},\therefore AC^{2}=BC^{2}+AB^{2},\therefore BC\perp AB,\therefore BM=\frac{1}{2}AC=\sqrt{3}，\because AF=FC,\therefore FM\perp AC，\therefore MF=\sqrt{CF^2-CM^2}=\sqrt{5-3}=\sqrt{2},又\because BF=\sqrt{BE^2+EF^2}=\sqrt{4+1}=\sqrt{5}，\therefore BF^{2}=MF^{2}+BM^{2}，即MF\perp BM.\because BM\cap AC=M，\therefore MF\perp 平面ABC.$
Subfield: Solid Geometry


Question: $如图①，在\triangle ABC中，AB\perp BC，且AB=2BC，将\triangle ABC沿中位线EF折起，使得AE\perp BE，连接AB，AC，M为AC的中点，连接ME，MF(如图②).$![Image](https://cdntk.53zaixian.com/tq/1693861405859053569/img/compression/M_4149470adf9bbb7d96d7c0792bc5c83c.jpg)
$(1)证明：MF\perp 平面ABC；$

Solution:
$证明：连接EC，设BC=2，则EF=1,AE=BE=2,AF=FC=\sqrt{5}，EC=2\sqrt{2}.\because AE\perp EF,AE\perp BE，EF\cap BE=E，\therefore AE\perp 平面BCFE，\because EC\subset 平面BCFE，\therefore AE\perp EC，连接BM，BF，\because AC=\sqrt{AE^2+EC^2}=\sqrt{4+8}=2\sqrt{3}，BC=2,AB=\sqrt{AE^2+BE^2}=2\sqrt{2},\therefore AC^{2}=BC^{2}+AB^{2},\therefore BC\perp AB,\therefore BM=\frac{1}{2}AC=\sqrt{3}，\because AF=FC,\therefore FM\perp AC，\therefore MF=\sqrt{CF^2-CM^2}=\sqrt{5-3}=\sqrt{2},又\because BF=\sqrt{BE^2+EF^2}=\sqrt{4+1}=\sqrt{5}，\therefore BF^{2}=MF^{2}+BM^{2}，即MF\perp BM.\because BM\cap AC=M，\therefore MF\perp 平面ABC.$

Final answer: 证明题
Subfield: Solid Geometry


Question: $如图①，在\triangle ABC中，AB\perp BC，且AB=2BC，将\triangle ABC沿中位线EF折起，使得AE\perp BE，连接AB，AC，M为AC的中点，连接ME，MF(如图②).$<img_1763>
$(2)求二面角E-MF-C的余弦值.$

Solution:
$由题意知，AE，BE，EF两两垂直.以点E为坐标原点，建立如图所示的空间直角坐标系，$<img_2233>$则E(0,0,0)，C(2,2,0),M(1,1,1),F(0,1,0),\therefore \overrightarrow{MF}=(-1,0,-1),\overrightarrow{EM}=(1,1,1),\overrightarrow{FC}=(2,1,0)，设平面EMF的法向量为n=(x_{1},y_{1},z_{1})，平面MFC的法向量为m=(x_{2},y_{2},z_{2})，则\left\{\begin{matrix}\overrightarrow{MF}\cdot \boldsymbol{n}=0,\\ \overrightarrow{EM}\cdot \boldsymbol{n}=0\end{matrix}\right.\Rightarrow \left\{\begin{matrix}-x_1-z_1=0,\\ x_1+y_1+z_1=0,\end{matrix}\right.令z_{1}=-1，则x_{1}=1,y_{1}=0，则n=(1,0,-1)是平面EMF的一个法向量.同理可得m=(1,-2,-1)为平面MFC的一个法向量，则\\cos  \langle m,n\rangle =\frac{\boldsymbol{m}\cdot \boldsymbol{n}}{|\boldsymbol{m}|\cdot |\boldsymbol{n}|}=\frac{2}{\sqrt{6}\times \sqrt{2}}=\frac{\sqrt{3}}{3}，结合题图可知二面角E-MF-C的平面角为钝角，故二面角E-MF-C的余弦值为-\frac{\sqrt{3}}{3}.$

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


Question: $如图①，在\triangle ABC中，AB\perp BC，且AB=2BC，将\triangle ABC沿中位线EF折起，使得AE\perp BE，连接AB，AC，M为AC的中点，连接ME，MF(如图②).$![Image](https://cdntk.53zaixian.com/tq/1693861405859053569/img/compression/M_4149470adf9bbb7d96d7c0792bc5c83c.jpg)
$(2)求二面角E-MF-C的余弦值.$

Solution:
$由题意知，AE，BE，EF两两垂直.以点E为坐标原点，建立如图所示的空间直角坐标系，$![Image](https://cdntk.53zaixian.com/tq/1693861406261706753/img/compression/M_388d57824e7633022cab63ef04c8062e.jpg)$则E(0,0,0)，C(2,2,0),M(1,1,1),F(0,1,0),\therefore \overrightarrow{MF}=(-1,0,-1),\overrightarrow{EM}=(1,1,1),\overrightarrow{FC}=(2,1,0)，设平面EMF的法向量为n=(x_{1},y_{1},z_{1})，平面MFC的法向量为m=(x_{2},y_{2},z_{2})，则\left\{\begin{matrix}\overrightarrow{MF}\cdot \boldsymbol{n}=0,\\ \overrightarrow{EM}\cdot \boldsymbol{n}=0\end{matrix}\right.\Rightarrow \left\{\begin{matrix}-x_1-z_1=0,\\ x_1+y_1+z_1=0,\end{matrix}\right.令z_{1}=-1，则x_{1}=1,y_{1}=0，则n=(1,0,-1)是平面EMF的一个法向量.同理可得m=(1,-2,-1)为平面MFC的一个法向量，则\\cos  \langle m,n\rangle =\frac{\boldsymbol{m}\cdot \boldsymbol{n}}{|\boldsymbol{m}|\cdot |\boldsymbol{n}|}=\frac{2}{\sqrt{6}\times \sqrt{2}}=\frac{\sqrt{3}}{3}，结合题图可知二面角E-MF-C的平面角为钝角，故二面角E-MF-C的余弦值为-\frac{\sqrt{3}}{3}.$

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


Question: $如图，在三棱柱ABC-A_{1}B_{1}C_{1}中，AB=AC=2，D为BC的中点，平面BB_{1}C_{1}C\perp 平面ABC.$![Image](https://cdntk.53zaixian.com/tq/1693861406593056770/img/compression/M_6280801aeecbd4dc436e6f61f456030a.jpg)
$(1)证明：AD\perp BB_{1}；$

Solution:
$证明：\because AB=AC，且D为BC的中点，\therefore AD\perp BC，又\because 平面BB_{1}C_{1}C\perp 平面ABC，平面BB_{1}C_{1}C\cap 平面ABC=BC，AD\subset 平面ABC，\therefore AD\perp 平面BB_{1}C_{1}C，\because BB_{1}\subset 平面BB_{1}C_{1}C，\therefore AD\perp BB_{1}.$

Final answer: 证明题
Subfield: Solid Geometry


Question: $如图，在三棱柱ABC-A_{1}B_{1}C_{1}中，AB=AC=2，D为BC的中点，平面BB_{1}C_{1}C\perp 平面ABC.$<img_2703>
$(1)证明：AD\perp BB_{1}；$

Solution:
$证明：\because AB=AC，且D为BC的中点，\therefore AD\perp BC，又\because 平面BB_{1}C_{1}C\perp 平面ABC，平面BB_{1}C_{1}C\cap 平面ABC=BC，AD\subset 平面ABC，\therefore AD\perp 平面BB_{1}C_{1}C，\because BB_{1}\subset 平面BB_{1}C_{1}C，\therefore AD\perp BB_{1}.$
Subfield: Solid Geometry


Question: $如图，在三棱柱ABC-A_{1}B_{1}C_{1}中，AB=AC=2，D为BC的中点，平面BB_{1}C_{1}C\perp 平面ABC.$![Image](https://cdntk.53zaixian.com/tq/1693861406593056770/img/compression/M_6280801aeecbd4dc436e6f61f456030a.jpg)
$(2)已知四边形BB_{1}C_{1}C是边长为2的菱形，且\angle B_{1}BC=60^\circ ，问在线段CC_{1}上是否存在一点E，使得平面EAD与平面EAC的夹角的余弦值为\frac{\sqrt{15}}{5}？若存在，求出CE的长度，若不存在，请说明理由.$

Solution:
$假设存在点E满足题意.连接B_{1}D，B_{1}C，\because 四边形BB_{1}C_{1}C是边长为2的菱形，且\angle B_{1}BC=60^\circ ，\therefore \triangle B_{1}BC为等边三角形.\because D为BC的中点，\therefore B_{1}D\perp BC.\because 平面BB_{1}C_{1}C\perp 平面ABC，平面BB_{1}C_{1}C\cap 平面ABC=BC，B_{1}D\subset 平面BB_{1}C_{1}C，\therefore B_{1}D\perp 平面ABC，\because AD\subset 平面ABC，\therefore B_{1}D\perp AD，\therefore B_{1}D，AD，BC两两垂直.以D为原点，BC，DA，DB_{1}所在直线分别为x轴、y轴、z轴，1为单位长度，建立空间直角坐标系，如图.$![Image](https://cdntk.53zaixian.com/tq/1693861406916018178/img/compression/M_c50cacfcc5c149806374e0978ac03d2b.jpg)$则D(0，0,0)，A(0,\sqrt{3},0)，C(1,0,0)，C_{1}(2,0,\sqrt{3})，\therefore \overrightarrow{CC_1}=(1,0,\sqrt{3})，\overrightarrow{AC}=(1,-\sqrt{3},0)，\overrightarrow{DA}=(0，\sqrt{3}，0).设\overrightarrow{CE}=\lambda \overrightarrow{CC_1}(0\leq \lambda \leq 1)，则\overrightarrow{AE}=\overrightarrow{AC}+\overrightarrow{CE}=(1+\lambda ,-\sqrt{3},\sqrt{3}\lambda ).设平面EAD的法向量为n=(x,y,z)，则\left\{\begin{matrix}{\boldsymbol{n}}\cdot \overrightarrow{AE}=0,\\ {\boldsymbol{n}}\cdot \overrightarrow{DA}=0\end{matrix}\right.\Rightarrow \left\{\begin{matrix}(1+\lambda )x-\sqrt{3}y+\sqrt{3}\lambda z=0,\\ \sqrt{3}y=0,\end{matrix}\right.令z=1+\lambda ，则x=-\sqrt{3}\lambda ，y=0,则n=(-\sqrt{3}\lambda ,0,1+\lambda )为平面EAD的一个法向量.设平面EAC的法向量为m=(x',y',z')，则\left\{\begin{matrix}{\boldsymbol{m}}\cdot \overrightarrow{AE}=0,\\ {\boldsymbol{m}}\cdot \overrightarrow{AC}=0\end{matrix}\right.\Rightarrow \left\{\begin{matrix}(1+\lambda )x'-\sqrt{3}y'+\sqrt{3}\lambda z'=0,\\ x'-\sqrt{3}y'=0,\end{matrix}\right.令x'=\sqrt{3},则y'=1，z'=-1，则m=(\sqrt{3},1,-1)为平面EAC的一个法向量.设平面EAD与平面EAC的夹角为\theta ，则\\cos   \theta =\frac{|{\boldsymbol{m}}\cdot {\boldsymbol{n}}|}{|{\boldsymbol{m}}||{\boldsymbol{n}}|}=\frac{|-3\lambda -1-\lambda |}{\sqrt{5}\times \sqrt{3\lambda ^2+{(1+\lambda )}^2}}=\frac{\sqrt{15}}{5}.解得\lambda =\frac{1}{2}(\lambda =-1舍去)，故点E为CC_{1}的中点，所以CE=1.所以在线段CC_{1}上存在点E满足题意，且CE=1.$

Final answer: $\frac{\sqrt{15}}{5},1$
Subfield: Solid Geometry


Question: $如图，在三棱柱ABC-A_{1}B_{1}C_{1}中，AB=AC=2，D为BC的中点，平面BB_{1}C_{1}C\perp 平面ABC.$<img_2703>
$(2)已知四边形BB_{1}C_{1}C是边长为2的菱形，且\angle B_{1}BC=60^\circ ，问在线段CC_{1}上是否存在一点E，使得平面EAD与平面EAC的夹角的余弦值为\frac{\sqrt{15}}{5}？若存在，求出CE的长度，若不存在，请说明理由.$

Solution:
$假设存在点E满足题意.连接B_{1}D，B_{1}C，\because 四边形BB_{1}C_{1}C是边长为2的菱形，且\angle B_{1}BC=60^\circ ，\therefore \triangle B_{1}BC为等边三角形.\because D为BC的中点，\therefore B_{1}D\perp BC.\because 平面BB_{1}C_{1}C\perp 平面ABC，平面BB_{1}C_{1}C\cap 平面ABC=BC，B_{1}D\subset 平面BB_{1}C_{1}C，\therefore B_{1}D\perp 平面ABC，\because AD\subset 平面ABC，\therefore B_{1}D\perp AD，\therefore B_{1}D，AD，BC两两垂直.以D为原点，BC，DA，DB_{1}所在直线分别为x轴、y轴、z轴，1为单位长度，建立空间直角坐标系，如图.$<img_1919>$则D(0，0,0)，A(0,\sqrt{3},0)，C(1,0,0)，C_{1}(2,0,\sqrt{3})，\therefore \overrightarrow{CC_1}=(1,0,\sqrt{3})，\overrightarrow{AC}=(1,-\sqrt{3},0)，\overrightarrow{DA}=(0，\sqrt{3}，0).设\overrightarrow{CE}=\lambda \overrightarrow{CC_1}(0\leq \lambda \leq 1)，则\overrightarrow{AE}=\overrightarrow{AC}+\overrightarrow{CE}=(1+\lambda ,-\sqrt{3},\sqrt{3}\lambda ).设平面EAD的法向量为n=(x,y,z)，则\left\{\begin{matrix}{\boldsymbol{n}}\cdot \overrightarrow{AE}=0,\\ {\boldsymbol{n}}\cdot \overrightarrow{DA}=0\end{matrix}\right.\Rightarrow \left\{\begin{matrix}(1+\lambda )x-\sqrt{3}y+\sqrt{3}\lambda z=0,\\ \sqrt{3}y=0,\end{matrix}\right.令z=1+\lambda ，则x=-\sqrt{3}\lambda ，y=0,则n=(-\sqrt{3}\lambda ,0,1+\lambda )为平面EAD的一个法向量.设平面EAC的法向量为m=(x',y',z')，则\left\{\begin{matrix}{\boldsymbol{m}}\cdot \overrightarrow{AE}=0,\\ {\boldsymbol{m}}\cdot \overrightarrow{AC}=0\end{matrix}\right.\Rightarrow \left\{\begin{matrix}(1+\lambda )x'-\sqrt{3}y'+\sqrt{3}\lambda z'=0,\\ x'-\sqrt{3}y'=0,\end{matrix}\right.令x'=\sqrt{3},则y'=1，z'=-1，则m=(\sqrt{3},1,-1)为平面EAC的一个法向量.设平面EAD与平面EAC的夹角为\theta ，则\\cos   \theta =\frac{|{\boldsymbol{m}}\cdot {\boldsymbol{n}}|}{|{\boldsymbol{m}}||{\boldsymbol{n}}|}=\frac{|-3\lambda -1-\lambda |}{\sqrt{5}\times \sqrt{3\lambda ^2+{(1+\lambda )}^2}}=\frac{\sqrt{15}}{5}.解得\lambda =\frac{1}{2}(\lambda =-1舍去)，故点E为CC_{1}的中点，所以CE=1.所以在线段CC_{1}上存在点E满足题意，且CE=1.$

Final answer: $\frac{\sqrt{15}}{5},1$
Subfield: Solid Geometry


Question: $如图，在四棱锥P-ABCD中，侧面PAD是边长为2的正三角形且与底面垂直，底面ABCD是菱形，且\angle ABC=60^\circ ，M为棱PC上的动点，且\frac{PM}{PC}=\lambda (\lambda \in [0,1]).$![Image](https://cdntk.53zaixian.com/tq/1693861407377391619/img/compression/M_958f18033f6decedc9ccbf28603a3da6.jpg)
$(1)求证：\triangle PBC为直角三角形；$

Solution:
$证明：取AD的中点O,连接OP，OC，AC，依题意可知\triangle PAD，\triangle ACD均为正三角形，所以OC\perp AD，OP\perp AD,又因为OC\cap OP=O，OC\subset 平面POC，OP\subset 平面POC，所以AD\perp 平面POC，又因为PC\subset 平面POC，所以AD\perp PC,因为BC\parallel AD,所以BC\perp PC，即\angle PCB=90^\circ ,从而\triangle PBC为直角三角形.$

Final answer: 证明题
Subfield: Solid Geometry


Question: $如图，在四棱锥P-ABCD中，侧面PAD是边长为2的正三角形且与底面垂直，底面ABCD是菱形，且\angle ABC=60^\circ ，M为棱PC上的动点，且\frac{PM}{PC}=\lambda (\lambda \in [0,1]).$<img_2157>
$(1)求证：\triangle PBC为直角三角形；$

Solution:
$证明：取AD的中点O,连接OP，OC，AC，依题意可知\triangle PAD，\triangle ACD均为正三角形，所以OC\perp AD，OP\perp AD,又因为OC\cap OP=O，OC\subset 平面POC，OP\subset 平面POC，所以AD\perp 平面POC，又因为PC\subset 平面POC，所以AD\perp PC,因为BC\parallel AD,所以BC\perp PC，即\angle PCB=90^\circ ,从而\triangle PBC为直角三角形.$