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


Question: $如图，等腰梯形ABCD中，AB\parallel CD，AD=AB=BC=1,CD=2,E为CD中点，以AE为折痕把\triangle ADE折起，使点D到达点P的位置(P∉平面ABCE).$<img_1534>
$(2)当平面PAE\perp 平面ABCE时，求二面角A-PE-C的余弦值.$

Solution:
$若平面PAE\perp 平面ABCE,即PO\perp 平面ABCE,以O为原点，OE，OB，OP所在直线为x,y,z轴，建立空间直角坐标系O-xyz,则P\left(0,0,\frac{\sqrt{3}}{2}\right)，E\left(\frac{1}{2},0,0\right)，C\left(1,\frac{\sqrt{3}}{2},0\right)，\therefore \overrightarrow{PE}=\left(\frac{1}{2},0,-\frac{\sqrt{3}}{2}\right)，\overrightarrow{EC}=\left(\frac{1}{2},\frac{\sqrt{3}}{2},0\right)，$<img_2476>$设平面PCE的一个法向量为n_{1}=(x,y,z),则\left\{\begin{matrix}\boldsymbol{n}_1\cdot \overrightarrow{PE}=0,\\ \boldsymbol{n}_1\cdot \overrightarrow{EC}=0,\end{matrix}\right.即\left\{\begin{matrix}\frac{1}{2}x-\frac{\sqrt{3}}{2}z=0,\\ \frac{1}{2}x+\frac{\sqrt{3}}{2}y=0,\end{matrix}\right.令x=\sqrt{3},得n_{1}=(\sqrt{3},-1,1),又OB\perp 平面PAE，\therefore \overrightarrow{OB}=n_{2}=\left(0,\frac{\sqrt{3}}{2},0\right)为平面PAE的一个法向量，设平面PAE与平面CPE夹角为\alpha ,则\\cos   \alpha =|\\cos  \langle n_{1},n_{2}\rangle |=\frac{\boldsymbol{n}_1\cdot \boldsymbol{n}_2}{|\boldsymbol{n}_1||\boldsymbol{n}_2|}=\frac{1}{\sqrt{5}}=\frac{\sqrt{5}}{5}，由图观察知，二面角A-PE-C是钝二面角，故二面角A-PE-C的余弦值为-\frac{\sqrt{5}}{5}.$

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


Question: $如图，等腰梯形ABCD中，AB\parallel CD，AD=AB=BC=1,CD=2,E为CD中点，以AE为折痕把\triangle ADE折起，使点D到达点P的位置(P∉平面ABCE).$![Image](https://cdntk.53zaixian.com/tq/1715283928091447298/img/compression/M_7c3c33302f258e56b6079208c0dcbfff.jpg)
$(2)当平面PAE\perp 平面ABCE时，求二面角A-PE-C的余弦值.$

Solution:
$若平面PAE\perp 平面ABCE,即PO\perp 平面ABCE,以O为原点，OE，OB，OP所在直线为x,y,z轴，建立空间直角坐标系O-xyz,则P\left(0,0,\frac{\sqrt{3}}{2}\right)，E\left(\frac{1}{2},0,0\right)，C\left(1,\frac{\sqrt{3}}{2},0\right)，\therefore \overrightarrow{PE}=\left(\frac{1}{2},0,-\frac{\sqrt{3}}{2}\right)，\overrightarrow{EC}=\left(\frac{1}{2},\frac{\sqrt{3}}{2},0\right)，$![Image](https://cdntk.53zaixian.com/tq/1715283928460546049/img/compression/M_af5087f738c9312cec89a50d54c71f51.jpg)$设平面PCE的一个法向量为n_{1}=(x,y,z),则\left\{\begin{matrix}\boldsymbol{n}_1\cdot \overrightarrow{PE}=0,\\ \boldsymbol{n}_1\cdot \overrightarrow{EC}=0,\end{matrix}\right.即\left\{\begin{matrix}\frac{1}{2}x-\frac{\sqrt{3}}{2}z=0,\\ \frac{1}{2}x+\frac{\sqrt{3}}{2}y=0,\end{matrix}\right.令x=\sqrt{3},得n_{1}=(\sqrt{3},-1,1),又OB\perp 平面PAE，\therefore \overrightarrow{OB}=n_{2}=\left(0,\frac{\sqrt{3}}{2},0\right)为平面PAE的一个法向量，设平面PAE与平面CPE夹角为\alpha ,则\\cos   \alpha =|\\cos  \langle n_{1},n_{2}\rangle |=\frac{\boldsymbol{n}_1\cdot \boldsymbol{n}_2}{|\boldsymbol{n}_1||\boldsymbol{n}_2|}=\frac{1}{\sqrt{5}}=\frac{\sqrt{5}}{5}，由图观察知，二面角A-PE-C是钝二面角，故二面角A-PE-C的余弦值为-\frac{\sqrt{5}}{5}.$

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


Question: $如图所示的五边形SBADC中四边形ABCD是矩形，BC=2AB,SB=SC,沿BC折叠成四棱锥S-ABCD，点M是BC的中点，SM=2.$<img_2677>
$(1)在四棱锥S-ABCD中，可以满足条件：①SA=\sqrt{6};②\\cos  \angle SBM=\frac{\sqrt{5}}{5}；③\\sin  \angle SAM=\frac{\sqrt{6}}{3}，请从中任选两个作为补充条件，证明:侧面SBC\perp 底面ABCD;(注:若选择不同的组合分别解答,则按第一个解答计分)$

Solution:
$方案一:选条件①②.因为在四棱锥S-ABCD中SB=SC，点M是BC的中点，SM=2，所以SM\perp BC，又因为在Rt\triangle SBM中，\\cos  \angle SBM=\frac{\sqrt{5}}{5}，所以BM=1.又因为四边形ABCD是矩形，BC=2AB， 所以BM=AB=1，AM=\sqrt{2}，由SA=\sqrt{6}，AM=\sqrt{2}，SM=2可得SA^{2}=AM^{2}+SM^{2},所以SM\perp AM.则由\left\{\begin{matrix}SM\bot BC,\\ SM\bot AM,\\ AM\bigcap BC=M\end{matrix}\right.可得SM\perp 底面ABCD,又因为SM\subset 侧面SBC,所以侧面SBC\perp 底面ABCD.方案二:选条件①③.因为在四棱锥S-ABCD中SB=SC，点M是BC的中点，SM=2，所以SM\perp BC.又因为在\triangle SAM中，SA=\sqrt{6},\\sin  \angle SAM=\frac{\sqrt{6}}{3}，SM=2,所以由正弦定理得\frac{SA}{\\\sin   \angle SMA}=\frac{SM}{\\\sin   \angle SAM}， 即\frac{\sqrt{6}}{\\\sin   \angle SMA}=\frac{2}{\frac{\sqrt{6}}{3}}，所以\\sin  \angle SMA=1，即\angle SMA=\frac{\pi }{2},所以SM\perp MA,则由\left\{\begin{matrix}SM\bot BC,\\ SM\bot AM,\\ AM\bigcap BC=M\end{matrix}\right.可得SM\perp 底面ABCD，又因为SM\subset 侧面SBC，所以侧面SBC\perp 底面ABCD.方案三:选条件②③.因为在四棱锥S-ABCD中SB=SC,点M是BC的中点，SM=2,所以SM\perp BC.又因为在Rt\triangle SBM中，\\cos  \angle SBM=\frac{\sqrt{5}}{5},所以BM=1.又因为四边形ABCD是矩形，BC=2AB，所以BM=AB=1, AM=\sqrt{2},又因为在\triangle SAM中，\\sin  \angle SAM=\frac{\sqrt{6}}{3},则\\cos  \angle SAM=\frac{\sqrt{3}}{3},设SA=x，由余弦定理SM^{2}=SA^{2}+AM^{2}-2SA\cdot AM\\cos  \angle SAM，所以有3x^{2}-2\sqrt{6}x-6=0,解得x_{1}=\sqrt{6}，x_{2}=-\frac{\sqrt{6}}{3}(舍)，所以SA=\sqrt{6}.由SA=\sqrt{6},AM=\sqrt{2},SM=2可得SA^{2}=AM^{2}+SM^{2},所以SM\perp AM.则由\left\{\begin{matrix}SM\bot BC,\\ SM\bot AM,\\ AM\bigcap BC=M\end{matrix}\right.可得SM\perp 底面ABCD,又因为SM\subset 侧面SBC,所以侧面SBC\perp 底面ABCD.$
Subfield: Solid Geometry


Question: $如图所示的五边形SBADC中四边形ABCD是矩形，BC=2AB,SB=SC,沿BC折叠成四棱锥S-ABCD，点M是BC的中点，SM=2.$![Image](https://cdntk.53zaixian.com/tq/1715283929051942914/img/compression/M_7b27a2f224d22d86fd55b694af38b247.jpg)
$(1)在四棱锥S-ABCD中，可以满足条件：①SA=\sqrt{6};②\\cos  \angle SBM=\frac{\sqrt{5}}{5}；③\\sin  \angle SAM=\frac{\sqrt{6}}{3}，请从中任选两个作为补充条件，证明:侧面SBC\perp 底面ABCD;(注:若选择不同的组合分别解答,则按第一个解答计分)$

Solution:
$方案一:选条件①②.因为在四棱锥S-ABCD中SB=SC，点M是BC的中点，SM=2，所以SM\perp BC，又因为在Rt\triangle SBM中，\\cos  \angle SBM=\frac{\sqrt{5}}{5}，所以BM=1.又因为四边形ABCD是矩形，BC=2AB， 所以BM=AB=1，AM=\sqrt{2}，由SA=\sqrt{6}，AM=\sqrt{2}，SM=2可得SA^{2}=AM^{2}+SM^{2},所以SM\perp AM.则由\left\{\begin{matrix}SM\bot BC,\\ SM\bot AM,\\ AM\bigcap BC=M\end{matrix}\right.可得SM\perp 底面ABCD,又因为SM\subset 侧面SBC,所以侧面SBC\perp 底面ABCD.方案二:选条件①③.因为在四棱锥S-ABCD中SB=SC，点M是BC的中点，SM=2，所以SM\perp BC.又因为在\triangle SAM中，SA=\sqrt{6},\\sin  \angle SAM=\frac{\sqrt{6}}{3}，SM=2,所以由正弦定理得\frac{SA}{\\\sin   \angle SMA}=\frac{SM}{\\\sin   \angle SAM}， 即\frac{\sqrt{6}}{\\\sin   \angle SMA}=\frac{2}{\frac{\sqrt{6}}{3}}，所以\\sin  \angle SMA=1，即\angle SMA=\frac{\pi }{2},所以SM\perp MA,则由\left\{\begin{matrix}SM\bot BC,\\ SM\bot AM,\\ AM\bigcap BC=M\end{matrix}\right.可得SM\perp 底面ABCD，又因为SM\subset 侧面SBC，所以侧面SBC\perp 底面ABCD.方案三:选条件②③.因为在四棱锥S-ABCD中SB=SC,点M是BC的中点，SM=2,所以SM\perp BC.又因为在Rt\triangle SBM中，\\cos  \angle SBM=\frac{\sqrt{5}}{5},所以BM=1.又因为四边形ABCD是矩形，BC=2AB，所以BM=AB=1, AM=\sqrt{2},又因为在\triangle SAM中，\\sin  \angle SAM=\frac{\sqrt{6}}{3},则\\cos  \angle SAM=\frac{\sqrt{3}}{3},设SA=x，由余弦定理SM^{2}=SA^{2}+AM^{2}-2SA\cdot AM\\cos  \angle SAM，所以有3x^{2}-2\sqrt{6}x-6=0,解得x_{1}=\sqrt{6}，x_{2}=-\frac{\sqrt{6}}{3}(舍)，所以SA=\sqrt{6}.由SA=\sqrt{6},AM=\sqrt{2},SM=2可得SA^{2}=AM^{2}+SM^{2},所以SM\perp AM.则由\left\{\begin{matrix}SM\bot BC,\\ SM\bot AM,\\ AM\bigcap BC=M\end{matrix}\right.可得SM\perp 底面ABCD,又因为SM\subset 侧面SBC,所以侧面SBC\perp 底面ABCD.$

Final answer: 证明题
Subfield: Solid Geometry


Question: $如图所示的五边形SBADC中四边形ABCD是矩形，BC=2AB,SB=SC,沿BC折叠成四棱锥S-ABCD，点M是BC的中点，SM=2.$![Image](https://cdntk.53zaixian.com/tq/1715283929051942914/img/compression/M_7b27a2f224d22d86fd55b694af38b247.jpg)
$(2)在(1)的条件下求直线SC与平面SAD所成角的正弦值.$

Solution:
$在(1)条件下知SM\perp 底面ABCD，且MD\perp AM，如图，以M为坐标原点，以MA所在直线为x轴，以MD所在直线为y轴，以MS所在直线为z轴建立空间直角坐标系，则S(0,0，2)，A(\sqrt{2},0,0)，D(0,\sqrt{2},0)， C\left(-\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2},0\right),$![Image](https://cdntk.53zaixian.com/tq/1715283929454596098/img/compression/M_96394cfc063d50f1115a40e987131d66.jpg)$设平面SAD的法向量为n=(x,y,z),则n\perp \overrightarrow{SD}，n\perp \overrightarrow{SA},故\left\{\begin{matrix}\sqrt{2}y-2z=0,\\ \sqrt{2}x-2z=0,\end{matrix}\right.令x=\sqrt{2}，得n=(\sqrt{2}，\sqrt{2}，1),而\overrightarrow{SC}=\left(-\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2},-2\right).设直线SC与平面SAD所成角为\theta ，则\\sin   \theta =\frac{|\boldsymbol{n}\cdot \overrightarrow{SC}|}{|\boldsymbol{n}|\cdot |\overrightarrow{SC}|}=\frac{2}{5}，所以直线SC与平面SAD所成角的正弦值为\frac{2}{5}.$

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


Question: $如图所示的五边形SBADC中四边形ABCD是矩形，BC=2AB,SB=SC,沿BC折叠成四棱锥S-ABCD，点M是BC的中点，SM=2.$<img_2677>
$(2)在(1)的条件下求直线SC与平面SAD所成角的正弦值.$

Solution:
$在(1)条件下知SM\perp 底面ABCD，且MD\perp AM，如图，以M为坐标原点，以MA所在直线为x轴，以MD所在直线为y轴，以MS所在直线为z轴建立空间直角坐标系，则S(0,0，2)，A(\sqrt{2},0,0)，D(0,\sqrt{2},0)， C\left(-\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2},0\right),$<img_1787>$设平面SAD的法向量为n=(x,y,z),则n\perp \overrightarrow{SD}，n\perp \overrightarrow{SA},故\left\{\begin{matrix}\sqrt{2}y-2z=0,\\ \sqrt{2}x-2z=0,\end{matrix}\right.令x=\sqrt{2}，得n=(\sqrt{2}，\sqrt{2}，1),而\overrightarrow{SC}=\left(-\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2},-2\right).设直线SC与平面SAD所成角为\theta ，则\\sin   \theta =\frac{|\boldsymbol{n}\cdot \overrightarrow{SC}|}{|\boldsymbol{n}|\cdot |\overrightarrow{SC}|}=\frac{2}{5}，所以直线SC与平面SAD所成角的正弦值为\frac{2}{5}.$

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


Question: $圆柱O_{1}O_{2}中,四边形DEFG为过轴O_{1}O_{2}的截面，DG= 4\sqrt{2}, DE=16,\triangle ABC为底面圆O的内接正三角形，AB\parallel DE.$<img_1560>
$(1)证明:CO_{2}\perp 平面ABFG；$

Solution:
$证明:连接CO_{1}并延长交AB于H ,连接O_{2}H，\because \triangle ABC为底面圆O_{1}的内接正三角形，\therefore CH\perp AB,\because AB\parallel DE，\therefore CH\perp DE,\because 四边形DEFG为圆柱O_{1}O_{2}的轴截面,\therefore O_{1}O_{2}\perp 圆面O_{1},DE\subset 圆面O_{1}，\therefore O_{1}O_{2}\perp DE，\because O_{1}O_{2}\cap CH=O_{1},\therefore DE\perp 平面CHO_{2},\because DE\parallel FG,\therefore FG\perp 平面CHO_{2},\therefore FG\perp CO_{2},\because DG=4\sqrt{2},DE=16,\therefore O_{1}C=8,O_{1}H=4,CH=12,O_{1}O_{2}=4\sqrt{2}，\therefore O_{2}C^{2}=O_{1}C^{2}+O_{1}O^2_2=96,O_{2}H^{2}=O_{1}H^{2}+O_{1}O^2_2=48,\therefore O_{2}C^{2}+O_{2}H^{2}=CH^{2},\therefore CO_{2}\perp O_{2}H,\because HO_{2}\cap FG=O_{2},\therefore CO_{2}\perp 平面ABFG.$<img_1669>
Subfield: Solid Geometry


Question: $圆柱O_{1}O_{2}中,四边形DEFG为过轴O_{1}O_{2}的截面，DG= 4\sqrt{2}, DE=16,\triangle ABC为底面圆O的内接正三角形，AB\parallel DE.$![Image](https://cdntk.53zaixian.com/tq/1715283930092130306/img/compression/M_e9a3d3eb9352013fcd1c6d2d10d69488.jpg)
$(1)证明:CO_{2}\perp 平面ABFG；$

Solution:
$证明:连接CO_{1}并延长交AB于H ,连接O_{2}H，\because \triangle ABC为底面圆O_{1}的内接正三角形，\therefore CH\perp AB,\because AB\parallel DE，\therefore CH\perp DE,\because 四边形DEFG为圆柱O_{1}O_{2}的轴截面,\therefore O_{1}O_{2}\perp 圆面O_{1},DE\subset 圆面O_{1}，\therefore O_{1}O_{2}\perp DE，\because O_{1}O_{2}\cap CH=O_{1},\therefore DE\perp 平面CHO_{2},\because DE\parallel FG,\therefore FG\perp 平面CHO_{2},\therefore FG\perp CO_{2},\because DG=4\sqrt{2},DE=16,\therefore O_{1}C=8,O_{1}H=4,CH=12,O_{1}O_{2}=4\sqrt{2}，\therefore O_{2}C^{2}=O_{1}C^{2}+O_{1}O^2_2=96,O_{2}H^{2}=O_{1}H^{2}+O_{1}O^2_2=48,\therefore O_{2}C^{2}+O_{2}H^{2}=CH^{2},\therefore CO_{2}\perp O_{2}H,\because HO_{2}\cap FG=O_{2},\therefore CO_{2}\perp 平面ABFG.$![Image](https://cdntk.53zaixian.com/tq/1715283929744003075/img/compression/M_89357bcbf8193761376c0280368bbe88.jpg)

Final answer: 证明题
Subfield: Solid Geometry


Question: $圆柱O_{1}O_{2}中,四边形DEFG为过轴O_{1}O_{2}的截面，DG= 4\sqrt{2}, DE=16,\triangle ABC为底面圆O的内接正三角形，AB\parallel DE.$![Image](https://cdntk.53zaixian.com/tq/1715283930092130306/img/compression/M_e9a3d3eb9352013fcd1c6d2d10d69488.jpg)
$(2)求平面CFD与平面ABFG所成角的正弦值.$

Solution:
$由(1)知O_{1}O_{2},CH,DE两两垂直,如图建立空间直角坐标系O_{1}-xyz,则C(0,8,0), F(8,0,4\sqrt{2}), D(-8,0,0),\overrightarrow{CF}=(8,-8,4\sqrt{2}),\overrightarrow{CD}=(-8,-8,0),设平面CFD的法向量为n=(x,y,z),则\left\{\begin{matrix}\boldsymbol{n}\cdot \overrightarrow{CF}=0,\\ \boldsymbol{n}\cdot \overrightarrow{CD}=0,\end{matrix}\right.\left\{\begin{matrix}8x-8y+4\sqrt{2}z=0,\\ -8x-8y=0,\end{matrix}\right.可取n=(-1，1，2\sqrt{2})，由(1)知平面ABFG的法向量可取\overrightarrow{O_2C}=n_{1}=(0,8,-4\sqrt{2}),\therefore \\cos  \langle n，n_{1}\rangle =\frac{\boldsymbol{n}\cdot \boldsymbol{n}_1}{|\boldsymbol{n}|\cdot |\boldsymbol{n}_1|}=-\frac{\sqrt{15}}{15}，\therefore 平面ABFG与平面CFD所成二面角的正弦值为\frac{\sqrt{210}}{15}.$

Final answer: $\frac{\sqrt{210}}{15}$
Subfield: Solid Geometry


Question: $圆柱O_{1}O_{2}中,四边形DEFG为过轴O_{1}O_{2}的截面，DG= 4\sqrt{2}, DE=16,\triangle ABC为底面圆O的内接正三角形，AB\parallel DE.$<img_1560>
$(2)求平面CFD与平面ABFG所成角的正弦值.$

Solution:
$由(1)知O_{1}O_{2},CH,DE两两垂直,如图建立空间直角坐标系O_{1}-xyz,则C(0,8,0), F(8,0,4\sqrt{2}), D(-8,0,0),\overrightarrow{CF}=(8,-8,4\sqrt{2}),\overrightarrow{CD}=(-8,-8,0),设平面CFD的法向量为n=(x,y,z),则\left\{\begin{matrix}\boldsymbol{n}\cdot \overrightarrow{CF}=0,\\ \boldsymbol{n}\cdot \overrightarrow{CD}=0,\end{matrix}\right.\left\{\begin{matrix}8x-8y+4\sqrt{2}z=0,\\ -8x-8y=0,\end{matrix}\right.可取n=(-1，1，2\sqrt{2})，由(1)知平面ABFG的法向量可取\overrightarrow{O_2C}=n_{1}=(0,8,-4\sqrt{2}),\therefore \\cos  \langle n，n_{1}\rangle =\frac{\boldsymbol{n}\cdot \boldsymbol{n}_1}{|\boldsymbol{n}|\cdot |\boldsymbol{n}_1|}=-\frac{\sqrt{15}}{15}，\therefore 平面ABFG与平面CFD所成二面角的正弦值为\frac{\sqrt{210}}{15}.$

Final answer: $\frac{\sqrt{210}}{15}$
Subfield: Solid Geometry


Question: $如图,在四棱锥P-ABCD中,底面ABCD是正方形,\triangle PAD为直角三角形,PB=PC,\angle ABP=\angle DCP,\triangle PBC的面积是\triangle PAD的面积的\sqrt{5}倍.$<img_2745>
$(1)证明:平面PAD\perp 平面ABCD;$

Solution:
$证明:如图,取AD的中点O，BC的中点F.连接OP,OF,PF.\because PB=PC,\angle ABP=\angle DCP,AB=CD,\therefore \triangle PBA\cong \triangle PCD.\therefore PA=PD,又\because \triangle PAD为直角三角形,\therefore AP\perp PD,\triangle PAD为等腰直角三角形，又\because OA=OD,\therefore OP\perp AD.\because PB=PC,BF=CF，\therefore PF\perp BC.又\because \triangle PBC的面积是\triangle PAD面积的\sqrt{5}倍,AD=BC，\therefore PF=\sqrt{5}OP.易知AD=2OP,\therefore OP^{2}+OF^{2}=OP^{2}+AD^{2}=OP^{2}+4OP^{2}=5OP^{2}=PF^{2},\therefore OP\perp OF,\because OP\perp AD，OF\perp AD,\therefore \angle POF为平面PAD和平面ABCD所成二面角的平面角，\therefore 平面PAD\perp 平面ABCD.$<img_2337>
Subfield: Solid Geometry


Question: $如图,在四棱锥P-ABCD中,底面ABCD是正方形,\triangle PAD为直角三角形,PB=PC,\angle ABP=\angle DCP,\triangle PBC的面积是\triangle PAD的面积的\sqrt{5}倍.$![Image](https://cdntk.53zaixian.com/tq/1715283930452840449/img/compression/M_cce8ec3f36990eeaa1903786b58f99d1.jpg)
$(1)证明:平面PAD\perp 平面ABCD;$

Solution:
$证明:如图,取AD的中点O，BC的中点F.连接OP,OF,PF.\because PB=PC,\angle ABP=\angle DCP,AB=CD,\therefore \triangle PBA\cong \triangle PCD.\therefore PA=PD,又\because \triangle PAD为直角三角形,\therefore AP\perp PD,\triangle PAD为等腰直角三角形，又\because OA=OD,\therefore OP\perp AD.\because PB=PC,BF=CF，\therefore PF\perp BC.又\because \triangle PBC的面积是\triangle PAD面积的\sqrt{5}倍,AD=BC，\therefore PF=\sqrt{5}OP.易知AD=2OP,\therefore OP^{2}+OF^{2}=OP^{2}+AD^{2}=OP^{2}+4OP^{2}=5OP^{2}=PF^{2},\therefore OP\perp OF,\because OP\perp AD，OF\perp AD,\therefore \angle POF为平面PAD和平面ABCD所成二面角的平面角，\therefore 平面PAD\perp 平面ABCD.$![Image](https://cdntk.53zaixian.com/tq/1715283931002294273/img/compression/M_2a659c6dd551ae7db408ca703922029c.jpg)

Final answer: 证明题
Subfield: Solid Geometry


Question: $如图,在四棱锥P-ABCD中,底面ABCD是正方形,\triangle PAD为直角三角形,PB=PC,\angle ABP=\angle DCP,\triangle PBC的面积是\triangle PAD的面积的\sqrt{5}倍.$![Image](https://cdntk.53zaixian.com/tq/1715283930452840449/img/compression/M_cce8ec3f36990eeaa1903786b58f99d1.jpg)
$(2)E为PC上的点,四棱锥E-ABCD的体积是四棱锥P-ABCD的体积的一半,求平面PAE与平面DAE的夹角的余弦值.$

Solution:
$由四棱锥E-ABCD的体积是四棱锥P-ABCD体积的一半,可得E为PC的中点.设AB=2，可得OA=OD=OP=1,AB=OF=2,由(1)可知OA,OF,OP两两垂直,以O为坐标原点，以\overrightarrow{OA},\overrightarrow{OF},\overrightarrow{OP}的方向分别为x,y,z轴正方向建立如图所示的空间直角坐标系，则O(0,0,0),A(1,0,0),D(-1,0,0),P(0,0,1),C(-1,2,0),E\left(-\frac{1}{2},1,\frac{1}{2}\right).设平面PAE的法向量为m=(x,y,z),由\overrightarrow{AP}=(-1,0,1)，\overrightarrow{PC}=(-1,2,-1),得\left\{\begin{matrix}\overrightarrow{AP}\cdot \boldsymbol{m}=-x+z=0,\\ \overrightarrow{PC}\cdot \boldsymbol{m}=-x+2y-z=0,\end{matrix}\right.取x=1,可得m=(1,1,1).设平面DAE的法向量为n=(a,b,c),由\overrightarrow{DA}=(2，0，0)，\overrightarrow{DE}=\left(\frac{1}{2},1,\frac{1}{2}\right)，得\left\{\begin{matrix}\overrightarrow{DA}\cdot \boldsymbol{n}=2a=0,\\ \overrightarrow{DE}\cdot \boldsymbol{n}=\frac{1}{2}a+b+\frac{1}{2}c=0,\end{matrix}\right.取b=-1,可得n=(0,-1,2),则\\cos  \langle m，n\rangle =\frac{1}{\sqrt{3}\times \sqrt{5}}=\frac{\sqrt{15}}{15}，故平面PAE与平面DAE夹角的余弦值为\frac{\sqrt{15}}{15}.$

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


Question: $如图,在四棱锥P-ABCD中,底面ABCD是正方形,\triangle PAD为直角三角形,PB=PC,\angle ABP=\angle DCP,\triangle PBC的面积是\triangle PAD的面积的\sqrt{5}倍.$<img_2745>
$(2)E为PC上的点,四棱锥E-ABCD的体积是四棱锥P-ABCD的体积的一半,求平面PAE与平面DAE的夹角的余弦值.$

Solution:
$由四棱锥E-ABCD的体积是四棱锥P-ABCD体积的一半,可得E为PC的中点.设AB=2，可得OA=OD=OP=1,AB=OF=2,由(1)可知OA,OF,OP两两垂直,以O为坐标原点，以\overrightarrow{OA},\overrightarrow{OF},\overrightarrow{OP}的方向分别为x,y,z轴正方向建立如图所示的空间直角坐标系，则O(0,0,0),A(1,0,0),D(-1,0,0),P(0,0,1),C(-1,2,0),E\left(-\frac{1}{2},1,\frac{1}{2}\right).设平面PAE的法向量为m=(x,y,z),由\overrightarrow{AP}=(-1,0,1)，\overrightarrow{PC}=(-1,2,-1),得\left\{\begin{matrix}\overrightarrow{AP}\cdot \boldsymbol{m}=-x+z=0,\\ \overrightarrow{PC}\cdot \boldsymbol{m}=-x+2y-z=0,\end{matrix}\right.取x=1,可得m=(1,1,1).设平面DAE的法向量为n=(a,b,c),由\overrightarrow{DA}=(2，0，0)，\overrightarrow{DE}=\left(\frac{1}{2},1,\frac{1}{2}\right)，得\left\{\begin{matrix}\overrightarrow{DA}\cdot \boldsymbol{n}=2a=0,\\ \overrightarrow{DE}\cdot \boldsymbol{n}=\frac{1}{2}a+b+\frac{1}{2}c=0,\end{matrix}\right.取b=-1,可得n=(0,-1,2),则\\cos  \langle m，n\rangle =\frac{1}{\sqrt{3}\times \sqrt{5}}=\frac{\sqrt{15}}{15}，故平面PAE与平面DAE夹角的余弦值为\frac{\sqrt{15}}{15}.$

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


Question: $如图，正四棱锥P-ABCD的底面边长和高均为2,E,F分别为PD,PB的中点.$<img_2171>
$(1)若点M是线段PC上的点,且PM=\frac{1}{3}PC,判断点M是否在平面AEF内,并证明你的结论;$

Solution:
$证明：连接AC、BD交于O，以O为坐标原点，OA、OB、OP所在直线为x，y，z轴建立空间直角坐标系.则A(\sqrt{2},0,0),B(0,\sqrt{2},0),P(0,0,2),C(-\sqrt{2},0,0),D(0,-\sqrt{2},0)，E\left(0,-\frac{\sqrt{2}}{2},1\right),F\left(0,\frac{\sqrt{2}}{2},1\right),\overrightarrow{AE}=\left(-\sqrt{2},-\frac{\sqrt{2}}{2},1\right),\overrightarrow{AF}=\left(-\sqrt{2},\frac{\sqrt{2}}{2},1\right),又\overrightarrow{PM}=\frac{1}{3}\overrightarrow{PC}，得\overrightarrow{AM}=\overrightarrow{AP}+\frac{1}{3}\overrightarrow{PC}=\left(-\frac{4}{3}\sqrt{2},0,\frac{4}{3}\right),\overrightarrow{AE}+\overrightarrow{AF}=(-2\sqrt{2},0,2),所以\overrightarrow{AM}=\frac{2}{3}\overrightarrow{AE}+\frac{2}{3}\overrightarrow{AF}，所以A、M、E、F四点共面，即点M在平面AEF内.$
Subfield: Solid Geometry


Question: $如图，正四棱锥P-ABCD的底面边长和高均为2,E,F分别为PD,PB的中点.$![Image](https://cdntk.53zaixian.com/tq/1715283958279782402/img/compression/M_639cce88f56232fd6d592334e5882d02.jpg)
$(1)若点M是线段PC上的点,且PM=\frac{1}{3}PC,判断点M是否在平面AEF内,并证明你的结论;$

Solution:
$证明：连接AC、BD交于O，以O为坐标原点，OA、OB、OP所在直线为x，y，z轴建立空间直角坐标系.则A(\sqrt{2},0,0),B(0,\sqrt{2},0),P(0,0,2),C(-\sqrt{2},0,0),D(0,-\sqrt{2},0)，E\left(0,-\frac{\sqrt{2}}{2},1\right),F\left(0,\frac{\sqrt{2}}{2},1\right),\overrightarrow{AE}=\left(-\sqrt{2},-\frac{\sqrt{2}}{2},1\right),\overrightarrow{AF}=\left(-\sqrt{2},\frac{\sqrt{2}}{2},1\right),又\overrightarrow{PM}=\frac{1}{3}\overrightarrow{PC}，得\overrightarrow{AM}=\overrightarrow{AP}+\frac{1}{3}\overrightarrow{PC}=\left(-\frac{4}{3}\sqrt{2},0,\frac{4}{3}\right),\overrightarrow{AE}+\overrightarrow{AF}=(-2\sqrt{2},0,2),所以\overrightarrow{AM}=\frac{2}{3}\overrightarrow{AE}+\frac{2}{3}\overrightarrow{AF}，所以A、M、E、F四点共面，即点M在平面AEF内.$

Final answer: 证明题
Subfield: Solid Geometry


Question: $如图，正四棱锥P-ABCD的底面边长和高均为2,E,F分别为PD,PB的中点.$![Image](https://cdntk.53zaixian.com/tq/1715283958279782402/img/compression/M_639cce88f56232fd6d592334e5882d02.jpg)
$(2)求直线PB与平面AEF所成角的正弦值.$

Solution:
$\overrightarrow{PB}=(0,\sqrt{2},-2)，设平面AEF的法向量n=(x,y,z)，由\left\{\begin{matrix}\boldsymbol{n}\cdot \overrightarrow{AE}=0,\\ \boldsymbol{n}\cdot \overrightarrow{AF}=0\end{matrix}\right.得\left\{\begin{matrix}-\sqrt{2}x-\frac{\sqrt{2}}{2}y+z=0,\\ -\sqrt{2}x+\frac{\sqrt{2}}{2}y+z=0,\end{matrix}\right.取x=1,则n=(1,0,\sqrt{2})，所以\\cos  \langle \overrightarrow{PB},n\rangle =\frac{-2\sqrt{2}}{\sqrt{6}\cdot \sqrt{3}}=-\frac{2}{3}，所以直线PB与平面AEF所成角的正弦值为\frac{2}{3}.$

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


Question: $如图，正四棱锥P-ABCD的底面边长和高均为2,E,F分别为PD,PB的中点.$<img_2171>
$(2)求直线PB与平面AEF所成角的正弦值.$

Solution:
$\overrightarrow{PB}=(0,\sqrt{2},-2)，设平面AEF的法向量n=(x,y,z)，由\left\{\begin{matrix}\boldsymbol{n}\cdot \overrightarrow{AE}=0,\\ \boldsymbol{n}\cdot \overrightarrow{AF}=0\end{matrix}\right.得\left\{\begin{matrix}-\sqrt{2}x-\frac{\sqrt{2}}{2}y+z=0,\\ -\sqrt{2}x+\frac{\sqrt{2}}{2}y+z=0,\end{matrix}\right.取x=1,则n=(1,0,\sqrt{2})，所以\\cos  \langle \overrightarrow{PB},n\rangle =\frac{-2\sqrt{2}}{\sqrt{6}\cdot \sqrt{3}}=-\frac{2}{3}，所以直线PB与平面AEF所成角的正弦值为\frac{2}{3}.$

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


Question: $如图，在直棱柱ABCD-A_{1}B_{1}C_{1}D_{1}中，底面ABCD是直角梯形，AB\parallel DC，AB\perp BC，AB=3DC=3，BC=6，点P在平面ADD_{1}A_{1}上，过点P和棱BB_{1}的平面把直棱柱分成体积相等的两部分.$![Image](https://cdntk.53zaixian.com/tq/1715283958682435587/img/compression/M_5dd78c490b1810849d47c89c9d9e96de.jpg)
$(1)求截面与直棱柱的侧面BCC_{1}B_{1}所成角的正切值；$

Solution:
$不妨设点P在棱A_{1}D_{1}上，过P作PQ\parallel DD_{1}，交AD于点Q，连接BQ，B_{1}P，如图所示，则截面为BB_{1}PQ，\because ABCD-A_{1}B_{1}C_{1}D_{1}为直棱柱，$![Image](https://cdntk.53zaixian.com/tq/1715283958967648258/img/compression/M_3d533c2079fef38832704cc881b65087.jpg)$\therefore BB_{1}\perp 平面A_{1}B_{1}C_{1}D_{1}，\therefore BB_{1}\perp PB_{1}，BB_{1}\perp B_{1}C_{1}，\therefore \angle PB_{1}C_{1}为截面与直棱柱的侧面BCC_{1}B_{1}所成角的平面角.过P作PH\perp A_{1}B_{1}，垂足为H，\because A_{1}B_{1}\perp B_{1}C_{1}，\therefore PH\parallel B_{1}C_{1}，\therefore \angle PB_{1}C_{1}=\angle B_{1}PH，由已知可得S_{四边形A_{1}B_{1}C_{1}D_{1}}=2S_{\triangle A_{1}B_{1}P}=12，\therefore PH=4，则A_{1}H=\frac{4}{3}，\therefore HB_{1}=\frac{5}{3}，\therefore \\tan  \angle PB_{1}C_{1}=\\tan  \angle B_{1}PH=\frac{HB_1}{PH}=\frac{5}{12}.$

Final answer: $\frac{5}{12}$
Subfield: Solid Geometry


Question: $如图，在直棱柱ABCD-A_{1}B_{1}C_{1}D_{1}中，底面ABCD是直角梯形，AB\parallel DC，AB\perp BC，AB=3DC=3，BC=6，点P在平面ADD_{1}A_{1}上，过点P和棱BB_{1}的平面把直棱柱分成体积相等的两部分.$<img_2176>
$(1)求截面与直棱柱的侧面BCC_{1}B_{1}所成角的正切值；$

Solution:
$不妨设点P在棱A_{1}D_{1}上，过P作PQ\parallel DD_{1}，交AD于点Q，连接BQ，B_{1}P，如图所示，则截面为BB_{1}PQ，\because ABCD-A_{1}B_{1}C_{1}D_{1}为直棱柱，$<img_2532>$\therefore BB_{1}\perp 平面A_{1}B_{1}C_{1}D_{1}，\therefore BB_{1}\perp PB_{1}，BB_{1}\perp B_{1}C_{1}，\therefore \angle PB_{1}C_{1}为截面与直棱柱的侧面BCC_{1}B_{1}所成角的平面角.过P作PH\perp A_{1}B_{1}，垂足为H，\because A_{1}B_{1}\perp B_{1}C_{1}，\therefore PH\parallel B_{1}C_{1}，\therefore \angle PB_{1}C_{1}=\angle B_{1}PH，由已知可得S_{四边形A_{1}B_{1}C_{1}D_{1}}=2S_{\triangle A_{1}B_{1}P}=12，\therefore PH=4，则A_{1}H=\frac{4}{3}，\therefore HB_{1}=\frac{5}{3}，\therefore \\tan  \angle PB_{1}C_{1}=\\tan  \angle B_{1}PH=\frac{HB_1}{PH}=\frac{5}{12}.$

Final answer: $\frac{5}{12}$
Subfield: Solid Geometry


Question: $如图，在直棱柱ABCD-A_{1}B_{1}C_{1}D_{1}中，底面ABCD是直角梯形，AB\parallel DC，AB\perp BC，AB=3DC=3，BC=6，点P在平面ADD_{1}A_{1}上，过点P和棱BB_{1}的平面把直棱柱分成体积相等的两部分.$<img_2176>
$(2)求棱DD_{1}到截面的距离.$

Solution:
$\because BB_{1}\perp 平面A_{1}B_{1}C_{1}D_{1}，BB_{1}\subset 平面BB_{1}PQ，\therefore 平面BB_{1}PQ\perp 平面A_{1}B_{1}C_{1}D_{1}，交线为B_{1}P.过D_{1}作D_{1}T\perp 直线B_{1}P，垂足为T，\therefore D_{1}T\perp 平面BB_{1}PQ，则D_{1}T的长度即为棱DD_{1}到截面的距离.连接B_{1}D_{1}，S_{\triangle B_{1}D_{1}P}=S_{四边形PB_{1}C_{1}D_{1}}-S_{\triangle B_{1}C_{1}D_{1}}=3，由(1)知B_{1}P=\frac{13}{3}，\therefore D_{1}T=\frac{18}{13}，\therefore 棱DD_{1}到截面的距离为\frac{18}{13}.$

Final answer: $\frac{18}{13}$
Subfield: Solid Geometry


Question: $如图，在直棱柱ABCD-A_{1}B_{1}C_{1}D_{1}中，底面ABCD是直角梯形，AB\parallel DC，AB\perp BC，AB=3DC=3，BC=6，点P在平面ADD_{1}A_{1}上，过点P和棱BB_{1}的平面把直棱柱分成体积相等的两部分.$![Image](https://cdntk.53zaixian.com/tq/1715283958682435587/img/compression/M_5dd78c490b1810849d47c89c9d9e96de.jpg)
$(2)求棱DD_{1}到截面的距离.$

Solution:
$\because BB_{1}\perp 平面A_{1}B_{1}C_{1}D_{1}，BB_{1}\subset 平面BB_{1}PQ，\therefore 平面BB_{1}PQ\perp 平面A_{1}B_{1}C_{1}D_{1}，交线为B_{1}P.过D_{1}作D_{1}T\perp 直线B_{1}P，垂足为T，\therefore D_{1}T\perp 平面BB_{1}PQ，则D_{1}T的长度即为棱DD_{1}到截面的距离.连接B_{1}D_{1}，S_{\triangle B_{1}D_{1}P}=S_{四边形PB_{1}C_{1}D_{1}}-S_{\triangle B_{1}C_{1}D_{1}}=3，由(1)知B_{1}P=\frac{13}{3}，\therefore D_{1}T=\frac{18}{13}，\therefore 棱DD_{1}到截面的距离为\frac{18}{13}.$

Final answer: $\frac{18}{13}$
Subfield: Solid Geometry


Question: $如图，在四棱锥P-ABCD中，PA\perp 底面ABCD，AB\perp AD，BC\parallel AD，PA=AB=BC=2，AD=4，E为棱PD的中点，\overrightarrow{PF}=\lambda \overrightarrow{PC}(\lambda 为常数，且0<\lambda <1).$<img_2149>
$(1)若直线BF\parallel 平面ACE，求实数\lambda 的值；$

Solution:
$因为PA\perp 底面ABCD，AB，AD\subset 平面ABCD，所以PA\perp AB，PA\perp AD.由题意可知，AB，AD，AP两两垂直，建立如图所示的空间直角坐标系O-xyz，则A(0，0，0)，B(2，0，0)，C(2，2，0)，D(0，4，0)，P(0，0，2)，E(0，2，1)，所以\overrightarrow{AC}=(2，2，0)，\overrightarrow{AE}=(0，2，1)，\overrightarrow{BP}=(-2，0，2)，\overrightarrow{PC}=(2，2，-2)，则\overrightarrow{PF}=\lambda \overrightarrow{PC}=(2\lambda ，2\lambda ，-2\lambda )，所以\overrightarrow{BF}=\overrightarrow{BP}+\overrightarrow{PF}=(2\lambda -2，2\lambda ，2-2\lambda ).设平面ACE的法向量为m=(x,y,z).由\left\{\begin{matrix}\overrightarrow{AC}\cdot \boldsymbol{m}=0,\\ \overrightarrow{AE}\cdot \boldsymbol{m}=0,\end{matrix}\right.得\left\{\begin{matrix}2x+2y=0,\\ 2y+z=0,\end{matrix}\right.令x=1，得m=(1,-1,2).因为BF\parallel 平面ACE，所以\overrightarrow{BF}\cdot m=2\lambda -2-2\lambda +4-4\lambda =0，解得\lambda =\frac{1}{2}.$<img_2101>

Final answer: $\frac{1}{2}$