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


Question: $如图，在四棱锥P-ABCD中，PA\perp AB，PC\perp CD，BC\parallel AD，\angle BAD=\frac{2\pi }{3}，PA=AB=BC=2，AD=4.$<img_2083>
$(1)证明：PA\perp 平面ABCD.$

Solution:
$证明：因为BC\parallel AD，\angle BAD=\frac{2\pi }{3}，所以\angle ABC=\pi -\angle BAD=\frac{\pi }{3}.因为AB=BC=2,所以\triangle ABC为等边三角形，所以AC=2，\angle CAD=\frac{\pi }{3}，在\triangle ACD中，由余弦定理得CD=\sqrt{AC^2+AD^2-2AC\cdot AD\\\cos   \angle CAD}=\sqrt{4+16-2\times 2\times 4\times \frac{1}{2}}=2\sqrt{3}，所以AC^{2}+CD^{2}=AD^{2}，所以CD\perp AC.因为CD\perp PC，且AC\cap PC=C，AC,PC\subset 平面PAC，所以CD\perp 平面PAC.因为PA\subset 平面PAC，所以CD\perp PA.因为PA\perp AB，且AB，CD相交，所以PA\perp 平面ABCD.$
Subfield: Solid Geometry


Question: $如图，在四棱锥P-ABCD中，PA\perp AB，PC\perp CD，BC\parallel AD，\angle BAD=\frac{2\pi }{3}，PA=AB=BC=2，AD=4.$![Image](https://cdntk.53zaixian.com/tq/1703664442340376577/img/compression/M_fe4e8603ce274e3c060cb72c1271f10c.jpg)
$(1)证明：PA\perp 平面ABCD.$

Solution:
$证明：因为BC\parallel AD，\angle BAD=\frac{2\pi }{3}，所以\angle ABC=\pi -\angle BAD=\frac{\pi }{3}.因为AB=BC=2,所以\triangle ABC为等边三角形，所以AC=2，\angle CAD=\frac{\pi }{3}，在\triangle ACD中，由余弦定理得CD=\sqrt{AC^2+AD^2-2AC\cdot AD\\\cos   \angle CAD}=\sqrt{4+16-2\times 2\times 4\times \frac{1}{2}}=2\sqrt{3}，所以AC^{2}+CD^{2}=AD^{2}，所以CD\perp AC.因为CD\perp PC，且AC\cap PC=C，AC,PC\subset 平面PAC，所以CD\perp 平面PAC.因为PA\subset 平面PAC，所以CD\perp PA.因为PA\perp AB，且AB，CD相交，所以PA\perp 平面ABCD.$

Final answer: 证明题
Subfield: Solid Geometry


Question: $如图，在四棱锥P-ABCD中，PA\perp AB，PC\perp CD，BC\parallel AD，\angle BAD=\frac{2\pi }{3}，PA=AB=BC=2，AD=4.$<img_2083>
$(2)若M为PD的中点，求P到平面MAC的距离.$

Solution:
$因为线段PD与平面MAC交于点M，且点M是线段PD的中点，所以P到平面MAC的距离与D到平面MAC的距离相等.因为M为PD的中点，PA\perp 平面ABCD，PA=2，所以点M到平面ACD的距离为\frac{1}{2}PA=1，因为CD\perp AC，AC=2，CD=2\sqrt{3}，所以\triangle ACD的面积为\frac{1}{2}\times 2\times 2\sqrt{3}=2\sqrt{3}.所以三棱锥M-ACD的体积为\frac{1}{3}\times 2\sqrt{3}\times 1=\frac{2\sqrt{3}}{3}.在\triangle MAC中，AC=2，AM=CM=\sqrt{5}，所以\triangle MAC的面积为\frac{1}{2}\times 2\times \sqrt{5-1}=2.设D到平面MAC的距离为d，则\frac{1}{3}\times d\times 2=\frac{2\sqrt{3}}{3}，所以d=\sqrt{3}.所以P到平面MAC的距离为\sqrt{3}.$

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


Question: $如图，在四棱锥P-ABCD中，PA\perp AB，PC\perp CD，BC\parallel AD，\angle BAD=\frac{2\pi }{3}，PA=AB=BC=2，AD=4.$![Image](https://cdntk.53zaixian.com/tq/1703664442340376577/img/compression/M_fe4e8603ce274e3c060cb72c1271f10c.jpg)
$(2)若M为PD的中点，求P到平面MAC的距离.$

Solution:
$因为线段PD与平面MAC交于点M，且点M是线段PD的中点，所以P到平面MAC的距离与D到平面MAC的距离相等.因为M为PD的中点，PA\perp 平面ABCD，PA=2，所以点M到平面ACD的距离为\frac{1}{2}PA=1，因为CD\perp AC，AC=2，CD=2\sqrt{3}，所以\triangle ACD的面积为\frac{1}{2}\times 2\times 2\sqrt{3}=2\sqrt{3}.所以三棱锥M-ACD的体积为\frac{1}{3}\times 2\sqrt{3}\times 1=\frac{2\sqrt{3}}{3}.在\triangle MAC中，AC=2，AM=CM=\sqrt{5}，所以\triangle MAC的面积为\frac{1}{2}\times 2\times \sqrt{5-1}=2.设D到平面MAC的距离为d，则\frac{1}{3}\times d\times 2=\frac{2\sqrt{3}}{3}，所以d=\sqrt{3}.所以P到平面MAC的距离为\sqrt{3}.$

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


Question: $如图，PA\perp 平面ABCD，四边形ABCD为矩形，PA=AB=1，AD=2，点F是PB的中点，点E在边BC上移动．$<img_1740>
$(1)求三棱锥E-PAD的体积；$

Solution:
$\because PA\perp 平面ABCD，\therefore P到平面ADE的距离为PA=1，\because 四边形ABCD为矩形，\therefore S_{\triangle ADE}=\frac{1}{2}AD\cdot AB=\frac{1}{2}\times 2\times 1=1，\therefore V_{E-PAD}=V_{P-ADE}=\frac{1}{3}PA\cdot S_{\triangle ADE}=\frac{1}{3}\times 1\times 1=\frac{1}{3}.$

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


Question: $如图，PA\perp 平面ABCD，四边形ABCD为矩形，PA=AB=1，AD=2，点F是PB的中点，点E在边BC上移动．$![Image](https://cdntk.53zaixian.com/tq/1703664444282339330/img/compression/M_5138667ce9a2a6e260529f000986894c.jpg)
$(1)求三棱锥E-PAD的体积；$

Solution:
$\because PA\perp 平面ABCD，\therefore P到平面ADE的距离为PA=1，\because 四边形ABCD为矩形，\therefore S_{\triangle ADE}=\frac{1}{2}AD\cdot AB=\frac{1}{2}\times 2\times 1=1，\therefore V_{E-PAD}=V_{P-ADE}=\frac{1}{3}PA\cdot S_{\triangle ADE}=\frac{1}{3}\times 1\times 1=\frac{1}{3}.$

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


Question: $如图，PA\perp 平面ABCD，四边形ABCD为矩形，PA=AB=1，AD=2，点F是PB的中点，点E在边BC上移动．$![Image](https://cdntk.53zaixian.com/tq/1703664444282339330/img/compression/M_5138667ce9a2a6e260529f000986894c.jpg)
$(2)证明：无论点E在边BC的何处，都有AF\perp PE．$

Solution:
$证明：\because PA\perp 平面ABCD，BC\subset 平面ABCD，\therefore PA\perp BC，\because 四边形ABCD为矩形，\therefore BC\perp AB，又PA\cap AB=A，PA,AB\subset 平面PAB，\therefore BC\perp 平面PAB，\because AF\subset 平面PAB，\therefore BC\perp AF，\because PA=AB，点F是PB的中点，\therefore AF\perp PB，\because PB\cap BC=B，PB,BC\subset 平面PBC，\therefore AF\perp 平面PBC，\because PE\subset 平面PBC，\therefore AF\perp PE．\therefore 无论点E在边BC的何处，都有AF\perp PE．$

Final answer: 证明题
Subfield: Solid Geometry


Question: $如图，PA\perp 平面ABCD，四边形ABCD为矩形，PA=AB=1，AD=2，点F是PB的中点，点E在边BC上移动．$<img_1740>
$(2)证明：无论点E在边BC的何处，都有AF\perp PE．$

Solution:
$证明：\because PA\perp 平面ABCD，BC\subset 平面ABCD，\therefore PA\perp BC，\because 四边形ABCD为矩形，\therefore BC\perp AB，又PA\cap AB=A，PA,AB\subset 平面PAB，\therefore BC\perp 平面PAB，\because AF\subset 平面PAB，\therefore BC\perp AF，\because PA=AB，点F是PB的中点，\therefore AF\perp PB，\because PB\cap BC=B，PB,BC\subset 平面PBC，\therefore AF\perp 平面PBC，\because PE\subset 平面PBC，\therefore AF\perp PE．\therefore 无论点E在边BC的何处，都有AF\perp PE．$
Subfield: Solid Geometry


Question: $已知三棱柱ABC-A_{1}B_{1}C_{1}中,BC=1,CC_{1}=BB_{1}=2,AB=\sqrt{2},\angle BCC_{1}=60^\circ ,AB\perp 平面BB_{1}C_{1}C.$![Image](https://cdntk.53zaixian.com/tq/1703664444802433025/img/compression/M_dbaaf78c8efba972c0ef8d32bfc09457.jpg)
$(1)求证:BC_{1}\perp 平面ABC；$

Solution:
$证明：\because BC=1,CC_{1}=2,\angle BCC_{1}=60^\circ ,\therefore 在\triangle BCC_{1}中,由余弦定理得BC_{1}=\sqrt{BC^2+CC^2_1-2BC\cdot CC_1\cdot \\\cos   \angle BCC_1}=\sqrt{3},\because BC^{2}+BC^2_1=CC^2_1,\therefore BC\perp BC_{1}.\because AB\perp 平面BB_{1}C_{1}C,BC_{1}\subset 平面BB_{1}C_{1}C，\therefore AB\perp BC_{1}.\because BC\cap AB=B,AB,BC\subset 平面ABC,\therefore BC_{1}\perp 平面ABC.$

Final answer: 证明题
Subfield: Solid Geometry


Question: $已知三棱柱ABC-A_{1}B_{1}C_{1}中,BC=1,CC_{1}=BB_{1}=2,AB=\sqrt{2},\angle BCC_{1}=60^\circ ,AB\perp 平面BB_{1}C_{1}C.$<img_2596>
$(1)求证:BC_{1}\perp 平面ABC；$

Solution:
$证明：\because BC=1,CC_{1}=2,\angle BCC_{1}=60^\circ ,\therefore 在\triangle BCC_{1}中,由余弦定理得BC_{1}=\sqrt{BC^2+CC^2_1-2BC\cdot CC_1\cdot \\\cos   \angle BCC_1}=\sqrt{3},\because BC^{2}+BC^2_1=CC^2_1,\therefore BC\perp BC_{1}.\because AB\perp 平面BB_{1}C_{1}C,BC_{1}\subset 平面BB_{1}C_{1}C，\therefore AB\perp BC_{1}.\because BC\cap AB=B,AB,BC\subset 平面ABC,\therefore BC_{1}\perp 平面ABC.$
Subfield: Solid Geometry


Question: $已知三棱柱ABC-A_{1}B_{1}C_{1}中,BC=1,CC_{1}=BB_{1}=2,AB=\sqrt{2},\angle BCC_{1}=60^\circ ,AB\perp 平面BB_{1}C_{1}C.$![Image](https://cdntk.53zaixian.com/tq/1703664444802433025/img/compression/M_dbaaf78c8efba972c0ef8d32bfc09457.jpg)
$(2)求三棱柱ABC-A_{1}B_{1}C_{1}的体积；$

Solution:
$易得S_{\triangle ABC}=\frac{1}{2}AB\cdot BC=\frac{1}{2}\times \sqrt{2}\times 1=\frac{\sqrt{2}}{2},由(1)知BC_{1}\perp 平面ABC,BC_{1}=\sqrt{3},\therefore 三棱柱ABC-A_{1}B_{1}C_{1}的体积V=S_{\triangle ABC}\cdot BC_{1}=\frac{\sqrt{2}}{2}\times \sqrt{3}=\frac{\sqrt{6}}{2}.$

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


Question: $已知三棱柱ABC-A_{1}B_{1}C_{1}中,BC=1,CC_{1}=BB_{1}=2,AB=\sqrt{2},\angle BCC_{1}=60^\circ ,AB\perp 平面BB_{1}C_{1}C.$<img_2596>
$(2)求三棱柱ABC-A_{1}B_{1}C_{1}的体积；$

Solution:
$易得S_{\triangle ABC}=\frac{1}{2}AB\cdot BC=\frac{1}{2}\times \sqrt{2}\times 1=\frac{\sqrt{2}}{2},由(1)知BC_{1}\perp 平面ABC,BC_{1}=\sqrt{3},\therefore 三棱柱ABC-A_{1}B_{1}C_{1}的体积V=S_{\triangle ABC}\cdot BC_{1}=\frac{\sqrt{2}}{2}\times \sqrt{3}=\frac{\sqrt{6}}{2}.$

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


Question: $已知三棱柱ABC-A_{1}B_{1}C_{1}中,BC=1,CC_{1}=BB_{1}=2,AB=\sqrt{2},\angle BCC_{1}=60^\circ ,AB\perp 平面BB_{1}C_{1}C.$![Image](https://cdntk.53zaixian.com/tq/1703664444802433025/img/compression/M_dbaaf78c8efba972c0ef8d32bfc09457.jpg)
$(3)试在棱CC_{1}(不包含端点C,C_{1})上确定一点E,使得EA\perp EB_{1}.$

Solution:
$如图,在棱CC_{1}(不包含端点C,C_{1})上取一点E,使得EA\perp EB_{1}，连接BE.$![Image](https://cdntk.53zaixian.com/tq/1703664444802433025/img/compression/M_1c9705fc1bfad48520b0973223b7baaa.jpg)$\because AB\perp 平面BB_{1}C_{1}C,B_{1}E\subset 平面BB_{1}C_{1}C,\therefore AB\perp B_{1}E,\because EA\perp EB_{1},AB\cap EA=A,AB,EA\subset 平面ABE,\therefore EB_{1}\perp 平面ABE.\because BE\subset 平面ABE,\therefore BE\perp EB_{1}.设CE=x(0在\triangle BCE中,由余弦定理得BE^{2}=BC^{2}+CE^{2}-2BC\cdot CE\cdot \\cos  \angle BCE=1+x^{2}-x,在\triangle B_{1}C_{1}E中,\angle B_{1}C_{1}E=120^\circ ,由余弦定理得EB^2_1=B_{1}C^2_1+C_{1}E^{2}-2B_{1}C_{1}\cdot C_{1}E\cdot \\cos  \angle B_{1}C_{1}E=x^{2}-5x+7,在Rt\triangle BEB_{1}中,由EB^2_1+BE^{2}=B_{1}B^{2},得(x^{2}-5x+7)+(1+x^{2}-x)=4,解得x=1或x=2(舍去).故当E为CC_{1}的中点时,EA\perp EB_{1}.$

Final answer: $当E为CC_{1}的中点时,EA\perp EB_1$
Subfield: Solid Geometry


Question: $如图，在三棱锥P-ABC中，\triangle PAC是正三角形，AC\perp BC，AC=BC，D是AB的中点．$![Image](https://cdntk.53zaixian.com/tq/1703664446761172994/img/compression/M_ba3cf9d5d5e02aa04c04640f6e37510d.jpg)
$(1)求证：AC\perp PD;$

Solution:
$证明：取AC的中点O，连接OP，OD，\because D为AB的中点，\therefore OD\parallel BC，\because AC\perp BC，\therefore AC\perp OD.\because \triangle PAC是正三角形，\therefore PO\perp AC.\because PO\cap OD=O，PO，OD\subset 平面POD,\therefore AC\perp 平面POD，\because PD\subset 平面POD，\therefore AC\perp PD.$![Image](https://cdntk.53zaixian.com/tq/1703664447109300226/img/compression/M_0d59aa25482fe5248ce47f5521a7e919.jpg)

Final answer: 证明题
Subfield: Solid Geometry


Question: $如图，在三棱锥P-ABC中，\triangle PAC是正三角形，AC\perp BC，AC=BC，D是AB的中点．$<img_2134>
$(1)求证：AC\perp PD;$

Solution:
$证明：取AC的中点O，连接OP，OD，\because D为AB的中点，\therefore OD\parallel BC，\because AC\perp BC，\therefore AC\perp OD.\because \triangle PAC是正三角形，\therefore PO\perp AC.\because PO\cap OD=O，PO，OD\subset 平面POD,\therefore AC\perp 平面POD，\because PD\subset 平面POD，\therefore AC\perp PD.$<img_1989>
Subfield: Solid Geometry


Question: $如图，在三棱锥P-ABC中，\triangle PAC是正三角形，AC\perp BC，AC=BC，D是AB的中点．$![Image](https://cdntk.53zaixian.com/tq/1703664446761172994/img/compression/M_ba3cf9d5d5e02aa04c04640f6e37510d.jpg)
$(2)若AC=BC=PD=2，求直线PC与平面PAB所成角的正弦值．$

Solution:
$解法一：取PA的中点N，连接ON，作OM\perp AB于点M，连接PM，作OH\perp PM于点H，连接NH，\because O，N分别为AC，PA的中点，\therefore ON\parallel PC，\therefore PC与平面PAB所成的角即为ON与平面PAB所成的角.\because \triangle PAC为等边三角形，O，D分别为AC，AB的中点，AC=BC=2，\therefore PO=\sqrt{3}，OD=1，又PD=2，\therefore PO^{2}+OD^{2}=PD^{2}，\therefore PO\perp OD，又PO\perp AC，AC\cap OD=O，\therefore PO\perp 平面ABC，\therefore PO\perp AB.又AB\perp OM，PO\cap OM=O，\therefore AB\perp 平面POM，\therefore AB\perp OH，又OH\perp PM，AB\cap PM=M，\therefore OH\perp 平面PAB，故\angle ONH即为PC与平面PAB所成的角.\because AC\perp BC，AC=BC=2，O为AC的中点，OM\perp AB，\therefore OM=\frac{1}{2}\cdot \frac{AC\cdot BC}{AB}=\frac{1}{2}\times \frac{2\times 2}{2\sqrt{2}}=\frac{\sqrt{2}}{2}，又PO\perp OM，\therefore PM=\sqrt{PO^2+OM^2}=\sqrt{3+\frac{1}{2}}=\frac{\sqrt{14}}{2}.在Rt\triangle OHN中，ON=1，OH=\frac{OM\cdot OP}{PM}=\frac{\frac{\sqrt{2}}{2}\times \sqrt{3}}{\frac{\sqrt{14}}{2}}=\frac{\sqrt{21}}{7}，\therefore \\sin  \angle ONH=\frac{OH}{ON}=\frac{\sqrt{21}}{7}，即直线PC与平面PAB所成角的正弦值为\frac{\sqrt{21}}{7}.解法二：同解法一得PO\perp 平面ABC，PO=\sqrt{3}，PM=\frac{\sqrt{14}}{2}，AB=2\sqrt{2}，易知V_{P-ABC}=V_{C-PAB}，设点C到平面PAB的距离为d，则\frac{1}{3}\times \frac{1}{2}\times 2\times 2\times \sqrt{3}=\frac{1}{3}\times \frac{1}{2}\times 2\sqrt{2}\times \frac{\sqrt{14}}{2}\times d，解得d=\frac{2\sqrt{21}}{7}，设直线PC与平面PAB所成的角为\theta ，则\\sin   \theta =\frac{d}{PC}=\frac{\frac{2\sqrt{21}}{7}}{2}=\frac{\sqrt{21}}{7}，即直线PC与平面PAB所成角的正弦值为\frac{\sqrt{21}}{7}.$

Final answer: $\frac{\sqrt{21}}{7}.$
Subfield: Solid Geometry


Question: $如图，在三棱锥P-ABC中，\triangle PAC是正三角形，AC\perp BC，AC=BC，D是AB的中点．$<img_2134>
$(2)若AC=BC=PD=2，求直线PC与平面PAB所成角的正弦值．$

Solution:
$解法一：取PA的中点N，连接ON，作OM\perp AB于点M，连接PM，作OH\perp PM于点H，连接NH，\because O，N分别为AC，PA的中点，\therefore ON\parallel PC，\therefore PC与平面PAB所成的角即为ON与平面PAB所成的角.\because \triangle PAC为等边三角形，O，D分别为AC，AB的中点，AC=BC=2，\therefore PO=\sqrt{3}，OD=1，又PD=2，\therefore PO^{2}+OD^{2}=PD^{2}，\therefore PO\perp OD，又PO\perp AC，AC\cap OD=O，\therefore PO\perp 平面ABC，\therefore PO\perp AB.又AB\perp OM，PO\cap OM=O，\therefore AB\perp 平面POM，\therefore AB\perp OH，又OH\perp PM，AB\cap PM=M，\therefore OH\perp 平面PAB，故\angle ONH即为PC与平面PAB所成的角.\because AC\perp BC，AC=BC=2，O为AC的中点，OM\perp AB，\therefore OM=\frac{1}{2}\cdot \frac{AC\cdot BC}{AB}=\frac{1}{2}\times \frac{2\times 2}{2\sqrt{2}}=\frac{\sqrt{2}}{2}，又PO\perp OM，\therefore PM=\sqrt{PO^2+OM^2}=\sqrt{3+\frac{1}{2}}=\frac{\sqrt{14}}{2}.在Rt\triangle OHN中，ON=1，OH=\frac{OM\cdot OP}{PM}=\frac{\frac{\sqrt{2}}{2}\times \sqrt{3}}{\frac{\sqrt{14}}{2}}=\frac{\sqrt{21}}{7}，\therefore \\sin  \angle ONH=\frac{OH}{ON}=\frac{\sqrt{21}}{7}，即直线PC与平面PAB所成角的正弦值为\frac{\sqrt{21}}{7}.解法二：同解法一得PO\perp 平面ABC，PO=\sqrt{3}，PM=\frac{\sqrt{14}}{2}，AB=2\sqrt{2}，易知V_{P-ABC}=V_{C-PAB}，设点C到平面PAB的距离为d，则\frac{1}{3}\times \frac{1}{2}\times 2\times 2\times \sqrt{3}=\frac{1}{3}\times \frac{1}{2}\times 2\sqrt{2}\times \frac{\sqrt{14}}{2}\times d，解得d=\frac{2\sqrt{21}}{7}，设直线PC与平面PAB所成的角为\theta ，则\\sin   \theta =\frac{d}{PC}=\frac{\frac{2\sqrt{21}}{7}}{2}=\frac{\sqrt{21}}{7}，即直线PC与平面PAB所成角的正弦值为\frac{\sqrt{21}}{7}.$

Final answer: $\frac{\sqrt{21}}{7}.$
Subfield: Solid Geometry


Question: $如图，在四棱锥P-ABCD中，AD\perp 平面PDC，AD\parallel BC，PD\perp PB，AD＝1，BC＝3，CD＝4，PD＝2.$![Image](https://cdntk.53zaixian.com/tq/1703664502313054211/img/compression/M_afe7c75830527c936bd7a4f0b985b53e.jpg)
$(1)求异面直线AP与BC所成角的余弦值；$

Solution:
$因为AD\parallel BC，所以\angle DAP(或其补角)即为异面直线AP与BC所成的角.因为AD\perp 平面PDC，PD\subset 平面PDC，所以AD\perp PD.在Rt\triangle PDA中，AP＝\sqrt{AD^2+PD^2}＝\sqrt{5}，故\\cos  \angle DAP＝\frac{AD}{AP}＝\frac{\sqrt{5}}{5}.所以异面直线AP与BC所成角的余弦值为\frac{\sqrt{5}}{5}.$

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


Question: $如图，在四棱锥P-ABCD中，AD\perp 平面PDC，AD\parallel BC，PD\perp PB，AD＝1，BC＝3，CD＝4，PD＝2.$<img_1972>
$(1)求异面直线AP与BC所成角的余弦值；$

Solution:
$因为AD\parallel BC，所以\angle DAP(或其补角)即为异面直线AP与BC所成的角.因为AD\perp 平面PDC，PD\subset 平面PDC，所以AD\perp PD.在Rt\triangle PDA中，AP＝\sqrt{AD^2+PD^2}＝\sqrt{5}，故\\cos  \angle DAP＝\frac{AD}{AP}＝\frac{\sqrt{5}}{5}.所以异面直线AP与BC所成角的余弦值为\frac{\sqrt{5}}{5}.$

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


Question: $如图，在四棱锥P-ABCD中，AD\perp 平面PDC，AD\parallel BC，PD\perp PB，AD＝1，BC＝3，CD＝4，PD＝2.$![Image](https://cdntk.53zaixian.com/tq/1703664502313054211/img/compression/M_afe7c75830527c936bd7a4f0b985b53e.jpg)
$(2)求证：PD\perp 平面PBC；$

Solution:
$证明：由(1)知AD\perp PD.因为BC\parallel AD，所以PD\perp BC.又PD\perp PB，PB\cap BC＝B，所以PD\perp 平面PBC.$

Final answer: 证明题
Subfield: Solid Geometry


Question: $如图，在四棱锥P-ABCD中，AD\perp 平面PDC，AD\parallel BC，PD\perp PB，AD＝1，BC＝3，CD＝4，PD＝2.$<img_1972>
$(2)求证：PD\perp 平面PBC；$

Solution:
$证明：由(1)知AD\perp PD.因为BC\parallel AD，所以PD\perp BC.又PD\perp PB，PB\cap BC＝B，所以PD\perp 平面PBC.$
Subfield: Solid Geometry


Question: $如图，在四棱锥P-ABCD中，AD\perp 平面PDC，AD\parallel BC，PD\perp PB，AD＝1，BC＝3，CD＝4，PD＝2.$![Image](https://cdntk.53zaixian.com/tq/1703664502313054211/img/compression/M_afe7c75830527c936bd7a4f0b985b53e.jpg)
$(3)求直线AB与平面PBC所成角的正弦值.$

Solution:
$过点D作DF\parallel AB，交BC于点F，连接PF，则DF与平面PBC所成的角等于AB与平面PBC所成的角.$![Image](https://cdntk.53zaixian.com/tq/1703664502686347265/img/compression/M_f1f78e00207f0640c283aa6d0cdecb19.jpg)$因为PD\perp 平面PBC，所以PF为DF在平面PBC上的射影，所以\angle DFP为直线DF和平面PBC所成的角.由AD\parallel BF，DF\parallel AB，可得四边形ADFB为平行四边形，故BF＝AD＝1.则CF＝BC-BF＝2.易得AD\perp DC，所以BC\perp DC.在Rt\triangle DCF中，DF＝\sqrt{CD^2+CF^2}＝2\sqrt{5}，在Rt\triangle DPF中，\\sin  \angle DFP＝\frac{PD}{DF}＝\frac{\sqrt{5}}{5}.所以直线AB与平面PBC所成角的正弦值为\frac{\sqrt{5}}{5}.$

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


Question: $如图，在四棱锥P-ABCD中，AD\perp 平面PDC，AD\parallel BC，PD\perp PB，AD＝1，BC＝3，CD＝4，PD＝2.$<img_1972>
$(3)求直线AB与平面PBC所成角的正弦值.$

Solution:
$过点D作DF\parallel AB，交BC于点F，连接PF，则DF与平面PBC所成的角等于AB与平面PBC所成的角.$<img_3023>$因为PD\perp 平面PBC，所以PF为DF在平面PBC上的射影，所以\angle DFP为直线DF和平面PBC所成的角.由AD\parallel BF，DF\parallel AB，可得四边形ADFB为平行四边形，故BF＝AD＝1.则CF＝BC-BF＝2.易得AD\perp DC，所以BC\perp DC.在Rt\triangle DCF中，DF＝\sqrt{CD^2+CF^2}＝2\sqrt{5}，在Rt\triangle DPF中，\\sin  \angle DFP＝\frac{PD}{DF}＝\frac{\sqrt{5}}{5}.所以直线AB与平面PBC所成角的正弦值为\frac{\sqrt{5}}{5}.$

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


Question: $在四棱锥P-ABCD中，四边形ABCD为菱形，PA=AB=2，PB=2\sqrt{2}，\angle ABC=60^\circ ，且平面PAC\perp 平面ABCD.$<img_2977>
$(1)证明：PA\perp 平面ABCD；$

Solution:
$证明：\because 四边形ABCD是菱形，\therefore BD\perp AC.\because 平面PAC\perp 平面ABCD，平面PAC\cap 平面ABCD=AC，BD\subset 平面ABCD，\therefore BD\perp 平面PAC，\because PA\subset 平面PAC，\therefore BD\perp PA.\because PA=AB=2，PB=2\sqrt{2}，\therefore PA^{2}+AB^{2}=PB^{2}，\therefore AB\perp PA.又AB\cap BD=B，AB，BD\subset 平面ABCD，\therefore PA\perp 平面ABCD.$
Subfield: Solid Geometry


Question: $在四棱锥P-ABCD中，四边形ABCD为菱形，PA=AB=2，PB=2\sqrt{2}，\angle ABC=60^\circ ，且平面PAC\perp 平面ABCD.$![Image](https://cdntk.53zaixian.com/tq/1703664461187903490/img/compression/M_0ffa15122394e0f6a9ce92d861ec3b27.jpg)
$(1)证明：PA\perp 平面ABCD；$

Solution:
$证明：\because 四边形ABCD是菱形，\therefore BD\perp AC.\because 平面PAC\perp 平面ABCD，平面PAC\cap 平面ABCD=AC，BD\subset 平面ABCD，\therefore BD\perp 平面PAC，\because PA\subset 平面PAC，\therefore BD\perp PA.\because PA=AB=2，PB=2\sqrt{2}，\therefore PA^{2}+AB^{2}=PB^{2}，\therefore AB\perp PA.又AB\cap BD=B，AB，BD\subset 平面ABCD，\therefore PA\perp 平面ABCD.$

Final answer: 证明题
Subfield: Solid Geometry


Question: $在四棱锥P-ABCD中，四边形ABCD为菱形，PA=AB=2，PB=2\sqrt{2}，\angle ABC=60^\circ ，且平面PAC\perp 平面ABCD.$![Image](https://cdntk.53zaixian.com/tq/1703664461187903490/img/compression/M_0ffa15122394e0f6a9ce92d861ec3b27.jpg)
$(2)若M为PC上一点，且PC\perp BM，求二面角M-BD-A的余弦值.$

Solution:
$设AC\cap BD=O，连接OM，由(1)知BD\perp 平面PAC，\because OM\subset 平面PAC，\therefore OM\perp BD，又AO\perp BD，\therefore \angle AOM是二面角M-BD-A的平面角.由BD\perp 平面PAC，知BD\perp PC，又BM\perp PC，BD\cap BM=B，\therefore PC\perp 平面BDM，又OM\subset 平面BDM，\therefore OM\perp PC.在Rt\triangle PAC中，PA=AC=2，\therefore \angle PCA=45^\circ ，\therefore \angle MOC=45^\circ ,\therefore \\cos  \angle AOM=\\cos  (180^\circ -\angle MOC)=\\cos   135^\circ =-\frac{\sqrt{2}}{2}.\therefore 二面角M-BD-A的余弦值为-\frac{\sqrt{2}}{2}.$![Image](https://cdntk.53zaixian.com/tq/1703664461187903490/img/compression/M_14e05f431faff69acde033e531533f05.jpg)

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


Question: $在四棱锥P-ABCD中，四边形ABCD为菱形，PA=AB=2，PB=2\sqrt{2}，\angle ABC=60^\circ ，且平面PAC\perp 平面ABCD.$<img_2977>
$(2)若M为PC上一点，且PC\perp BM，求二面角M-BD-A的余弦值.$

Solution:
$设AC\cap BD=O，连接OM，由(1)知BD\perp 平面PAC，\because OM\subset 平面PAC，\therefore OM\perp BD，又AO\perp BD，\therefore \angle AOM是二面角M-BD-A的平面角.由BD\perp 平面PAC，知BD\perp PC，又BM\perp PC，BD\cap BM=B，\therefore PC\perp 平面BDM，又OM\subset 平面BDM，\therefore OM\perp PC.在Rt\triangle PAC中，PA=AC=2，\therefore \angle PCA=45^\circ ，\therefore \angle MOC=45^\circ ,\therefore \\cos  \angle AOM=\\cos  (180^\circ -\angle MOC)=\\cos   135^\circ =-\frac{\sqrt{2}}{2}.\therefore 二面角M-BD-A的余弦值为-\frac{\sqrt{2}}{2}.$<img_2359>

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


Question: $如图，在直角梯形ABCD中，AB\parallel DC，\angle ABC=90^\circ ，AB=2DC=2BC，E为AB的中点，沿DE将\triangle ADE折起，使得点A到点P的位置，且PE\perp EB，M为PB的中点，N是BC上的动点(与点B，C不重合)．$![Image](https://cdntk.53zaixian.com/tq/1703664461842214913/img/compression/M_0ee5dab92abf2957a7edc89170e72f08.jpg)
$(2)是否存在点N，使得二面角B-EN-M的正切值为\sqrt{5}？若存在，请确定点N的位置；若不存在，请说明理由．$

Solution:
$假设存在点N满足题意，如图，过M作MQ\perp EB于Q，因为PE\perp EB，所以PE\parallel MQ，由(1)知PE\perp 平面EBCD，所以MQ\perp 平面EBCD，因为EN\subset 平面EBCD，所以MQ\perp EN.过Q作QR\perp EN于R，连接MR，因为MQ\cap QR=Q，所以EN\perp 平面MQR，因为MR\subset 平面MQR，所以EN\perp MR，所以\angle MRQ为二面角B-EN-M的平面角.设PE=EB=BC=2，则MQ=1，EQ=1,设BN=x(0易知Rt\triangle EBN∽Rt\triangle ERQ，所以\frac{BN}{RQ}=\frac{EN}{EQ}，所以\frac{x}{RQ}=\frac{\sqrt{2^2+x^2}}{1}，得RQ=\frac{x}{\sqrt{4+x^2}}，所以\\tan  \angle MRQ=\frac{MQ}{RQ}=\frac{\sqrt{x^2+4}}{x}=\sqrt{5}，解得x=1，即N为BC的中点.故存在点N使得二面角B-EN-M的正切值为\sqrt{5}，N为BC的中点.$![Image](https://cdntk.53zaixian.com/tq/1703664462186147841/img/compression/M_67e11517e559a68aa92e34dc2338949c.jpg)

Final answer: $存在点N，N为BC的中点。$
Subfield: Solid Geometry


Question: $如图，直三棱柱ABC-DEF的底面是边长为2的正三角形，侧棱AD=1，P是线段CF的延长线上的一点，平面PAB分别与DF，EF相交于M，N.$<img_1883>
$(1)求证：MN\parallel 平面CDE；$

Solution:
$证明：由题意得AB\parallel DE，AB\not\subset 平面DEF，DE\subset 平面DEF，\therefore AB\parallel 平面DEF.\because 平面PAB\cap 平面DEF=MN，AB\subset 平面PAB，\therefore AB\parallel MN，\therefore DE\parallel MN，又MN\not\subset 平面CDE，DE\subset 平面CDE，\therefore MN\parallel 平面CDE.$
Subfield: Solid Geometry


Question: $如图，直三棱柱ABC-DEF的底面是边长为2的正三角形，侧棱AD=1，P是线段CF的延长线上的一点，平面PAB分别与DF，EF相交于M，N.$![Image](https://cdntk.53zaixian.com/tq/1703664462601383940/img/compression/M_f07cde00df83c231bc0a8bfd704f575e.jpg)
$(1)求证：MN\parallel 平面CDE；$

Solution:
$证明：由题意得AB\parallel DE，AB\not\subset 平面DEF，DE\subset 平面DEF，\therefore AB\parallel 平面DEF.\because 平面PAB\cap 平面DEF=MN，AB\subset 平面PAB，\therefore AB\parallel MN，\therefore DE\parallel MN，又MN\not\subset 平面CDE，DE\subset 平面CDE，\therefore MN\parallel 平面CDE.$

Final answer: 证明题
Subfield: Solid Geometry


Question: $如图，直三棱柱ABC-DEF的底面是边长为2的正三角形，侧棱AD=1，P是线段CF的延长线上的一点，平面PAB分别与DF，EF相交于M，N.$![Image](https://cdntk.53zaixian.com/tq/1703664462601383940/img/compression/M_f07cde00df83c231bc0a8bfd704f575e.jpg)
$(2)当PF为何值时，平面PAB\perp 平面CDE?$

Solution:
$分别取线段AB，DE的中点G，H，连接CG，GH，CH，PG,则GH\parallel CP，\therefore P，C，G，H四点共面.易得Rt\triangle PCA\cong Rt\triangle PCB，\therefore PA=PB，\therefore PG\perp AB.\because AB\parallel DE，\therefore PG\perp DE.若PG\perp CH，则PG\perp 平面CDE，则平面PAB\perp 平面CDE.此时\angle CPG=\angle HCG，则\triangle PCG∽\triangle CGH,\therefore \frac{PC}{CG}=\frac{CG}{GH}.\because \triangle ABC是边长为2的正三角形，\therefore CG=2\\sin   60^\circ =\sqrt{3}.又GH=1，\therefore PC=\frac{CG^2}{GH}=3,\therefore PF=PC-FC=2，\therefore 当PF=2时，平面PAB\perp 平面CDE.$![Image](https://cdntk.53zaixian.com/tq/1703664462974676993/img/compression/M_eed915ad5eb4fe695d75e9cd5d248e9c.jpg)

Final answer: $当PF=2时，平面PAB\perp 平面CDE$
Subfield: Solid Geometry


Question: $如图，直三棱柱ABC-DEF的底面是边长为2的正三角形，侧棱AD=1，P是线段CF的延长线上的一点，平面PAB分别与DF，EF相交于M，N.$<img_1883>
$(2)平面PAB\perp 平面CDE, 求PF的值$

Solution:
$分别取线段AB，DE的中点G，H，连接CG，GH，CH，PG,则GH\parallel CP，\therefore P，C，G，H四点共面.易得Rt\triangle PCA\cong Rt\triangle PCB，\therefore PA=PB，\therefore PG\perp AB.\because AB\parallel DE，\therefore PG\perp DE.若PG\perp CH，则PG\perp 平面CDE，则平面PAB\perp 平面CDE.此时\angle CPG=\angle HCG，则\triangle PCG∽\triangle CGH,\therefore \frac{PC}{CG}=\frac{CG}{GH}.\because \triangle ABC是边长为2的正三角形，\therefore CG=2\\sin   60^\circ =\sqrt{3}.又GH=1，\therefore PC=\frac{CG^2}{GH}=3,\therefore PF=PC-FC=2，\therefore 当PF=2时，平面PAB\perp 平面CDE.$<img_2640>



Final answer: $2$
Subfield: Solid Geometry


Question: $如图，在四棱锥P-ABCD中，底面ABCD是边长为2的菱形且\angle ABC=\frac{\pi }{3}，PB=PA=4，PC=\sqrt{6}．$![Image](https://cdntk.53zaixian.com/tq/1703664463234723843/img/compression/M_2037bafbe98dd8e609d06d4defaf99b9.jpg)
$(1)求PD的值；$

Solution:
$取线段AB的中点E，连接CE，PE，因为四边形ABCD是边长为2的菱形，所以BC=2，BE=1.在\triangle BCE中，由余弦定理可得CE^{2}=BC^{2}+BE^{2}-2BC\cdot BE\cdot \\cos  \frac{\pi }{3}=3，\therefore BE^{2}+CE^{2}=BC^{2}，\therefore BE\perp CE，即CE\perp AB.\because PB=PA，E是AB的中点，\therefore PE\perp AB.\because PE\cap CE=E，PE，CE\subset 平面PCE，\therefore AB\perp 平面PCE.\because PC\subset 平面PCE，\therefore PC\perp AB，\because CD\parallel AB，\therefore PC\perp CD.\therefore PD=\sqrt{PC^2+CD^2}=\sqrt{6+4}=\sqrt{10}.$

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


Question: $如图，在四棱锥P-ABCD中，底面ABCD是边长为2的菱形且\angle ABC=\frac{\pi }{3}，PB=PA=4，PC=\sqrt{6}．$<img_2087>
$(1)求PD的值；$

Solution:
$取线段AB的中点E，连接CE，PE，因为四边形ABCD是边长为2的菱形，所以BC=2，BE=1.在\triangle BCE中，由余弦定理可得CE^{2}=BC^{2}+BE^{2}-2BC\cdot BE\cdot \\cos  \frac{\pi }{3}=3，\therefore BE^{2}+CE^{2}=BC^{2}，\therefore BE\perp CE，即CE\perp AB.\because PB=PA，E是AB的中点，\therefore PE\perp AB.\because PE\cap CE=E，PE，CE\subset 平面PCE，\therefore AB\perp 平面PCE.\because PC\subset 平面PCE，\therefore PC\perp AB，\because CD\parallel AB，\therefore PC\perp CD.\therefore PD=\sqrt{PC^2+CD^2}=\sqrt{6+4}=\sqrt{10}.$

Final answer: $\sqrt{10}$