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


Question: $如图，在四棱锥P-ABCD中，底面ABCD是矩形.已知AB=3,AD=2,PA=2,PD=2\sqrt{2},\angle PAB=60^\circ .$<img_2605>
$(2)求异面直线PC与AD所成角的正切值；$

Solution:
$易知BC\parallel AD，所以\angle PCB(或其补角)是异面直线PC与AD所成的角.在\triangle PAB中，由余弦定理得PB=\sqrt{PA^2+AB^2-2PA\cdot AB\cdot \\\cos   \angle PAB}=\sqrt{7}，由(1)知AD\perp 平面PAB,又PB\subset 平面PAB，BC\parallel AD，所以BC\perp PB，所以\\tan  \angle PCB=\frac{PB}{BC}=\frac{\sqrt{7}}{2}，所以异面直线PC与AD所成角的正切值为\frac{\sqrt{7}}{2}.$

Final answer: $\frac{\sqrt{7}}{2}$
Subfield: Solid Geometry


Question: $如图，在四棱锥P-ABCD中，底面ABCD是矩形.已知AB=3,AD=2,PA=2,PD=2\sqrt{2},\angle PAB=60^\circ .$<img_2605>
$(3)求二面角A-PB-D的正切值.$

Solution:
$过点A作AH\perp PB于H，连接HD，如图.$<img_3015>$由(1)知AD\perp 平面PAB，又PB\subset 平面PAB，所以AD\perp PB，又AH\perp PB，AH\cap AD=A，AH，AD\subset 平面AHD，所以PB\perp 平面AHD，又HD\subset 平面AHD，所以PB\perp HD，故\angle AHD是二面角A-PB-D的平面角.由题意可得\frac{1}{2}\cdot PB\cdot AH=\frac{1}{2}\cdot AP\cdot AB\cdot \\sin  \angle PAB，由(2)知PB=\sqrt{7}，所以AH=\frac{3\sqrt{21}}{7}.故\\tan  \angle AHD=\frac{AD}{AH}=\frac{2}{\frac{3\sqrt{21}}{7}}=\frac{2\sqrt{21}}{9}，所以二面角A-PB-D的正切值为\frac{2\sqrt{21}}{9}.$

Final answer: $\frac{2\sqrt{21}}{9}$
Subfield: Solid Geometry


Question: $如图，在四棱锥P-ABCD中，底面ABCD是矩形.已知AB=3,AD=2,PA=2,PD=2\sqrt{2},\angle PAB=60^\circ .$![Image](https://cdntk.53zaixian.com/tq/1650310188403490819/img/compression/M_9b670173eef297d194513e0dcf7a3a15.jpg)
$(3)求二面角A-PB-D的正切值.$

Solution:
$过点A作AH\perp PB于H，连接HD，如图.$![Image](https://cdntk.53zaixian.com/tq/1650310188948750338/img/compression/M_30b7223369347e2c5e44eb820c8d65ca.jpg)$由(1)知AD\perp 平面PAB，又PB\subset 平面PAB，所以AD\perp PB，又AH\perp PB，AH\cap AD=A，AH，AD\subset 平面AHD，所以PB\perp 平面AHD，又HD\subset 平面AHD，所以PB\perp HD，故\angle AHD是二面角A-PB-D的平面角.由题意可得\frac{1}{2}\cdot PB\cdot AH=\frac{1}{2}\cdot AP\cdot AB\cdot \\sin  \angle PAB，由(2)知PB=\sqrt{7}，所以AH=\frac{3\sqrt{21}}{7}.故\\tan  \angle AHD=\frac{AD}{AH}=\frac{2}{\frac{3\sqrt{21}}{7}}=\frac{2\sqrt{21}}{9}，所以二面角A-PB-D的正切值为\frac{2\sqrt{21}}{9}.$

Final answer: $\frac{2\sqrt{21}}{9}$
Subfield: Solid Geometry


Question: $如图，三棱台ABC-A_{1}B_{1}C_{1}中，BC\perp CC_{1},AB=BC=CA=2B_{1}C_{1},CC_{1}=\sqrt{3}B_{1}C_{1}.$![Image](https://cdntk.53zaixian.com/tq/1650310189527564290/img/compression/M_abc657017d66a0a8ad3ddda03b9b7547.jpg)
$(1)求证：AB_{1}\perp BC；$

Solution:
$证明：取BC的中点D，连接AD,B_{1}D.因为BC=2B_{1}C_{1}，所以DC=B_{1}C_{1}，又BC\perp CC_{1}，DC\parallel B_{1}C_{1}，所以四边形DCC_{1}B_{1}为矩形，故B_{1}D\perp BC.因为AB=BC=CA，所以AD\perp BC，又B_{1}D\cap AD=D，B_{1}D,AD\subset 平面B_{1}AD，所以BC\perp 平面B_{1}AD，又AB_{1}\subset 平面B_{1}AD，所以AB_{1}\perp BC.$

Final answer: 证明题
Subfield: Solid Geometry


Question: $如图，三棱台ABC-A_{1}B_{1}C_{1}中，BC\perp CC_{1},AB=BC=CA=2B_{1}C_{1},CC_{1}=\sqrt{3}B_{1}C_{1}.$<img_2689>
$(1)求证：AB_{1}\perp BC；$

Solution:
$证明：取BC的中点D，连接AD,B_{1}D.因为BC=2B_{1}C_{1}，所以DC=B_{1}C_{1}，又BC\perp CC_{1}，DC\parallel B_{1}C_{1}，所以四边形DCC_{1}B_{1}为矩形，故B_{1}D\perp BC.因为AB=BC=CA，所以AD\perp BC，又B_{1}D\cap AD=D，B_{1}D,AD\subset 平面B_{1}AD，所以BC\perp 平面B_{1}AD，又AB_{1}\subset 平面B_{1}AD，所以AB_{1}\perp BC.$
Subfield: Solid Geometry


Question: $如图，三棱台ABC-A_{1}B_{1}C_{1}中，BC\perp CC_{1},AB=BC=CA=2B_{1}C_{1},CC_{1}=\sqrt{3}B_{1}C_{1}.$<img_2689>
$(2)若二面角C_{1}-BC-A的平面角的大小为60^\circ ，求直线AC_{1}与平面BCC_{1}B_{1}所成角的正弦值.$

Solution:
$由(1)知BD\perp 平面B_{1}AD，故二面角C_{1}-BC-A的平面角为\angle ADB_{1}，结合题图知\angle ADB_{1}=60^\circ .设BC_{1}\cap B_{1}D=E，连接AE,DC_{1}，易知四边形B_{1}C_{1}DB为平行四边形，所以E为B_{1}D的中点，设B_{1}C_{1}=2，则BD=DC=2，CC_{1}=\sqrt{3}B_{1}C_{1}=2\sqrt{3}，B_{1}E=ED=\sqrt{3}，C_{1}E=\sqrt{B_1C^2_1+B_1E^2}=\sqrt{7},AD=2\sqrt{3}，故在\triangle AED中，由余弦定理得AE=\sqrt{ED^2+AD^2-2ED\cdot AD\cdot \\\cos    60^\circ }=3.故AE^{2}+ED^{2}=AD^{2}，所以AE\perp ED.因为BD\perp 平面B_{1}AD，AE\subset 平面B_{1}AD，所以BD\perp AE，又BD\cap ED=D，BD,ED\subset 平面BCC_{1}B_{1}，所以AE\perp 平面BCC_{1}B_{1}，故直线AC_{1}与平面BCC_{1}B_{1}所成的角为\angle AC_{1}E.易得\\sin  \angle AC_{1}E=\frac{AE}{AC_1}=\frac{3}{\sqrt{3^2+(\sqrt{7})^2}}=\frac{3}{4}，故直线AC_{1}与平面BCC_{1}B_{1}所成角的正弦值为\frac{3}{4}.$<img_3111>

Final answer: $\frac{3}{4}$
Subfield: Solid Geometry


Question: $如图，三棱台ABC-A_{1}B_{1}C_{1}中，BC\perp CC_{1},AB=BC=CA=2B_{1}C_{1},CC_{1}=\sqrt{3}B_{1}C_{1}.$![Image](https://cdntk.53zaixian.com/tq/1650310189527564290/img/compression/M_abc657017d66a0a8ad3ddda03b9b7547.jpg)
$(2)若二面角C_{1}-BC-A的平面角的大小为60^\circ ，求直线AC_{1}与平面BCC_{1}B_{1}所成角的正弦值.$

Solution:
$由(1)知BD\perp 平面B_{1}AD，故二面角C_{1}-BC-A的平面角为\angle ADB_{1}，结合题图知\angle ADB_{1}=60^\circ .设BC_{1}\cap B_{1}D=E，连接AE,DC_{1}，易知四边形B_{1}C_{1}DB为平行四边形，所以E为B_{1}D的中点，设B_{1}C_{1}=2，则BD=DC=2，CC_{1}=\sqrt{3}B_{1}C_{1}=2\sqrt{3}，B_{1}E=ED=\sqrt{3}，C_{1}E=\sqrt{B_1C^2_1+B_1E^2}=\sqrt{7},AD=2\sqrt{3}，故在\triangle AED中，由余弦定理得AE=\sqrt{ED^2+AD^2-2ED\cdot AD\cdot \\\cos    60^\circ }=3.故AE^{2}+ED^{2}=AD^{2}，所以AE\perp ED.因为BD\perp 平面B_{1}AD，AE\subset 平面B_{1}AD，所以BD\perp AE，又BD\cap ED=D，BD,ED\subset 平面BCC_{1}B_{1}，所以AE\perp 平面BCC_{1}B_{1}，故直线AC_{1}与平面BCC_{1}B_{1}所成的角为\angle AC_{1}E.易得\\sin  \angle AC_{1}E=\frac{AE}{AC_1}=\frac{3}{\sqrt{3^2+(\sqrt{7})^2}}=\frac{3}{4}，故直线AC_{1}与平面BCC_{1}B_{1}所成角的正弦值为\frac{3}{4}.$![Image](https://cdntk.53zaixian.com/tq/1650310190039269378/img/compression/M_e2a8c5733ebe09521d48bb24712c6c28.jpg)

Final answer: $\frac{3}{4}$
Subfield: Solid Geometry


Question: $在正六棱柱ABCDEF-A_{1}B_{1}C_{1}D_{1}E_{1}F_{1}中，各棱长都为a，O为B_{1}E的中点.$![Image](https://cdntk.53zaixian.com/tq/1650310190664220673/img/compression/M_bb771b35aff32c2cc2a3ea8a25da1112.jpg)
$(1)求A_{1}E与平面BCC_{1}B_{1}所成角的正切值；$

Solution:
$过A_{1}作A_{1}P\perp E_{1}F_{1}，交E_{1}F_{1}的延长线于点P，连接PE，可得A_{1}P\perp 平面EFF_{1}E_{1}，进而可得\angle A_{1}EP为A_{1}E与平面BCC_{1}B_{1}所成的角，结合条件即得所求正切值；$

Final answer: $A_{1}E与平面BCC_{1}B_{1}所成角的正切值为\frac{A_{1}P}{EP}。$
Subfield: Solid Geometry


Question: $在正六棱柱ABCDEF-A_{1}B_{1}C_{1}D_{1}E_{1}F_{1}中，各棱长都为a，O为B_{1}E的中点.$![Image](https://cdntk.53zaixian.com/tq/1650310190664220673/img/compression/M_bb771b35aff32c2cc2a3ea8a25da1112.jpg)
$(2)求平面ACO与平面EDD_{1}E_{1}所成的锐二面角的正弦值.$

Solution:
$延长AC交ED的延长线于点G，过C作CH\perp DG，过H作HK\perp D_{1}G于K，则可得\angle CKH为平面ACO与平面EDD_{1}E_{1}所成的锐二面角的平面角，进而即得所求正弦值.$

Final answer: $\sin (\angle CKH)$
Subfield: Solid Geometry


Question: $如图所示，正四棱锥E-ABCD中，EA=5，AB=6，F为线段BC的中点，EO是正四棱锥的高.$![Image](https://cdntk.53zaixian.com/tq/1650310148935782404/img/compression/M_f038038af357fcc011861d3615e51130.jpg)
$(1)求正四棱锥E-ABCD的表面积；$

Solution:
$由题可知EB=EC，四边形ABCD是正方形，又F为线段BC的中点，所以EF\perp BC，BF=\frac{1}{2}BC=\frac{1}{2}AB=3，又EB=EA=5，所以EF=\sqrt[　]{{5}^{2}-{3}^{2}}=4，(3分)所以正四棱锥E-ABCD的表面积S=\left(\frac{1}{2}\times 6\times 4\right)\times 4+6\times 6=48+36=84.(5分)$

Final answer: $84$
Subfield: Solid Geometry


Question: $如图所示，正四棱锥E-ABCD中，EA=5，AB=6，F为线段BC的中点，EO是正四棱锥的高.$<img_2795>
$(1)求正四棱锥E-ABCD的表面积；$

Solution:
$由题可知EB=EC，四边形ABCD是正方形，又F为线段BC的中点，所以EF\perp BC，BF=\frac{1}{2}BC=\frac{1}{2}AB=3，又EB=EA=5，所以EF=\sqrt[　]{{5}^{2}-{3}^{2}}=4，(3分)所以正四棱锥E-ABCD的表面积S=\left(\frac{1}{2}\times 6\times 4\right)\times 4+6\times 6=48+36=84.(5分)$

Final answer: $84$
Subfield: Solid Geometry


Question: $如图所示，正四棱锥E-ABCD中，EA=5，AB=6，F为线段BC的中点，EO是正四棱锥的高.$<img_2795>
$(2)求正四棱锥E-ABCD的体积.$

Solution:
$由(1)知EF=4，因为EO是正四棱锥E-ABCD的高，所以EO\perp OF，易知OF=\frac{1}{2}AB=3，所以EO=\sqrt[　]{EF^{2}-OF^{2}}=\sqrt[　]{{4}^{2}-{3}^{2}}=\sqrt{7}，(8分)所以正四棱锥E-ABCD的体积V=\frac{1}{3}\times 6\times 6\times \sqrt[　]{7}=12\sqrt{7}.(10分)$

Final answer: $12\sqrt 7$
Subfield: Solid Geometry


Question: $如图所示，正四棱锥E-ABCD中，EA=5，AB=6，F为线段BC的中点，EO是正四棱锥的高.$![Image](https://cdntk.53zaixian.com/tq/1650310148935782404/img/compression/M_f038038af357fcc011861d3615e51130.jpg)
$(2)求正四棱锥E-ABCD的体积.$

Solution:
$由(1)知EF=4，因为EO是正四棱锥E-ABCD的高，所以EO\perp OF，易知OF=\frac{1}{2}AB=3，所以EO=\sqrt[　]{EF^{2}-OF^{2}}=\sqrt[　]{{4}^{2}-{3}^{2}}=\sqrt{7}，(8分)所以正四棱锥E-ABCD的体积V=\frac{1}{3}\times 6\times 6\times \sqrt[　]{7}=12\sqrt{7}.(10分)$

Final answer: $12\sqrt 7$
Subfield: Solid Geometry


Question: $如图，已知四棱锥P-ABCD中，底面ABCD是正方形，\triangle PAD是正三角形，CD\perp 平面PAD，E、F、G、O分别是PC、PD、BC、AD的中点.$<img_3121>
$(1)求证：PO\perp 平面ABCD；$

Solution:
$证明：因为CD\perp 平面PAD，CD\subset 平面ABCD，所以平面PAD\perp 平面ABCD，(2分)因为\triangle PAD是正三角形，O为AD的中点，所以PO\perp AD，(4分)又平面PAD\cap 平面ABCD=AD，PO\subset 平面PAD，所以PO\perp 平面ABCD.(6分)$
Subfield: Solid Geometry


Question: $如图，已知四棱锥P-ABCD中，底面ABCD是正方形，\triangle PAD是正三角形，CD\perp 平面PAD，E、F、G、O分别是PC、PD、BC、AD的中点.$![Image](https://cdntk.53zaixian.com/tq/1650310150387011586/img/compression/M_19694cd4625c8a805cf79337be8865b9.jpg)
$(1)求证：PO\perp 平面ABCD；$

Solution:
$证明：因为CD\perp 平面PAD，CD\subset 平面ABCD，所以平面PAD\perp 平面ABCD，(2分)因为\triangle PAD是正三角形，O为AD的中点，所以PO\perp AD，(4分)又平面PAD\cap 平面ABCD=AD，PO\subset 平面PAD，所以PO\perp 平面ABCD.(6分)$

Final answer: 证明题
Subfield: Solid Geometry


Question: $如图，已知四棱锥P-ABCD中，底面ABCD是正方形，\triangle PAD是正三角形，CD\perp 平面PAD，E、F、G、O分别是PC、PD、BC、AD的中点.$![Image](https://cdntk.53zaixian.com/tq/1650310150387011586/img/compression/M_19694cd4625c8a805cf79337be8865b9.jpg)
$(2)求平面EFG与平面ABCD夹角的大小.$

Solution:
$如图，连接OF，OG，$![Image](https://cdntk.53zaixian.com/tq/1650310150990991362/img/compression/M_76d4cc30249e7b3e31eae8f8bc778d00.jpg)$因为E、F、G、O分别是PC、PD、BC、AD的中点，所以EF\parallel CD\parallel OG，OF\parallel PA，所以E、F、G、O四点共面.(8分)因为CD\perp 平面PAD，所以OG\perp 平面PAD，(9分)又OD\subset 平面PAD，OF\subset 平面PAD，所以OG\perp OD，OG\perp OF.所以平面EFG与平面ABCD的夹角为\angle FOD，(10分)因为\triangle PAD是正三角形，OF\parallel PA，所以\angle FOD=\angle PAD=60^\circ .所以平面EFG与平面ABCD夹角的大小为60^\circ .(12分)$

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


Question: $如图，已知四棱锥P-ABCD中，底面ABCD是正方形，\triangle PAD是正三角形，CD\perp 平面PAD，E、F、G、O分别是PC、PD、BC、AD的中点.$<img_3121>
$(2)求平面EFG与平面ABCD夹角的大小.$

Solution:
$如图，连接OF，OG，$<img_2146>$因为E、F、G、O分别是PC、PD、BC、AD的中点，所以EF\parallel CD\parallel OG，OF\parallel PA，所以E、F、G、O四点共面.(8分)因为CD\perp 平面PAD，所以OG\perp 平面PAD，(9分)又OD\subset 平面PAD，OF\subset 平面PAD，所以OG\perp OD，OG\perp OF.所以平面EFG与平面ABCD的夹角为\angle FOD，(10分)因为\triangle PAD是正三角形，OF\parallel PA，所以\angle FOD=\angle PAD=60^\circ .所以平面EFG与平面ABCD夹角的大小为60^\circ .(12分)$

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


Question: $如图所示，在三棱柱ABC-A_{1}B_{1}C_{1}中，A_{1}A\perp 平面ABC，\angle ACB=90^\circ ，M是AB的中点，AC=CB=CC_{1}=2.$![Image](https://cdntk.53zaixian.com/tq/1650310151540445187/img/compression/M_a88ed9a5b7a37a74c91a0c5203e9e460.jpg)
$(1)求证：平面A_{1}CM\perp 平面ABB_{1}A_{1}；$

Solution:
$证明：由A_{1}A\perp 平面ABC，CM\subset 平面ABC，得A_{1}A\perp CM.由AC=CB，M是AB的中点，得AB\perp CM，(3分)又A_{1}A\cap AB=A，A_{1}A，AB\subset 平面ABB_{1}A_{1}，\therefore CM\perp 平面ABB_{1}A_{1}，又CM\subset 平面A_{1}CM，\therefore 平面A_{1}CM\perp 平面ABB_{1}A_{1}.(6分)$

Final answer: 证明题
Subfield: Solid Geometry


Question: $如图所示，在三棱柱ABC-A_{1}B_{1}C_{1}中，A_{1}A\perp 平面ABC，\angle ACB=90^\circ ，M是AB的中点，AC=CB=CC_{1}=2.$<img_1543>
$(1)求证：平面A_{1}CM\perp 平面ABB_{1}A_{1}；$

Solution:
$证明：由A_{1}A\perp 平面ABC，CM\subset 平面ABC，得A_{1}A\perp CM.由AC=CB，M是AB的中点，得AB\perp CM，(3分)又A_{1}A\cap AB=A，A_{1}A，AB\subset 平面ABB_{1}A_{1}，\therefore CM\perp 平面ABB_{1}A_{1}，又CM\subset 平面A_{1}CM，\therefore 平面A_{1}CM\perp 平面ABB_{1}A_{1}.(6分)$
Subfield: Solid Geometry


Question: $如图所示，在三棱柱ABC-A_{1}B_{1}C_{1}中，A_{1}A\perp 平面ABC，\angle ACB=90^\circ ，M是AB的中点，AC=CB=CC_{1}=2.$<img_1543>
$(2)求点M到平面A_{1}CB_{1}的距离.$

Solution:
$如图，取A_{1}B_{1}的中点N，连接MN，$<img_2665>$设M到平面A_{1}CB_{1}的距离为h，由题意知A_{1}C=CB_{1}=A_{1}B_{1}=2CM=2\sqrt{2}，MN=2，\therefore S_{\triangle A_{1}CB_{1}}=\frac{1}{2}\times (2\sqrt{2})^{2}\times \\sin   60^\circ =2\sqrt{3}，(8分)S_{\triangle A_{1}MB_{1}}=\frac{1}{2}\times 2\sqrt{2}\times 2=2\sqrt{2}，(9分)又V_{C-A_{1}MB_{1}}=V_{M-A_{1}CB_{1}}，即\frac{1}{3}CM\cdot S_{\triangle A_{1}MB_{1}}=\frac{1}{3}h\cdot S_{\triangle A_{1}CB_{1}}，\therefore 点M到平面A_{1}CB_{1}的距离h=\frac{CM\cdot S_{\vartriangle A_1MB_1}}{S_{\vartriangle A_1CB_1}}=\frac{2\sqrt{3}}{3}.(12分)$

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


Question: $如图所示，在三棱柱ABC-A_{1}B_{1}C_{1}中，A_{1}A\perp 平面ABC，\angle ACB=90^\circ ，M是AB的中点，AC=CB=CC_{1}=2.$![Image](https://cdntk.53zaixian.com/tq/1650310151540445187/img/compression/M_a88ed9a5b7a37a74c91a0c5203e9e460.jpg)
$(2)求点M到平面A_{1}CB_{1}的距离.$

Solution:
$如图，取A_{1}B_{1}的中点N，连接MN，$![Image](https://cdntk.53zaixian.com/tq/1650310151884378113/img/compression/M_fffe6cde22211376cc6217646b0894f1.jpg)$设M到平面A_{1}CB_{1}的距离为h，由题意知A_{1}C=CB_{1}=A_{1}B_{1}=2CM=2\sqrt{2}，MN=2，\therefore S_{\triangle A_{1}CB_{1}}=\frac{1}{2}\times (2\sqrt{2})^{2}\times \\sin   60^\circ =2\sqrt{3}，(8分)S_{\triangle A_{1}MB_{1}}=\frac{1}{2}\times 2\sqrt{2}\times 2=2\sqrt{2}，(9分)又V_{C-A_{1}MB_{1}}=V_{M-A_{1}CB_{1}}，即\frac{1}{3}CM\cdot S_{\triangle A_{1}MB_{1}}=\frac{1}{3}h\cdot S_{\triangle A_{1}CB_{1}}，\therefore 点M到平面A_{1}CB_{1}的距离h=\frac{CM\cdot S_{\vartriangle A_1MB_1}}{S_{\vartriangle A_1CB_1}}=\frac{2\sqrt{3}}{3}.(12分)$

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


Question: $如图，四边形ABCD是边长为3的正方形，DE\perp 平面ABCD，AF\perp 平面ABCD，DE=3，AF=1.$![Image](https://cdntk.53zaixian.com/tq/1650310153448853506/img/compression/M_7cee04b770fb4bd5b813f675982682cb.jpg)
$(1)证明：平面ABF\parallel 平面DCE；$

Solution:
$证明：\because 四边形ABCD是正方形，\therefore AB\parallel CD，又AB\not\subset 平面DCE，CD\subset 平面DCE，\therefore AB\parallel 平面DCE.(2分)\because DE\perp 平面ABCD，AF\perp 平面ABCD，\therefore DE\parallel AF，又AF\not\subset 平面DCE，DE\subset 平面DCE，\therefore AF\parallel 平面DCE，(4分)又AF\cap AB=A，AF\subset 平面ABF，AB\subset 平面ABF，\therefore 平面ABF\parallel 平面DCE.(6分)$

Final answer: 证明题
Subfield: Solid Geometry


Question: $如图，四边形ABCD是边长为3的正方形，DE\perp 平面ABCD，AF\perp 平面ABCD，DE=3，AF=1.$<img_3082>
$(1)证明：平面ABF\parallel 平面DCE；$

Solution:
$证明：\because 四边形ABCD是正方形，\therefore AB\parallel CD，又AB\not\subset 平面DCE，CD\subset 平面DCE，\therefore AB\parallel 平面DCE.(2分)\because DE\perp 平面ABCD，AF\perp 平面ABCD，\therefore DE\parallel AF，又AF\not\subset 平面DCE，DE\subset 平面DCE，\therefore AF\parallel 平面DCE，(4分)又AF\cap AB=A，AF\subset 平面ABF，AB\subset 平面ABF，\therefore 平面ABF\parallel 平面DCE.(6分)$
Subfield: Solid Geometry


Question: $如图，四边形ABCD是边长为3的正方形，DE\perp 平面ABCD，AF\perp 平面ABCD，DE=3，AF=1.$![Image](https://cdntk.53zaixian.com/tq/1650310153448853506/img/compression/M_7cee04b770fb4bd5b813f675982682cb.jpg)
$(2)在DE上是否存在一点G，使平面FBG将多面体ABCDEF分成上、下两部分，且体积比为3:11？若存在，求出点G的位置；若不存在，请说明理由.$

Solution:
$假设在DE上存在一点G，使平面FBG将多面体ABCDEF分成上、下两部分，且体积比为3:11.如图，过点G作MG\parallel BF，交EC于点M，连接BG，BM，GF，BD，V_{多面体ABCDEF}=V_{B-ADEF}+V_{B-CDE}=\frac{1}{3}\times 3\times \frac{(1+3)\times 3}{2}+\frac{1}{3}\times 3\times \frac{3\times 3}{2}=\frac{21}{2}.
(8分)
设EG=t，t＞0，则V_{多面体GFBME}=V_{B-EFG}+V_{B-EGM}=\frac{21}{2}\times \frac{3}{14}=\frac{9}{4}.设点M到DE的距离为h，过F作FN\parallel AD，交DE于点N，连接NC，则四边形ADNF为矩形，\therefore DN=AF=1，FN=AD，又BC\xlongequal{//} AD，\therefore FN\xlongequal{//} BC，\therefore 四边形FBCN是平行四边形，\therefore BF\parallel NC，\therefore GM\parallel NC，则\frac{h}{3}=\frac{EG}{EN}=\frac{t}{3-1}，
(9分)
\therefore h=\frac{3}{2}t，\therefore S_{\triangle EGM}=\frac{3}{4}t^{2}，\therefore \frac{1}{3}\times 3\times \frac{3}{4}t^{2}+\frac{1}{3}\times 3\times \frac{3}{2}t=\frac{9}{4}，解得t=1或t=-3(舍去)，
(11分)
\therefore 在DE上存在一点G，使平面FBG将多面体ABCDEF分成上、下两部分，且体积比为3:11，点G为DE的一个三等分点，EG=1.
(12分)
$![Image](https://cdntk.53zaixian.com/tq/1650310153947975682/img/compression/M_308e698272916235fc1f17cdc3312731.jpg)

Final answer: $在DE上存在一点G，使平面FBG将多面体ABCDEF分成上、下两部分，且体积比为3:11，点G为DE的一个三等分点，EG=1.$
Subfield: Solid Geometry


Question: $如图，在直三棱柱ABC-A_{1}B_{1}C_{1}中，AC=2，BC=2\sqrt{3}，AC\perp BC，D是线段AB上的动点.$<img_1751>
$(1)当D是AB的中点时，证明：AC_{1}\parallel 平面B_{1}CD；$

Solution:
$证明：如图，连接BC_{1}，交B_{1}C于E，则E是BC_{1}的中点，连接DE，又\because D是AB的中点，\therefore DE\parallel AC_{1}，又DE\subset 平面B_{1}CD，AC_{1}\not\subset 平面B_{1}CD，\therefore AC_{1}\parallel 平面B_{1}CD.
(6分)
$<img_2762>
Subfield: Solid Geometry


Question: $如图，在直三棱柱ABC-A_{1}B_{1}C_{1}中，AC=2，BC=2\sqrt{3}，AC\perp BC，D是线段AB上的动点.$![Image](https://cdntk.53zaixian.com/tq/1650310211502215170/img/compression/M_05ebb69fe628c7f5d40422cbf6765b15.jpg)
$(1)当D是AB的中点时，证明：AC_{1}\parallel 平面B_{1}CD；$

Solution:
$证明：如图，连接BC_{1}，交B_{1}C于E，则E是BC_{1}的中点，连接DE，又\because D是AB的中点，\therefore DE\parallel AC_{1}，又DE\subset 平面B_{1}CD，AC_{1}\not\subset 平面B_{1}CD，\therefore AC_{1}\parallel 平面B_{1}CD.
(6分)
$![Image](https://cdntk.53zaixian.com/tq/1650310211502215170/img/compression/M_e370e73214ca044f4a76d0cb31d8f2b8.jpg)

Final answer: 证明题
Subfield: Solid Geometry


Question: $如图，在直三棱柱ABC-A_{1}B_{1}C_{1}中，AC=2，BC=2\sqrt{3}，AC\perp BC，D是线段AB上的动点.$<img_1751>
$(2)若CD\perp AB，证明：平面ABB_{1}A_{1}\perp 平面B_{1}CD.$

Solution:
$证明：\because AA_{1}\perp 平面ABC，CD\subset 平面ABC，\therefore AA_{1}\perp CD，又CD\perp AB，AA_{1}\cap AB=A，AB，AA_{1}\subset 平面ABB_{1}A_{1}，\therefore CD\perp 平面ABB_{1}A_{1}，
(10分)
又CD\subset 平面B_{1}CD，\therefore 平面ABB_{1}A_{1}\perp 平面B_{1}CD.
(12分)$
Subfield: Solid Geometry


Question: $如图，在直三棱柱ABC-A_{1}B_{1}C_{1}中，AC=2，BC=2\sqrt{3}，AC\perp BC，D是线段AB上的动点.$![Image](https://cdntk.53zaixian.com/tq/1650310211502215170/img/compression/M_05ebb69fe628c7f5d40422cbf6765b15.jpg)
$(2)若CD\perp AB，证明：平面ABB_{1}A_{1}\perp 平面B_{1}CD.$

Solution:
$证明：\because AA_{1}\perp 平面ABC，CD\subset 平面ABC，\therefore AA_{1}\perp CD，又CD\perp AB，AA_{1}\cap AB=A，AB，AA_{1}\subset 平面ABB_{1}A_{1}，\therefore CD\perp 平面ABB_{1}A_{1}，
(10分)
又CD\subset 平面B_{1}CD，\therefore 平面ABB_{1}A_{1}\perp 平面B_{1}CD.
(12分)$

Final answer: 证明题
Subfield: Solid Geometry


Question: $如图，在正三棱柱A_{1}B_{1}C_{1}-ABC中，O为AB_{1}与A_{1}B的交点，M为BC_{1}的中点，\overrightarrow{EF}=2\overrightarrow{AE}=2\overrightarrow{FC_1}.$![Image](https://cdntk.53zaixian.com/tq/1650310230091370497/img/compression/M_131eb93fdbbd8ef7b385d9b61f9aaa0a.jpg)
$(1)证明：OM\parallel 平面A_{1}B_{1}C_{1}；$

Solution:
$证明：由题意得O为A_{1}B的中点，又\because M为BC_{1}的中点，\therefore OM为\triangle A_{1}BC_{1}的中位线，\therefore OM\parallel A_{1}C_{1}，(2分)又\because A_{1}C_{1}\subset 平面A_{1}B_{1}C_{1}，OM\not\subset 平面A_{1}B_{1}C_{1}，\therefore OM\parallel 平面A_{1}B_{1}C_{1}.(4分)$

Final answer: 证明题
Subfield: Solid Geometry


Question: $如图，在正三棱柱A_{1}B_{1}C_{1}-ABC中，O为AB_{1}与A_{1}B的交点，M为BC_{1}的中点，\overrightarrow{EF}=2\overrightarrow{AE}=2\overrightarrow{FC_1}.$<img_3060>
$(1)证明：OM\parallel 平面A_{1}B_{1}C_{1}；$

Solution:
$证明：由题意得O为A_{1}B的中点，又\because M为BC_{1}的中点，\therefore OM为\triangle A_{1}BC_{1}的中位线，\therefore OM\parallel A_{1}C_{1}，(2分)又\because A_{1}C_{1}\subset 平面A_{1}B_{1}C_{1}，OM\not\subset 平面A_{1}B_{1}C_{1}，\therefore OM\parallel 平面A_{1}B_{1}C_{1}.(4分)$
Subfield: Solid Geometry


Question: $如图，在正三棱柱A_{1}B_{1}C_{1}-ABC中，O为AB_{1}与A_{1}B的交点，M为BC_{1}的中点，\overrightarrow{EF}=2\overrightarrow{AE}=2\overrightarrow{FC_1}.$<img_3060>
$(2)若G为线段FC上一动点，在平面AA_{1}B_{1}B上是否存在一点N，使得NG\parallel 平面A_{1}OE恒成立？若存在，请找出点N位置并证明NG\parallel 平面A_{1}OE；若不存在，请说明理由.$

Solution:
$存在点N，当N在AB_{1}的延长线上，且AB_{1}=2B_{1}N时满足条件.(5分)证明如下：如图，延长AB_{1}至N，使得AB_{1}=2B_{1}N，连接FN，CN.$<img_2631>$\because \left\{\begin{matrix}A_1A=CC_1，\\ \angle A_1AE=\angle CC_1F，\\ AE=C_1F，\end{matrix}\right.\therefore \triangle A_{1}AE\cong \triangle CC_{1}F，(7分)\therefore \angle A_{1}EA=\angle CFC_{1}，\therefore \angle CFE=\angle A_{1}EF，\therefore A_{1}E\parallel CF.又\because A_{1}E\subset 平面A_{1}OE，CF\not\subset 平面A_{1}OE，\therefore CF\parallel 平面A_{1}OE.(8分)由\overrightarrow{EF}=2\overrightarrow{AE}=2\overrightarrow{FC_1}及AB_{1}=2B_{1}N，得\frac{AO}{AN}=\frac{AE}{AF}=\frac{1}{3}，\therefore \triangle AEO∽\triangle AFN，\therefore EO\parallel FN.又\because EO\subset 平面A_{1}OE，FN\not\subset 平面A_{1}OE，\therefore FN\parallel 平面A_{1}OE.(10分)又\because FN，CF\subset 平面FCN且FN\cap CF=F，\therefore 平面A_{1}OE\parallel 平面FCN.(11分)易知NG\subset 平面FCN，\therefore NG\parallel 平面A_{1}OE.(12分)$



Final answer: $存在点N，当N在AB_{1}的延长线上，且AB_{1}=2B_{1}N时满足条件$
Subfield: Solid Geometry


Question: $如图，在正三棱柱A_{1}B_{1}C_{1}-ABC中，O为AB_{1}与A_{1}B的交点，M为BC_{1}的中点，\overrightarrow{EF}=2\overrightarrow{AE}=2\overrightarrow{FC_1}.$![Image](https://cdntk.53zaixian.com/tq/1650310230091370497/img/compression/M_131eb93fdbbd8ef7b385d9b61f9aaa0a.jpg)
$(2)若G为线段FC上一动点，在平面AA_{1}B_{1}B上是否存在一点N，使得NG\parallel 平面A_{1}OE恒成立？若存在，请找出点N位置并证明NG\parallel 平面A_{1}OE；若不存在，请说明理由.$

Solution:
$存在点N，当N在AB_{1}的延长线上，且AB_{1}=2B_{1}N时满足条件.(5分)证明如下：如图，延长AB_{1}至N，使得AB_{1}=2B_{1}N，连接FN，CN.$![Image](https://cdntk.53zaixian.com/tq/1650310230804402177/img/compression/M_442dc0efa8409077c49246cd27427cc2.jpg)$\because \left\{\begin{matrix}A_1A=CC_1，\\ \angle A_1AE=\angle CC_1F，\\ AE=C_1F，\end{matrix}\right.\therefore \triangle A_{1}AE\cong \triangle CC_{1}F，(7分)\therefore \angle A_{1}EA=\angle CFC_{1}，\therefore \angle CFE=\angle A_{1}EF，\therefore A_{1}E\parallel CF.又\because A_{1}E\subset 平面A_{1}OE，CF\not\subset 平面A_{1}OE，\therefore CF\parallel 平面A_{1}OE.(8分)由\overrightarrow{EF}=2\overrightarrow{AE}=2\overrightarrow{FC_1}及AB_{1}=2B_{1}N，得\frac{AO}{AN}=\frac{AE}{AF}=\frac{1}{3}，\therefore \triangle AEO∽\triangle AFN，\therefore EO\parallel FN.又\because EO\subset 平面A_{1}OE，FN\not\subset 平面A_{1}OE，\therefore FN\parallel 平面A_{1}OE.(10分)又\because FN，CF\subset 平面FCN且FN\cap CF=F，\therefore 平面A_{1}OE\parallel 平面FCN.(11分)易知NG\subset 平面FCN，\therefore NG\parallel 平面A_{1}OE.(12分)$

Final answer: $存在点N，当N在AB_{1}的延长线上，且AB_{1}=2B_{1}N时满足条件$
Subfield: Solid Geometry


Question: $如图，在三棱柱ABC-A_{1}B_{1}C_{1}中，AB\perp BC,AB=BC=2，D为AC中点，四边形BCC_{1}B_{1}为正方形.$![Image](https://cdntk.53zaixian.com/tq/1623874893693214722/img/compression/M_988b5fccf5685e30662ed1784267102c.jpg)
$(1)求证：B_{1}C\parallel 平面A_{1}BD；$

Solution:
$证明：连接AB_{1}与A_{1}B交于点O,连接OD，因为O、D分别是\triangle AB_{1}C的边AB_{1}与AC的中点，所以OD\parallel B_{1}C，又OD\subset 平面A_{1}BD，B_{1}C\not\subset 平面A_{1}BD，所以B_{1}C\parallel 平面A_{1}BD.$

Final answer: 证明题
Subfield: Solid Geometry


Question: $如图，在三棱柱ABC-A_{1}B_{1}C_{1}中，AB\perp BC,AB=BC=2，D为AC中点，四边形BCC_{1}B_{1}为正方形.$<img_3097>
$(1)求证：B_{1}C\parallel 平面A_{1}BD；$

Solution:
$证明：连接AB_{1}与A_{1}B交于点O,连接OD，因为O、D分别是\triangle AB_{1}C的边AB_{1}与AC的中点，所以OD\parallel B_{1}C，又OD\subset 平面A_{1}BD，B_{1}C\not\subset 平面A_{1}BD，所以B_{1}C\parallel 平面A_{1}BD.$
Subfield: Solid Geometry


Question: $如图，在三棱柱ABC-A_{1}B_{1}C_{1}中，AB\perp BC,AB=BC=2，D为AC中点，四边形BCC_{1}B_{1}为正方形.$![Image](https://cdntk.53zaixian.com/tq/1623874893693214722/img/compression/M_988b5fccf5685e30662ed1784267102c.jpg)
$(2)再从条件①、条件②这两个条件中选择一个作为已知，求直线AB与平面A_{1}BD所成角的正弦值.条件①：AB\perp B_{1}C；条件②：A_{1}B=B_{1}C.注：如果选择条件①和条件②分别解答，按第一个解答计分.$

Solution:
$若选条件①：因为AB\perp BC，AB\perp B_{1}C，且BC\cap B_{1}C=C，BC,B_{1}C\subset 平面BCC_{1}B_{1}，所以AB\perp 平面BCC_{1}B_{1}，又因为BB_{1}\subset 平面BCC_{1}B_{1}，所以AB\perp BB_{1}，以B为原点，以BC,BA,BB_{1}所在直线分别为x轴、y轴和z轴建立空间直角坐标系，如图所示，$![Image](https://cdntk.53zaixian.com/tq/1623874894242668545/img/compression/M_37660af580ff4f82b55dac8ce72d203d.jpg)$因为AB=BC=2且四边形BCC_{1}B_{1}为正方形，则A(0,2,0),B(0,0,0),D(1,1,0),A_{1}(0,2,2)，所以\overrightarrow{BA}=(0,2,0),\overrightarrow{BD}=(1,1,0),\overrightarrow{BA_1}=(0,2,2)，设平面A_{1}BD的法向量为m=(x,y,z)，则\left\{\begin{matrix}\boldsymbol{m} \cdot \overrightarrow{BD}=x+y=0,\\ \boldsymbol{m} \cdot \overrightarrow{BA_1}=2y+2z=0,\end{matrix}\right.令y=-1，得x=1,z=1，所以m=(1,-1,1)，设直线AB与平面A_{1}BD所成角为\theta ，则\\sin   \theta =|\\cos  \langle m,\overrightarrow{BA}\rangle |=\frac{|\boldsymbol{m} \cdot \overrightarrow{BA}|}{|\boldsymbol{m}| \cdot |\overrightarrow{BA}|}=\frac{2}{\sqrt{3} \cdot 2}=\frac{\sqrt{3}}{3}，即直线AB与平面A_{1}BD所成角的正弦值为\frac{\sqrt{3}}{3}.若选条件②：由四边形BCC_{1}B_{1}为正方形且BC=2，可得B_{1}C=2\sqrt{2}，所以A_{1}B=B_{1}C=2\sqrt{2}，A_{1}B^2_1+BB^2_1=8=A_{1}B^{2}，所以A_{1}B_{1}\perp BB_{1}，所以BB_{1}\perp BA，以B为原点，以BC,BA,BB_{1}所在直线分别为x轴、y轴和z轴建立空间直角坐标系，以下同选条件①.$

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


Question: $如图，在三棱柱ABC-A_{1}B_{1}C_{1}中，AB\perp BC,AB=BC=2，D为AC中点，四边形BCC_{1}B_{1}为正方形.$<img_3097>
$(2)再从条件①、条件②这两个条件中选择一个作为已知，求直线AB与平面A_{1}BD所成角的正弦值.条件①：AB\perp B_{1}C；条件②：A_{1}B=B_{1}C.注：如果选择条件①和条件②分别解答，按第一个解答计分.$

Solution:
$若选条件①：因为AB\perp BC，AB\perp B_{1}C，且BC\cap B_{1}C=C，BC,B_{1}C\subset 平面BCC_{1}B_{1}，所以AB\perp 平面BCC_{1}B_{1}，又因为BB_{1}\subset 平面BCC_{1}B_{1}，所以AB\perp BB_{1}，以B为原点，以BC,BA,BB_{1}所在直线分别为x轴、y轴和z轴建立空间直角坐标系，如图所示，$<img_2753>$因为AB=BC=2且四边形BCC_{1}B_{1}为正方形，则A(0,2,0),B(0,0,0),D(1,1,0),A_{1}(0,2,2)，所以\overrightarrow{BA}=(0,2,0),\overrightarrow{BD}=(1,1,0),\overrightarrow{BA_1}=(0,2,2)，设平面A_{1}BD的法向量为m=(x,y,z)，则\left\{\begin{matrix}\boldsymbol{m} \cdot \overrightarrow{BD}=x+y=0,\\ \boldsymbol{m} \cdot \overrightarrow{BA_1}=2y+2z=0,\end{matrix}\right.令y=-1，得x=1,z=1，所以m=(1,-1,1)，设直线AB与平面A_{1}BD所成角为\theta ，则\\sin   \theta =|\\cos  \langle m,\overrightarrow{BA}\rangle |=\frac{|\boldsymbol{m} \cdot \overrightarrow{BA}|}{|\boldsymbol{m}| \cdot |\overrightarrow{BA}|}=\frac{2}{\sqrt{3} \cdot 2}=\frac{\sqrt{3}}{3}，即直线AB与平面A_{1}BD所成角的正弦值为\frac{\sqrt{3}}{3}.若选条件②：由四边形BCC_{1}B_{1}为正方形且BC=2，可得B_{1}C=2\sqrt{2}，所以A_{1}B=B_{1}C=2\sqrt{2}，A_{1}B^2_1+BB^2_1=8=A_{1}B^{2}，所以A_{1}B_{1}\perp BB_{1}，所以BB_{1}\perp BA，以B为原点，以BC,BA,BB_{1}所在直线分别为x轴、y轴和z轴建立空间直角坐标系，以下同选条件①.$

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