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


Question: $如图，四棱锥P-ABCD中，底面ABCD是边长为2的菱形，\angle DAB=60^\circ ，\angle ADP=90^\circ ，平面ADP\perp 平面ABCD，点F为棱PD的中点.$![Image](https://cdntk.53zaixian.com/tq/1706233566166462465/img/compression/M_7371af061860f58f2019ce4db996f5db.jpg)
$(2)当二面角D-FC-B的余弦值为\frac{\sqrt{2}}{4}时，求直线PB与平面ABCD所成的角.$

Solution:
$设PD=a，$![Image](https://cdntk.53zaixian.com/tq/1706233566397149185/img/compression/M_6f20dfd7d9717a8eb9cdda6e8a850307.jpg)$由(1)知PD\perp 平面ABCD，则\angle PBD就是直线PB与平面ABCD所成的角.由题意得，\triangle BDC为等边三角形，过B作BH\perp CD于H，则H为CD的中点，\because PD\perp 平面ABCD,BH\subset 平面ABCD，\therefore PD\perp BH，又PD\cap CD=D,PD，CD\subset 平面PDC，\therefore BH\perp 平面PDC.过H作HG\perp FC于G，连接BG，则CF\perp BG，\angle BGH就是二面角D-FC-B的平面角.(平面PDC的垂线比较明显，用垂线法找二面角D-FC-B的平面角)\because \\cos  \angle BGH=\frac{\sqrt{2}}{4},\therefore \\tan  \angle BGH=\sqrt{7}，易得BH=\sqrt{3},\therefore GH=\frac{\sqrt{21}}{7}.\because \\sin  \angle GCH=\frac{GH}{HC}=\frac{FD}{FC},\therefore \frac{\frac{\sqrt{21}}{7}}{1}=\frac{\frac{a}{2}}{\sqrt{\left(\frac{a}{2}\right)^2+2^2}}，\therefore a=2\sqrt{3}，\therefore \\tan  \angle PBD=\frac{PD}{BD}=\frac{2\sqrt{3}}{2}=\sqrt{3}，\therefore \angle PBD=60^\circ ，即直线PB与平面ABCD所成的角为60^\circ .$

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


Question: $如图，四棱锥P-ABCD中，底面ABCD是边长为2的菱形，\angle DAB=60^\circ ，\angle ADP=90^\circ ，平面ADP\perp 平面ABCD，点F为棱PD的中点.$<img_1644>
$(2)当二面角D-FC-B的余弦值为\frac{\sqrt{2}}{4}时，求直线PB与平面ABCD所成的角.$

Solution:
$设PD=a，$<img_2590>$由(1)知PD\perp 平面ABCD，则\angle PBD就是直线PB与平面ABCD所成的角.由题意得，\triangle BDC为等边三角形，过B作BH\perp CD于H，则H为CD的中点，\because PD\perp 平面ABCD,BH\subset 平面ABCD，\therefore PD\perp BH，又PD\cap CD=D,PD，CD\subset 平面PDC，\therefore BH\perp 平面PDC.过H作HG\perp FC于G，连接BG，则CF\perp BG，\angle BGH就是二面角D-FC-B的平面角.(平面PDC的垂线比较明显，用垂线法找二面角D-FC-B的平面角)\because \\cos  \angle BGH=\frac{\sqrt{2}}{4},\therefore \\tan  \angle BGH=\sqrt{7}，易得BH=\sqrt{3},\therefore GH=\frac{\sqrt{21}}{7}.\because \\sin  \angle GCH=\frac{GH}{HC}=\frac{FD}{FC},\therefore \frac{\frac{\sqrt{21}}{7}}{1}=\frac{\frac{a}{2}}{\sqrt{\left(\frac{a}{2}\right)^2+2^2}}，\therefore a=2\sqrt{3}，\therefore \\tan  \angle PBD=\frac{PD}{BD}=\frac{2\sqrt{3}}{2}=\sqrt{3}，\therefore \angle PBD=60^\circ ，即直线PB与平面ABCD所成的角为60^\circ .$

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


Question: $已知三棱锥P-ABC的底面ABC是边长为2的等边三角形，PA\perp 平面ABC，\angle PCA=45^\circ ，点M为线段PC上一动点.$![Image](https://cdntk.53zaixian.com/tq/1707197310870700033/img/compression/M_fd48d820957f9916b0df81bab1ab3a27.jpg)
$(1)当点M为PC的中点时，证明：BM\perp AC；$

Solution:
$证明：设E为AC的中点，连接ME,BE,因为点M为PC的中点，所以ME\parallel PA,又PA\perp 平面ABC，所以ME\perp 平面ABC，因为AC\subset 平面ABC，所以ME\perp AC.因为底面ABC是等边三角形，所以BE\perp AC,又ME\cap BE=E,ME,BE\subset 平面MBE，所以AC\perp 平面MBE，又BM\subset 平面MBE，所以BM\perp AC.$

Final answer: 证明题
Subfield: Solid Geometry


Question: $已知三棱锥P-ABC的底面ABC是边长为2的等边三角形，PA\perp 平面ABC，\angle PCA=45^\circ ，点M为线段PC上一动点.$<img_2341>
$(1)当点M为PC的中点时，证明：BM\perp AC；$

Solution:
$证明：设E为AC的中点，连接ME,BE,因为点M为PC的中点，所以ME\parallel PA,又PA\perp 平面ABC，所以ME\perp 平面ABC，因为AC\subset 平面ABC，所以ME\perp AC.因为底面ABC是等边三角形，所以BE\perp AC,又ME\cap BE=E,ME,BE\subset 平面MBE，所以AC\perp 平面MBE，又BM\subset 平面MBE，所以BM\perp AC.$
Subfield: Solid Geometry


Question: $已知三棱锥P-ABC的底面ABC是边长为2的等边三角形，PA\perp 平面ABC，\angle PCA=45^\circ ，点M为线段PC上一动点.$<img_2341>
$(2)当平面ABC与平面ABM所成二面角的平面角为60^\circ 时，求此时MC的长度.$

Solution:
$如图，过点M作MN\perp AC,垂足为N,因为PA\perp 平面ABC，AC\subset 平面ABC，所以PA\perp AC,所以PA\parallel MN,所以MN\perp 平面ABC，又AB\subset 平面ABC，所以MN\perp AB,$<img_3080>$过点N作NF\perp AB,垂足为F,连接MF,因为MN\cap NF=N,MN,NF\subset 平面MNF，所以AB\perp 平面MNF，又MF\subset 平面MNF,所以AB\perp MF,又NF\subset 平面ABC,MF\subset 平面ABM,所以\angle MFN为平面ABC与平面ABM所成二面角的平面角，即\angle MFN=60^\circ ，设NF=x，则MN=\sqrt{3}x，因为\angle PCA=45^\circ ，所以NC=\sqrt{3}x,MC=\sqrt{6}x,又底面ABC是等边三角形，所以\angle FAN=60^\circ ，所以AN=\frac{x}{\\\sin   60^{\circ}}=\frac{2\sqrt{3}}{3}x，又AC=2,所以\frac{2\sqrt{3}}{3}x+\sqrt{3}x=2,所以x=\frac{2\sqrt{3}}{5}，则MC=\sqrt{6}x=\frac{6\sqrt{2}}{5}，又CP=2\sqrt{2},所以MC=\frac{3}{5}CP.$





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


Question: $已知三棱锥P-ABC的底面ABC是边长为2的等边三角形，PA\perp 平面ABC，\angle PCA=45^\circ ，点M为线段PC上一动点.$![Image](https://cdntk.53zaixian.com/tq/1707197310870700033/img/compression/M_fd48d820957f9916b0df81bab1ab3a27.jpg)
$(2)当平面ABC与平面ABM所成二面角的平面角为60^\circ 时，试确定点M的位置.$

Solution:
$如图，过点M作MN\perp AC,垂足为N,因为PA\perp 平面ABC，AC\subset 平面ABC，所以PA\perp AC,所以PA\parallel MN,所以MN\perp 平面ABC，又AB\subset 平面ABC，所以MN\perp AB,$![Image](https://cdntk.53zaixian.com/tq/1707197310870700033/img/compression/M_19b70776afa31a63f5c1d329b6e97f7e.jpg)$过点N作NF\perp AB,垂足为F,连接MF,因为MN\cap NF=N,MN,NF\subset 平面MNF，所以AB\perp 平面MNF，又MF\subset 平面MNF,所以AB\perp MF,又NF\subset 平面ABC,MF\subset 平面ABM,所以\angle MFN为平面ABC与平面ABM所成二面角的平面角，即\angle MFN=60^\circ ，设NF=x，则MN=\sqrt{3}x，因为\angle PCA=45^\circ ，所以NC=\sqrt{3}x,MC=\sqrt{6}x,又底面ABC是等边三角形，所以\angle FAN=60^\circ ，所以AN=\frac{x}{\\\sin   60^{\circ}}=\frac{2\sqrt{3}}{3}x，又AC=2,所以\frac{2\sqrt{3}}{3}x+\sqrt{3}x=2,所以x=\frac{2\sqrt{3}}{5}，则MC=\sqrt{6}x=\frac{6\sqrt{2}}{5}，又CP=2\sqrt{2},所以MC=\frac{3}{5}CP.$

Final answer: $MC=6\sqrt 2/5
MC=3/5CP$
Subfield: Solid Geometry


Question: $如图所示，在四棱锥V-ABCD中，底面ABCD是正方形，侧面VAD是正三角形，平面VAD\perp 底面ABCD.$![Image](https://cdntk.53zaixian.com/tq/1707197316235214850/img/compression/M_8a7390fc56b129e4de37ac4cbdedcf5f.jpg)
$(1)求证：AB\perp 平面VAD；$

Solution:
$证明：\because 底面ABCD是正方形，\therefore AB\perp AD，又平面VAD\perp 底面ABCD，平面VAD\cap 底面ABCD=AD，AB\subset 底面ABCD，\therefore AB\perp 平面VAD.$

Final answer: 证明题
Subfield: Solid Geometry


Question: $如图所示，在四棱锥V-ABCD中，底面ABCD是正方形，侧面VAD是正三角形，平面VAD\perp 底面ABCD.$<img_1910>
$(1)求证：AB\perp 平面VAD；$

Solution:
$证明：\because 底面ABCD是正方形，\therefore AB\perp AD，又平面VAD\perp 底面ABCD，平面VAD\cap 底面ABCD=AD，AB\subset 底面ABCD，\therefore AB\perp 平面VAD.$
Subfield: Solid Geometry


Question: $如图所示，在四棱锥V-ABCD中，底面ABCD是正方形，侧面VAD是正三角形，平面VAD\perp 底面ABCD.$![Image](https://cdntk.53zaixian.com/tq/1707197316235214850/img/compression/M_8a7390fc56b129e4de37ac4cbdedcf5f.jpg)
$(2)求平面VAD与平面VDB所成的二面角的正切值.$

Solution:
$取VD的中点E，连接AE，BE，$![Image](https://cdntk.53zaixian.com/tq/1707197316533010433/img/compression/M_c9adc29f5dafc1e9f2ff1ccc6ee49f46.jpg)$\because \triangle VAD是正三角形，\therefore AE\perp VD，AE=\frac{\sqrt{3}}{2}AD，\because AB\perp 平面VAD，VD\subset 平面VAD，\therefore AB\perp VD，又AB\cap AE=A，AB，AE\subset 平面ABE，\therefore VD\perp 平面ABE，\because BE\subset 平面ABE，\therefore VD\perp BE，\therefore \angle AEB就是平面VAD与平面VDB所成的二面角的平面角，在Rt\triangle BAE中，\\tan  \angle AEB=\frac{AB}{AE}=\frac{AD}{\frac{\sqrt{3}}{2}AD}=\frac{2\sqrt{3}}{3}，\therefore 平面VAD与平面VDB所成的二面角的正切值为\frac{2\sqrt{3}}{3}.$

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


Question: $如图所示，在四棱锥V-ABCD中，底面ABCD是正方形，侧面VAD是正三角形，平面VAD\perp 底面ABCD.$<img_1910>
$(2)求平面VAD与平面VDB所成的二面角的正切值.$

Solution:
$取VD的中点E，连接AE，BE，$<img_2223>$\because \triangle VAD是正三角形，\therefore AE\perp VD，AE=\frac{\sqrt{3}}{2}AD，\because AB\perp 平面VAD，VD\subset 平面VAD，\therefore AB\perp VD，又AB\cap AE=A，AB，AE\subset 平面ABE，\therefore VD\perp 平面ABE，\because BE\subset 平面ABE，\therefore VD\perp BE，\therefore \angle AEB就是平面VAD与平面VDB所成的二面角的平面角，在Rt\triangle BAE中，\\tan  \angle AEB=\frac{AB}{AE}=\frac{AD}{\frac{\sqrt{3}}{2}AD}=\frac{2\sqrt{3}}{3}，\therefore 平面VAD与平面VDB所成的二面角的正切值为\frac{2\sqrt{3}}{3}.$

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


Question: $如图，在正方体ABCD-A_{1}B_{1}C_{1}D_{1}中，E为棱AA_{1}的中点，过点B，E，D_{1}的平面与棱CC_{1}交于点F.$![Image](https://cdntk.53zaixian.com/tq/1707197288069955586/img/compression/M_aea1fd049898e7433cd50a2262d5d7c8.jpg)
$(1)求证：四边形BFD_{1}E为平行四边形；$

Solution:
$证明：在正方体ABCD-A_{1}B_{1}C_{1}D_{1}中，因为平面ABB_{1}A_{1}\parallel 平面DCC_{1}D_{1}，平面BFD_{1}E\cap 平面ABB_{1}A_{1}=BE,平面BFD_{1}E\cap 平面DCC_{1}D_{1}=D_{1}F，所以BE\parallel D_{1}F.同理可证ED_{1}\parallel FB，所以四边形BFD_{1}E为平行四边形.$

Final answer: 证明题
Subfield: Solid Geometry


Question: $如图，在正方体ABCD-A_{1}B_{1}C_{1}D_{1}中，E为棱AA_{1}的中点，过点B，E，D_{1}的平面与棱CC_{1}交于点F.$<img_2625>
$(1)求证：四边形BFD_{1}E为平行四边形；$

Solution:
$证明：在正方体ABCD-A_{1}B_{1}C_{1}D_{1}中，因为平面ABB_{1}A_{1}\parallel 平面DCC_{1}D_{1}，平面BFD_{1}E\cap 平面ABB_{1}A_{1}=BE,平面BFD_{1}E\cap 平面DCC_{1}D_{1}=D_{1}F，所以BE\parallel D_{1}F.同理可证ED_{1}\parallel FB，所以四边形BFD_{1}E为平行四边形.$
Subfield: Solid Geometry


Question: $如图，在正方体ABCD-A_{1}B_{1}C_{1}D_{1}中，E为棱AA_{1}的中点，过点B，E，D_{1}的平面与棱CC_{1}交于点F.$![Image](https://cdntk.53zaixian.com/tq/1707197288069955586/img/compression/M_aea1fd049898e7433cd50a2262d5d7c8.jpg)
$(2)试确定点F的位置.$

Solution:
$由(1)知四边形BFD_{1}E为平行四边形，所以FB=ED_{1}，因为CB=A_{1}D_{1},\angle BCF=\angle D_{1}A_{1}E，所以\triangle BCF\cong \triangle D_{1}A_{1}E，所以CF=EA_{1}，又E是AA_{1}的中点，所以F是CC_{1}的中点.$

Final answer: $F是CC_{1}的中点$
Subfield: Solid Geometry


Question: $如图所示，三棱柱ABC-A_{1}B_{1}C_{1}的各棱长均为4，侧棱垂直于底面，E,F,G,H分别为AB,AC,A_{1}C_{1},A_{1}B_{1}的中点.求：$<img_2721>
$(1)B_{1}C_{1}与平面BCGH的距离；$

Solution:
$分别取B_{1}C_{1}，BC的中点D_{1},D,连接A_{1}D_{1}交HG于M，连接DD_{1},DM,如图所示.由题意易得HG\perp A_{1}D_{1},HG\perp DD_{1},\because A_{1}D_{1}\cap DD_{1}=D_{1}，A_{1}D_{1}，DD_{1}\subset 平面MD_{1}D，\therefore HG\perp 平面MD_{1}D.过点D_{1}作D_{1}P\perp MD，垂足为P，易知 D_{1}P\perp HG,又HG\cap MD=M,\therefore D_{1}P\perp 平面BCGH,\therefore D_{1}P的长就是点D_{1}与平面BCGH的距离.易知HG\parallel B_{1}C_{1}，HG\subset 平面BCGH,B_{1}C_{1}\not\subset 平面BCGH,\therefore B_{1}C_{1}\parallel 平面BCGH,\therefore D_{1}到平面BCGH的距离和B_{1}C_{1}与平面BCGH的距离相等,\therefore D_{1}P的长即为所求.在Rt\triangle DD_{1}M中，DD_{1}=4,D_{1}M=\sqrt{3},\therefore MD=\sqrt{19},\therefore D_{1}P=\frac{DD_1\cdot D_1M}{MD}=\frac{4\sqrt{3}}{\sqrt{19}}=\frac{4\sqrt{57}}{19}.\therefore B_{1}C_{1}与平面BCGH的距离是\frac{4\sqrt{57}}{19}.$<img_2827>

Final answer: $\frac{4\sqrt{57}}{19}$
Subfield: Solid Geometry


Question: $如图所示，三棱柱ABC-A_{1}B_{1}C_{1}的各棱长均为4，侧棱垂直于底面，E,F,G,H分别为AB,AC,A_{1}C_{1},A_{1}B_{1}的中点.求：$![Image](https://cdntk.53zaixian.com/tq/1707197294097170434/img/compression/M_899bf44086019995aea5981f6640aab5.jpg)
$(1)B_{1}C_{1}与平面BCGH的距离；$

Solution:
$分别取B_{1}C_{1}，BC的中点D_{1},D,连接A_{1}D_{1}交HG于M，连接DD_{1},DM,如图所示.由题意易得HG\perp A_{1}D_{1},HG\perp DD_{1},\because A_{1}D_{1}\cap DD_{1}=D_{1}，A_{1}D_{1}，DD_{1}\subset 平面MD_{1}D，\therefore HG\perp 平面MD_{1}D.过点D_{1}作D_{1}P\perp MD，垂足为P，易知 D_{1}P\perp HG,又HG\cap MD=M,\therefore D_{1}P\perp 平面BCGH,\therefore D_{1}P的长就是点D_{1}与平面BCGH的距离.易知HG\parallel B_{1}C_{1}，HG\subset 平面BCGH,B_{1}C_{1}\not\subset 平面BCGH,\therefore B_{1}C_{1}\parallel 平面BCGH,\therefore D_{1}到平面BCGH的距离和B_{1}C_{1}与平面BCGH的距离相等,\therefore D_{1}P的长即为所求.在Rt\triangle DD_{1}M中，DD_{1}=4,D_{1}M=\sqrt{3},\therefore MD=\sqrt{19},\therefore D_{1}P=\frac{DD_1\cdot D_1M}{MD}=\frac{4\sqrt{3}}{\sqrt{19}}=\frac{4\sqrt{57}}{19}.\therefore B_{1}C_{1}与平面BCGH的距离是\frac{4\sqrt{57}}{19}.$![Image](https://cdntk.53zaixian.com/tq/1707197294403354626/img/compression/M_96fee8ea4650cd861b799df7998c55c6.jpg)

Final answer: $\frac{4\sqrt{57}}{19}$
Subfield: Solid Geometry


Question: $如图所示，三棱柱ABC-A_{1}B_{1}C_{1}的各棱长均为4，侧棱垂直于底面，E,F,G,H分别为AB,AC,A_{1}C_{1},A_{1}B_{1}的中点.求：$![Image](https://cdntk.53zaixian.com/tq/1707197294097170434/img/compression/M_899bf44086019995aea5981f6640aab5.jpg)
$(2)平面A_{1}EF与平面BCGH之间的距离.$

Solution:
$由题意易知A_{1}E\parallel HB,\because A_{1}E\not\subset 平面BCGH,HB\subset 平面BCGH,\therefore A_{1}E\parallel 平面BCGH,同理，A_{1}F\parallel 平面BCGH,又A_{1}E\cap A_{1}F=A_{1},A_{1}E,A_{1}F\subset 平面A_{1}EF,\therefore 平面A_{1}EF\parallel 平面BCGH,\therefore 平面A_{1}EF与平面BCGH间的距离等于点A_{1}到平面BCGH的距离.又H,G分别是A_{1}B_{1},A_{1}C_{1}的中点,\therefore 点A_{1}到平面BCGH的距离等于B_{1}C_{1}到平面BCGH的距离.\therefore 平面A_{1}EF与平面BCGH之间的距离为\frac{4\sqrt{57}}{19}.$

Final answer: $\frac{4\sqrt{57}}{19}$
Subfield: Solid Geometry


Question: $如图所示，三棱柱ABC-A_{1}B_{1}C_{1}的各棱长均为4，侧棱垂直于底面，E,F,G,H分别为AB,AC,A_{1}C_{1},A_{1}B_{1}的中点.求：$<img_2721>
$(2)平面A_{1}EF与平面BCGH之间的距离.$

Solution:
$由题意易知A_{1}E\parallel HB,\because A_{1}E\not\subset 平面BCGH,HB\subset 平面BCGH,\therefore A_{1}E\parallel 平面BCGH,同理，A_{1}F\parallel 平面BCGH,又A_{1}E\cap A_{1}F=A_{1},A_{1}E,A_{1}F\subset 平面A_{1}EF,\therefore 平面A_{1}EF\parallel 平面BCGH,\therefore 平面A_{1}EF与平面BCGH间的距离等于点A_{1}到平面BCGH的距离.又H,G分别是A_{1}B_{1},A_{1}C_{1}的中点,\therefore 点A_{1}到平面BCGH的距离等于B_{1}C_{1}到平面BCGH的距离.\therefore 平面A_{1}EF与平面BCGH之间的距离为\frac{4\sqrt{57}}{19}.$

Final answer: $\frac{4\sqrt{57}}{19}$
Subfield: Solid Geometry


Question: $如图，三棱柱ABC-A_{1}B_{1}C_{1}中，底面ABC为正三角形，侧棱与底面垂直，它的所有棱长都为2，E，F，G分别为B_{1}C_{1}，A_{1}B_{1}，AB的中点.$<img_2947>
$(1)求证：平面A_{1}C_{1}G\parallel 平面BEF；$

Solution:
$证明：\because E，F分别为B_{1}C_{1}，A_{1}B_{1}的中点，\therefore EF\parallel A_{1}C_{1}，\because A_{1}C_{1}\subset 平面A_{1}C_{1}G，EF\not\subset 平面A_{1}C_{1}G，\therefore EF\parallel 平面A_{1}C_{1}G，\because A_{1}B_{1\xlongequal{//} }AB，F，G分别为A_{1}B_{1}，AB的中点，\therefore A_{1}F\xlongequal{//} BG，\therefore 四边形A_{1}GBF为平行四边形，\therefore BF\parallel A_{1}G，\because A_{1}G\subset 平面A_{1}C_{1}G，BF\not\subset 平面A_{1}C_{1}G，\therefore BF\parallel 平面A_{1}C_{1}G，又EF\cap BF=F，EF，BF\subset 平面BEF，\therefore 平面A_{1}C_{1}G\parallel 平面BEF.$
Subfield: Solid Geometry


Question: $如图，三棱柱ABC-A_{1}B_{1}C_{1}中，底面ABC为正三角形，侧棱与底面垂直，它的所有棱长都为2，E，F，G分别为B_{1}C_{1}，A_{1}B_{1}，AB的中点.$![Image](https://cdntk.53zaixian.com/tq/1707197297167400962/img/compression/M_ebf7d9062d9e8452b8eda891e30cb120.jpg)
$(1)求证：平面A_{1}C_{1}G\parallel 平面BEF；$

Solution:
$证明：\because E，F分别为B_{1}C_{1}，A_{1}B_{1}的中点，\therefore EF\parallel A_{1}C_{1}，\because A_{1}C_{1}\subset 平面A_{1}C_{1}G，EF\not\subset 平面A_{1}C_{1}G，\therefore EF\parallel 平面A_{1}C_{1}G，\because A_{1}B_{1\xlongequal{//} }AB，F，G分别为A_{1}B_{1}，AB的中点，\therefore A_{1}F\xlongequal{//} BG，\therefore 四边形A_{1}GBF为平行四边形，\therefore BF\parallel A_{1}G，\because A_{1}G\subset 平面A_{1}C_{1}G，BF\not\subset 平面A_{1}C_{1}G，\therefore BF\parallel 平面A_{1}C_{1}G，又EF\cap BF=F，EF，BF\subset 平面BEF，\therefore 平面A_{1}C_{1}G\parallel 平面BEF.$

Final answer: 证明题
Subfield: Solid Geometry


Question: $如图，三棱柱ABC-A_{1}B_{1}C_{1}中，底面ABC为正三角形，侧棱与底面垂直，它的所有棱长都为2，E，F，G分别为B_{1}C_{1}，A_{1}B_{1}，AB的中点.$![Image](https://cdntk.53zaixian.com/tq/1707197297167400962/img/compression/M_ebf7d9062d9e8452b8eda891e30cb120.jpg)
$(2)求C_{1}G与平面BCC_{1}B_{1}所成角的正弦值.$

Solution:
$在平面ABC内，过点G作GH\perp BC，垂足为H，连接C_{1}H.\because CC_{1}\perp 平面ABC，GH\subset 平面ABC，\therefore CC_{1}\perp GH.又BC\cap CC_{1}=C，BC，CC_{1}\subset 平面BCC_{1}B_{1}，\therefore GH\perp 平面BCC_{1}B_{1}.\therefore \angle GC_{1}H即为C_{1}G与平面BCC_{1}B_{1}所成的角.易知BG=1，\angle GBH=60^\circ ，\angle BHG=90^\circ ，\therefore BH=\frac{1}{2}，GH=\frac{\sqrt{3}}{2}，则CH=\frac{3}{2}.又CC_{1}\perp BC，CC_{1}=2，\therefore C_{1}H=\sqrt{CH^2+CC^2_1}=\sqrt{\left(\frac{3}{2}\right)^2+2^2}=\frac{5}{2}.又GH\perp C_{1}H，\therefore C_{1}G=\sqrt{GH^2+C_1H^2}=\sqrt{\left(\frac{\sqrt{3}}{2}\right)^2+\left(\frac{5}{2}\right)^2}=\sqrt{7}，\therefore \\sin  \angle GC_{1}H=\frac{GH}{C_1G}=\frac{\frac{\sqrt{3}}{2}}{\sqrt{7}}=\frac{\sqrt{21}}{14}，故C_{1}G与平面BCC_{1}B_{1}所成角的正弦值为\frac{\sqrt{21}}{14}.$![Image](https://cdntk.53zaixian.com/tq/1707197297494556673/img/compression/M_acd539af6a9e409783cd263693ada0de.jpg)

Final answer: $\frac{\sqrt{21}}{14}$
Subfield: Solid Geometry


Question: $如图，三棱柱ABC-A_{1}B_{1}C_{1}中，底面ABC为正三角形，侧棱与底面垂直，它的所有棱长都为2，E，F，G分别为B_{1}C_{1}，A_{1}B_{1}，AB的中点.$<img_2947>
$(2)求C_{1}G与平面BCC_{1}B_{1}所成角的正弦值.$

Solution:
$在平面ABC内，过点G作GH\perp BC，垂足为H，连接C_{1}H.\because CC_{1}\perp 平面ABC，GH\subset 平面ABC，\therefore CC_{1}\perp GH.又BC\cap CC_{1}=C，BC，CC_{1}\subset 平面BCC_{1}B_{1}，\therefore GH\perp 平面BCC_{1}B_{1}.\therefore \angle GC_{1}H即为C_{1}G与平面BCC_{1}B_{1}所成的角.易知BG=1，\angle GBH=60^\circ ，\angle BHG=90^\circ ，\therefore BH=\frac{1}{2}，GH=\frac{\sqrt{3}}{2}，则CH=\frac{3}{2}.又CC_{1}\perp BC，CC_{1}=2，\therefore C_{1}H=\sqrt{CH^2+CC^2_1}=\sqrt{\left(\frac{3}{2}\right)^2+2^2}=\frac{5}{2}.又GH\perp C_{1}H，\therefore C_{1}G=\sqrt{GH^2+C_1H^2}=\sqrt{\left(\frac{\sqrt{3}}{2}\right)^2+\left(\frac{5}{2}\right)^2}=\sqrt{7}，\therefore \\sin  \angle GC_{1}H=\frac{GH}{C_1G}=\frac{\frac{\sqrt{3}}{2}}{\sqrt{7}}=\frac{\sqrt{21}}{14}，故C_{1}G与平面BCC_{1}B_{1}所成角的正弦值为\frac{\sqrt{21}}{14}.$<img_2082>

Final answer: $\frac{\sqrt{21}}{14}$
Subfield: Solid Geometry


Question: $如图，在三棱台ABC-DEF中，AB=BC=CA=2DF=2，FC=1，\angle ACF=\angle BCF=90^\circ ， G为线段AC的中点，H为线段BC上的点，BD\parallel 平面FGH.$![Image](https://cdntk.53zaixian.com/tq/1706233578451578883/img/compression/M_cd5fde6007e875833577316e4334d01e.jpg)
$(1)求证：H为线段BC的中点；$

Solution:
$证明：连接CD，设CD\cap FG=O，连接HO，DG，因为BD\parallel 平面FGH，BD\subset 平面CBD，且平面CBD\cap 平面FGH=HO，所以BD\parallel HO，易得四边形DFCG是正方形，所以O是CD的中点，所以H是BC的中点.$

Final answer: 证明题
Subfield: Solid Geometry


Question: $如图，在三棱台ABC-DEF中，AB=BC=CA=2DF=2，FC=1，\angle ACF=\angle BCF=90^\circ ， G为线段AC的中点，H为线段BC上的点，BD\parallel 平面FGH.$<img_2901>
$(1)求证：H为线段BC的中点；$

Solution:
$证明：连接CD，设CD\cap FG=O，连接HO，DG，因为BD\parallel 平面FGH，BD\subset 平面CBD，且平面CBD\cap 平面FGH=HO，所以BD\parallel HO，易得四边形DFCG是正方形，所以O是CD的中点，所以H是BC的中点.$
Subfield: Solid Geometry


Question: $如图，在三棱台ABC-DEF中，AB=BC=CA=2DF=2，FC=1，\angle ACF=\angle BCF=90^\circ ， G为线段AC的中点，H为线段BC上的点，BD\parallel 平面FGH.$![Image](https://cdntk.53zaixian.com/tq/1706233578451578883/img/compression/M_cd5fde6007e875833577316e4334d01e.jpg)
$(2)求三棱台ABC-DEF的表面积.$

Solution:
$因为AB=BC=CA=2，所以\triangle ABC为等边三角形，所以\triangle DEF也为等边三角形，且EF=DE=DF=1，则S_{\triangle DEF}=\frac{\sqrt{3}}{4}\times DE^{2}=\frac{\sqrt{3}}{4}，S_{\triangle ABC}=\frac{\sqrt{3}}{4}\times AB^{2}=\sqrt{3}，又因为\angle ACF=\angle BCF=90^\circ ，所以侧面ADFC和侧面EFCB均为直角梯形，且FC=1，设侧面ADFC的面积为S_{1}，则S_{1}=\frac{1}{2}\times (1+2)\times 1=\frac{3}{2}，连接EH，易得侧面ADEB为等腰梯形，其中DE=1,AB=2，且AD=BE=\sqrt{BH^2+EH^2}=\sqrt{2},过点E作EM\perp AB，垂足为M，可得EM=\sqrt{BE^2-BM^2}=\sqrt{(\sqrt{2})^2-\left(\frac{1}{2}\right)^2}=\frac{\sqrt{7}}{2}，设侧面ABED的面积为S_{2}，则S_{2}=\frac{1}{2}\times (1+2)\times \frac{\sqrt{7}}{2}=\frac{3\sqrt{7}}{4}，所以三棱台的表面积S=\frac{\sqrt{3}}{4}+\sqrt{3}+2\times \frac{3}{2}+\frac{3\sqrt{7}}{4}=\frac{5\sqrt{3}}{4}+\frac{3\sqrt{7}}{4}+3.$![Image](https://cdntk.53zaixian.com/tq/1706233578774540289/img/compression/M_0df88739f60bb1f3fcfeff1e049bf998.jpg)

Final answer: $\frac{5\sqrt{3}}{4}+\frac{3\sqrt{7}}{4}+3$
Subfield: Solid Geometry


Question: $如图，在三棱台ABC-DEF中，AB=BC=CA=2DF=2，FC=1，\angle ACF=\angle BCF=90^\circ ， G为线段AC的中点，H为线段BC上的点，BD\parallel 平面FGH.$<img_2901>
$(2)求三棱台ABC-DEF的表面积.$

Solution:
$因为AB=BC=CA=2，所以\triangle ABC为等边三角形，所以\triangle DEF也为等边三角形，且EF=DE=DF=1，则S_{\triangle DEF}=\frac{\sqrt{3}}{4}\times DE^{2}=\frac{\sqrt{3}}{4}，S_{\triangle ABC}=\frac{\sqrt{3}}{4}\times AB^{2}=\sqrt{3}，又因为\angle ACF=\angle BCF=90^\circ ，所以侧面ADFC和侧面EFCB均为直角梯形，且FC=1，设侧面ADFC的面积为S_{1}，则S_{1}=\frac{1}{2}\times (1+2)\times 1=\frac{3}{2}，连接EH，易得侧面ADEB为等腰梯形，其中DE=1,AB=2，且AD=BE=\sqrt{BH^2+EH^2}=\sqrt{2},过点E作EM\perp AB，垂足为M，可得EM=\sqrt{BE^2-BM^2}=\sqrt{(\sqrt{2})^2-\left(\frac{1}{2}\right)^2}=\frac{\sqrt{7}}{2}，设侧面ABED的面积为S_{2}，则S_{2}=\frac{1}{2}\times (1+2)\times \frac{\sqrt{7}}{2}=\frac{3\sqrt{7}}{4}，所以三棱台的表面积S=\frac{\sqrt{3}}{4}+\sqrt{3}+2\times \frac{3}{2}+\frac{3\sqrt{7}}{4}=\frac{5\sqrt{3}}{4}+\frac{3\sqrt{7}}{4}+3.$<img_2331>

Final answer: $\frac{5\sqrt{3}}{4}+\frac{3\sqrt{7}}{4}+3$
Subfield: Solid Geometry


Question: $如图,在\triangle ABC中,\angle ACB=90^\circ ,\angle ABC=30^\circ ,BC=\sqrt{3},在三角形内挖去一个半圆，圆心O在边BC上,半圆与AC，AB分别相切于点C，M,与BC交于另一点N,将\triangle ABC绕直线BC旋转一周得到一个旋转体.$![Image](https://cdntk.53zaixian.com/tq/1706233594708447234/img/compression/M_4f736dda32d26817a79fd8893d0c3015.jpg)
$(1)求该旋转体中间空心球的表面积的大小;$

Solution:
$如图，连接OM，则OM\perp AB，$![Image](https://cdntk.53zaixian.com/tq/1706233595572473857/img/compression/M_877eb1d74efce694783a19f495f73265.jpg)$设OM=r,则OB=\sqrt{3}-r，在\triangle BMO中，\\sin  \angle MBO=\frac{r}{\sqrt{3}-r}=\frac{1}{2}\Rightarrow r=\frac{\sqrt{3}}{3}，\therefore S_{球面}=4\pi r^{2}=\frac{4\pi }{3}.$

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


Question: $如图,在\triangle ABC中,\angle ACB=90^\circ ,\angle ABC=30^\circ ,BC=\sqrt{3},在三角形内挖去一个半圆，圆心O在边BC上,半圆与AC，AB分别相切于点C，M,与BC交于另一点N,将\triangle ABC绕直线BC旋转一周得到一个旋转体.$<img_1888>
$(1)求该旋转体中间空心球的表面积的大小;$

Solution:
$如图，连接OM，则OM\perp AB，$<img_2970>$设OM=r,则OB=\sqrt{3}-r，在\triangle BMO中，\\sin  \angle MBO=\frac{r}{\sqrt{3}-r}=\frac{1}{2}\Rightarrow r=\frac{\sqrt{3}}{3}，\therefore S_{球面}=4\pi r^{2}=\frac{4\pi }{3}.$

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


Question: $如图,在\triangle ABC中,\angle ACB=90^\circ ,\angle ABC=30^\circ ,BC=\sqrt{3},在三角形内挖去一个半圆，圆心O在边BC上,半圆与AC，AB分别相切于点C，M,与BC交于另一点N,将\triangle ABC绕直线BC旋转一周得到一个旋转体.$<img_1888>
$(2)求图中阴影部分绕直线BC旋转一周所得旋转体的体积.$

Solution:
$\because \angle ACB=90^\circ ,\angle ABC=30^\circ ,BC=\sqrt{3},\therefore AC=1，将阴影部分绕直线BC旋转一周得到一个圆锥，里面挖去一个内切球，\therefore 所求体积V=V_{圆锥}-V_{球}=\frac{1}{3}\pi \times AC^{2}\times BC-\frac{4}{3}\pi r^{3}=\frac{1}{3}\pi \times \sqrt{3}-\frac{4}{3}\pi \times \frac{\sqrt{3}}{9}=\frac{5\sqrt{3}}{27}\pi .$

Final answer: $\frac{5\sqrt{3}}{27}\pi$
Subfield: Solid Geometry


Question: $如图,在\triangle ABC中,\angle ACB=90^\circ ,\angle ABC=30^\circ ,BC=\sqrt{3},在三角形内挖去一个半圆，圆心O在边BC上,半圆与AC，AB分别相切于点C，M,与BC交于另一点N,将\triangle ABC绕直线BC旋转一周得到一个旋转体.$![Image](https://cdntk.53zaixian.com/tq/1706233594708447234/img/compression/M_4f736dda32d26817a79fd8893d0c3015.jpg)
$(2)求图中阴影部分绕直线BC旋转一周所得旋转体的体积.$

Solution:
$\because \angle ACB=90^\circ ,\angle ABC=30^\circ ,BC=\sqrt{3},\therefore AC=1，将阴影部分绕直线BC旋转一周得到一个圆锥，里面挖去一个内切球，\therefore 所求体积V=V_{圆锥}-V_{球}=\frac{1}{3}\pi \times AC^{2}\times BC-\frac{4}{3}\pi r^{3}=\frac{1}{3}\pi \times \sqrt{3}-\frac{4}{3}\pi \times \frac{\sqrt{3}}{9}=\frac{5\sqrt{3}}{27}\pi .$

Final answer: $\frac{5\sqrt{3}}{27}\pi$
Subfield: Solid Geometry


Question: $如图，正方体ABCD-A_{1}B_{1}C_{1}D_{1}的棱长为2，E，F分别是棱AB，BC上的点，且AE=BF=x.$![Image](https://cdntk.53zaixian.com/tq/1706233627403046913/img/compression/M_3a51c1de3645b5559f1324a6eb4ce68e.jpg)
$(1)当x为何值时，三棱锥B_{1}-BEF的体积最大？$

Solution:
$在正方体ABCD-A_{1}B_{1}C_{1}D_{1}中，BB_{1}\perp 平面ABCD，于是有V_{B1-BEF}=\frac{1}{3}S_{\triangle BEF}\cdot BB_{1}=\frac{1}{3}\times \frac{1}{2}(2-x)x\times 2=\frac{1}{3}\times (2-x)x=\frac{1}{3}(-x^{2}+2x)=\frac{1}{3}[-(x-1)^{2}+1]，0将三棱锥B_{1}-BEF的体积表示成关于x的二次函数，体现了函数思想.易知当x=1时，三棱锥B_{1}-BEF的体积取得最大值，为\frac{1}{3}.$

Final answer: $1$
Subfield: Solid Geometry


Question: $如图，正方体ABCD-A_{1}B_{1}C_{1}D_{1}的棱长为2，E，F分别是棱AB，BC上的点，且AE=BF=x.$<img_2208>
$(1)当x为何值时，三棱锥B_{1}-BEF的体积最大？$

Solution:
$在正方体ABCD-A_{1}B_{1}C_{1}D_{1}中，BB_{1}\perp 平面ABCD，于是有V_{B1-BEF}=\frac{1}{3}S_{\triangle BEF}\cdot BB_{1}=\frac{1}{3}\times \frac{1}{2}(2-x)x\times 2=\frac{1}{3}\times (2-x)x=\frac{1}{3}(-x^{2}+2x)=\frac{1}{3}[-(x-1)^{2}+1]，0将三棱锥B_{1}-BEF的体积表示成关于x的二次函数，体现了函数思想.易知当x=1时，三棱锥B_{1}-BEF的体积取得最大值，为\frac{1}{3}.$

Final answer: $1$
Subfield: Solid Geometry


Question: $如图，正方体ABCD-A_{1}B_{1}C_{1}D_{1}的棱长为2，E，F分别是棱AB，BC上的点，且AE=BF=x.$<img_2208>
$(2)求三棱锥B_{1}-BEF的体积最大时，二面角B_{1}-EF-B的正切值；$

Solution:
$由(1)知，当E，F分别为AB，BC的中点时，三棱锥B_{1}-BEF的体积最大，取EF的中点O，连接OB,OB_{1}，$<img_1696>$于是得BO\perp EF，B_{1}E=\sqrt{BE^2+B_1B^2}=\sqrt{BF^2+B_1B^2}=B_{1}F，因此B_{1}O\perp EF，则\angle B_{1}OB是二面角B_{1}-EF-B的平面角.在Rt\triangle BEF中，BO=\frac{1}{2}EF=\frac{\sqrt{2}}{2}，在Rt\triangle BB_{1}O中，\\tan  \angle B_{1}OB=\frac{BB_1}{BO}=2\sqrt{2}，即三棱锥B_{1}-BEF的体积最大时，二面角B_{1}-EF-B的正切值为2\sqrt{2}.$

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


Question: $如图，正方体ABCD-A_{1}B_{1}C_{1}D_{1}的棱长为2，E，F分别是棱AB，BC上的点，且AE=BF=x.$![Image](https://cdntk.53zaixian.com/tq/1706233627403046913/img/compression/M_3a51c1de3645b5559f1324a6eb4ce68e.jpg)
$(2)求三棱锥B_{1}-BEF的体积最大时，二面角B_{1}-EF-B的正切值；$

Solution:
$由(1)知，当E，F分别为AB，BC的中点时，三棱锥B_{1}-BEF的体积最大，取EF的中点O，连接OB,OB_{1}，$![Image](https://cdntk.53zaixian.com/tq/1706233627780534273/img/compression/M_ee1338ae47285097b6fe60af9e0d7f24.jpg)$于是得BO\perp EF，B_{1}E=\sqrt{BE^2+B_1B^2}=\sqrt{BF^2+B_1B^2}=B_{1}F，因此B_{1}O\perp EF，则\angle B_{1}OB是二面角B_{1}-EF-B的平面角.在Rt\triangle BEF中，BO=\frac{1}{2}EF=\frac{\sqrt{2}}{2}，在Rt\triangle BB_{1}O中，\\tan  \angle B_{1}OB=\frac{BB_1}{BO}=2\sqrt{2}，即三棱锥B_{1}-BEF的体积最大时，二面角B_{1}-EF-B的正切值为2\sqrt{2}.$

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


Question: $如图，正方体ABCD-A_{1}B_{1}C_{1}D_{1}的棱长为2，E，F分别是棱AB，BC上的点，且AE=BF=x.$<img_2208>
$(3)求异面直线A_{1}E与B_{1}F所成角的取值范围.$

Solution:
$在AD上取点H，使AH=BF=AE，连接A_{1}H,EH,FH，如图，$<img_2259>$显然HF=AB=A_{1}B_{1}，HF\parallel AB\parallel A_{1}B_{1}，即四边形A_{1}B_{1}FH是平行四边形，则A_{1}H\parallel B_{1}F，故\angle HA_{1}E(或其补角)即为异面直线A_{1}E与B_{1}F所成的角.在Rt\triangle A_{1}AH中，A_{1}H=\sqrt{4+x^2}，在Rt\triangle A_{1}AE中，A_{1}E=\sqrt{4+x^2}，在Rt\triangle HAE中，HE=\sqrt{x^2+x^2}=\sqrt{2}x，将异面直线所成角的余弦值表示成关于x的函数，通过变量x的范围求异面直线所成角的取值范围，体现了函数思想.在\triangle HA_{1}E中，由余弦定理的推论得\\cos  \angle HA_{1}E=\frac{A_1H^2+A_1E^2-EH^2}{2A_1H\cdot A_1E}=\frac{4}{4+x^2}，而0<x\leq 2，则4<x^{2}+4\leq 8，即\frac{1}{2}\leq \frac{4}{x^2+4}<1，即\frac{1}{2}\leq \\cos  \angle HA_{1}E<1，则0<\angle HA_{1}E\leq \frac{\pi }{3}，所以异面直线A_{1}E与B_{1}F所成角的取值范围为\left(0,\frac{\pi }{3}\right].$

Final answer: $(0,\frac{\pi }{3}]$