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


Question: $如图，线段AA_{1}是圆柱OO_{1}的母线，BC是圆柱下底面\odot O的直径.$![Image](https://cdntk.53zaixian.com/tq/1715549479019962371/img/compression/M_90b3cf5f52739697c5be4aba769b586e.jpg)
$(2)若BC=AA_{1}=2，\angle ABC=30^\circ ，求点A到平面A_{1}BC的距离.$

Solution:
$Rt\triangle BAC中，BC=2，\angle ABC=30^\circ ，则AC=1,AB=\sqrt{3}，可得S_{\triangle ABC}=\frac{1}{2}AC\cdot AB=\frac{\sqrt{3}}{2}，Rt\triangle BAA_{1}中，AA_{1}=2,AB=\sqrt{3}，则A_{1}B=\sqrt{7}，Rt\triangle CAA_{1}中，AA_{1}=2,AC=1，则A_{1}C=\sqrt{5}，\triangle A_{1}BC中，\\cos  \angle A_{1}BC=\frac{A_1B^2+BC^2-A_1C^2}{2A_1B\cdot BC}=\frac{6}{4\sqrt{7}}=\frac{3\sqrt{7}}{14},则\\sin  \angle A_{1}BC=\frac{\sqrt{133}}{14}，S_{\triangle A_{1}BC}=\frac{1}{2}A_{1}B\cdot BC\cdot \\sin  \angle A_{1}BC=\frac{\sqrt{19}}{2}，设点A到平面A_{1}BC的距离为d，由等体积法得V_{A_{1}-ABC}=V_{A-A_{1}BC}，即\frac{1}{3}S_{\triangle ABC}\cdot AA_{1}=\frac{1}{3}S_{\triangle A_{1}BC}\cdot d，解得d=\frac{2\sqrt{57}}{19}.所以点A到平面A_{1}BC的距离为\frac{2\sqrt{57}}{19}.$

Final answer: $\frac{2\sqrt{57}}{19}$
Subfield: Solid Geometry


Question: $如图，线段AA_{1}是圆柱OO_{1}的母线，BC是圆柱下底面\odot O的直径.$<img_2273>
$(2)若BC=AA_{1}=2，\angle ABC=30^\circ ，求点A到平面A_{1}BC的距离.$

Solution:
$Rt\triangle BAC中，BC=2，\angle ABC=30^\circ ，则AC=1,AB=\sqrt{3}，可得S_{\triangle ABC}=\frac{1}{2}AC\cdot AB=\frac{\sqrt{3}}{2}，Rt\triangle BAA_{1}中，AA_{1}=2,AB=\sqrt{3}，则A_{1}B=\sqrt{7}，Rt\triangle CAA_{1}中，AA_{1}=2,AC=1，则A_{1}C=\sqrt{5}，\triangle A_{1}BC中，\\cos  \angle A_{1}BC=\frac{A_1B^2+BC^2-A_1C^2}{2A_1B\cdot BC}=\frac{6}{4\sqrt{7}}=\frac{3\sqrt{7}}{14},则\\sin  \angle A_{1}BC=\frac{\sqrt{133}}{14}，S_{\triangle A_{1}BC}=\frac{1}{2}A_{1}B\cdot BC\cdot \\sin  \angle A_{1}BC=\frac{\sqrt{19}}{2}，设点A到平面A_{1}BC的距离为d，由等体积法得V_{A_{1}-ABC}=V_{A-A_{1}BC}，即\frac{1}{3}S_{\triangle ABC}\cdot AA_{1}=\frac{1}{3}S_{\triangle A_{1}BC}\cdot d，解得d=\frac{2\sqrt{57}}{19}.所以点A到平面A_{1}BC的距离为\frac{2\sqrt{57}}{19}.$

Final answer: $\frac{2\sqrt{57}}{19}$
Subfield: Solid Geometry


Question: $如图,在正方体ABCD-A_{1}B_{1}C_{1}D_{1}中,E,F,G分别为所在棱的中点,Q,H分别为正方形ADD_{1}A_{1}和正方形ABCD的中心,连接EF,EG,FG,D_{1}Q,CH,QH,CD_{1}.$<img_1641>
$(1)证明:平面EFG\parallel 平面CD_{1}QH;$

Solution:
$证明:如图，连接A_{1}C_{1},BC_{1},由E,F,G分别为所在棱的中点，得A_{1}C_{1}\parallel GF,EF\parallel BC_{1}.连接AH,AQ，CQ.由H为正方形ABCD的中心可得A,H,C三点共线,同理A,Q,D_{1}三点共线，$<img_2909>$\therefore A\in 平面CD_{1}QH.又知A_{1}C_{1}\parallel AC,所以GF\parallel AC,又AC\subset 平面ACQ,GF\not\subset 平面ACQ,所以GF\parallel 平面ACQ.同理,EF\parallel 平面ACQ.又GF\cap EF=F,所以平面EFG\parallel 平面ACQ,即平面EFG\parallel 平面CD_{1}QH.$
Subfield: Solid Geometry


Question: $如图,在正方体ABCD-A_{1}B_{1}C_{1}D_{1}中,E,F,G分别为所在棱的中点,Q,H分别为正方形ADD_{1}A_{1}和正方形ABCD的中心,连接EF,EG,FG,D_{1}Q,CH,QH,CD_{1}.$![Image](https://cdntk.53zaixian.com/tq/1715549479581999107/img/compression/M_45a8dcec9b9427d41a370f6a7e73567f.jpg)
$(1)证明:平面EFG\parallel 平面CD_{1}QH;$

Solution:
$证明:如图，连接A_{1}C_{1},BC_{1},由E,F,G分别为所在棱的中点，得A_{1}C_{1}\parallel GF,EF\parallel BC_{1}.连接AH,AQ，CQ.由H为正方形ABCD的中心可得A,H,C三点共线,同理A,Q,D_{1}三点共线，$![Image](https://cdntk.53zaixian.com/tq/1715549479875600385/img/compression/M_e15837e6ef6ca9de497c56034737a0f4.jpg)$\therefore A\in 平面CD_{1}QH.又知A_{1}C_{1}\parallel AC,所以GF\parallel AC,又AC\subset 平面ACQ,GF\not\subset 平面ACQ,所以GF\parallel 平面ACQ.同理,EF\parallel 平面ACQ.又GF\cap EF=F,所以平面EFG\parallel 平面ACQ,即平面EFG\parallel 平面CD_{1}QH.$

Final answer: 证明题
Subfield: Solid Geometry


Question: $如图,在正方体ABCD-A_{1}B_{1}C_{1}D_{1}中,E,F,G分别为所在棱的中点,Q,H分别为正方形ADD_{1}A_{1}和正方形ABCD的中心,连接EF,EG,FG,D_{1}Q,CH,QH,CD_{1}.$<img_1641>
$(2)在线段CD上是否存在一点P,使得DQ\parallel 平面D_{1}PH?若存在,写出P点的位置并给出证明;若不存在,请说明理由.$

Solution:
$线段CD上存在一点P,当DP=\frac{1}{3}DC时,满足DQ\parallel 平面D_{1}PH.证明如下:连接PH并延长交AB于点M,连接D_{1}M,则平面D_{1}PH与平面D_{1}PM为同一平面.由H为AC的中点,得\triangle AMH\cong \triangle CPH.则AM=\frac{2}{3}AB=\frac{2}{3}CD,取线段D_{1}M的中点N,连接QN,NP.因为Q,N分别为AD_{1},D_{1}M的中点,所以QN=\frac{1}{2}AM=\frac{1}{3}AB=\frac{1}{3}DC且QN\parallel AM,又DP\parallel AM且DP=\frac{1}{3}DC，所以QN\parallel DP且QN=DP,所以四边形QDPN为平行四边形,故QD\parallel NP,又QD\not\subset 平面D_{1}PH,NP\subset 平面D_{1}PH,所以DQ\parallel 平面D_{1}PH.$
Subfield: Solid Geometry


Question: $如图,在正方体ABCD-A_{1}B_{1}C_{1}D_{1}中,E,F,G分别为所在棱的中点,Q,H分别为正方形ADD_{1}A_{1}和正方形ABCD的中心,连接EF,EG,FG,D_{1}Q,CH,QH,CD_{1}.$![Image](https://cdntk.53zaixian.com/tq/1715549479581999107/img/compression/M_45a8dcec9b9427d41a370f6a7e73567f.jpg)
$(2)在线段CD上是否存在一点P,使得DQ\parallel 平面D_{1}PH?若存在,写出P点的位置并给出证明;若不存在,请说明理由.$

Solution:
$线段CD上存在一点P,当DP=\frac{1}{3}DC时,满足DQ\parallel 平面D_{1}PH.证明如下:连接PH并延长交AB于点M,连接D_{1}M,则平面D_{1}PH与平面D_{1}PM为同一平面.由H为AC的中点,得\triangle AMH\cong \triangle CPH.则AM=\frac{2}{3}AB=\frac{2}{3}CD,取线段D_{1}M的中点N,连接QN,NP.因为Q,N分别为AD_{1},D_{1}M的中点,所以QN=\frac{1}{2}AM=\frac{1}{3}AB=\frac{1}{3}DC且QN\parallel AM,又DP\parallel AM且DP=\frac{1}{3}DC，所以QN\parallel DP且QN=DP,所以四边形QDPN为平行四边形,故QD\parallel NP,又QD\not\subset 平面D_{1}PH,NP\subset 平面D_{1}PH,所以DQ\parallel 平面D_{1}PH.$

Final answer: 证明题
Subfield: Solid Geometry


Question: $小明同学参加综合实践活动,设计了一个封闭的包装盒.包装盒如图所示:底面ABCD是边长为8(单位:cm)的正方形，\triangle EAB，\triangle FBC，\triangle GCD，\triangle HDA均为正三角形,且它们所在的平面都与平面ABCD垂直.$![Image](https://cdntk.53zaixian.com/tq/1715549427211685890/img/compression/M_cc58899ee4709d31ed155b5bf73410b9.jpg)
$(1)证明：EF\parallel 平面ABCD；$

Solution:
$取AB、BC、CD、DA的中点M、N、P、Q，连接EM、FN、GP、HQ、MN、NP、PQ、QM.证明：在正三角形ABE中，M为AB的中点，所以EM\perp AB.又平面ABE\cap 平面ABCD=AB,且平面ABE\perp 平面ABCD,所以EM\perp 平面ABCD.同理FN\perp 平面ABCD，所以EM\parallel FN，又EM=FN,所以四边形EMNF为平行四边形，所以EF\parallel MN.又MN\subset 平面ABCD,且EF\not\subset 平面ABCD,所以EF\parallel 平面ABCD.
$

Final answer: 证明题
Subfield: Solid Geometry


Question: $小明同学参加综合实践活动,设计了一个封闭的包装盒.包装盒如图所示:底面ABCD是边长为8(单位:cm)的正方形，\triangle EAB，\triangle FBC，\triangle GCD，\triangle HDA均为正三角形,且它们所在的平面都与平面ABCD垂直.$<img_2697>
$(1)证明：EF\parallel 平面ABCD；$

Solution:
$取AB、BC、CD、DA的中点M、N、P、Q，连接EM、FN、GP、HQ、MN、NP、PQ、QM.证明：在正三角形ABE中，M为AB的中点，所以EM\perp AB.又平面ABE\cap 平面ABCD=AB,且平面ABE\perp 平面ABCD,所以EM\perp 平面ABCD.同理FN\perp 平面ABCD，所以EM\parallel FN，又EM=FN,所以四边形EMNF为平行四边形，所以EF\parallel MN.又MN\subset 平面ABCD,且EF\not\subset 平面ABCD,所以EF\parallel 平面ABCD.
$
Subfield: Solid Geometry


Question: $小明同学参加综合实践活动,设计了一个封闭的包装盒.包装盒如图所示:底面ABCD是边长为8(单位:cm)的正方形，\triangle EAB，\triangle FBC，\triangle GCD，\triangle HDA均为正三角形,且它们所在的平面都与平面ABCD垂直.$<img_2697>
$(2)求该包装盒的容积(不计包装盒材料的厚度).$

Solution:
$
如图，可将包装盒分割为长方体MNPQ-EFGH和四个全等的四棱锥.$<img_1711>$易得MN=4\sqrt{2} cm，EM=4\sqrt{3} cm.所以V_{长方体MNPQ-EFGH}=(4\sqrt{2})^{2}\times 4\sqrt{3}=128\sqrt{3} cm^{3},V_{四棱锥B-MNFE}=\frac{1}{3}\times 4\sqrt{3}\times 4\sqrt{2}\times 2\sqrt{2}=\frac{64\sqrt{3}}{3} cm^{3},所以该包装盒的容积为128\sqrt{3}+4\times \frac{64\sqrt{3}}{3}=\frac{640\sqrt{3}}{3} cm^{3}.$

Final answer: $640\sqrt 3/3$ $cm^3$
Subfield: Solid Geometry


Question: $小明同学参加综合实践活动,设计了一个封闭的包装盒.包装盒如图所示:底面ABCD是边长为8(单位:cm)的正方形，\triangle EAB，\triangle FBC，\triangle GCD，\triangle HDA均为正三角形,且它们所在的平面都与平面ABCD垂直.$![Image](https://cdntk.53zaixian.com/tq/1715549427211685890/img/compression/M_cc58899ee4709d31ed155b5bf73410b9.jpg)
$(2)求该包装盒的容积(不计包装盒材料的厚度).$

Solution:
$
如图，可将包装盒分割为长方体MNPQ-EFGH和四个全等的四棱锥.$![Image](https://cdntk.53zaixian.com/tq/1715549427211685890/img/compression/M_44be4e1d34b1deb80ca5d761edc3cf53.jpg)$易得MN=4\sqrt{2} cm，EM=4\sqrt{3} cm.所以V_{长方体MNPQ-EFGH}=(4\sqrt{2})^{2}\times 4\sqrt{3}=128\sqrt{3} cm^{3},V_{四棱锥B-MNFE}=\frac{1}{3}\times 4\sqrt{3}\times 4\sqrt{2}\times 2\sqrt{2}=\frac{64\sqrt{3}}{3} cm^{3},所以该包装盒的容积为128\sqrt{3}+4\times \frac{64\sqrt{3}}{3}=\frac{640\sqrt{3}}{3} cm^{3}.$

Final answer: $640\sqrt 3/3$ $cm^3$
Subfield: Solid Geometry


Question: $已知直三棱柱ABC-A_{1}B_{1}C_{1}中，侧面AA_{1}B_{1}B为正方形，AB=BC=2，E，F分别为AC和CC_{1}的中点，BF\perp A_{1}B_{1}.
$<img_2863>
$(1)求三棱锥F-EBC的体积；$

Solution:
$\because 侧面四边形AA_{1}B_{1}B为正方形，\therefore A_{1}B_{1}\perp BB_{1}，\because BF\perp A_{1}B_{1}且BB_{1}\cap BF=B，BB_{1},BF\subset 平面BB_{1}C_{1}C，\therefore A_{1}B_{1}\perp 平面BB_{1}C_{1}C，(线面垂直的判定定理)又\because AB\parallel A_{1}B_{1}，\therefore AB\perp 平面BB_{1}C_{1}C，又\because BC\subset 平面BB_{1}C_{1}C，\therefore AB\perp BC，(线面垂直的性质定理)\because AB=BC=2，E为AC的中点，\therefore S_{\triangle EBC}=\frac{1}{2}S_{\triangle ABC}=\frac{1}{2}\times \frac{1}{2}\times 2\times 2=1.由直三棱柱知CF\perp 平面ABC.\because F为CC_{1}的中点，\therefore CF=\frac{1}{2}CC_{1}=\frac{1}{2}BB_{1}=\frac{1}{2}AB=1，\therefore V_{F-EBC}=\frac{1}{3}S_{\triangle EBC}\cdot CF=\frac{1}{3}\times 1\times 1=\frac{1}{3}.$

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


Question: $已知直三棱柱ABC-A_{1}B_{1}C_{1}中，侧面AA_{1}B_{1}B为正方形，AB=BC=2，E，F分别为AC和CC_{1}的中点，BF\perp A_{1}B_{1}.
$![Image](https://cdntk.53zaixian.com/tq/1715549487395860481/img/compression/M_3dc56b5f4ffc27d7a10428af0a27f077.jpg)
$(1)求三棱锥F-EBC的体积；$

Solution:
$\because 侧面四边形AA_{1}B_{1}B为正方形，\therefore A_{1}B_{1}\perp BB_{1}，\because BF\perp A_{1}B_{1}且BB_{1}\cap BF=B，BB_{1},BF\subset 平面BB_{1}C_{1}C，\therefore A_{1}B_{1}\perp 平面BB_{1}C_{1}C，(线面垂直的判定定理)又\because AB\parallel A_{1}B_{1}，\therefore AB\perp 平面BB_{1}C_{1}C，又\because BC\subset 平面BB_{1}C_{1}C，\therefore AB\perp BC，(线面垂直的性质定理)\because AB=BC=2，E为AC的中点，\therefore S_{\triangle EBC}=\frac{1}{2}S_{\triangle ABC}=\frac{1}{2}\times \frac{1}{2}\times 2\times 2=1.由直三棱柱知CF\perp 平面ABC.\because F为CC_{1}的中点，\therefore CF=\frac{1}{2}CC_{1}=\frac{1}{2}BB_{1}=\frac{1}{2}AB=1，\therefore V_{F-EBC}=\frac{1}{3}S_{\triangle EBC}\cdot CF=\frac{1}{3}\times 1\times 1=\frac{1}{3}.$

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


Question: $如图，D为圆锥的顶点，O是圆锥底面的圆心，\triangle ABC是底面的内接正三角形，P为DO上一点，\angle APC=90^\circ .$<img_1990>
$(1)证明：平面PAB\perp 平面PAC；$

Solution:
$证明：由题设可知，PA=PB=PC.由于\triangle ABC是正三角形，故可得\triangle PAC\cong \triangle PAB，\triangle PAC\cong \triangle PBC.又\angle APC=90^\circ ，故\angle APB=90^\circ ，\angle BPC=90^\circ .从而PB\perp PA，PB\perp PC，故PB\perp 平面PAC，所以平面PAB\perp 平面PAC.$
Subfield: Solid Geometry


Question: $如图，D为圆锥的顶点，O是圆锥底面的圆心，\triangle ABC是底面的内接正三角形，P为DO上一点，\angle APC=90^\circ .$![Image](https://cdntk.53zaixian.com/tq/1715549489019056130/img/compression/M_65ea453f50f947a562a670181b855bc0.jpg)
$(1)证明：平面PAB\perp 平面PAC；$

Solution:
$证明：由题设可知，PA=PB=PC.由于\triangle ABC是正三角形，故可得\triangle PAC\cong \triangle PAB，\triangle PAC\cong \triangle PBC.又\angle APC=90^\circ ，故\angle APB=90^\circ ，\angle BPC=90^\circ .从而PB\perp PA，PB\perp PC，故PB\perp 平面PAC，所以平面PAB\perp 平面PAC.$

Final answer: 证明题
Subfield: Solid Geometry


Question: $如图，D为圆锥的顶点，O是圆锥底面的圆心，\triangle ABC是底面的内接正三角形，P为DO上一点，\angle APC=90^\circ .$<img_1990>
$(2)设DO=\sqrt{2}，圆锥的侧面积为\sqrt{3}\pi ，求三棱锥P-ABC的体积.$

Solution:
$设圆锥的底面半径为r，母线长为l.由题设可得rl=\sqrt{3}，l^{2}-r^{2}=2.解得r=1，l=\sqrt{3}.从而AB=\sqrt{3}.由(1)可得PA^{2}+PB^{2}=AB^{2}，故PA=PB=PC=\frac{\sqrt{6}}{2}.所以三棱锥P-ABC的体积为\frac{1}{3}\times \frac{1}{2}\times PA\times PB\times PC=\frac{1}{3}\times \frac{1}{2}\times \left(\frac{\sqrt{6}}{2}\right)^3=\frac{\sqrt{6}}{8}.$

Final answer: $\frac{\sqrt{6}}{8}$
Subfield: Solid Geometry


Question: $如图，D为圆锥的顶点，O是圆锥底面的圆心，\triangle ABC是底面的内接正三角形，P为DO上一点，\angle APC=90^\circ .$![Image](https://cdntk.53zaixian.com/tq/1715549489019056130/img/compression/M_65ea453f50f947a562a670181b855bc0.jpg)
$(2)设DO=\sqrt{2}，圆锥的侧面积为\sqrt{3}\pi ，求三棱锥P-ABC的体积.$

Solution:
$设圆锥的底面半径为r，母线长为l.由题设可得rl=\sqrt{3}，l^{2}-r^{2}=2.解得r=1，l=\sqrt{3}.从而AB=\sqrt{3}.由(1)可得PA^{2}+PB^{2}=AB^{2}，故PA=PB=PC=\frac{\sqrt{6}}{2}.所以三棱锥P-ABC的体积为\frac{1}{3}\times \frac{1}{2}\times PA\times PB\times PC=\frac{1}{3}\times \frac{1}{2}\times \left(\frac{\sqrt{6}}{2}\right)^3=\frac{\sqrt{6}}{8}.$

Final answer: $\frac{\sqrt{6}}{8}$
Subfield: Solid Geometry


Question: $如图，四面体ABCD中，AD\perp CD，AD=CD，\angle ADB=\angle BDC，E为AC的中点.$![Image](https://cdntk.53zaixian.com/tq/1715549490273153026/img/compression/M_c9b4b059a249c6edf1e2d0e1b4e0b200.jpg)
$(2)设AB=BD=2，\angle ACB=60^\circ ，点F在BD上，当\triangle AFC的面积最小时，求三棱锥F-ABC的体积.$

Solution:
$解法一：由(1)可知AB=BC且\angle ACB=60^\circ ,\therefore \triangle ABC为等边三角形，\therefore AC=AB=2.又AD=DC,AD\perp CD,\therefore AD=DC=\sqrt{2},连接EF,由(1)知AC\perp 平面BED,\because EF\subset 平面BED,\therefore AC\perp EF,\therefore S_{\triangle ACF}=\frac{1}{2}AC\times EF=EF,在Rt\triangle ADC中，可得DE=1,在\triangle ABC中，可得BE=\sqrt{3},又BD=2,\therefore BD^{2}=DE^{2}+BE^{2},\therefore \triangle BED为直角三角形，且\angle EBD=30^\circ ,\therefore EF的最小值为Rt\triangle BED斜边上的高h,且h=BE\\sin  \angle EBD=\frac{\sqrt{3}}{2},\because AC\perp 平面BEF,\therefore V_{F-ABC}=\frac{1}{3}S_{\triangle BEF}\times AC=\frac{1}{3}\times \left(\frac{1}{2}BF\cdot h\right)\times AC=\frac{1}{3}\times \left(\frac{1}{2}BE\\\cos    30^{\circ}\cdot h\right)\times AC=\frac{1}{3}\times \left(\frac{1}{2}\times \sqrt{3}\times \frac{\sqrt{3}}{2}\times \frac{\sqrt{3}}{2}\right)\times 2=\frac{\sqrt{3}}{4}.解法二：连接EF，结合(1)易得AF=CF，又E为AC的中点，\therefore EF\perp AC，易知当EF\perp BD时，EF最短，此时S_{\triangle AFC}最小.由(1)知AB=BC，又AB=2，\angle ACB=60^\circ ，\therefore \triangle ABC为等边三角形，\therefore AC=2，BE=\sqrt{3}，易得DE=1，\therefore DE^{2}+BE^{2}=4=BD^{2}，\therefore DE\perp BE，\therefore EF_{min}=\frac{\sqrt{3}\times 1}{2}=\frac{\sqrt{3}}{2}，过F作FH\perp BE于H，则FH\parallel DE，FH\perp AC，又AC\cap BE=E，AC，BE\subset 平面ABC，\therefore FH\perp 平面ABC，\because \\cos  \angle FEB=\frac{EF}{BE}=\frac{1}{2}，\therefore \angle FEB=60^\circ ，\therefore FH=EF\cdot \\sin   60^\circ =\frac{\sqrt{3}}{2}\times \frac{\sqrt{3}}{2}=\frac{3}{4}，\therefore V_{F-ABC}=\frac{1}{3}\cdot S_{\triangle ABC}\cdot FH=\frac{1}{3}\times \frac{1}{2}\times 2\times 2\times \frac{\sqrt{3}}{2}\times \frac{3}{4}=\frac{\sqrt{3}}{4}.$![Image](https://cdntk.53zaixian.com/tq/1715549490575142914/img/compression/M_ee980d7f8dd05cb096679e831045abae.jpg)

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


Question: $如图，四面体ABCD中，AD\perp CD，AD=CD，\angle ADB=\angle BDC，E为AC的中点.$<img_2285>
$(2)设AB=BD=2，\angle ACB=60^\circ ，点F在BD上，当\triangle AFC的面积最小时，求三棱锥F-ABC的体积.$

Solution:
$解法一：由(1)可知AB=BC且\angle ACB=60^\circ ,\therefore \triangle ABC为等边三角形，\therefore AC=AB=2.又AD=DC,AD\perp CD,\therefore AD=DC=\sqrt{2},连接EF,由(1)知AC\perp 平面BED,\because EF\subset 平面BED,\therefore AC\perp EF,\therefore S_{\triangle ACF}=\frac{1}{2}AC\times EF=EF,在Rt\triangle ADC中，可得DE=1,在\triangle ABC中，可得BE=\sqrt{3},又BD=2,\therefore BD^{2}=DE^{2}+BE^{2},\therefore \triangle BED为直角三角形，且\angle EBD=30^\circ ,\therefore EF的最小值为Rt\triangle BED斜边上的高h,且h=BE\\sin  \angle EBD=\frac{\sqrt{3}}{2},\because AC\perp 平面BEF,\therefore V_{F-ABC}=\frac{1}{3}S_{\triangle BEF}\times AC=\frac{1}{3}\times \left(\frac{1}{2}BF\cdot h\right)\times AC=\frac{1}{3}\times \left(\frac{1}{2}BE\\\cos    30^{\circ}\cdot h\right)\times AC=\frac{1}{3}\times \left(\frac{1}{2}\times \sqrt{3}\times \frac{\sqrt{3}}{2}\times \frac{\sqrt{3}}{2}\right)\times 2=\frac{\sqrt{3}}{4}.解法二：连接EF，结合(1)易得AF=CF，又E为AC的中点，\therefore EF\perp AC，易知当EF\perp BD时，EF最短，此时S_{\triangle AFC}最小.由(1)知AB=BC，又AB=2，\angle ACB=60^\circ ，\therefore \triangle ABC为等边三角形，\therefore AC=2，BE=\sqrt{3}，易得DE=1，\therefore DE^{2}+BE^{2}=4=BD^{2}，\therefore DE\perp BE，\therefore EF_{min}=\frac{\sqrt{3}\times 1}{2}=\frac{\sqrt{3}}{2}，过F作FH\perp BE于H，则FH\parallel DE，FH\perp AC，又AC\cap BE=E，AC，BE\subset 平面ABC，\therefore FH\perp 平面ABC，\because \\cos  \angle FEB=\frac{EF}{BE}=\frac{1}{2}，\therefore \angle FEB=60^\circ ，\therefore FH=EF\cdot \\sin   60^\circ =\frac{\sqrt{3}}{2}\times \frac{\sqrt{3}}{2}=\frac{3}{4}，\therefore V_{F-ABC}=\frac{1}{3}\cdot S_{\triangle ABC}\cdot FH=\frac{1}{3}\times \frac{1}{2}\times 2\times 2\times \frac{\sqrt{3}}{2}\times \frac{3}{4}=\frac{\sqrt{3}}{4}.$<img_1710>

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


Question: $如图，在三棱锥A-BCD中，平面ABD\perp 平面BCD，AB=AD，O为BD的中点.$![Image](https://cdntk.53zaixian.com/tq/1715549490906492931/img/compression/M_abb2bf59ade97eb7cba8d3eb8fddc1f1.jpg)
$(1)证明：OA\perp CD；$

Solution:
$证明：在\triangle ABD中，\because AB=AD，O为BD的中点，\therefore AO\perp BD，又\because 平面ABD\perp 平面BCD，平面ABD\cap 平面BCD=BD，AO\subset 平面ABD，\therefore AO\perp 平面BCD，又CD\subset 平面BCD，\therefore AO\perp CD.$

Final answer: 证明题
Subfield: Solid Geometry


Question: $如图，在三棱锥A-BCD中，平面ABD\perp 平面BCD，AB=AD，O为BD的中点.$<img_3030>
$(1)证明：OA\perp CD；$

Solution:
$证明：在\triangle ABD中，\because AB=AD，O为BD的中点，\therefore AO\perp BD，又\because 平面ABD\perp 平面BCD，平面ABD\cap 平面BCD=BD，AO\subset 平面ABD，\therefore AO\perp 平面BCD，又CD\subset 平面BCD，\therefore AO\perp CD.$
Subfield: Solid Geometry


Question: $如图，在三棱锥A-BCD中，平面ABD\perp 平面BCD，AB=AD，O为BD的中点.$<img_3030>
$(2)若\triangle OCD是边长为1的等边三角形，点E在棱AD上，DE=2EA，且二面角E-BC-D的大小为45^\circ ，求三棱锥A-BCD的体积.$

Solution:
$在\triangle ABD中，过E作EN\parallel AO交BD于N，则由AO\perp 平面BCD得EN\perp 平面BCD，\therefore EN\perp BC，\because OB=OD=OC=1，\therefore \angle BCD=90^\circ ，即DC\perp BC.$<img_2711>$在\triangle BCD中，过N作NM\parallel CD交BC于M，则NM\perp BC.连接EM，\because BC\perp EN，BC\perp NM，EN\cap NM=N，\therefore BC\perp 平面EMN，\therefore EM\perp BC，\therefore \angle EMN为二面角E-BC-D的平面角，又知二面角E-BC-D的大小为45^\circ ，\therefore \angle EMN=45^\circ ，\therefore \triangle EMN为等腰直角三角形，又由DE=2EA得DN=2NO，\therefore MN=\frac{2}{3}CD=\frac{2}{3}=EN=ND，\therefore AO=OD=1，\therefore V_{A-BCD}=\frac{1}{3}S_{\triangle BCD}\cdot AO=\frac{1}{3}\times \frac{1}{2}\times 1\times \sqrt{3}\times 1=\frac{\sqrt{3}}{6}.故三棱锥A-BCD的体积为\frac{\sqrt{3}}{6}.$

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


Question: $如图，在三棱锥A-BCD中，平面ABD\perp 平面BCD，AB=AD，O为BD的中点.$![Image](https://cdntk.53zaixian.com/tq/1715549490906492931/img/compression/M_abb2bf59ade97eb7cba8d3eb8fddc1f1.jpg)
$(2)若\triangle OCD是边长为1的等边三角形，点E在棱AD上，DE=2EA，且二面角E-BC-D的大小为45^\circ ，求三棱锥A-BCD的体积.$

Solution:
$在\triangle ABD中，过E作EN\parallel AO交BD于N，则由AO\perp 平面BCD得EN\perp 平面BCD，\therefore EN\perp BC，\because OB=OD=OC=1，\therefore \angle BCD=90^\circ ，即DC\perp BC.$![Image](https://cdntk.53zaixian.com/tq/1715549491191705602/img/compression/M_6228825b5f31057a2414e56144b8b9ce.jpg)$在\triangle BCD中，过N作NM\parallel CD交BC于M，则NM\perp BC.连接EM，\because BC\perp EN，BC\perp NM，EN\cap NM=N，\therefore BC\perp 平面EMN，\therefore EM\perp BC，\therefore \angle EMN为二面角E-BC-D的平面角，又知二面角E-BC-D的大小为45^\circ ，\therefore \angle EMN=45^\circ ，\therefore \triangle EMN为等腰直角三角形，又由DE=2EA得DN=2NO，\therefore MN=\frac{2}{3}CD=\frac{2}{3}=EN=ND，\therefore AO=OD=1，\therefore V_{A-BCD}=\frac{1}{3}S_{\triangle BCD}\cdot AO=\frac{1}{3}\times \frac{1}{2}\times 1\times \sqrt{3}\times 1=\frac{\sqrt{3}}{6}.故三棱锥A-BCD的体积为\frac{\sqrt{3}}{6}.$

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


Question: $如图所示，圆锥的高PO=\sqrt{3}，底面圆O的半径为1，延长直径AB到点C，使得BC=1，分别过点A,C作底面圆O的切线，两切线相交于点E，点D是切线CE与圆O的切点.$<img_3127>
$(1)证明：平面PDE\perp 平面POD；$

Solution:
$证明：由题意知，PO\perp 平面ABD，又D是切线CE与圆O的切点，所以CE\subset 平面ABD，且OD\perp CE，则PO\perp CE，又PO\cap OD=O,PO,OD\subset 平面POD，所以CE\perp 平面POD，又CE\subset 平面PDE，所以平面PDE\perp 平面POD.$
Subfield: Solid Geometry


Question: $如图所示，圆锥的高PO=\sqrt{3}，底面圆O的半径为1，延长直径AB到点C，使得BC=1，分别过点A,C作底面圆O的切线，两切线相交于点E，点D是切线CE与圆O的切点.$![Image](https://cdntk.53zaixian.com/tq/1715549502428246020/img/compression/M_bf82d1c1b723866a569f59ea9f913d44.jpg)
$(1)证明：平面PDE\perp 平面POD；$

Solution:
$证明：由题意知，PO\perp 平面ABD，又D是切线CE与圆O的切点，所以CE\subset 平面ABD，且OD\perp CE，则PO\perp CE，又PO\cap OD=O,PO,OD\subset 平面POD，所以CE\perp 平面POD，又CE\subset 平面PDE，所以平面PDE\perp 平面POD.$

Final answer: 证明题
Subfield: Solid Geometry


Question: $如图所示，圆锥的高PO=\sqrt{3}，底面圆O的半径为1，延长直径AB到点C，使得BC=1，分别过点A,C作底面圆O的切线，两切线相交于点E，点D是切线CE与圆O的切点.$![Image](https://cdntk.53zaixian.com/tq/1715549502428246020/img/compression/M_bf82d1c1b723866a569f59ea9f913d44.jpg)
$(2)若点E到平面PAD的距离为d_{1}，求d_{1}的值.$

Solution:
$因为OD\perp CE，OD=1，OC=2，所以CD=\sqrt{3}，\angle OCD=30^\circ 又AE\perp AC，CA=3，所以AE=\sqrt{3},CE=2\sqrt{3}，所以ED=\sqrt{3},\angle AED=60^\circ ，所以AD=\sqrt{3}，且\triangle ADE的面积为\frac{1}{2}AE\cdot ED\cdot \\sin   60^\circ =\frac{1}{2}\times \sqrt{3}\times \sqrt{3}\times \frac{\sqrt{3}}{2}=\frac{3\sqrt{3}}{4}，因为PO=\sqrt{3},OA=OD=1,\angle POD=\angle POA=90^\circ ，所以PA=PD=2，所以\triangle PAD为等腰三角形，其底边AD上的高为\sqrt{4-\frac{3}{4}}=\frac{\sqrt{13}}{2}，所以\triangle PAD的面积为\frac{1}{2}\times \sqrt{3}\times \frac{\sqrt{13}}{2}=\frac{\sqrt{39}}{4}，因为V_{P-ADE}=V_{E-PAD}，所以\frac{1}{3}S_{\triangle ADE}\cdot PO=\frac{1}{3}S_{\triangle PAD}\cdot d_{1}，所以\frac{1}{3}\times \frac{3\sqrt{3}}{4}\times \sqrt{3}=\frac{1}{3}\times \frac{\sqrt{39}}{4}\times d_{1}，所以d_{1}=\frac{3\sqrt{39}}{13}.$![Image](https://cdntk.53zaixian.com/tq/1715549503011254273/img/compression/M_830124a75fecfc657349e492b0c0c0b8.jpg)

Final answer: $d_{1}=\frac{3\sqrt{39}}{13}$
Subfield: Solid Geometry


Question: $如图所示，圆锥的高PO=\sqrt{3}，底面圆O的半径为1，延长直径AB到点C，使得BC=1，分别过点A,C作底面圆O的切线，两切线相交于点E，点D是切线CE与圆O的切点.$<img_3127>
$(2)若点E到平面PAD的距离为d_{1}，求d_{1}的值.$

Solution:
$因为OD\perp CE，OD=1，OC=2，所以CD=\sqrt{3}，\angle OCD=30^\circ 又AE\perp AC，CA=3，所以AE=\sqrt{3},CE=2\sqrt{3}，所以ED=\sqrt{3},\angle AED=60^\circ ，所以AD=\sqrt{3}，且\triangle ADE的面积为\frac{1}{2}AE\cdot ED\cdot \\sin   60^\circ =\frac{1}{2}\times \sqrt{3}\times \sqrt{3}\times \frac{\sqrt{3}}{2}=\frac{3\sqrt{3}}{4}，因为PO=\sqrt{3},OA=OD=1,\angle POD=\angle POA=90^\circ ，所以PA=PD=2，所以\triangle PAD为等腰三角形，其底边AD上的高为\sqrt{4-\frac{3}{4}}=\frac{\sqrt{13}}{2}，所以\triangle PAD的面积为\frac{1}{2}\times \sqrt{3}\times \frac{\sqrt{13}}{2}=\frac{\sqrt{39}}{4}，因为V_{P-ADE}=V_{E-PAD}，所以\frac{1}{3}S_{\triangle ADE}\cdot PO=\frac{1}{3}S_{\triangle PAD}\cdot d_{1}，所以\frac{1}{3}\times \frac{3\sqrt{3}}{4}\times \sqrt{3}=\frac{1}{3}\times \frac{\sqrt{39}}{4}\times d_{1}，所以d_{1}=\frac{3\sqrt{39}}{13}.$<img_2056>

Final answer: $d_{1}=\frac{3\sqrt{39}}{13}$
Subfield: Solid Geometry


Question: $在\triangle ABC中，\angle ACB=45^\circ ，BC=3，过点A作AD\perp BC，交线段BC于点D(如图1)，沿AD将\triangle ABD折起，使\angle BDC=90^\circ (如图2)，点E，M分别为棱BC，AC的中点.
$![Image](https://cdntk.53zaixian.com/tq/1715549503518765057/img/compression/M_d5588d2fc5a447ab167bfa60b3ad6306.jpg)$
图1
$![Image](https://cdntk.53zaixian.com/tq/1715549503518765057/img/compression/M_6eb5a7185ee7d8db4f55ea72f1ff80d8.jpg)$
图2
$
$(1)求证：CD\perp ME；$

Solution:
$证明：在\triangle ABC中，M，E分别为AC，BC的中点，则ME\parallel AB，折叠前AD\perp BC，则折叠后AD\perp CD，又\angle BDC=90^\circ ，所以CD\perp BD，且AD\cap BD=D，AD\subset 平面ADB，BD\subset 平面ADB，所以CD\perp 平面ADB，又AB\subset 平面ADB，所以CD\perp AB，因为ME\parallel AB，所以CD\perp ME.$

Final answer: 证明题
Subfield: Solid Geometry


Question: $在\triangle ABC中，\angle ACB=45^\circ ，BC=3，过点A作AD\perp BC，交线段BC于点D(如图1)，沿AD将\triangle ABD折起，使\angle BDC=90^\circ (如图2)，点E，M分别为棱BC，AC的中点.
$<img_2451>$
图1
$<img_2565>$
图2
$
$(1)求证：CD\perp ME；$

Solution:
$证明：在\triangle ABC中，M，E分别为AC，BC的中点，则ME\parallel AB，折叠前AD\perp BC，则折叠后AD\perp CD，又\angle BDC=90^\circ ，所以CD\perp BD，且AD\cap BD=D，AD\subset 平面ADB，BD\subset 平面ADB，所以CD\perp 平面ADB，又AB\subset 平面ADB，所以CD\perp AB，因为ME\parallel AB，所以CD\perp ME.$
Subfield: Solid Geometry


Question: $在\triangle ABC中，\angle ACB=45^\circ ，BC=3，过点A作AD\perp BC，交线段BC于点D(如图1)，沿AD将\triangle ABD折起，使\angle BDC=90^\circ (如图2)，点E，M分别为棱BC，AC的中点.
$<img_2451>$
图1
$<img_2565>$
图2
$
$(2)求三棱锥A-BCD体积的最大值.$

Solution:
$设BD=x(0因为AD\perp CD，AD\perp BD，且CD\cap BD=D，CD\subset 平面BDC，BD\subset 平面BDC，所以AD\perp 平面BDC，所以AD为三棱锥A-BCD的高，在\triangle ADC中，\angle ACD=45^\circ ,\angle ADC=90^\circ ，所以AD=CD=3-x，所以三棱锥体积为V=\frac{1}{3}\times \frac{1}{2}x(3-x)^{2}=\frac{1}{6}x(3-x)^{2}(0则V'=\frac{1}{2}(3-x)(1-x)，令V'=0，解得x=1或x=3(舍去)，令V'>0，解得0所以V=\frac{1}{6}x(3-x)^{2}在(0,1)上单调递增，在(1,3)上单调递减，故当x=1，即BD=1，CD=2时，V取得最大值，V_{max}=\frac{1}{6}\times 1\times (3-1)^{2}=\frac{2}{3}.$

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


Question: $在\triangle ABC中，\angle ACB=45^\circ ，BC=3，过点A作AD\perp BC，交线段BC于点D(如图1)，沿AD将\triangle ABD折起，使\angle BDC=90^\circ (如图2)，点E，M分别为棱BC，AC的中点.
$![Image](https://cdntk.53zaixian.com/tq/1715549503518765057/img/compression/M_d5588d2fc5a447ab167bfa60b3ad6306.jpg)$
图1
$![Image](https://cdntk.53zaixian.com/tq/1715549503518765057/img/compression/M_6eb5a7185ee7d8db4f55ea72f1ff80d8.jpg)$
图2
$
$(2)求三棱锥A-BCD体积的最大值.$

Solution:
$设BD=x(0因为AD\perp CD，AD\perp BD，且CD\cap BD=D，CD\subset 平面BDC，BD\subset 平面BDC，所以AD\perp 平面BDC，所以AD为三棱锥A-BCD的高，在\triangle ADC中，\angle ACD=45^\circ ,\angle ADC=90^\circ ，所以AD=CD=3-x，所以三棱锥体积为V=\frac{1}{3}\times \frac{1}{2}x(3-x)^{2}=\frac{1}{6}x(3-x)^{2}(0则V'=\frac{1}{2}(3-x)(1-x)，令V'=0，解得x=1或x=3(舍去)，令V'>0，解得0所以V=\frac{1}{6}x(3-x)^{2}在(0,1)上单调递增，在(1,3)上单调递减，故当x=1，即BD=1，CD=2时，V取得最大值，V_{max}=\frac{1}{6}\times 1\times (3-1)^{2}=\frac{2}{3}.$

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


Question: $在四棱锥P-ABCD中，AB\parallel CD，CD=2AB，AC与BD相交于点M，点N在线段AP上，AN=\lambda AP(\lambda >0)，且MN\parallel 平面PCD.
$![Image](https://cdntk.53zaixian.com/tq/1715549504181465089/img/compression/M_ad6f5313566b7aee7f6b280b39f0590b.jpg)
$(1)求实数\lambda 的值；$

Solution:
$因为AB\parallel CD,所以\frac{AM}{MC}=\frac{AB}{CD}=\frac{1}{2},则\frac{AM}{AC}=\frac{1}{3}.因为MN\parallel 平面PCD，MN\subset 平面PAC，平面PAC\cap 平面PCD=PC，所以MN\parallel PC.所以\frac{AN}{AP}=\frac{AM}{AC}=\frac{1}{3}，即\lambda =\frac{1}{3}.$

Final answer: $\frac{1}{3}$