di-zhang-fdu/OlympiadExtended huggingface dataset:
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)
$(2)若\overrightarrow{BH}=\lambda \overrightarrow{BP}，是否存在\lambda ，使得平面CDH\perp 平面PAB？若存在，求出\lambda 的值；若不存在，请说明理由．$

Solution:
$过点C作CM\perp PE于点M，\because AB\perp 平面PCE，AB\subset 平面PAB，\therefore 平面PAB\perp 平面PCE.\because 平面PAB\cap 平面PCE=PE，CM\subset 平面PCE，CM\perp PE，\therefore CM\perp 平面PAB.过点M作HN\parallel AB，分别交PA,PB于点N,H，连接CH，DN，\because CD\parallel AB，\therefore HN\parallel CD，\therefore C,D,N,H四点共面.\because CM\subset 平面CDNH，\therefore 平面CDNH\perp 平面PAB.\because PA=PB=4，AE=1，PE\perp AB，\therefore PE=\sqrt{PA^2-AE^2}=\sqrt{15}.由(1)知CE=\sqrt{3}，在\triangle PCE中，由余弦定理的推论可得\\cos  \angle PCE=\frac{PC^2+CE^2-PE^2}{2PC\cdot CE}=-\frac{\sqrt{2}}{2}，\therefore \\sin  \angle PCE=\sqrt{1-\\\cos   ^2\angle PCE}=\frac{\sqrt{2}}{2}，\because S_{\triangle PCE}=\frac{1}{2}PC\cdot CE\\sin  \angle PCE=\frac{1}{2}CM\cdot PE，\therefore CM=\frac{PC\cdot CE\\\sin   \angle PCE}{PE}=\frac{\sqrt{15}}{5},\therefore EM=\sqrt{CE^2-CM^2}=\frac{2\sqrt{15}}{5}，又HN\parallel AB，\therefore \lambda =\frac{BH}{BP}=\frac{EM}{PE}=\frac{2}{5}.故存在\lambda =\frac{2}{5}满足题意.$![Image](https://cdntk.53zaixian.com/tq/1703664463570268162/img/compression/M_a766c079901da1b1addd00296714c29b.jpg)

Final answer: $\frac{2}{5}$
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>
$(2)若\overrightarrow{BH}=\lambda \overrightarrow{BP}，是否存在\lambda ，使得平面CDH\perp 平面PAB？若存在，求出\lambda 的值；若不存在，请说明理由．$

Solution:
$过点C作CM\perp PE于点M，\because AB\perp 平面PCE，AB\subset 平面PAB，\therefore 平面PAB\perp 平面PCE.\because 平面PAB\cap 平面PCE=PE，CM\subset 平面PCE，CM\perp PE，\therefore CM\perp 平面PAB.过点M作HN\parallel AB，分别交PA,PB于点N,H，连接CH，DN，\because CD\parallel AB，\therefore HN\parallel CD，\therefore C,D,N,H四点共面.\because CM\subset 平面CDNH，\therefore 平面CDNH\perp 平面PAB.\because PA=PB=4，AE=1，PE\perp AB，\therefore PE=\sqrt{PA^2-AE^2}=\sqrt{15}.由(1)知CE=\sqrt{3}，在\triangle PCE中，由余弦定理的推论可得\\cos  \angle PCE=\frac{PC^2+CE^2-PE^2}{2PC\cdot CE}=-\frac{\sqrt{2}}{2}，\therefore \\sin  \angle PCE=\sqrt{1-\\\cos   ^2\angle PCE}=\frac{\sqrt{2}}{2}，\because S_{\triangle PCE}=\frac{1}{2}PC\cdot CE\\sin  \angle PCE=\frac{1}{2}CM\cdot PE，\therefore CM=\frac{PC\cdot CE\\\sin   \angle PCE}{PE}=\frac{\sqrt{15}}{5},\therefore EM=\sqrt{CE^2-CM^2}=\frac{2\sqrt{15}}{5}，又HN\parallel AB，\therefore \lambda =\frac{BH}{BP}=\frac{EM}{PE}=\frac{2}{5}.故存在\lambda =\frac{2}{5}满足题意.$<img_1831>

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


Question: $如图，在正三棱柱ABC-A_{1}B_{1}C_{1}中，各棱长均相等，D是BC的中点.$![Image](https://cdntk.53zaixian.com/tq/1703958631244120066/img/compression/M_c40d7f03059539f6c98aa1b6808092cc.jpg)
$(1)求证：AD\perp C_{1}D；$

Solution:
$证明：\because AB=AC，D是BC的中点，\therefore AD\perp BC.\because BB_{1}\perp 平面ABC，AD\subset 平面ABC，\therefore AD\perp BB_{1}.(2分)\because BC\cap BB_{1}=B,BC,BB_{1}\subset 平面BB_{1}C_{1}C，\therefore AD\perp 平面BB_{1}C_{1}C，(3分)又C_{1}D\subset 平面BB_{1}C_{1}C，\therefore AD\perp C_{1}D.(4分)$

Final answer: 证明题
Subfield: Solid Geometry


Question: $如图，在正三棱柱ABC-A_{1}B_{1}C_{1}中，各棱长均相等，D是BC的中点.$<img_3078>
$(1)求证：AD\perp C_{1}D；$

Solution:
$证明：\because AB=AC，D是BC的中点，\therefore AD\perp BC.\because BB_{1}\perp 平面ABC，AD\subset 平面ABC，\therefore AD\perp BB_{1}.(2分)\because BC\cap BB_{1}=B,BC,BB_{1}\subset 平面BB_{1}C_{1}C，\therefore AD\perp 平面BB_{1}C_{1}C，(3分)又C_{1}D\subset 平面BB_{1}C_{1}C，\therefore AD\perp C_{1}D.(4分)$
Subfield: Solid Geometry


Question: $如图，在正三棱柱ABC-A_{1}B_{1}C_{1}中，各棱长均相等，D是BC的中点.$![Image](https://cdntk.53zaixian.com/tq/1703958631244120066/img/compression/M_c40d7f03059539f6c98aa1b6808092cc.jpg)
$(2)求证：A_{1}B\parallel 平面AC_{1}D；$

Solution:
$证明：连接A_{1}C，交AC_{1}于O，连接OD,$![Image](https://cdntk.53zaixian.com/tq/1703958631671939073/img/compression/M_3b4740d592074ee214ef55ed0f4ba702.jpg)$易得O为A_{1}C的中点，\because D为BC的中点，\therefore A_{1}B\parallel OD，(6分)\because OD\subset 平面ADC_{1}，A_{1}B\not\subset 平面ADC_{1}，\therefore A_{1}B\parallel 平面AC_{1}D.(8分)$

Final answer: 证明题
Subfield: Solid Geometry


Question: $如图，在正三棱柱ABC-A_{1}B_{1}C_{1}中，各棱长均相等，D是BC的中点.$<img_3078>
$(2)求证：A_{1}B\parallel 平面AC_{1}D；$

Solution:
$证明：连接A_{1}C，交AC_{1}于O，连接OD,$<img_1827>$易得O为A_{1}C的中点，\because D为BC的中点，\therefore A_{1}B\parallel OD，(6分)\because OD\subset 平面ADC_{1}，A_{1}B\not\subset 平面ADC_{1}，\therefore A_{1}B\parallel 平面AC_{1}D.(8分)$
Subfield: Solid Geometry


Question: $如图，在正三棱柱ABC-A_{1}B_{1}C_{1}中，各棱长均相等，D是BC的中点.$![Image](https://cdntk.53zaixian.com/tq/1703958631244120066/img/compression/M_c40d7f03059539f6c98aa1b6808092cc.jpg)
$(3)求异面直线A_{1}B与AC_{1}所成角的余弦值．$

Solution:
$由(2)知A_{1}B\parallel OD，\therefore \angle AOD(或其补角)为异面直线A_{1}B与AC_{1}所成的角，(10分)设AB=2a，则AD=\sqrt{3}a，AO=OD=\sqrt{2}a，由余弦定理的推论得\\cos  \angle AOD=\frac{OA^2+OD^2-AD^2}{2OA\cdot OD}=\frac{2a^2+2a^2-3a^2}{2\times \sqrt{2}a\times \sqrt{2}a}=\frac{1}{4}，所以异面直线A_{1}B与AC_{1}所成角的余弦值为\frac{1}{4}．(12分)$

Final answer: $\frac{1}{4}$
Subfield: Solid Geometry


Question: $如图，在正三棱柱ABC-A_{1}B_{1}C_{1}中，各棱长均相等，D是BC的中点.$<img_3078>
$(3)求异面直线A_{1}B与AC_{1}所成角的余弦值．$

Solution:
$由(2)知A_{1}B\parallel OD，\therefore \angle AOD(或其补角)为异面直线A_{1}B与AC_{1}所成的角，(10分)设AB=2a，则AD=\sqrt{3}a，AO=OD=\sqrt{2}a，由余弦定理的推论得\\cos  \angle AOD=\frac{OA^2+OD^2-AD^2}{2OA\cdot OD}=\frac{2a^2+2a^2-3a^2}{2\times \sqrt{2}a\times \sqrt{2}a}=\frac{1}{4}，所以异面直线A_{1}B与AC_{1}所成角的余弦值为\frac{1}{4}．(12分)$

Final answer: $\frac{1}{4}$
Subfield: Solid Geometry


Question: $如图，已知三棱柱ABC-A_{1}B_{1}C_{1}的棱长均为2，\angle A_{1}AC=60^\circ ，A_{1}B=\sqrt{6}．$<img_2594>
$(1)证明：平面A_{1}ACC_{1}\perp 平面ABC；$

Solution:
$证明：取AC的中点D，连接A_{1}D，BD，$<img_1658>$\because D为AC的中点，AA_{1}=AC=2,\therefore AD=\frac{1}{2}AA_{1}=1，又\angle A_{1}AC=60^\circ ，\therefore A_{1}D\perp AC，A_{1}D=\sqrt{3}.(2分)\because AB=AC=BC=2，D为AC的中点，\therefore BD=\sqrt{3}.\because A_{1}B=\sqrt{6}，\therefore A_{1}D^{2}+BD^{2}=A_{1}B^{2}，\therefore A_{1}D\perp BD.又BD\cap AC=D，BD，AC\subset 平面ABC，\therefore A_{1}D\perp 平面ABC.(4分)\because A_{1}D\subset 平面A_{1}ACC_{1}，\therefore 平面A_{1}ACC_{1}\perp 平面ABC.(6分)$
Subfield: Solid Geometry


Question: $如图，已知三棱柱ABC-A_{1}B_{1}C_{1}的棱长均为2，\angle A_{1}AC=60^\circ ，A_{1}B=\sqrt{6}．$![Image](https://cdntk.53zaixian.com/tq/1703958632053620738/img/compression/M_fa7643f7ec1e092f030e51a23b937588.jpg)
$(1)证明：平面A_{1}ACC_{1}\perp 平面ABC；$

Solution:
$证明：取AC的中点D，连接A_{1}D，BD，$![Image](https://cdntk.53zaixian.com/tq/1703958632338833409/img/compression/M_7aad6e7ca45c111046eecefb4f7b263d.jpg)$\because D为AC的中点，AA_{1}=AC=2,\therefore AD=\frac{1}{2}AA_{1}=1，又\angle A_{1}AC=60^\circ ，\therefore A_{1}D\perp AC，A_{1}D=\sqrt{3}.(2分)\because AB=AC=BC=2，D为AC的中点，\therefore BD=\sqrt{3}.\because A_{1}B=\sqrt{6}，\therefore A_{1}D^{2}+BD^{2}=A_{1}B^{2}，\therefore A_{1}D\perp BD.又BD\cap AC=D，BD，AC\subset 平面ABC，\therefore A_{1}D\perp 平面ABC.(4分)\because A_{1}D\subset 平面A_{1}ACC_{1}，\therefore 平面A_{1}ACC_{1}\perp 平面ABC.(6分)$

Final answer: 证明题
Subfield: Solid Geometry


Question: $如图，已知三棱柱ABC-A_{1}B_{1}C_{1}的棱长均为2，\angle A_{1}AC=60^\circ ，A_{1}B=\sqrt{6}．$![Image](https://cdntk.53zaixian.com/tq/1703958632053620738/img/compression/M_fa7643f7ec1e092f030e51a23b937588.jpg)
$(2)求直线CC_{1}到平面A_{1}ABB_{1}的距离．$

Solution:
$\because CC_{1}\parallel AA_{1}，CC_{1}\not\subset 平面A_{1}ABB_{1}，AA_{1}\subset 平面A_{1}ABB_{1}，\therefore CC_{1}\parallel 平面A_{1}ABB_{1}，则直线CC_{1}到平面A_{1}ABB_{1}的距离即点C到平面A_{1}ABB_{1}的距离.(8分)连接A_{1}C,利用等体积法得V_{A_{1}-ABC}=V_{C-A_{1}AB},(9分)在\triangle A_{1}AB中，AA_{1}=AB=2，A_{1}B=\sqrt{6}，\therefore \\cos  \angle A_{1}AB=\frac{AA^2_1+AB^2-A_1B^2}{2AA_1\cdot AB}=\frac{4+4-6}{2\times 2\times 2}=\frac{1}{4}，\therefore \\sin  \angle A_{1}AB=\frac{\sqrt{15}}{4}，\therefore S_{\triangle A_{1}AB}=\frac{1}{2}\times 2\times 2\times \frac{\sqrt{15}}{4}=\frac{\sqrt{15}}{2}.(10分)设点C到平面A_{1}ABB_{1}的距离为d，则\frac{1}{3}S_{\triangle ABC}\cdot A_{1}D=\frac{1}{3}S_{\triangle A_{1}AB}\cdot d，即\frac{1}{3}\times \frac{1}{2}\times 2\times \sqrt{3}\times \sqrt{3}=\frac{1}{3}\times \frac{\sqrt{15}}{2}d，解得d=\frac{2\sqrt{15}}{5}.故直线CC_{1}到平面A_{1}ABB_{1}的距离为\frac{2\sqrt{15}}{5}.(12分)$

Final answer: $\frac{2\sqrt{15}}{5}$
Subfield: Solid Geometry


Question: $如图，已知三棱柱ABC-A_{1}B_{1}C_{1}的棱长均为2，\angle A_{1}AC=60^\circ ，A_{1}B=\sqrt{6}．$<img_2594>
$(2)求直线CC_{1}到平面A_{1}ABB_{1}的距离．$

Solution:
$\because CC_{1}\parallel AA_{1}，CC_{1}\not\subset 平面A_{1}ABB_{1}，AA_{1}\subset 平面A_{1}ABB_{1}，\therefore CC_{1}\parallel 平面A_{1}ABB_{1}，则直线CC_{1}到平面A_{1}ABB_{1}的距离即点C到平面A_{1}ABB_{1}的距离.(8分)连接A_{1}C,利用等体积法得V_{A_{1}-ABC}=V_{C-A_{1}AB},(9分)在\triangle A_{1}AB中，AA_{1}=AB=2，A_{1}B=\sqrt{6}，\therefore \\cos  \angle A_{1}AB=\frac{AA^2_1+AB^2-A_1B^2}{2AA_1\cdot AB}=\frac{4+4-6}{2\times 2\times 2}=\frac{1}{4}，\therefore \\sin  \angle A_{1}AB=\frac{\sqrt{15}}{4}，\therefore S_{\triangle A_{1}AB}=\frac{1}{2}\times 2\times 2\times \frac{\sqrt{15}}{4}=\frac{\sqrt{15}}{2}.(10分)设点C到平面A_{1}ABB_{1}的距离为d，则\frac{1}{3}S_{\triangle ABC}\cdot A_{1}D=\frac{1}{3}S_{\triangle A_{1}AB}\cdot d，即\frac{1}{3}\times \frac{1}{2}\times 2\times \sqrt{3}\times \sqrt{3}=\frac{1}{3}\times \frac{\sqrt{15}}{2}d，解得d=\frac{2\sqrt{15}}{5}.故直线CC_{1}到平面A_{1}ABB_{1}的距离为\frac{2\sqrt{15}}{5}.(12分)$

Final answer: $\frac{2\sqrt{15}}{5}$
Subfield: Solid Geometry


Question: $如图，P为圆锥的顶点，AB，CD为底面圆中两条互相垂直的直径，E为PB的中点.$![Image](https://cdntk.53zaixian.com/tq/1703958632678572035/img/compression/M_c4aead622ad19ddc1b38e85769acd417.jpg)
$(1)证明：平面PAB\perp 平面PCD；$

Solution:
$证明：设AB与CD交于点O，连接PO，$![Image](https://cdntk.53zaixian.com/tq/1703958632988950530/img/compression/M_2d4e3aeba235d5bae87264a4f7fcc809.jpg)$因为AB，CD为底面圆中两条互相垂直的直径，所以O为底面圆的圆心，所以PO为圆锥的高，所以PO\perp CD，(2分)又AB\perp CD，AB\cap PO=O，AB,PO\subset 平面PAB，所以CD\perp 平面PAB.(4分)因为CD\subset 平面PCD，所以平面PAB\perp 平面PCD.(6分)$

Final answer: 证明题
Subfield: Solid Geometry


Question: $如图，P为圆锥的顶点，AB，CD为底面圆中两条互相垂直的直径，E为PB的中点.$<img_1771>
$(1)证明：平面PAB\perp 平面PCD；$

Solution:
$证明：设AB与CD交于点O，连接PO，$<img_2688>$因为AB，CD为底面圆中两条互相垂直的直径，所以O为底面圆的圆心，所以PO为圆锥的高，所以PO\perp CD，(2分)又AB\perp CD，AB\cap PO=O，AB,PO\subset 平面PAB，所以CD\perp 平面PAB.(4分)因为CD\subset 平面PCD，所以平面PAB\perp 平面PCD.(6分)$
Subfield: Solid Geometry


Question: $如图，P为圆锥的顶点，AB，CD为底面圆中两条互相垂直的直径，E为PB的中点.$![Image](https://cdntk.53zaixian.com/tq/1703958632678572035/img/compression/M_c4aead622ad19ddc1b38e85769acd417.jpg)
$(2)若AB=4，且直线CE与平面PCD所成角的正切值为\frac{\sqrt{5}}{5}，求该圆锥的体积.$

Solution:
$过E作EF\perp PO于F，则EF\parallel AB,因为E为PB的中点，所以F为PO的中点.连接CF，由(1)知平面PAB\perp 平面PCD，因为平面PAB\cap 平面PCD=PO，EF\subset 平面PAB，EF\perp PO,所以EF\perp 平面PCD，所以\angle ECF为直线CE与平面PCD所成的角，则\\tan  \angle ECF=\frac{\sqrt{5}}{5}.(9分)因为AB=4，所以EF=\frac{1}{2}OB=\frac{1}{4}AB=1，所以\\tan  \angle ECF=\frac{EF}{CF}=\frac{1}{CF}=\frac{\sqrt{5}}{5}，解得CF=\sqrt{5}，所以PO=2OF=2\times \sqrt{5-4}=2，故该圆锥的体积为\frac{1}{3}\times \pi \times 2^{2}\times 2=\frac{8\pi }{3}. (12分)$

Final answer: $\frac{8\pi }{3}$
Subfield: Solid Geometry


Question: $如图，P为圆锥的顶点，AB，CD为底面圆中两条互相垂直的直径，E为PB的中点.$<img_1771>
$(2)若AB=4，且直线CE与平面PCD所成角的正切值为\frac{\sqrt{5}}{5}，求该圆锥的体积.$

Solution:
$过E作EF\perp PO于F，则EF\parallel AB,因为E为PB的中点，所以F为PO的中点.连接CF，由(1)知平面PAB\perp 平面PCD，因为平面PAB\cap 平面PCD=PO，EF\subset 平面PAB，EF\perp PO,所以EF\perp 平面PCD，所以\angle ECF为直线CE与平面PCD所成的角，则\\tan  \angle ECF=\frac{\sqrt{5}}{5}.(9分)因为AB=4，所以EF=\frac{1}{2}OB=\frac{1}{4}AB=1，所以\\tan  \angle ECF=\frac{EF}{CF}=\frac{1}{CF}=\frac{\sqrt{5}}{5}，解得CF=\sqrt{5}，所以PO=2OF=2\times \sqrt{5-4}=2，故该圆锥的体积为\frac{1}{3}\times \pi \times 2^{2}\times 2=\frac{8\pi }{3}. (12分)$

Final answer: $\frac{8\pi }{3}$
Subfield: Solid Geometry


Question: $如图，在四边形ABCD中，AB\perp AD,AD\parallel BC,AD=6,BC=2AB=4，E，F分别在BC，AD上，EF\parallel AB，现将四边形ABEF沿EF折起，使BE\perp EC.$![Image](https://cdntk.53zaixian.com/tq/1703958634217881602/img/compression/M_bfaa738ab17c66930740ebf1cdbdeea3.jpg)
$(1)若BE=3，在折叠后的线段AD上是否存在一点P，使得CP\parallel 平面ABEF？若存在，求出\frac{AP}{PD}的值；若不存在，说明理由．$

Solution:
$假设存在满足条件的点P.如图，过点P作PM\parallel FD，交AF于点M，连接ME，$![Image](https://cdntk.53zaixian.com/tq/1703958634578591746/img/compression/M_15442069702de23ba92589a6ce73097a.jpg)$\because CE\parallel FD,\therefore MP\parallel EC,\therefore M,P,C,E四点共面.\because CP\parallel 平面ABEF，CP\subset 平面CEMP,平面ABEF\cap 平面CEMP=ME,\therefore CP\parallel ME,\therefore 四边形CEMP为平行四边形，(2分)\therefore MP=CE=4-BE=1,易得FD=6-3=3，由MP\parallel FD可得\frac{AP}{AD}=\frac{MP}{FD}=\frac{1}{3},\therefore \frac{AP}{PD}=\frac{1}{2}.故存在点P使得CP\parallel 平面ABEF，此时，\frac{AP}{PD}=\frac{1}{2}.(4分)$

Final answer: $\frac{1}{2}$
Subfield: Solid Geometry


Question: $如图，在四边形ABCD中，AB\perp AD,AD\parallel BC,AD=6,BC=2AB=4，E，F分别在BC，AD上，EF\parallel AB，现将四边形ABEF沿EF折起，使BE\perp EC.$<img_1964>
$(1)若BE=3，在折叠后的线段AD上是否存在一点P，使得CP\parallel 平面ABEF？若存在，求出\frac{AP}{PD}的值；若不存在，说明理由．$

Solution:
$假设存在满足条件的点P.如图，过点P作PM\parallel FD，交AF于点M，连接ME，$<img_2981>$\because CE\parallel FD,\therefore MP\parallel EC,\therefore M,P,C,E四点共面.\because CP\parallel 平面ABEF，CP\subset 平面CEMP,平面ABEF\cap 平面CEMP=ME,\therefore CP\parallel ME,\therefore 四边形CEMP为平行四边形，(2分)\therefore MP=CE=4-BE=1,易得FD=6-3=3，由MP\parallel FD可得\frac{AP}{AD}=\frac{MP}{FD}=\frac{1}{3},\therefore \frac{AP}{PD}=\frac{1}{2}.故存在点P使得CP\parallel 平面ABEF，此时，\frac{AP}{PD}=\frac{1}{2}.(4分)$

Final answer: $\frac{1}{2}$
Subfield: Solid Geometry


Question: $如图，在四边形ABCD中，AB\perp AD,AD\parallel BC,AD=6,BC=2AB=4，E，F分别在BC，AD上，EF\parallel AB，现将四边形ABEF沿EF折起，使BE\perp EC.$![Image](https://cdntk.53zaixian.com/tq/1703958634217881602/img/compression/M_bfaa738ab17c66930740ebf1cdbdeea3.jpg)
$(2)求三棱锥A-CDF的体积的最大值，并求出此时点F到平面ACD的距离．$

Solution:
$由题意得BE\perp EF,又BE\perp EC,EC\cap EF=E,EC,EF\subset 平面CEFD,\therefore BE\perp 平面CEFD,又AF\parallel BE,\therefore AF\perp 平面CEFD.设BE=x(0故V_{A-CDF}=\frac{1}{3}\times \frac{1}{2}\times 2\times (6-x)\times x=-\frac{1}{3}(x-3)^{2}+3，\therefore 当x=3时，V_{A-CDF}取得最大值，最大值为3，(6分)此时EC=1，AF=3，FD=3，DC=2\sqrt{2}，\therefore AD=\sqrt{AF^2+FD^2}=3\sqrt{2}，AC=\sqrt{EF^2+EC^2+AF^2}=\sqrt{14}，在\triangle ACD中，由余弦定理的推论得\\cos  \angle ADC=\frac{18+8-14}{2\times 3\sqrt{2}\times 2\sqrt{2}}=\frac{1}{2}，\therefore \\sin  \angle ADC=\frac{\sqrt{3}}{2}，\therefore S_{\triangle ACD}=\frac{1}{2}DC\cdot AD\cdot \\sin  \angle ADC=3\sqrt{3}.(9分)设F到平面ACD的距离为h，由V_{F-ACD}=V_{A-CDF}，得\frac{1}{3}S_{\triangle ACD}\cdot h=3,解得h=\sqrt{3}．综上，三棱锥A-CDF的体积的最大值为3，此时点F到平面ACD的距离为\sqrt{3}.(12分)$

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


Question: $如图，在四边形ABCD中，AB\perp AD,AD\parallel BC,AD=6,BC=2AB=4，E，F分别在BC，AD上，EF\parallel AB，现将四边形ABEF沿EF折起，使BE\perp EC.$<img_1964>
$(2)求三棱锥A-CDF的体积的最大值，并求出此时点F到平面ACD的距离．$

Solution:
$由题意得BE\perp EF,又BE\perp EC,EC\cap EF=E,EC,EF\subset 平面CEFD,\therefore BE\perp 平面CEFD,又AF\parallel BE,\therefore AF\perp 平面CEFD.设BE=x(0故V_{A-CDF}=\frac{1}{3}\times \frac{1}{2}\times 2\times (6-x)\times x=-\frac{1}{3}(x-3)^{2}+3，\therefore 当x=3时，V_{A-CDF}取得最大值，最大值为3，(6分)此时EC=1，AF=3，FD=3，DC=2\sqrt{2}，\therefore AD=\sqrt{AF^2+FD^2}=3\sqrt{2}，AC=\sqrt{EF^2+EC^2+AF^2}=\sqrt{14}，在\triangle ACD中，由余弦定理的推论得\\cos  \angle ADC=\frac{18+8-14}{2\times 3\sqrt{2}\times 2\sqrt{2}}=\frac{1}{2}，\therefore \\sin  \angle ADC=\frac{\sqrt{3}}{2}，\therefore S_{\triangle ACD}=\frac{1}{2}DC\cdot AD\cdot \\sin  \angle ADC=3\sqrt{3}.(9分)设F到平面ACD的距离为h，由V_{F-ACD}=V_{A-CDF}，得\frac{1}{3}S_{\triangle ACD}\cdot h=3,解得h=\sqrt{3}．综上，三棱锥A-CDF的体积的最大值为3，此时点F到平面ACD的距离为\sqrt{3}.(12分)$

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


Question: $如图所示，在四棱锥C-ABED中，四边形ABED是正方形，点G，F分别是线段EC，BD的中点.$![Image](https://cdntk.53zaixian.com/tq/1703958665205784578/img/compression/M_6f23ac4201c3a1c2d3433fb69a74e72b.jpg)
$(1)求证：GF\parallel 平面ABC；$

Solution:
$证明：连接AE，\because 四边形ABED是正方形，且F是BD的中点，\therefore F也是AE的中点.\because G是EC的中点，\therefore GF\parallel AC.(3分)\because GF\not\subset 平面ABC，AC\subset 平面ABC，\therefore GF\parallel 平面ABC.(6分)$![Image](https://cdntk.53zaixian.com/tq/1703958665511968769/img/compression/M_6c70ed7a9aed07305d24d51a36be443f.jpg)

Final answer: 证明题
Subfield: Solid Geometry


Question: $如图所示，在四棱锥C-ABED中，四边形ABED是正方形，点G，F分别是线段EC，BD的中点.$<img_1646>
$(1)求证：GF\parallel 平面ABC；$

Solution:
$证明：连接AE，\because 四边形ABED是正方形，且F是BD的中点，\therefore F也是AE的中点.\because G是EC的中点，\therefore GF\parallel AC.(3分)\because GF\not\subset 平面ABC，AC\subset 平面ABC，\therefore GF\parallel 平面ABC.(6分)$<img_2355>
Subfield: Solid Geometry


Question: $如图所示，在四棱锥C-ABED中，四边形ABED是正方形，点G，F分别是线段EC，BD的中点.$![Image](https://cdntk.53zaixian.com/tq/1703958665205784578/img/compression/M_6f23ac4201c3a1c2d3433fb69a74e72b.jpg)
$(2)若AD\perp 平面ABC，且AB=2AC，求异面直线GF与CD所成角的余弦值.$

Solution:
$由(1)可知GF\parallel AC，\therefore \angle DCA(或其补角)是异面直线GF与CD所成的角.(8分)\because AD\perp 平面ABC，AC\subset 平面ABC，\therefore AD\perp AC.\because AB=2AC，四边形ABED是正方形，\therefore AD=2AC，设AC=1，则AD=2，CD=\sqrt{1^2+2^2}=\sqrt{5}，(10分)\therefore \\cos  \angle DCA=\frac{AC}{CD}=\frac{\sqrt{5}}{5}，\therefore 异面直线GF与CD所成角的余弦值为\frac{\sqrt{5}}{5}.(12分)$

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


Question: $如图所示，在四棱锥C-ABED中，四边形ABED是正方形，点G，F分别是线段EC，BD的中点.$<img_1646>
$(2)若AD\perp 平面ABC，且AB=2AC，求异面直线GF与CD所成角的余弦值.$

Solution:
$由(1)可知GF\parallel AC，\therefore \angle DCA(或其补角)是异面直线GF与CD所成的角.(8分)\because AD\perp 平面ABC，AC\subset 平面ABC，\therefore AD\perp AC.\because AB=2AC，四边形ABED是正方形，\therefore AD=2AC，设AC=1，则AD=2，CD=\sqrt{1^2+2^2}=\sqrt{5}，(10分)\therefore \\cos  \angle DCA=\frac{AC}{CD}=\frac{\sqrt{5}}{5}，\therefore 异面直线GF与CD所成角的余弦值为\frac{\sqrt{5}}{5}.(12分)$

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


Question: $下图为长方体ABCD-A_{1}B_{1}C_{1}D_{1}.$![Image](https://cdntk.53zaixian.com/tq/1704021339654078466/img/compression/M_46fdcf4f7e0ac7fc03ab4a007c063b6f.jpg)
$(1)这个长方体是棱柱吗？如果是，是几棱柱？为什么？$

Solution:
$是棱柱，并且是四棱柱，因为长方体有相对的两个面是互相平行的四边形(作底面)，其余各面都是矩形(作侧面)，且相邻侧面的公共边互相平行，符合棱柱的定义，又因为底面是四边形，所以长方体是四棱柱.$

Final answer: $是四棱柱$
Subfield: Solid Geometry


Question: $下图为长方体ABCD-A_{1}B_{1}C_{1}D_{1}.$![Image](https://cdntk.53zaixian.com/tq/1704021339654078466/img/compression/M_46fdcf4f7e0ac7fc03ab4a007c063b6f.jpg)
$(2)用平面BCNM把这个长方体分成两部分，各部分形成的几何体还是棱柱吗？如果是，请说明是几棱柱，并用符号表示；如果不是，请说明理由.$

Solution:
$是.截面BCNM的右上方部分是三棱柱BB_{1}M-CC_{1}N，左下方部分是四棱柱ABMA_{1}-DCND_{1}.$

Final answer: $是，三棱柱BB_{1}M-CC_{1}N，四棱柱ABMA_{1}-DCND_{1}$
Subfield: Solid Geometry


Question: $如图，在正方体ABCD-A_{1}B_{1}C_{1}D_{1}中，AC，BD交于点O，判断下列命题是否正确，并说明理由.$![Image](https://cdntk.53zaixian.com/tq/1704021365516541954/img/compression/M_7b5ab9159a622c8f377747dc4b0d2fa5.jpg)
$(1)由点A，O，C可以确定一个平面；$

Solution:
$不正确.因为点A，O，C在同一条直线上，故不能确定一个平面.$

Final answer: $不正确.因为点A，O，C在同一条直线上，故不能确定一个平面.$
Subfield: Solid Geometry


Question: $如图，已知正四棱柱ABCP－A'B'C'P'，点Q，R分别在棱A'B',B'C'上.$![Image](https://cdntk.53zaixian.com/tq/1704021368335114243/img/compression/M_0ba5fc2359cc513235b769f35c0471f5.jpg)
$(1)请在正四棱柱ABCP-A'B'C'P'中，画出经过P，Q，R三点的截面(不用证明)；$

Solution:
$作直线QR分别交P'A',P'C'的延长线于点M,N，连接PM交AA'于点S，连接PN交CC'于点T，连接SQ,TR，如图所示，五边形PSQRT即为所求.$![Image](https://cdntk.53zaixian.com/tq/1704021368704212993/img/compression/M_9c80949a80a7daeabd7f6134db3e65ab.jpg)

Final answer: 证明题
Subfield: Solid Geometry


Question: $如图，已知正四棱柱ABCP－A'B'C'P'，点Q，R分别在棱A'B',B'C'上.$<img_2662>
$(1)请在正四棱柱ABCP-A'B'C'P'中，画出经过P，Q，R三点的截面(不用证明)；$

Solution:
$作直线QR分别交P'A',P'C'的延长线于点M,N，连接PM交AA'于点S，连接PN交CC'于点T，连接SQ,TR，如图所示，五边形PSQRT即为所求.$<img_2796>
Subfield: Solid Geometry


Question: $如图，已知正四棱柱ABCP－A'B'C'P'，点Q，R分别在棱A'B',B'C'上.$<img_2662>
$(2)若Q，R分别为A'B'，B'C'的中点，证明：AQ，CR，BB'三线共点．$

Solution:
$证明：在正四棱柱ABCP-A'B'C'P'中，侧面BCC'B'为矩形，所以BC\parallel B'C'.又R为B'C'的中点，所以BB'与CR必相交于一点，设该点为O，易知\frac{OB'}{OB}=\frac{1}{2}，同理，BB'与AQ也相交于一点，设该点为O'，易知\frac{O'B'}{O'B}=\frac{1}{2}，所以O，O'重合，所以AQ，CR，BB'三线共点.$<img_1517>
Subfield: Solid Geometry


Question: $如图，已知正四棱柱ABCP－A'B'C'P'，点Q，R分别在棱A'B',B'C'上.$![Image](https://cdntk.53zaixian.com/tq/1704021368335114243/img/compression/M_0ba5fc2359cc513235b769f35c0471f5.jpg)
$(2)若Q，R分别为A'B'，B'C'的中点，证明：AQ，CR，BB'三线共点．$

Solution:
$证明：在正四棱柱ABCP-A'B'C'P'中，侧面BCC'B'为矩形，所以BC\parallel B'C'.又R为B'C'的中点，所以BB'与CR必相交于一点，设该点为O，易知\frac{OB'}{OB}=\frac{1}{2}，同理，BB'与AQ也相交于一点，设该点为O'，易知\frac{O'B'}{O'B}=\frac{1}{2}，所以O，O'重合，所以AQ，CR，BB'三线共点.$![Image](https://cdntk.53zaixian.com/tq/1704021369035563010/img/compression/M_7dc0dbc6a5024a9bc6893ae060f6ca06.jpg)

Final answer: 证明题
Subfield: Solid Geometry


Question: $已知正方体ABCD-A_{1}B_{1}C_{1}D_{1}中,E,F分别是棱CC_{1},AA_{1}的中点．$![Image](https://cdntk.53zaixian.com/tq/1704021369782149122/img/compression/M_d9d1facb9b4d3982c53417b31b53fb83.jpg)
$(1)画出平面BED_{1}F与平面ABCD的交线，并说明理由；$

Solution:
$如图所示，延长D_{1}F交DA的延长线于点P，连接PB，则直线PB即为平面BED_{1}F与平面ABCD的交线.$![Image](https://cdntk.53zaixian.com/tq/1704021369350135811/img/compression/M_e7914f71bc185ced674cc0b7e0547af6.jpg)$理由如下：\because P\in AD，P\in D_{1}F，DA\subset 平面ABCD，D_{1}F\subset 平面BED_{1}F，\therefore P\in 平面ABCD，P\in 平面BED_{1}F，即P为平面ABCD和平面BED_{1}F的公共点，又\because B为平面ABCD和平面BED_{1}F的公共点，\therefore 直线PB为平面BED_{1}F与平面ABCD的交线．$

Final answer: $直线PB为平面BED_1F与平面ABCD的交线。$
Subfield: Solid Geometry


Question: $已知正方体ABCD-A_{1}B_{1}C_{1}D_{1}中,E,F分别是棱CC_{1},AA_{1}的中点．$<img_1628>
$(2)设H为直线B_{1}D与平面BED_{1}F的交点，求证：B，H，D_{1}三点共线．$

Solution:
$证明：连接BD，B_{1}D_{1}，\because BB_{1}\parallel DD_{1}，且BB_{1}=DD_{1}，\therefore 四边形BB_{1}D_{1}D为平行四边形.\because H为直线B_{1}D与平面BED_{1}F的交点，\therefore H\in B_{1}D，又\because B_{1}D\subset 平面BB_{1}D_{1}D，\therefore H\in 平面BB_{1}D_{1}D，又\because H\in 平面BED_{1}F，平面BED_{1}F\cap 平面BB_{1}D_{1}D=BD_{1}，\therefore H\in BD_{1}，\therefore B，H，D_{1}三点共线．$
Subfield: Solid Geometry


Question: $已知正方体ABCD-A_{1}B_{1}C_{1}D_{1}中,E,F分别是棱CC_{1},AA_{1}的中点．$![Image](https://cdntk.53zaixian.com/tq/1704021369782149122/img/compression/M_d9d1facb9b4d3982c53417b31b53fb83.jpg)
$(2)设H为直线B_{1}D与平面BED_{1}F的交点，求证：B，H，D_{1}三点共线．$

Solution:
$证明：连接BD，B_{1}D_{1}，\because BB_{1}\parallel DD_{1}，且BB_{1}=DD_{1}，\therefore 四边形BB_{1}D_{1}D为平行四边形.\because H为直线B_{1}D与平面BED_{1}F的交点，\therefore H\in B_{1}D，又\because B_{1}D\subset 平面BB_{1}D_{1}D，\therefore H\in 平面BB_{1}D_{1}D，又\because H\in 平面BED_{1}F，平面BED_{1}F\cap 平面BB_{1}D_{1}D=BD_{1}，\therefore H\in BD_{1}，\therefore B，H，D_{1}三点共线．$

Final answer: 证明题
Subfield: Solid Geometry


Question: $如图，圆柱OO'的轴截面为正方形ABCD，EF是圆柱上异于AD，BC的母线．$![Image](https://cdntk.53zaixian.com/tq/1704021431136043011/img/compression/M_3f80b4d9a9f636d36917b85ba74f4e24.jpg)
$(1)请作出平面BDE与圆O'所在平面的交线l，并判断l与平面BEF的位置关系，要求说明作法及理由；$

Solution:
$交线l如图所示．$![Image](https://cdntk.53zaixian.com/tq/1704021431513530369/img/compression/M_4ff29b785639b3717511e19dc8e00a8e.jpg)$作法：在平面BDE中过D点作直线l\parallel BE，则直线l就是所求作的交线．理由：在圆柱OO'中，EF,BC是母线，\therefore EF\parallel BC，EF=BC，\therefore 四边形EFCB是平行四边形，\therefore EB\parallel FC，又\because EB\not\subset 平面FCD，CF\subset 平面FCD，\therefore EB\parallel 平面FCD.\because 交线l=平面FCD\cap 平面DBE，\therefore l\parallel EB，\therefore 过D作直线l\parallel EB，则直线l就是所求作的交线．\because l\parallel EB,l\not\subset 平面BEF，EB\subset 平面BEF，\therefore l\parallel 平面BEF.$

Final answer: $交线l与平面BEF平行。$
Subfield: Solid Geometry


Question: $如图，圆柱OO'的轴截面为正方形ABCD，EF是圆柱上异于AD，BC的母线．$<img_2702>
$(2)M，N分别是DE，BF的中点，证明：MN\parallel 平面ABE．$

Solution:
$证明：如图，取EF的中点G，连接MG，NG，$<img_2743>$\because M，G分别是DE，EF的中点，\therefore MG\parallel DF，\because MG\not\subset 平面DFC，DF\subset 平面DFC，\therefore MG\parallel 平面DFC，\therefore MG\parallel 平面ABE，同理可证GN\parallel 平面ABE，\because MG\cap GN=G，MG，GN\subset 平面MGN，\therefore 平面MGN\parallel 平面ABE，又\because MN\subset 平面MGN，\therefore MN\parallel 平面ABE.$
Subfield: Solid Geometry


Question: $如图，圆柱OO'的轴截面为正方形ABCD，EF是圆柱上异于AD，BC的母线．$![Image](https://cdntk.53zaixian.com/tq/1704021431136043011/img/compression/M_3f80b4d9a9f636d36917b85ba74f4e24.jpg)
$(2)M，N分别是DE，BF的中点，证明：MN\parallel 平面ABE．$

Solution:
$证明：如图，取EF的中点G，连接MG，NG，$![Image](https://cdntk.53zaixian.com/tq/1704021431916183554/img/compression/M_6b19f687724d02932efe62620b556dcd.jpg)$\because M，G分别是DE，EF的中点，\therefore MG\parallel DF，\because MG\not\subset 平面DFC，DF\subset 平面DFC，\therefore MG\parallel 平面DFC，\therefore MG\parallel 平面ABE，同理可证GN\parallel 平面ABE，\because MG\cap GN=G，MG，GN\subset 平面MGN，\therefore 平面MGN\parallel 平面ABE，又\because MN\subset 平面MGN，\therefore MN\parallel 平面ABE.$

Final answer: 证明题
Subfield: Solid Geometry


Question: $在棱长为1的正方体ABCD－A_{1}B_{1}C_{1}D_{1}中，M，N，P分别为C_{1}D_{1}，AD，CC_{1}的中点.$![Image](https://cdntk.53zaixian.com/tq/1704021398381604865/img/compression/M_2f2b6baa7c3b635cad07f39057204685.jpg)
$(1)求证：MP\parallel 平面BC_{1}A_{1}；$

Solution:
$证明：连接D_{1}C,如图所示，$![Image](https://cdntk.53zaixian.com/tq/1704021397932814339/img/compression/M_79a25b410c92d371a14e80cf893337bf.jpg)$因为M，P分别为C_{1}D_{1}，C_{1}C的中点，所以MP\parallel D_{1}C，又D_{1}C\parallel A_{1}B，所以MP\parallel A_{1}B,又MP\not\subset 平面BC_{1}A_{1}，A_{1}B\subset 平面BC_{1}A_{1}， 所以MP\parallel 平面BC_{1}A_{1}.$

Final answer: 证明题