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


Question: $如图①，在直三棱柱ABC－A_{1}B_{1}C_{1}中，M为A_{1}B_{1}的中点，已知AB=AC=1，\angle ABC=\frac{\pi }{4}，AA_{1}=2.$![Image](https://cdntk.53zaixian.com/tq/1706913511910637570/img/compression/M_632c3736b592088fc19c5d1e9fdf0d7c.jpg)
$(1)求三棱锥A_{1}－AC_{1}M的体积；$

Solution:
$由题设，易知\triangle A_{1}C_{1}M是直角三角形，\therefore S_{\triangle A_{1}C_{1}M}=\frac{1}{2}A_{1}C_{1}\cdot A_{1}M=\frac{1}{4}，则V_{A_{1}－AC_{1}M}=V_{A－A_{1}C_{1}M}=\frac{1}{3}\cdot S_{\triangle A_{1}C_{1}M}\cdot 2=\frac{1}{6}.$

Final answer: $\frac{1}{6}$
Subfield: Solid Geometry


Question: $如图①，在直三棱柱ABC－A_{1}B_{1}C_{1}中，M为A_{1}B_{1}的中点，已知AB=AC=1，\angle ABC=\frac{\pi }{4}，AA_{1}=2.$![Image](https://cdntk.53zaixian.com/tq/1706913511910637570/img/compression/M_632c3736b592088fc19c5d1e9fdf0d7c.jpg)
$(2)证明：B_{1}C\parallel 平面AC_{1}M；$

Solution:
$证明：连接A_{1}C，交AC_{1}于点N，连接MN,则点N为A_{1}C的中点，\because M为A_{1}B_{1}的中点，\therefore B_{1}C\parallel MN，又MN\subset 平面AC_{1}M，B_{1}C\not\subset 平面AC_{1}M，\therefore B_{1}C\parallel 平面AC_{1}M.$![Image](https://cdntk.53zaixian.com/tq/1706913512489451521/img/compression/M_0ca9674066d4578784847fc6b9e02d34.jpg)

Final answer: 证明题
Subfield: Solid Geometry


Question: $如图①，在直三棱柱ABC－A_{1}B_{1}C_{1}中，M为A_{1}B_{1}的中点，已知AB=AC=1，\angle ABC=\frac{\pi }{4}，AA_{1}=2.$<img_2731>
$(2)证明：B_{1}C\parallel 平面AC_{1}M；$

Solution:
$证明：连接A_{1}C，交AC_{1}于点N，连接MN,则点N为A_{1}C的中点，\because M为A_{1}B_{1}的中点，\therefore B_{1}C\parallel MN，又MN\subset 平面AC_{1}M，B_{1}C\not\subset 平面AC_{1}M，\therefore B_{1}C\parallel 平面AC_{1}M.$<img_1544>
Subfield: Solid Geometry


Question: $如图①，在直三棱柱ABC－A_{1}B_{1}C_{1}中，M为A_{1}B_{1}的中点，已知AB=AC=1，\angle ABC=\frac{\pi }{4}，AA_{1}=2.$![Image](https://cdntk.53zaixian.com/tq/1706913511910637570/img/compression/M_632c3736b592088fc19c5d1e9fdf0d7c.jpg)
$(3)将两块形状与该直三棱柱完全相同的木料按如图②所示的两种方案沿阴影面进行切割，将木料一分为二，留下体积较大的一块木料.根据你所学的知识，请判断采用哪一种方案会使留下的木料表面积较大，求出这个较大的表面积，并说明理由.$

Solution:
$按方案1切割：V_{B_{1}－ABC}=\frac{1}{3}V_{A_{1}B_{1}C_{1}－ABC}，故体积较大的一块是四棱锥B_{1}－ACC_{1}A_{1}，其表面积S_{1}=S_{四边形ACC_{1}A_{1}}+S_{\triangle CC_{1}B_{1}}+S_{\triangle ACB_{1}}+S_{\triangle AA_{1}B_{1}}+S_{\triangle A_{1}C_{1}B_{1}}，在\triangle ACB_{1}中，AC=1,AB_{1}=\sqrt{AA^2_1+A_1B^2_1}=\sqrt{5},B_{1}C=\sqrt{A_1B^2_1+A_1C^2_1+CC^2_1}=\sqrt{6}，\therefore AC^{2}+AB^2_1=B_{1}C^{2}，则\triangle ACB_{1}是以\angle CAB_{1}为直角的直角三角形，\therefore S_{\triangle ACB_{1}}=\frac{1}{2}\times 1\times \sqrt{5}=\frac{\sqrt{5}}{2}，\therefore S_{1}=2+\sqrt{2}+\frac{\sqrt{5}}{2}+1+\frac{1}{2}=\frac{7}{2}+\sqrt{2}+\frac{\sqrt{5}}{2}；按方案2切割：V_{A_{1}－ABC}=\frac{1}{3}V_{A_{1}B_{1}C_{1}－ABC}，故体积较大的一块是四棱锥A_{1}－BCC_{1}B_{1}，其表面积S_{2}=S_{BCC_{1}B_{1}}+S_{\triangle CC_{1}A_{1}}+S_{\triangle A_{1}B_{1}B}+S_{\triangle A_{1}BC}+S_{\triangle A_{1}B_{1}C_{1}}，在\triangle BCA_{1}中，易知BC=\sqrt{2},A_{1}C=A_{1}B=\sqrt{5}，则\triangle BCA_{1}为等腰三角形，\therefore S_{\triangle BCA_{1}}=\frac{1}{2}\times \sqrt{2}\times \sqrt{(\sqrt{5})^2-\left(\frac{\sqrt{2}}{2}\right)^2}=\frac{3}{2}，\therefore S_{2}=1+2\sqrt{2}+1+\frac{3}{2}+\frac{1}{2}=4+2\sqrt{2}，综上，S_{2}－S_{1}=4+2\sqrt{2}－\frac{7}{2}－\sqrt{2}－\frac{\sqrt{5}}{2}=\frac{1+2\sqrt{2}-\sqrt{5}}{2}>0，\therefore S_{2}>S_{1}，\therefore 按方案2会使留下的木料表面积较大，这个较大的表面积为4+2\sqrt{2}.$

Final answer: $按方案2会使留下的木料表面积较大，这个较大的表面积为4+2\sqrt 2.$
Subfield: Solid Geometry


Question: $如图①，在直三棱柱ABC－A_{1}B_{1}C_{1}中，M为A_{1}B_{1}的中点，已知AB=AC=1，\angle ABC=\frac{\pi }{4}，AA_{1}=2.$<img_2731>
$(3)将两块形状与该直三棱柱完全相同的木料按如图②所示的两种方案沿阴影面进行切割，将木料一分为二，留下体积较大的一块木料.根据你所学的知识，请判断采用哪一种方案会使留下的木料表面积较大，求出这个较大的表面积，并说明理由.$

Solution:
$按方案1切割：V_{B_{1}－ABC}=\frac{1}{3}V_{A_{1}B_{1}C_{1}－ABC}，故体积较大的一块是四棱锥B_{1}－ACC_{1}A_{1}，其表面积S_{1}=S_{四边形ACC_{1}A_{1}}+S_{\triangle CC_{1}B_{1}}+S_{\triangle ACB_{1}}+S_{\triangle AA_{1}B_{1}}+S_{\triangle A_{1}C_{1}B_{1}}，在\triangle ACB_{1}中，AC=1,AB_{1}=\sqrt{AA^2_1+A_1B^2_1}=\sqrt{5},B_{1}C=\sqrt{A_1B^2_1+A_1C^2_1+CC^2_1}=\sqrt{6}，\therefore AC^{2}+AB^2_1=B_{1}C^{2}，则\triangle ACB_{1}是以\angle CAB_{1}为直角的直角三角形，\therefore S_{\triangle ACB_{1}}=\frac{1}{2}\times 1\times \sqrt{5}=\frac{\sqrt{5}}{2}，\therefore S_{1}=2+\sqrt{2}+\frac{\sqrt{5}}{2}+1+\frac{1}{2}=\frac{7}{2}+\sqrt{2}+\frac{\sqrt{5}}{2}；按方案2切割：V_{A_{1}－ABC}=\frac{1}{3}V_{A_{1}B_{1}C_{1}－ABC}，故体积较大的一块是四棱锥A_{1}－BCC_{1}B_{1}，其表面积S_{2}=S_{BCC_{1}B_{1}}+S_{\triangle CC_{1}A_{1}}+S_{\triangle A_{1}B_{1}B}+S_{\triangle A_{1}BC}+S_{\triangle A_{1}B_{1}C_{1}}，在\triangle BCA_{1}中，易知BC=\sqrt{2},A_{1}C=A_{1}B=\sqrt{5}，则\triangle BCA_{1}为等腰三角形，\therefore S_{\triangle BCA_{1}}=\frac{1}{2}\times \sqrt{2}\times \sqrt{(\sqrt{5})^2-\left(\frac{\sqrt{2}}{2}\right)^2}=\frac{3}{2}，\therefore S_{2}=1+2\sqrt{2}+1+\frac{3}{2}+\frac{1}{2}=4+2\sqrt{2}，综上，S_{2}－S_{1}=4+2\sqrt{2}－\frac{7}{2}－\sqrt{2}－\frac{\sqrt{5}}{2}=\frac{1+2\sqrt{2}-\sqrt{5}}{2}>0，\therefore S_{2}>S_{1}，\therefore 按方案2会使留下的木料表面积较大，这个较大的表面积为4+2\sqrt{2}.$



Final answer: $4+2\sqrt{2}$
Subfield: Solid Geometry


Question: $如图，在多面体ABCDE中，平面ACD\perp 平面ABC，AC\perp BC，BC=2AC=4，DA=DC=3，F是BC的中点，EF\perp 平面ABC，EF=2\sqrt{2}.$![Image](https://cdntk.53zaixian.com/tq/1704026547703459844/img/compression/M_eec94297c9a249fcc99e0035821a53ac.jpg)
$(1)证明：A,B,E,D四点共面；$

Solution:
$证明：如图，取AC的中点M，连接DM，MF，\because DA=DC=3，AC=2，M为AC的中点，\therefore DM\perp AC，且DM=2\sqrt{2}.(2分)\because 平面ACD\perp 平面ABC，平面ACD\cap 平面ABC=AC，DM\subset 平面ACD，\therefore DM\perp 平面ABC.又EF\perp 平面ABC，\therefore DM\parallel EF，且DM=EF=2\sqrt{2}，\therefore 四边形DEFM是平行四边形，\therefore DE\parallel MF，(4分)在\triangle ABC中，\because M，F是AC，BC的中点，\therefore MF\parallel AB，\therefore DE\parallel AB，\therefore A,B,E,D四点共面.(6分)$![Image](https://cdntk.53zaixian.com/tq/1704026547997061121/img/compression/M_0356dc0c6aa04989b853b210b9fa47a7.jpg)

Final answer: 证明题
Subfield: Solid Geometry


Question: $如图，在多面体ABCDE中，平面ACD\perp 平面ABC，AC\perp BC，BC=2AC=4，DA=DC=3，F是BC的中点，EF\perp 平面ABC，EF=2\sqrt{2}.$<img_2202>
$(1)证明：A,B,E,D四点共面；$

Solution:
$证明：如图，取AC的中点M，连接DM，MF，\because DA=DC=3，AC=2，M为AC的中点，\therefore DM\perp AC，且DM=2\sqrt{2}.(2分)\because 平面ACD\perp 平面ABC，平面ACD\cap 平面ABC=AC，DM\subset 平面ACD，\therefore DM\perp 平面ABC.又EF\perp 平面ABC，\therefore DM\parallel EF，且DM=EF=2\sqrt{2}，\therefore 四边形DEFM是平行四边形，\therefore DE\parallel MF，(4分)在\triangle ABC中，\because M，F是AC，BC的中点，\therefore MF\parallel AB，\therefore DE\parallel AB，\therefore A,B,E,D四点共面.(6分)$<img_2155>
Subfield: Solid Geometry


Question: $如图，在多面体ABCDE中，平面ACD\perp 平面ABC，AC\perp BC，BC=2AC=4，DA=DC=3，F是BC的中点，EF\perp 平面ABC，EF=2\sqrt{2}.$<img_2202>
$(2)求三棱锥　　　　的体积.从①B-ACD；②A-BCE；③B-CDE这三个条件中任选一个，补充在上面问题中并作答.注：如果选择多个条件分别解答，按第一个解答计分.$

Solution:
$选①：\because 平面ACD\perp 平面ABC，平面ACD\cap 平面ABC=AC,AC\perp BC,BC\subset 平面ABC，\therefore BC\perp 平面ACD，\therefore V_{B-ACD}=\frac{1}{3}S_{\triangle ACD}\times BC.(8分)由(1)知，DM\perp AC,且DM=2\sqrt{2}，\therefore S_{\triangle ACD}=\frac{1}{2}\times AC\times DM=\frac{1}{2}\times 2\times 2\sqrt{2}=2\sqrt{2}，(10分)\therefore V_{B-ACD}=\frac{1}{3}\times 2\sqrt{2}\times 4=\frac{8\sqrt{2}}{3}.(12分)选②：由题图可知，V_{A-BCE}=V_{E-ABC}.\because EF\perp 平面ABC,\therefore V_{E-ABC}=\frac{1}{3}\times S_{\triangle ABC}\times EF.(8分)\because AC\perp BC,\therefore S_{\triangle ABC}=\frac{1}{2}\times AC\times BC=\frac{1}{2}\times 2\times 4=4,\therefore V_{E-ABC}=\frac{1}{3}\times 4\times 2\sqrt{2}=\frac{8\sqrt{2}}{3},即V_{A-BCE}=\frac{8\sqrt{2}}{3}.(12分)选③:由(1)知DM\parallel EF，\because DM\not\subset 平面BCE，EF\subset 平面BCE，\therefore DM\parallel 平面BCE，\therefore 点D到平面BCE的距离等于点M到平面BCE的距离，则三棱锥D-BCE与三棱锥M-BCE的体积相等，(8分)\because AC\perp BC，AC=2，BC=4，M为AC的中点，\therefore S_{\triangle BCM}=\frac{1}{2}CM\cdot BC=2,(10分)又EF\perp 平面ABC，且EF=2\sqrt{2}，\therefore V_{B-CDE}=V_{D-BCE}=V_{M-BCE}=V_{E-BCM}=\frac{1}{3}S_{\triangle BCM}\cdot EF=\frac{4\sqrt{2}}{3}.(12分)$

Final answer: $\frac{8\sqrt 2}{3}, \frac{8\sqrt 2}{3}, \frac{4\sqrt 2}{3}$
Subfield: Solid Geometry


Question: $如图，在多面体ABCDE中，平面ACD\perp 平面ABC，AC\perp BC，BC=2AC=4，DA=DC=3，F是BC的中点，EF\perp 平面ABC，EF=2\sqrt{2}.$![Image](https://cdntk.53zaixian.com/tq/1704026547703459844/img/compression/M_eec94297c9a249fcc99e0035821a53ac.jpg)
$(2)求三棱锥　　　　的体积.从①B-ACD；②A-BCE；③B-CDE这三个条件中任选一个，补充在上面问题中并作答.注：如果选择多个条件分别解答，按第一个解答计分.$

Solution:
$选①：\because 平面ACD\perp 平面ABC，平面ACD\cap 平面ABC=AC,AC\perp BC,BC\subset 平面ABC，\therefore BC\perp 平面ACD，\therefore V_{B-ACD}=\frac{1}{3}S_{\triangle ACD}\times BC.(8分)由(1)知，DM\perp AC,且DM=2\sqrt{2}，\therefore S_{\triangle ACD}=\frac{1}{2}\times AC\times DM=\frac{1}{2}\times 2\times 2\sqrt{2}=2\sqrt{2}，(10分)\therefore V_{B-ACD}=\frac{1}{3}\times 2\sqrt{2}\times 4=\frac{8\sqrt{2}}{3}.(12分)选②：由题图可知，V_{A-BCE}=V_{E-ABC}.\because EF\perp 平面ABC,\therefore V_{E-ABC}=\frac{1}{3}\times S_{\triangle ABC}\times EF.(8分)\because AC\perp BC,\therefore S_{\triangle ABC}=\frac{1}{2}\times AC\times BC=\frac{1}{2}\times 2\times 4=4,\therefore V_{E-ABC}=\frac{1}{3}\times 4\times 2\sqrt{2}=\frac{8\sqrt{2}}{3},即V_{A-BCE}=\frac{8\sqrt{2}}{3}.(12分)选③:由(1)知DM\parallel EF，\because DM\not\subset 平面BCE，EF\subset 平面BCE，\therefore DM\parallel 平面BCE，\therefore 点D到平面BCE的距离等于点M到平面BCE的距离，则三棱锥D-BCE与三棱锥M-BCE的体积相等，(8分)\because AC\perp BC，AC=2，BC=4，M为AC的中点，\therefore S_{\triangle BCM}=\frac{1}{2}CM\cdot BC=2,(10分)又EF\perp 平面ABC，且EF=2\sqrt{2}，\therefore V_{B-CDE}=V_{D-BCE}=V_{M-BCE}=V_{E-BCM}=\frac{1}{3}S_{\triangle BCM}\cdot EF=\frac{4\sqrt{2}}{3}.(12分)$

Final answer: $\frac{8\sqrt 2}{3}, \frac{8\sqrt 2}{3}, \frac{4\sqrt 2}{3}$
Subfield: Solid Geometry


Question: $如图，在正三棱柱ABC-A_{1}B_{1}C_{1}中，E为AC的中点.$<img_2569>
$(1)求证：AB_{1}\parallel 平面BEC_{1}；$

Solution:
$证明：如图所示，连接B_{1}C，交BC_{1}于点O，易知O为B_{1}C的中点，连接EO.$<img_1845>$\because E为AC的中点，\therefore AB_{1}\parallel EO.(3分)又\because AB_{1}\not\subset 平面BEC_{1}，EO\subset 平面BEC_{1}，\therefore AB_{1}\parallel 平面BEC_{1}.(5分)$
Subfield: Solid Geometry


Question: $如图，在正三棱柱ABC-A_{1}B_{1}C_{1}中，E为AC的中点.$![Image](https://cdntk.53zaixian.com/tq/1704026548332605443/img/compression/M_fdf06be77802e639d5e767aa2e2e9a7b.jpg)
$(1)求证：AB_{1}\parallel 平面BEC_{1}；$

Solution:
$证明：如图所示，连接B_{1}C，交BC_{1}于点O，易知O为B_{1}C的中点，连接EO.$![Image](https://cdntk.53zaixian.com/tq/1704026548731064322/img/compression/M_89b784173e4f2f234bf2cc93632b727d.jpg)$\because E为AC的中点，\therefore AB_{1}\parallel EO.(3分)又\because AB_{1}\not\subset 平面BEC_{1}，EO\subset 平面BEC_{1}，\therefore AB_{1}\parallel 平面BEC_{1}.(5分)$

Final answer: 证明题
Subfield: Solid Geometry


Question: $如图，在正三棱柱ABC-A_{1}B_{1}C_{1}中，E为AC的中点.$![Image](https://cdntk.53zaixian.com/tq/1704026548332605443/img/compression/M_fdf06be77802e639d5e767aa2e2e9a7b.jpg)
$(2)若BB_{1}=BA，求异面直线AB_{1}与EC_{1}所成角的余弦值.$

Solution:
$由(1)知AB_{1}\parallel EO，\therefore \angle C_{1}EO(或其补角)为异面直线AB_{1}与EC_{1}所成的角.(6分)设BB_{1}=BA=a，则C_{1}B=\sqrt{2}a，C_{1}O=\frac{1}{2}C_{1}B=\frac{\sqrt{2}}{2}a，C_{1}E=\sqrt{C_1C^2+CE^2}=\frac{\sqrt{5}}{2}a.　(7分)\because 三棱柱ABC-A_{1}B_{1}C_{1}是正三棱柱，\therefore 易得BE\perp 平面AA_{1}C_{1}C，\because EC_{1}\subset 平面AA_{1}C_{1}C，\therefore BE\perp EC_{1}.(9分)在Rt\triangle BEC_{1}中，EO=\frac{1}{2}C_{1}B=\frac{\sqrt{2}}{2}a，在\triangle OEC_{1}中，\\cos  \angle C_{1}EO=\frac{EC^2_1+EO^2-OC^2_1}{2\times EC_1\times EO}=\frac{\sqrt{10}}{4}，\therefore 异面直线AB_{1}与EC_{1}所成角的余弦值为\frac{\sqrt{10}}{4}.(12分)$

Final answer: $\frac{\sqrt{10}}{4}$
Subfield: Solid Geometry


Question: $如图，在正三棱柱ABC-A_{1}B_{1}C_{1}中，E为AC的中点.$<img_2569>
$(2)若BB_{1}=BA，求异面直线AB_{1}与EC_{1}所成角的余弦值.$

Solution:
$由(1)知AB_{1}\parallel EO，\therefore \angle C_{1}EO(或其补角)为异面直线AB_{1}与EC_{1}所成的角.(6分)设BB_{1}=BA=a，则C_{1}B=\sqrt{2}a，C_{1}O=\frac{1}{2}C_{1}B=\frac{\sqrt{2}}{2}a，C_{1}E=\sqrt{C_1C^2+CE^2}=\frac{\sqrt{5}}{2}a.　(7分)\because 三棱柱ABC-A_{1}B_{1}C_{1}是正三棱柱，\therefore 易得BE\perp 平面AA_{1}C_{1}C，\because EC_{1}\subset 平面AA_{1}C_{1}C，\therefore BE\perp EC_{1}.(9分)在Rt\triangle BEC_{1}中，EO=\frac{1}{2}C_{1}B=\frac{\sqrt{2}}{2}a，在\triangle OEC_{1}中，\\cos  \angle C_{1}EO=\frac{EC^2_1+EO^2-OC^2_1}{2\times EC_1\times EO}=\frac{\sqrt{10}}{4}，\therefore 异面直线AB_{1}与EC_{1}所成角的余弦值为\frac{\sqrt{10}}{4}.(12分)$

Final answer: $\frac{\sqrt{10}}{4}$
Subfield: Solid Geometry


Question: $如图，在正三棱柱ABC－A_{1}B_{1}C_{1}中，BA=BB_{1}=2，点D是棱AA_{1}的中点.$![Image](https://cdntk.53zaixian.com/tq/1704026549045637123/img/compression/M_4015d3cf3fe707cb64783ff7e3e73edc.jpg)
$(1)求证：BD\perp B_{1}C；$

Solution:
$证明：设AB的中点为M，连接B_{1}M，CM，如图所示：$![Image](https://cdntk.53zaixian.com/tq/1704026549418930178/img/compression/M_6403369d014cd0b3c331ec00fc7cd2c4.jpg)$根据题意，易证得\triangle B_{1}BM\cong \triangle BAD，\therefore \angle ABD=\angle BB_{1}M,又\angle BB_{1}M+\angle BMB_{1}=90^\circ ，\therefore \angle ABD+\angle BMB_{1}=90^\circ ，\therefore BD\perp B_{1}M.\because AA_{1}\perp 平面ABC，CM\subset 平面ABC，\therefore AA_{1}\perp CM，(2分)又AC=BC，M为AB的中点，\therefore CM\perp AB，\because AB\cap AA_{1}=A，AB，AA_{1}\subset 平面ABB_{1}A_{1},\therefore CM\perp 平面ABB_{1}A_{1}，\because BD\subset 平面ABB_{1}A_{1}，\therefore BD\perp CM，(4分)\because B_{1}M\cap CM=M，B_{1}M\subset 平面B_{1}CM，CM\subset 平面B_{1}CM，\therefore BD\perp 平面B_{1}CM，\because B_{1}C\subset 平面B_{1}CM，\therefore BD\perp B_{1}C.(6分)$

Final answer: 证明题
Subfield: Solid Geometry


Question: $如图，在正三棱柱ABC－A_{1}B_{1}C_{1}中，BA=BB_{1}=2，点D是棱AA_{1}的中点.$<img_2441>
$(1)求证：BD\perp B_{1}C；$

Solution:
$证明：设AB的中点为M，连接B_{1}M，CM，如图所示：$<img_2472>$根据题意，易证得\triangle B_{1}BM\cong \triangle BAD，\therefore \angle ABD=\angle BB_{1}M,又\angle BB_{1}M+\angle BMB_{1}=90^\circ ，\therefore \angle ABD+\angle BMB_{1}=90^\circ ，\therefore BD\perp B_{1}M.\because AA_{1}\perp 平面ABC，CM\subset 平面ABC，\therefore AA_{1}\perp CM，(2分)又AC=BC，M为AB的中点，\therefore CM\perp AB，\because AB\cap AA_{1}=A，AB，AA_{1}\subset 平面ABB_{1}A_{1},\therefore CM\perp 平面ABB_{1}A_{1}，\because BD\subset 平面ABB_{1}A_{1}，\therefore BD\perp CM，(4分)\because B_{1}M\cap CM=M，B_{1}M\subset 平面B_{1}CM，CM\subset 平面B_{1}CM，\therefore BD\perp 平面B_{1}CM，\because B_{1}C\subset 平面B_{1}CM，\therefore BD\perp B_{1}C.(6分)$
Subfield: Solid Geometry


Question: $如图，在正三棱柱ABC－A_{1}B_{1}C_{1}中，BA=BB_{1}=2，点D是棱AA_{1}的中点.$![Image](https://cdntk.53zaixian.com/tq/1704026549045637123/img/compression/M_4015d3cf3fe707cb64783ff7e3e73edc.jpg)
$(2)求点B到平面DCB_{1}的距离.$

Solution:
$设点B到平面DCB_{1}的距离为h，\because V_{B－DCB_{1}}=\frac{1}{3}S_{\triangle DCB_{1}}\cdot h=V_{C－BDB_{1}}，又B_{1}D=DC=\sqrt{2^2+1^2}=\sqrt{5}，B_{1}C=\sqrt{2^2+2^2}=2\sqrt{2}，\therefore \triangle DCB_{1}的边B_{1}C上的高为\sqrt{(\sqrt{5})^2-\left(\frac{2\sqrt{2}}{2}\right)^2}=\sqrt{3},\therefore S_{\triangle DCB_{1}}=\frac{1}{2}\times 2\sqrt{2}\times \sqrt{3}=\sqrt{6},(8分)由(1)知，CM\perp 平面ABB_{1}A_{1}，\therefore 三棱锥C－BDB_{1}的高为CM=\frac{\sqrt{3}}{2}\times 2=\sqrt{3}，又S_{\triangle BDB_{1}}=\frac{1}{2}\times 2\times 2=2，\therefore V_{C－BDB_{1}}=\frac{1}{3}\times 2\times \sqrt{3}=\frac{2\sqrt{3}}{3}，(10分)\therefore \frac{1}{3}\times \sqrt{6}\times h=\frac{2\sqrt{3}}{3}，\therefore h=\sqrt{2}，故点B到平面DCB_{1}的距离为\sqrt{2}.(12分)$

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


Question: $如图，在正三棱柱ABC－A_{1}B_{1}C_{1}中，BA=BB_{1}=2，点D是棱AA_{1}的中点.$<img_2441>
$(2)求点B到平面DCB_{1}的距离.$

Solution:
$设点B到平面DCB_{1}的距离为h，\because V_{B－DCB_{1}}=\frac{1}{3}S_{\triangle DCB_{1}}\cdot h=V_{C－BDB_{1}}，又B_{1}D=DC=\sqrt{2^2+1^2}=\sqrt{5}，B_{1}C=\sqrt{2^2+2^2}=2\sqrt{2}，\therefore \triangle DCB_{1}的边B_{1}C上的高为\sqrt{(\sqrt{5})^2-\left(\frac{2\sqrt{2}}{2}\right)^2}=\sqrt{3},\therefore S_{\triangle DCB_{1}}=\frac{1}{2}\times 2\sqrt{2}\times \sqrt{3}=\sqrt{6},(8分)由(1)知，CM\perp 平面ABB_{1}A_{1}，\therefore 三棱锥C－BDB_{1}的高为CM=\frac{\sqrt{3}}{2}\times 2=\sqrt{3}，又S_{\triangle BDB_{1}}=\frac{1}{2}\times 2\times 2=2，\therefore V_{C－BDB_{1}}=\frac{1}{3}\times 2\times \sqrt{3}=\frac{2\sqrt{3}}{3}，(10分)\therefore \frac{1}{3}\times \sqrt{6}\times h=\frac{2\sqrt{3}}{3}，\therefore h=\sqrt{2}，故点B到平面DCB_{1}的距离为\sqrt{2}.(12分)$

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


Question: $如图，在四棱锥P－ABCD中，\triangle PAB是等边三角形，CB\perp 平面PAB，AD\parallel BC，且PB=BC=2AD=2，F为PC的中点.$<img_1929>
$(1)求证：FD\parallel 平面PAB；$

Solution:
$证明：取PB的中点H，连接FH，AH，如图所示：$<img_2119>$\because F为PC的中点，\therefore FH是\triangle PBC的中位线，\therefore FH=\frac{1}{2}CB，FH\parallel CB,又AD\parallel CB,AD=\frac{1}{2}BC，\therefore FH\parallel AD,FH=AD,\therefore 四边形ADFH为平行四边形,\therefore FD\parallel AH.(2分)又FD\not\subset 平面ABP，AH\subset 平面ABP，\therefore FD\parallel 平面PAB.(3分)$
Subfield: Solid Geometry


Question: $如图，在四棱锥P－ABCD中，\triangle PAB是等边三角形，CB\perp 平面PAB，AD\parallel BC，且PB=BC=2AD=2，F为PC的中点.$![Image](https://cdntk.53zaixian.com/tq/1704026549804806146/img/compression/M_0b5b2e5b39cc2bc695f2de1455479c8a.jpg)
$(1)求证：FD\parallel 平面PAB；$

Solution:
$证明：取PB的中点H，连接FH，AH，如图所示：$![Image](https://cdntk.53zaixian.com/tq/1704026550090018818/img/compression/M_06466a351dfba2d0086f4f37a15ea2b4.jpg)$\because F为PC的中点，\therefore FH是\triangle PBC的中位线，\therefore FH=\frac{1}{2}CB，FH\parallel CB,又AD\parallel CB,AD=\frac{1}{2}BC，\therefore FH\parallel AD,FH=AD,\therefore 四边形ADFH为平行四边形,\therefore FD\parallel AH.(2分)又FD\not\subset 平面ABP，AH\subset 平面ABP，\therefore FD\parallel 平面PAB.(3分)$

Final answer: 证明题
Subfield: Solid Geometry


Question: $如图，在四棱锥P－ABCD中，\triangle PAB是等边三角形，CB\perp 平面PAB，AD\parallel BC，且PB=BC=2AD=2，F为PC的中点.$![Image](https://cdntk.53zaixian.com/tq/1704026549804806146/img/compression/M_0b5b2e5b39cc2bc695f2de1455479c8a.jpg)
$(2)求证：平面PBC\perp 平面ADF；$

Solution:
$证明：由(1)知，四边形ADFH为平行四边形，\therefore A，D，F，H共面.\because \triangle PAB是等边三角形，H为PB的中点，\therefore PB\perp AH，(4分)又CB\perp 平面PAB，FH\parallel CB，\therefore FH\perp 平面PAB，\therefore FH\perp PB.(5分)又FH\cap AH=H,FH,AH\subset 平面ADF,\therefore PB\perp 平面ADF.又PB\subset 平面PBC，\therefore 平面PBC\perp 平面ADF.(7分)$

Final answer: 证明题
Subfield: Solid Geometry


Question: $如图，在四棱锥P－ABCD中，\triangle PAB是等边三角形，CB\perp 平面PAB，AD\parallel BC，且PB=BC=2AD=2，F为PC的中点.$<img_1929>
$(2)求证：平面PBC\perp 平面ADF；$

Solution:
$证明：由(1)知，四边形ADFH为平行四边形，\therefore A，D，F，H共面.\because \triangle PAB是等边三角形，H为PB的中点，\therefore PB\perp AH，(4分)又CB\perp 平面PAB，FH\parallel CB，\therefore FH\perp 平面PAB，\therefore FH\perp PB.(5分)又FH\cap AH=H,FH,AH\subset 平面ADF,\therefore PB\perp 平面ADF.又PB\subset 平面PBC，\therefore 平面PBC\perp 平面ADF.(7分)$
Subfield: Solid Geometry


Question: $如图，在四棱锥P－ABCD中，\triangle PAB是等边三角形，CB\perp 平面PAB，AD\parallel BC，且PB=BC=2AD=2，F为PC的中点.$<img_1929>
$(3)求直线AB与平面PDC所成角的正弦值.$

Solution:
$取BC的中点G，连接DG，FG，则DG\parallel AB，且DG=AB=2，\therefore 直线AB与平面PDC所成的角即为DG与平面PDC所成的角.(8分)\because FG\parallel PB，DG\parallel AB，DG\cap FG=G，AB\cap PB=B,DG,FG\subset 平面DFG,AB,PB\subset 平面PAB,\therefore 平面PAB\parallel 平面DFG，又CB\perp 平面PAB，\therefore CG\perp 平面DFG.设G到平面CDF的距离为GO，由等体积法得，V_{G－CFD}=V_{C－DFG}，\therefore \frac{1}{3}S_{\triangle CDF}\cdot GO=\frac{1}{3}S_{\triangle DFG}\cdot CG，又易知\triangle DFG,\triangle CDF为直角三角形，且CF=\sqrt{2},DF=\sqrt{3},FG=1,\therefore GO=\frac{S_{\vartriangle DFG}\cdot CG}{S_{\vartriangle CDF}}=\frac{\frac{1}{2}DF\cdot FG\cdot CG}{\frac{1}{2}DF\cdot CF}=\frac{\frac{1}{2}\times \sqrt{3}\times 1\times 1}{\frac{1}{2}\times \sqrt{3}\times \sqrt{2}}=\frac{\sqrt{2}}{2}.(10分)设DG与平面PDC所成的角为\angle GDO，\therefore \\sin  \angle GDO=\frac{GO}{DG}=\frac{\frac{\sqrt{2}}{2}}{2}=\frac{\sqrt{2}}{4}，\therefore 直线AB与平面PDC所成角的正弦值为\frac{\sqrt{2}}{4}.(12分)$

Final answer: $\frac{\sqrt{2}}{4}$
Subfield: Solid Geometry


Question: $如图，在四棱锥P－ABCD中，\triangle PAB是等边三角形，CB\perp 平面PAB，AD\parallel BC，且PB=BC=2AD=2，F为PC的中点.$![Image](https://cdntk.53zaixian.com/tq/1704026549804806146/img/compression/M_0b5b2e5b39cc2bc695f2de1455479c8a.jpg)
$(3)求直线AB与平面PDC所成角的正弦值.$

Solution:
$取BC的中点G，连接DG，FG，则DG\parallel AB，且DG=AB=2，\therefore 直线AB与平面PDC所成的角即为DG与平面PDC所成的角.(8分)\because FG\parallel PB，DG\parallel AB，DG\cap FG=G，AB\cap PB=B,DG,FG\subset 平面DFG,AB,PB\subset 平面PAB,\therefore 平面PAB\parallel 平面DFG，又CB\perp 平面PAB，\therefore CG\perp 平面DFG.设G到平面CDF的距离为GO，由等体积法得，V_{G－CFD}=V_{C－DFG}，\therefore \frac{1}{3}S_{\triangle CDF}\cdot GO=\frac{1}{3}S_{\triangle DFG}\cdot CG，又易知\triangle DFG,\triangle CDF为直角三角形，且CF=\sqrt{2},DF=\sqrt{3},FG=1,\therefore GO=\frac{S_{\vartriangle DFG}\cdot CG}{S_{\vartriangle CDF}}=\frac{\frac{1}{2}DF\cdot FG\cdot CG}{\frac{1}{2}DF\cdot CF}=\frac{\frac{1}{2}\times \sqrt{3}\times 1\times 1}{\frac{1}{2}\times \sqrt{3}\times \sqrt{2}}=\frac{\sqrt{2}}{2}.(10分)设DG与平面PDC所成的角为\angle GDO，\therefore \\sin  \angle GDO=\frac{GO}{DG}=\frac{\frac{\sqrt{2}}{2}}{2}=\frac{\sqrt{2}}{4}，\therefore 直线AB与平面PDC所成角的正弦值为\frac{\sqrt{2}}{4}.(12分)$

Final answer: $\frac{\sqrt{2}}{4}$
Subfield: Solid Geometry


Question: $如图甲，在矩形ABCD中，E是CD的中点，AB=2，BC=\sqrt{2}，以AE，BE为折痕将\triangle ADE与\triangle BCE折起，使D，C重合(仍记为D)，如图乙(以\triangle ABD所在平面为底面).$![Image](https://cdntk.53zaixian.com/tq/1704026576604352514/img/compression/M_05a4c076c8e94c3d357875aecd507a09.jpg)
$(1)探索折叠形成的几何体中直线DE的几何性质并证明(写出一条即可，不含DE\perp DA，DE\perp DB)；$

Solution:
$性质1：DE\perp 平面ABD.证明如下：(1分)翻折前，DE\perp DA，CE\perp BC，翻折后垂直关系不变，则DE\perp DA，DE\perp DB，又DA\cap DB=D，DA,DB\subset 平面ABD,所以DE\perp 平面ABD.(4分)性质2：DE\perp AB.证明如下：(1分)与性质1证明方法相同，得到DE\perp 平面ABD，又AB\subset 平面ABD，所以DE\perp AB.(4分)性质3：DE与平面ABD内任一直线都垂直.证明如下：(1分)与性质1证明方法相同，得到DE\perp 平面ABD，从而DE与平面ABD内任一直线都垂直.(4分)性质4：直线DE与平面ABE所成的角等于\frac{\pi }{4}.证明如下：(1分)如图①，取AB的中点F，连接DF，EF，$![Image](https://cdntk.53zaixian.com/tq/1704026576969256961/img/compression/M_5d50dd29db143ba78c97282f04998828.jpg)$图①由DA=DB，得DF\perp AB，与性质2证明方法相同，得DE\perp AB，因为DE\cap DF=D，DE,DF\subset 平面DEF,所以AB\perp 平面DEF，因为AB\subset 平面ABE，所以平面DEF\perp 平面ABE.作DH\perp EF于H，则DH\perp 平面ABE，所以\angle DEF即为直线DE与平面ABE所成的角，易知DE=1，EF=\sqrt{2}，所以\\cos  \angle DEF=\frac{DE}{EF}=\frac{1}{\sqrt{2}}=\frac{\sqrt{2}}{2}，所以\angle DEF=\frac{\pi }{4}.(4分)写出其中一条即可.$

Final answer: 证明题
Subfield: Solid Geometry


Question: $如图甲，在矩形ABCD中，E是CD的中点，AB=2，BC=\sqrt{2}，以AE，BE为折痕将\triangle ADE与\triangle BCE折起，使D，C重合(仍记为D)，如图乙(以\triangle ABD所在平面为底面).$<img_2432>
$(1)探索折叠形成的几何体中直线DE的几何性质并证明(写出一条即可，不含DE\perp DA，DE\perp DB)；$

Solution:
$性质1：DE\perp 平面ABD.证明如下：(1分)翻折前，DE\perp DA，CE\perp BC，翻折后垂直关系不变，则DE\perp DA，DE\perp DB，又DA\cap DB=D，DA,DB\subset 平面ABD,所以DE\perp 平面ABD.(4分)性质2：DE\perp AB.证明如下：(1分)与性质1证明方法相同，得到DE\perp 平面ABD，又AB\subset 平面ABD，所以DE\perp AB.(4分)性质3：DE与平面ABD内任一直线都垂直.证明如下：(1分)与性质1证明方法相同，得到DE\perp 平面ABD，从而DE与平面ABD内任一直线都垂直.(4分)性质4：直线DE与平面ABE所成的角等于\frac{\pi }{4}.证明如下：(1分)如图①，取AB的中点F，连接DF，EF，$<img_1613>$图①由DA=DB，得DF\perp AB，与性质2证明方法相同，得DE\perp AB，因为DE\cap DF=D，DE,DF\subset 平面DEF,所以AB\perp 平面DEF，因为AB\subset 平面ABE，所以平面DEF\perp 平面ABE.作DH\perp EF于H，则DH\perp 平面ABE，所以\angle DEF即为直线DE与平面ABE所成的角，易知DE=1，EF=\sqrt{2}，所以\\cos  \angle DEF=\frac{DE}{EF}=\frac{1}{\sqrt{2}}=\frac{\sqrt{2}}{2}，所以\angle DEF=\frac{\pi }{4}.(4分)写出其中一条即可.$
Subfield: Solid Geometry


Question: $如图甲，在矩形ABCD中，E是CD的中点，AB=2，BC=\sqrt{2}，以AE，BE为折痕将\triangle ADE与\triangle BCE折起，使D，C重合(仍记为D)，如图乙(以\triangle ABD所在平面为底面).$<img_2432>
$(2)求翻折后几何体E-ABD外接球的体积.$

Solution:
$解法一：易知AD=BD=\sqrt{2}，AB=2，则AB^{2}=AD^{2}+BD^{2},所以\triangle ABD是等腰直角三角形，如图②，取AB的中点P，连接DP，$<img_2807>$图②则P是\triangle ABD的外心，(6分)设几何体E-ABD外接球的球心是O，则OP\perp 平面ABD.(8分)作OM\perp DE于M，则M是DE的中点，易得四边形OPDM是矩形，OP=DM=\frac{1}{2}，DP=\frac{1}{2}AB=1，则几何体E-ABD外接球的半径R=\sqrt{OP^2+PD^2}=\sqrt{\frac{1}{4}+1}=\frac{\sqrt{5}}{2}，(10分)则几何体E-ABD外接球的体积V=\frac{4}{3}\pi R^{3}=\frac{5\sqrt{5}}{6}\pi .(12分)解法二：易知DA，DB，DE两两垂直，外接球就是以DA，DB，DE为相邻的棱的长方体的外接球，(7分)设外接球的半径为R，则(2R)^{2}=DA^{2}+DB^{2}+DE^{2}=2+2+1=5，解得R=\frac{\sqrt{5}}{2}(负值舍去)，(10分)故外接球的体积V=\frac{4}{3}\pi R^{3}=\frac{5\sqrt{5}}{6}\pi .(12分)$

Final answer: $\frac{5\sqrt 5}{6\pi}$
Subfield: Solid Geometry


Question: $如图甲，在矩形ABCD中，E是CD的中点，AB=2，BC=\sqrt{2}，以AE，BE为折痕将\triangle ADE与\triangle BCE折起，使D，C重合(仍记为D)，如图乙(以\triangle ABD所在平面为底面).$![Image](https://cdntk.53zaixian.com/tq/1704026576604352514/img/compression/M_05a4c076c8e94c3d357875aecd507a09.jpg)
$(2)求翻折后几何体E-ABD外接球的体积.$

Solution:
$解法一：易知AD=BD=\sqrt{2}，AB=2，则AB^{2}=AD^{2}+BD^{2},所以\triangle ABD是等腰直角三角形，如图②，取AB的中点P，连接DP，$![Image](https://cdntk.53zaixian.com/tq/1704026577292218369/img/compression/M_b078bafcfefb7d64c3733634da8c76c2.jpg)$图②则P是\triangle ABD的外心，(6分)设几何体E-ABD外接球的球心是O，则OP\perp 平面ABD.(8分)作OM\perp DE于M，则M是DE的中点，易得四边形OPDM是矩形，OP=DM=\frac{1}{2}，DP=\frac{1}{2}AB=1，则几何体E-ABD外接球的半径R=\sqrt{OP^2+PD^2}=\sqrt{\frac{1}{4}+1}=\frac{\sqrt{5}}{2}，(10分)则几何体E-ABD外接球的体积V=\frac{4}{3}\pi R^{3}=\frac{5\sqrt{5}}{6}\pi .(12分)解法二：易知DA，DB，DE两两垂直，外接球就是以DA，DB，DE为相邻的棱的长方体的外接球，(7分)设外接球的半径为R，则(2R)^{2}=DA^{2}+DB^{2}+DE^{2}=2+2+1=5，解得R=\frac{\sqrt{5}}{2}(负值舍去)，(10分)故外接球的体积V=\frac{4}{3}\pi R^{3}=\frac{5\sqrt{5}}{6}\pi .(12分)$

Final answer: $\frac{5\sqrt 5}{6\pi}$
Subfield: Solid Geometry


Question: $如图，正四棱柱ABCD-A'B'C'D'.$![Image](https://cdntk.53zaixian.com/tq/1702600539585576962/img/compression/M_595e9aedcb5c4e848e0f36d34da76673.jpg)
$(1)请在正四棱柱ABCD-A'B'C'D'中，画出经过D，Q，R三点的截面(不必给出证明)；$

Solution:
$作直线QR分别交D'A',D'C'的延长线于M,N，连接MD交AA'于S，连接DN交CC'于点T，连接SQ,TR，如图,五边形DSQRT即为所求.$![Image](https://cdntk.53zaixian.com/tq/1702600539954675714/img/compression/M_7ffd63f61156dcbdafbcddbe679b30e6.jpg)

Final answer: 证明题