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


Question: $已知\triangle ABC的内角A，B，C的对边分别为a，b，c，c=1且2c\\sin   A\\cos   B=a\\sin   A－b\\sin   B+\frac{1}{4}b\\sin   C.
$![Image](https://cdntk.53zaixian.com/tq/1703968917644656642/img/compression/M_1ace49af54bd63b7ae303577fa477a5c.jpg)
$若E，F分别为边AB，AC上的动点，线段EF交BC边上的中线AD于点G，且\triangle AEF的面积为\triangle ABC面积的一半，求\overrightarrow{AG}\cdot \overrightarrow{EF}的最小值.$

Solution:
$设|\overrightarrow{AE}|=x(0\because \triangle AEF的面积为\triangle ABC面积的一半，\therefore \frac{1}{2}xy\\sin  \angle BAC=\frac{1}{2}\times \frac{1}{2}\times |\overrightarrow{AB}|\times |\overrightarrow{AC}|\\sin  \angle BAC，即xy=2.设\overrightarrow{AG}=\lambda \overrightarrow{AD}，则\overrightarrow{AG}=\frac{\lambda }{2}\overrightarrow{AB}+\frac{\lambda }{2}\overrightarrow{AC}.(7分)\because E，G，F三点共线，\therefore 设\overrightarrow{AG}=\mu \overrightarrow{AE}+(1-\mu )\overrightarrow{AF}，则\overrightarrow{AG}=x\mu \overrightarrow{AB}+\frac{y(1-\mu )}{4}\overrightarrow{AC}.\therefore \left\{\begin{matrix}x\mu =\frac{\lambda }{2},\\ \frac{y(1-\mu )}{4}=\frac{\lambda }{2},\end{matrix}\right.\therefore \mu =\frac{y}{4x+y}，\therefore \overrightarrow{AG}=\frac{xy}{4x+y}\overrightarrow{AB}+\frac{xy}{4x+y}\overrightarrow{AC}=\frac{2}{4x+y}\overrightarrow{AB}+\frac{2}{4x+y}\overrightarrow{AC}.(9分)又\overrightarrow{EF}=\overrightarrow{AF}-\overrightarrow{AE}=\frac{y}{4}\overrightarrow{AC}-x\overrightarrow{AB}，\therefore \overrightarrow{AG}\cdot \overrightarrow{EF}=\left(\frac{2}{4x+y}\overrightarrow{AB}+\frac{2}{4x+y}\overrightarrow{AC}\right)\cdot \left(\frac{y}{4}\overrightarrow{AC}-x\overrightarrow{AB}\right)=\frac{2}{4x+y}\left[\right.\frac{y}{4}\overrightarrow{AC}^{2}-x\overrightarrow{AB}^{2}+\left(\frac{y}{4}-x\right)\overrightarrow{AC}\cdot \overrightarrow{AB}\left]\right.=\frac{9y-6x}{4x+y}，(10分)又xy=2，\therefore \overrightarrow{AG}\cdot \overrightarrow{EF}=\frac{18-6x^2}{4x^2+2}=\frac{21}{4x^2+2}-\frac{3}{2}，\because 0<x\leq 1，0<y\leq 4，x=\frac{2}{y}，\therefore \frac{1}{2}\leq x\leq 1.\therefore 当x=1时，\overrightarrow{AG}\cdot \overrightarrow{EF}取得最小值，为2.(12分)$

Final answer: $2$
Subfield: Plane Geometry


Question: $已知\triangle ABC的内角A，B，C的对边分别为a，b，c，c=1且2c\\sin   A\\cos   B=a\\sin   A－b\\sin   B+\frac{1}{4}b\\sin   C.
$<img_2839>
$若E，F分别为边AB，AC上的动点，线段EF交BC边上的中线AD于点G，且\triangle AEF的面积为\triangle ABC面积的一半，求\overrightarrow{AG}\cdot \overrightarrow{EF}的最小值.$

Solution:
$设|\overrightarrow{AE}|=x(0\because \triangle AEF的面积为\triangle ABC面积的一半，\therefore \frac{1}{2}xy\\sin  \angle BAC=\frac{1}{2}\times \frac{1}{2}\times |\overrightarrow{AB}|\times |\overrightarrow{AC}|\\sin  \angle BAC，即xy=2.设\overrightarrow{AG}=\lambda \overrightarrow{AD}，则\overrightarrow{AG}=\frac{\lambda }{2}\overrightarrow{AB}+\frac{\lambda }{2}\overrightarrow{AC}.(7分)\because E，G，F三点共线，\therefore 设\overrightarrow{AG}=\mu \overrightarrow{AE}+(1-\mu )\overrightarrow{AF}，则\overrightarrow{AG}=x\mu \overrightarrow{AB}+\frac{y(1-\mu )}{4}\overrightarrow{AC}.\therefore \left\{\begin{matrix}x\mu =\frac{\lambda }{2},\\ \frac{y(1-\mu )}{4}=\frac{\lambda }{2},\end{matrix}\right.\therefore \mu =\frac{y}{4x+y}，\therefore \overrightarrow{AG}=\frac{xy}{4x+y}\overrightarrow{AB}+\frac{xy}{4x+y}\overrightarrow{AC}=\frac{2}{4x+y}\overrightarrow{AB}+\frac{2}{4x+y}\overrightarrow{AC}.(9分)又\overrightarrow{EF}=\overrightarrow{AF}-\overrightarrow{AE}=\frac{y}{4}\overrightarrow{AC}-x\overrightarrow{AB}，\therefore \overrightarrow{AG}\cdot \overrightarrow{EF}=\left(\frac{2}{4x+y}\overrightarrow{AB}+\frac{2}{4x+y}\overrightarrow{AC}\right)\cdot \left(\frac{y}{4}\overrightarrow{AC}-x\overrightarrow{AB}\right)=\frac{2}{4x+y}\left[\right.\frac{y}{4}\overrightarrow{AC}^{2}-x\overrightarrow{AB}^{2}+\left(\frac{y}{4}-x\right)\overrightarrow{AC}\cdot \overrightarrow{AB}\left]\right.=\frac{9y-6x}{4x+y}，(10分)又xy=2，\therefore \overrightarrow{AG}\cdot \overrightarrow{EF}=\frac{18-6x^2}{4x^2+2}=\frac{21}{4x^2+2}-\frac{3}{2}，\because 0<x\leq 1，0<y\leq 4，x=\frac{2}{y}，\therefore \frac{1}{2}\leq x\leq 1.\therefore 当x=1时，\overrightarrow{AG}\cdot \overrightarrow{EF}取得最小值，为2.(12分)$

Final answer: $2$
Subfield: Plane Geometry


Question: $如图①所示，在等腰梯形ABCD中，AB\parallel CD，AB=3，CD=1，BC=2，E,F分别为腰AD,BC的中点，H,M分别为线段EF,AB的中点，将四边形CDEF沿EF折起，使平面EFC'D'\perp 平面ABFE，如图②所示.$<img_1622>
$求证：C'E\perp 平面D'MH;$

Solution:
$证明：连接C'H，在等腰梯形ABCD中，H,M分别为线段EF,AB的中点，E,F分别为腰AD,BC的中点，所以MH\perp EF.因为平面EFC'D'\perp 平面ABFE，平面EFC'D'\cap 平面ABFE=EF，所以MH\perp 平面EFC'D'，(2分)又C'E\subset 平面EFC'D'，所以MH\perp C'E.因为AB\parallel CD，AB=3，CD=1，E,F分别为腰AD,BC的中点，所以EF\parallel CD，即EF\parallel C'D',EF=2.又H是EF的中点，所以EH\parallel C'D',EH=C'D'，所以四边形C'D'EH是平行四边形.(4分)根据题意，得ED'=C'D'=1,所以四边形C'D'EH是菱形，所以C'E\perp D'H，又MH\cap D'H=H，MH,D'H\subset 平面D'MH，所以C'E\perp 平面D'MH.(6分)$
Subfield: Plane Geometry


Question: $如图①所示，在等腰梯形ABCD中，AB\parallel CD，AB=3，CD=1，BC=2，E,F分别为腰AD,BC的中点，H,M分别为线段EF,AB的中点，将四边形CDEF沿EF折起，使平面EFC'D'\perp 平面ABFE，如图②所示.$![Image](https://cdntk.53zaixian.com/tq/1703969007085015042/img/compression/M_52af94056aab5bd8f476254398e8e4c8.jpg)
$求证：C'E\perp 平面D'MH;$

Solution:
$证明：连接C'H，在等腰梯形ABCD中，H,M分别为线段EF,AB的中点，E,F分别为腰AD,BC的中点，所以MH\perp EF.因为平面EFC'D'\perp 平面ABFE，平面EFC'D'\cap 平面ABFE=EF，所以MH\perp 平面EFC'D'，(2分)又C'E\subset 平面EFC'D'，所以MH\perp C'E.因为AB\parallel CD，AB=3，CD=1，E,F分别为腰AD,BC的中点，所以EF\parallel CD，即EF\parallel C'D',EF=2.又H是EF的中点，所以EH\parallel C'D',EH=C'D'，所以四边形C'D'EH是平行四边形.(4分)根据题意，得ED'=C'D'=1,所以四边形C'D'EH是菱形，所以C'E\perp D'H，又MH\cap D'H=H，MH,D'H\subset 平面D'MH，所以C'E\perp 平面D'MH.(6分)$

Final answer: 证明题
Subfield: Plane Geometry


Question: $如图①所示，在等腰梯形ABCD中，AB\parallel CD，AB=3，CD=1，BC=2，E,F分别为腰AD,BC的中点，H,M分别为线段EF,AB的中点，将四边形CDEF沿EF折起，使平面EFC'D'\perp 平面ABFE，如图②所示.$![Image](https://cdntk.53zaixian.com/tq/1703969007085015042/img/compression/M_52af94056aab5bd8f476254398e8e4c8.jpg)
$若N为线段C'D'的中点，在直线BF上是否存在点Q，使得NQ\parallel 平面D'MH？若存在，求出线段NQ的长度;若不存在，请说明理由.$

Solution:
$在直线BF上存在点Q，使得NQ\parallel 平面D'MH.如图，在线段MB上取点P，使得MP=\frac{1}{2}，连接CP，交EF于点L，连接NL，$![Image](https://cdntk.53zaixian.com/tq/1703969007387004930/img/compression/M_557d812a4e59a148be00604441046f62.jpg)$则CL\parallel HM,NL\parallel D'H，CP\perp EF.(8分)又CL\subset 平面NLC，HM\not\subset 平面NLC.所以HM\parallel 平面NLC，同理可得D'H\parallel 平面NLC.又HM\cap D'H=H，HM,D'H\subset 平面D'HM，所以平面NLC\parallel 平面D'HM，所以NC\parallel 平面D'HM，所以点C就是所求的点Q.(10分)在Rt\triangle CLF中，CL=\sqrt{1^2-{\left(\frac{1}{2}\right)}^2}=\frac{\sqrt{3}}{2}，易知NL=C'F=CF=1，所以NQ=NC=\sqrt{NL^2+CL^2}=\sqrt{1+\left(\frac{\sqrt{3}}{2}\right)^2}=\frac{\sqrt{7}}{2}.故线段NQ的长度为\frac{\sqrt{7}}{2}.(12分)$

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


Question: $如图①所示，在等腰梯形ABCD中，AB\parallel CD，AB=3，CD=1，BC=2，E,F分别为腰AD,BC的中点，H,M分别为线段EF,AB的中点，将四边形CDEF沿EF折起，使平面EFC'D'\perp 平面ABFE，如图②所示.$<img_1622>
$若N为线段C'D'的中点，在直线BF上是否存在点Q，使得NQ\parallel 平面D'MH？若存在，求出线段NQ的长度;若不存在，请说明理由.$

Solution:
$在直线BF上存在点Q，使得NQ\parallel 平面D'MH.如图，在线段MB上取点P，使得MP=\frac{1}{2}，连接CP，交EF于点L，连接NL，$<img_2334>$则CL\parallel HM,NL\parallel D'H，CP\perp EF.(8分)又CL\subset 平面NLC，HM\not\subset 平面NLC.所以HM\parallel 平面NLC，同理可得D'H\parallel 平面NLC.又HM\cap D'H=H，HM,D'H\subset 平面D'HM，所以平面NLC\parallel 平面D'HM，所以NC\parallel 平面D'HM，所以点C就是所求的点Q.(10分)在Rt\triangle CLF中，CL=\sqrt{1^2-{\left(\frac{1}{2}\right)}^2}=\frac{\sqrt{3}}{2}，易知NL=C'F=CF=1，所以NQ=NC=\sqrt{NL^2+CL^2}=\sqrt{1+\left(\frac{\sqrt{3}}{2}\right)^2}=\frac{\sqrt{7}}{2}.故线段NQ的长度为\frac{\sqrt{7}}{2}.(12分)$

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


Question: $如图，在平行四边形ABCD中，\angle BAD=60^\circ ，\overrightarrow{BE}=\frac{1}{2}\overrightarrow{BC}，\overrightarrow{CF}=2\overrightarrow{FD}.$<img_1732>
$若\overrightarrow{EF}=x\overrightarrow{AB}+y\overrightarrow{AD}，求3x+2y的值；$

Solution:
$在平行四边形ABCD中，\overrightarrow{BE}=\frac{1}{2}\overrightarrow{BC}，\overrightarrow{CF}=2\overrightarrow{FD},所以\overrightarrow{EF}=\overrightarrow{AF}-\overrightarrow{AE}=\overrightarrow{AD}+\frac{1}{3}\overrightarrow{AB}-\left(\overrightarrow{AB}+\frac{1}{2}\overrightarrow{AD}\right)=-\frac{2}{3}\overrightarrow{AB}+\frac{1}{2}\overrightarrow{AD}，(2分)又\overrightarrow{EF}=x\overrightarrow{AB}+y\overrightarrow{AD}，\therefore x=-\frac{2}{3}，y=\frac{1}{2}，\therefore 3x+2y=-1.(5分)$

Final answer: $-1$
Subfield: Plane Geometry


Question: $如图，在平行四边形ABCD中，\angle BAD=60^\circ ，\overrightarrow{BE}=\frac{1}{2}\overrightarrow{BC}，\overrightarrow{CF}=2\overrightarrow{FD}.$![Image](https://cdntk.53zaixian.com/tq/1703969059209940995/img/compression/M_01720f3ef7c934bd555262879684f64a.jpg)
$若\overrightarrow{EF}=x\overrightarrow{AB}+y\overrightarrow{AD}，求3x+2y的值；$

Solution:
$在平行四边形ABCD中，\overrightarrow{BE}=\frac{1}{2}\overrightarrow{BC}，\overrightarrow{CF}=2\overrightarrow{FD},所以\overrightarrow{EF}=\overrightarrow{AF}-\overrightarrow{AE}=\overrightarrow{AD}+\frac{1}{3}\overrightarrow{AB}-\left(\overrightarrow{AB}+\frac{1}{2}\overrightarrow{AD}\right)=-\frac{2}{3}\overrightarrow{AB}+\frac{1}{2}\overrightarrow{AD}，(2分)又\overrightarrow{EF}=x\overrightarrow{AB}+y\overrightarrow{AD}，\therefore x=-\frac{2}{3}，y=\frac{1}{2}，\therefore 3x+2y=-1.(5分)$

Final answer: $-1$
Subfield: Plane Geometry


Question: $如图，在平行四边形ABCD中，\angle BAD=60^\circ ，\overrightarrow{BE}=\frac{1}{2}\overrightarrow{BC}，\overrightarrow{CF}=2\overrightarrow{FD}.$![Image](https://cdntk.53zaixian.com/tq/1703969059209940995/img/compression/M_01720f3ef7c934bd555262879684f64a.jpg)
$若|\overrightarrow{AB}|=6，\overrightarrow{AC}\cdot \overrightarrow{EF}=-18，求边AD的长.$

Solution:
$设AD=x，则\overrightarrow{AC}\cdot \overrightarrow{EF}=(\overrightarrow{AB}+\overrightarrow{AD})\cdot \left(-\frac{2}{3}\overrightarrow{AB}+\frac{1}{2}\overrightarrow{AD}\right)=-\frac{2}{3}\overrightarrow{AB}^{2}+\frac{1}{2}\overrightarrow{AD}^{2}-\frac{1}{6}\overrightarrow{AB}\cdot \overrightarrow{AD}=-\frac{2}{3}|\overrightarrow{AB}|^{2}+\frac{1}{2}|\overrightarrow{AD}|^{2}-\frac{1}{6}|\overrightarrow{AB}||\overrightarrow{AD}|\\cos  \angle BAD=\frac{1}{2}x^{2}-\frac{1}{2}x-24=-18，(8分)\therefore x^{2}-x-12=0，\therefore x=4或x=-3(舍去)，即AD=4.(10分)$

Final answer: $4$
Subfield: Plane Geometry


Question: $如图，在平行四边形ABCD中，\angle BAD=60^\circ ，\overrightarrow{BE}=\frac{1}{2}\overrightarrow{BC}，\overrightarrow{CF}=2\overrightarrow{FD}.$<img_1732>
$若|\overrightarrow{AB}|=6，\overrightarrow{AC}\cdot \overrightarrow{EF}=-18，求边AD的长.$

Solution:
$设AD=x，则\overrightarrow{AC}\cdot \overrightarrow{EF}=(\overrightarrow{AB}+\overrightarrow{AD})\cdot \left(-\frac{2}{3}\overrightarrow{AB}+\frac{1}{2}\overrightarrow{AD}\right)=-\frac{2}{3}\overrightarrow{AB}^{2}+\frac{1}{2}\overrightarrow{AD}^{2}-\frac{1}{6}\overrightarrow{AB}\cdot \overrightarrow{AD}=-\frac{2}{3}|\overrightarrow{AB}|^{2}+\frac{1}{2}|\overrightarrow{AD}|^{2}-\frac{1}{6}|\overrightarrow{AB}||\overrightarrow{AD}|\\cos  \angle BAD=\frac{1}{2}x^{2}-\frac{1}{2}x-24=-18，(8分)\therefore x^{2}-x-12=0，\therefore x=4或x=-3(舍去)，即AD=4.(10分)$

Final answer: $4$
Subfield: Plane Geometry


Question: $如图，EA\perp 平面ABCD，EA\parallel FC，AC=EA=2FC=2，四边形ABCD是菱形.$<img_3016>
$证明：FA\perp 平面EBD；$

Solution:
$证明：设AC\cap BD=O，连接OE，\because EA\perp 平面ABCD，EA\parallel FC，\therefore FC\perp 平面ABCD，又BD\subset 平面ABCD，\therefore FC\perp BD，\angle EAC=\angle ACF=90^\circ ，又AC=EA=2FC，O为AC的中点，\therefore FC=OA，\therefore \triangle EAO\cong \triangle ACF，则\angle EOA=\angle AFC，又\angle AFC+\angle FAC=90^\circ ，\therefore \angle AOE+\angle FAC=90^\circ ，故OE\perp AF.(3分)\because FC\perp BD，AC\perp BD，FC\cap AC=C，FC,AC\subset 平面FAC，\therefore BD\perp 平面FAC，又AF\subset 平面FAC，\therefore BD\perp AF，又OE\cap BD=O,OE,BD\subset 平面EBD，\therefore AF\perp 平面EBD.(6分)$
Subfield: Plane Geometry


Question: $如图，EA\perp 平面ABCD，EA\parallel FC，AC=EA=2FC=2，四边形ABCD是菱形.$![Image](https://cdntk.53zaixian.com/tq/1703969060489203714/img/compression/M_1cc149769bfcf44224dcbdccbcba54d7.jpg)
$证明：FA\perp 平面EBD；$

Solution:
$证明：设AC\cap BD=O，连接OE，\because EA\perp 平面ABCD，EA\parallel FC，\therefore FC\perp 平面ABCD，又BD\subset 平面ABCD，\therefore FC\perp BD，\angle EAC=\angle ACF=90^\circ ，又AC=EA=2FC，O为AC的中点，\therefore FC=OA，\therefore \triangle EAO\cong \triangle ACF，则\angle EOA=\angle AFC，又\angle AFC+\angle FAC=90^\circ ，\therefore \angle AOE+\angle FAC=90^\circ ，故OE\perp AF.(3分)\because FC\perp BD，AC\perp BD，FC\cap AC=C，FC,AC\subset 平面FAC，\therefore BD\perp 平面FAC，又AF\subset 平面FAC，\therefore BD\perp AF，又OE\cap BD=O,OE,BD\subset 平面EBD，\therefore AF\perp 平面EBD.(6分)$

Final answer: 证明题
Subfield: Plane Geometry


Question: $如图，EA\perp 平面ABCD，EA\parallel FC，AC=EA=2FC=2，四边形ABCD是菱形.$<img_3016>
$若菱形ABCD的边长为\sqrt{10}，求直线AB与平面EBD所成角的正弦值.$

Solution:
$由(1)可得AF\perp 平面EBD，设垂足为P，则P为OE与AF的交点，连接BP，则\angle ABP为直线AB与平面EBD所成的角，\\sin  \angle ABP=\\cos  \angle PAB，(9分)易求得AF=\sqrt{FC^2+AC^2}=\sqrt{5}，BF=\sqrt{BC^2+FC^2}=\sqrt{11}，则\\sin  \angle ABP=\\cos  \angle PAB=\\cos  \angle FAB=\frac{AF^2+AB^2-FB^2}{2AF\cdot AB}=\frac{5+10-11}{2\times \sqrt{5}\times \sqrt{10}}=\frac{\sqrt{2}}{5}.故直线AB与平面EBD所成角的正弦值为\frac{\sqrt{2}}{5}.(12分)$<img_2471>

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


Question: $如图，EA\perp 平面ABCD，EA\parallel FC，AC=EA=2FC=2，四边形ABCD是菱形.$![Image](https://cdntk.53zaixian.com/tq/1703969060489203714/img/compression/M_1cc149769bfcf44224dcbdccbcba54d7.jpg)
$若菱形ABCD的边长为\sqrt{10}，求直线AB与平面EBD所成角的正弦值.$

Solution:
$由(1)可得AF\perp 平面EBD，设垂足为P，则P为OE与AF的交点，连接BP，则\angle ABP为直线AB与平面EBD所成的角，\\sin  \angle ABP=\\cos  \angle PAB，(9分)易求得AF=\sqrt{FC^2+AC^2}=\sqrt{5}，BF=\sqrt{BC^2+FC^2}=\sqrt{11}，则\\sin  \angle ABP=\\cos  \angle PAB=\\cos  \angle FAB=\frac{AF^2+AB^2-FB^2}{2AF\cdot AB}=\frac{5+10-11}{2\times \sqrt{5}\times \sqrt{10}}=\frac{\sqrt{2}}{5}.故直线AB与平面EBD所成角的正弦值为\frac{\sqrt{2}}{5}.(12分)$![Image](https://cdntk.53zaixian.com/tq/1703969061810409474/img/compression/M_0db7fc1fe6b591aa33036c8523dfe206.jpg)

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


Question: $如图所示，已知AB\perp 平面ACD，DE\perp 平面ACD，\triangle ACD为等边三角形，AD=DE=2AB，F为CD的中点.$<img_2673>
$证明：AF\parallel 平面BCE；$

Solution:
$证明：取CE的中点M，连接MF，BM，\because F是CD的中点，\therefore MF是\triangle CDE的中位线，\therefore MF\parallel DE，MF=\frac{1}{2}DE.又DE=2AB，\therefore AB=MF，(1分)\because AB\perp 平面ACD，DE\perp 平面ACD，\therefore DE\parallel AB，\therefore AB\parallel MF，\therefore 四边形ABMF是平行四边形，\therefore AF\parallel BM，(2分)\because AF\not\subset 平面BCE，BM\subset 平面BCE，\therefore AF\parallel 平面BCE.(3分)$
Subfield: Plane Geometry


Question: $如图所示，已知AB\perp 平面ACD，DE\perp 平面ACD，\triangle ACD为等边三角形，AD=DE=2AB，F为CD的中点.$![Image](https://cdntk.53zaixian.com/tq/1703969062108205058/img/compression/M_59f683658acafdca5c37b06fb3902d57.jpg)
$证明：AF\parallel 平面BCE；$

Solution:
$证明：取CE的中点M，连接MF，BM，\because F是CD的中点，\therefore MF是\triangle CDE的中位线，\therefore MF\parallel DE，MF=\frac{1}{2}DE.又DE=2AB，\therefore AB=MF，(1分)\because AB\perp 平面ACD，DE\perp 平面ACD，\therefore DE\parallel AB，\therefore AB\parallel MF，\therefore 四边形ABMF是平行四边形，\therefore AF\parallel BM，(2分)\because AF\not\subset 平面BCE，BM\subset 平面BCE，\therefore AF\parallel 平面BCE.(3分)$

Final answer: 证明题
Subfield: Plane Geometry


Question: $如图所示，已知AB\perp 平面ACD，DE\perp 平面ACD，\triangle ACD为等边三角形，AD=DE=2AB，F为CD的中点.$<img_2673>
$证明：平面BCE\perp 平面CDE；$

Solution:
$证明：\because AB\perp 平面ACD，AF\subset 平面ACD，\therefore AB\perp AF，\therefore 四边形ABMF是矩形，\therefore BM\perp MF．\because \triangle ACD是正三角形，F是CD的中点，\therefore CD\perp AF．\because BM\parallel AF，\therefore CD\perp BM，(5分)\because MF\cap CD=F，MF，CD\subset 平面CDE，\therefore BM\perp 平面CDE，\because BM\subset 平面BCE，\therefore 平面BCE\perp 平面CDE.(7分)$
Subfield: Plane Geometry


Question: $如图所示，已知AB\perp 平面ACD，DE\perp 平面ACD，\triangle ACD为等边三角形，AD=DE=2AB，F为CD的中点.$![Image](https://cdntk.53zaixian.com/tq/1703969062108205058/img/compression/M_59f683658acafdca5c37b06fb3902d57.jpg)
$证明：平面BCE\perp 平面CDE；$

Solution:
$证明：\because AB\perp 平面ACD，AF\subset 平面ACD，\therefore AB\perp AF，\therefore 四边形ABMF是矩形，\therefore BM\perp MF．\because \triangle ACD是正三角形，F是CD的中点，\therefore CD\perp AF．\because BM\parallel AF，\therefore CD\perp BM，(5分)\because MF\cap CD=F，MF，CD\subset 平面CDE，\therefore BM\perp 平面CDE，\because BM\subset 平面BCE，\therefore 平面BCE\perp 平面CDE.(7分)$

Final answer: 证明题
Subfield: Plane Geometry


Question: $江西某中学校园内有块扇形空地OPQ，经测量其半径为60 m，圆心角为\frac{\pi }{3}，学校准备在此扇形空地上修建一所矩形室内篮球场ABCD，初步设计方案1如图1所示．$<img_2716>$图1$
$取弧PQ的中点E，连接OE，设\angle BOE=\alpha ，试用\alpha 表示方案1中矩形ABCD的面积，并求其最大值；$

Solution:
$如图所示，设OE交AD于点M，交BC于点N，显然矩形ABCD关于OE所在直线对称，点M、N分别为AD、BC的中点，\angle BOE=\alpha ，0<\alpha <\frac{\pi }{6}．在Rt\triangle ONB中，BN=60\\sin   \alpha ，ON=60\\cos   \alpha .OM=\frac{DM}{\\\tan   \frac{\pi }{6}}=\sqrt{3}DM=\sqrt{3}CN=60\sqrt{3}\\sin   \alpha ，\therefore MN=ON-OM=60\\cos   \alpha -60\sqrt{3}\\sin   \alpha ，即AB=60\\cos   \alpha -60\sqrt{3}\\sin   \alpha ，又BC=2CN=120\\sin   \alpha ，故矩形ABCD的面积S=AB\cdot BC=3 600(\\cos   \alpha -\sqrt{3}\\sin   \alpha )\cdot 2\\sin   \alpha =3 600(2\\sin   \alpha \\cos   \alpha -2\sqrt{3}\\sin  ^{2}\alpha )=3 600[\\sin   2\alpha -\sqrt{3}(1-\\cos   2\alpha )]=3 600(\\sin   2\alpha +\sqrt{3}\\cos   2\alpha -\sqrt{3})=7 200\\sin  \left(2\alpha +\frac{\pi }{3}\right)-3 600\sqrt{3}．\because 0<\alpha <\frac{\pi }{6}，\therefore 0<2\alpha <\frac{\pi }{3}，\therefore \frac{\pi }{3}<2\alpha +\frac{\pi }{3}<\frac{2\pi }{3}．故当2\alpha +\frac{\pi }{3}=\frac{\pi }{2}，即\alpha =\frac{\pi }{12}时，S取得最大值3 600(2-\sqrt{3})，\therefore 矩形ABCD面积的最大值为3 600(2-\sqrt{3})m^{2}．$<img_2712>

Final answer: $3 600(2-\sqrt{3})$ $m^{2}$
Subfield: Plane Geometry


Question: $江西某中学校园内有块扇形空地OPQ，经测量其半径为60 m，圆心角为\frac{\pi }{3}，学校准备在此扇形空地上修建一所矩形室内篮球场ABCD，初步设计方案1如图1所示．$![Image](https://cdntk.53zaixian.com/tq/1632611650396233731/img/compression/M_5839487b9709daa6fc03df22d74a3e13.jpg)$图1$
$取弧PQ的中点E，连接OE，设\angle BOE=\alpha ，试用\alpha 表示方案1中矩形ABCD的面积，并求其最大值；$

Solution:
$如图所示，设OE交AD于点M，交BC于点N，显然矩形ABCD关于OE所在直线对称，点M、N分别为AD、BC的中点，\angle BOE=\alpha ，0<\alpha <\frac{\pi }{6}．在Rt\triangle ONB中，BN=60\\sin   \alpha ，ON=60\\cos   \alpha .OM=\frac{DM}{\\\tan   \frac{\pi }{6}}=\sqrt{3}DM=\sqrt{3}CN=60\sqrt{3}\\sin   \alpha ，\therefore MN=ON-OM=60\\cos   \alpha -60\sqrt{3}\\sin   \alpha ，即AB=60\\cos   \alpha -60\sqrt{3}\\sin   \alpha ，又BC=2CN=120\\sin   \alpha ，故矩形ABCD的面积S=AB\cdot BC=3 600(\\cos   \alpha -\sqrt{3}\\sin   \alpha )\cdot 2\\sin   \alpha =3 600(2\\sin   \alpha \\cos   \alpha -2\sqrt{3}\\sin  ^{2}\alpha )=3 600[\\sin   2\alpha -\sqrt{3}(1-\\cos   2\alpha )]=3 600(\\sin   2\alpha +\sqrt{3}\\cos   2\alpha -\sqrt{3})=7 200\\sin  \left(2\alpha +\frac{\pi }{3}\right)-3 600\sqrt{3}．\because 0<\alpha <\frac{\pi }{6}，\therefore 0<2\alpha <\frac{\pi }{3}，\therefore \frac{\pi }{3}<2\alpha +\frac{\pi }{3}<\frac{2\pi }{3}．故当2\alpha +\frac{\pi }{3}=\frac{\pi }{2}，即\alpha =\frac{\pi }{12}时，S取得最大值3 600(2-\sqrt{3})，\therefore 矩形ABCD面积的最大值为3 600(2-\sqrt{3})m^{2}．$![Image](https://cdntk.53zaixian.com/tq/1632611651277037570/img/compression/M_e044160cffcec5f0a3082dfc7d6ea9a1.jpg)

Final answer: $3 600(2-\sqrt{3})$ $m^{2}$
Subfield: Plane Geometry


Question: $江西某中学校园内有块扇形空地OPQ，经测量其半径为60 m，圆心角为\frac{\pi }{3}，学校准备在此扇形空地上修建一所矩形室内篮球场ABCD，初步设计方案1如图1所示．$<img_2716>$图1$
$你有没有更好的设计方案2来获得更大的篮球场面积？若有，在图2中画出来，并证明你的结论．　$<img_1896>$图2$

Solution:
$如图所示，在半径OP上截取线段AB为矩形的一边，作矩形ABCD，设\angle BOC=\theta ，0<\theta <\frac{\pi }{3}，可得CB=60\\sin   \theta ，OB=60\\cos   \theta ，则OA=CB\\tan  \frac{\pi }{6}=20\sqrt{3}\\sin   \theta ，故矩形ABCD的面积为S=(OB-OA)\cdot CB=(60\\cos   \theta -20\sqrt{3}\\sin   \theta )\times 60\\sin   \theta =3 600\left(\\\sin    \theta \\\cos    \theta -\frac{\sqrt{3}}{3}\\\sin   ^2\theta \right)=1 800\left(\\\sin    2\theta +\frac{\sqrt{3}}{3}\\\cos    2\theta \right)-600\sqrt{3}=1 200\sqrt{3}\left(\frac{\sqrt{3}}{2}\\\sin    2\theta +\frac{1}{2}\\\cos    2\theta \right)-600\sqrt{3}=1 200\sqrt{3}\\sin  \left(2\theta +\frac{\pi }{6}\right)-600\sqrt{3}，由0<\theta <\frac{\pi }{3}，可得\frac{\pi }{6}<2\theta +\frac{\pi }{6}<\frac{5\pi }{6}，\therefore 当2\theta +\frac{\pi }{6}=\frac{\pi }{2}，即\theta =\frac{\pi }{6}时，S有最大值600\sqrt{3}，即篮球场面积的最大值为600\sqrt{3} m^{2}，现将两种方案的最大值进行比较大小，\because 3 600(2-\sqrt{3})-600\sqrt{3}=600(12-7\sqrt{3})<0，\therefore 方案2的篮球场面积更大．$<img_2383>
Subfield: Plane Geometry


Question: $江西某中学校园内有块扇形空地OPQ，经测量其半径为60 m，圆心角为\frac{\pi }{3}，学校准备在此扇形空地上修建一所矩形室内篮球场ABCD，初步设计方案1如图1所示．$![Image](https://cdntk.53zaixian.com/tq/1632611650396233731/img/compression/M_5839487b9709daa6fc03df22d74a3e13.jpg)$图1$
$你有没有更好的设计方案2来获得更大的篮球场面积？若有，在图2中画出来，并证明你的结论．　$![Image](https://cdntk.53zaixian.com/tq/1632611651855851522/img/compression/M_4c2cab4c8f84b1745c6133a1ecb2b92b.jpg)$图2$

Solution:
$如图所示，在半径OP上截取线段AB为矩形的一边，作矩形ABCD，设\angle BOC=\theta ，0<\theta <\frac{\pi }{3}，可得CB=60\\sin   \theta ，OB=60\\cos   \theta ，则OA=CB\\tan  \frac{\pi }{6}=20\sqrt{3}\\sin   \theta ，故矩形ABCD的面积为S=(OB-OA)\cdot CB=(60\\cos   \theta -20\sqrt{3}\\sin   \theta )\times 60\\sin   \theta =3 600\left(\\\sin    \theta \\\cos    \theta -\frac{\sqrt{3}}{3}\\\sin   ^2\theta \right)=1 800\left(\\\sin    2\theta +\frac{\sqrt{3}}{3}\\\cos    2\theta \right)-600\sqrt{3}=1 200\sqrt{3}\left(\frac{\sqrt{3}}{2}\\\sin    2\theta +\frac{1}{2}\\\cos    2\theta \right)-600\sqrt{3}=1 200\sqrt{3}\\sin  \left(2\theta +\frac{\pi }{6}\right)-600\sqrt{3}，由0<\theta <\frac{\pi }{3}，可得\frac{\pi }{6}<2\theta +\frac{\pi }{6}<\frac{5\pi }{6}，\therefore 当2\theta +\frac{\pi }{6}=\frac{\pi }{2}，即\theta =\frac{\pi }{6}时，S有最大值600\sqrt{3}，即篮球场面积的最大值为600\sqrt{3} m^{2}，现将两种方案的最大值进行比较大小，\because 3 600(2-\sqrt{3})-600\sqrt{3}=600(12-7\sqrt{3})<0，\therefore 方案2的篮球场面积更大．$![Image](https://cdntk.53zaixian.com/tq/1632611651855851522/img/compression/M_835dba3ebaf9aa6f5cce2281b0a5b42b.jpg)

Final answer: 证明题
Subfield: Plane Geometry


Question: $在平面四边形ABCD中，对角线AC与BD交于点E,\angle ABD=45^\circ ,AE=EC,DE=2BE,AB=6，AD=3\sqrt{2}．$<img_3062>
$求AC的长；$

Solution:
$在\triangle ABD中，由余弦定理，得AD^{2}=AB^{2}+BD^{2}-2AB\cdot BD\cdot \\cos  \angle ABD，因为AB=6，AD=3\sqrt{2}，\angle ABD=45^\circ ,所以18=36+BD^{2}-2\times 6\times BD\times \\cos   45^\circ ，化简得BD^{2}-6\sqrt{2}BD+18=0，解得BD=3\sqrt{2}，因为BD^{2}+AD^{2}=AB^{2}，所以\angle ADB=90^\circ ．又DE=2BE，所以DE=2\sqrt{2}，所以AE^{2}=DE^{2}+AD^{2}=(2\sqrt{2})^{2}+(3\sqrt{2})^{2}=26，则AE=\sqrt{26}，又AE=EC，所以AC=2\sqrt{26}．$

Final answer: $2\sqrt{26}$
Subfield: Plane Geometry


Question: $在平面四边形ABCD中，对角线AC与BD交于点E,\angle ABD=45^\circ ,AE=EC,DE=2BE,AB=6，AD=3\sqrt{2}．$![Image](https://cdntk.53zaixian.com/tq/1623609001336000514/img/compression/M_d86890274e7fa0fecd13e7af80e6dd36.jpg)
$求AC的长；$

Solution:
$在\triangle ABD中，由余弦定理，得AD^{2}=AB^{2}+BD^{2}-2AB\cdot BD\cdot \\cos  \angle ABD，因为AB=6，AD=3\sqrt{2}，\angle ABD=45^\circ ,所以18=36+BD^{2}-2\times 6\times BD\times \\cos   45^\circ ，化简得BD^{2}-6\sqrt{2}BD+18=0，解得BD=3\sqrt{2}，因为BD^{2}+AD^{2}=AB^{2}，所以\angle ADB=90^\circ ．又DE=2BE，所以DE=2\sqrt{2}，所以AE^{2}=DE^{2}+AD^{2}=(2\sqrt{2})^{2}+(3\sqrt{2})^{2}=26，则AE=\sqrt{26}，又AE=EC，所以AC=2\sqrt{26}．$

Final answer: $2\sqrt{26}$
Subfield: Plane Geometry


Question: $在平面四边形ABCD中，对角线AC与BD交于点E,\angle ABD=45^\circ ,AE=EC,DE=2BE,AB=6，AD=3\sqrt{2}．$<img_3062>
$求\sin  \angle ADC的值．$

Solution:
$由\angle ADB=90^\circ ,AE=\sqrt{26},DE=2\sqrt{2},AD=3\sqrt{2}，得\\sin  \angle EAD=\frac{DE}{AE}=\frac{2\sqrt{2}}{\sqrt{26}},\\cos  \angle EAD=\frac{AD}{AE}=\frac{3\sqrt{2}}{\sqrt{26}}．在\triangle ACD中，由余弦定理，得CD^{2}=AD^{2}+AC^{2}-2AD\cdot AC\cdot \\cos  \angle EAD=50，则CD=5\sqrt{2}．在\triangle ACD中，由正弦定理，得\frac{AC}{\\\sin   \angle ADC}=\frac{CD}{\\\sin   \angle EAD}，则\\sin  \angle ADC=\frac{2\sqrt{26} \times \frac{2\sqrt{2}}{\sqrt{26}}}{5\sqrt{2}}=\frac{4}{5}.$

Final answer: $\frac{4}{5}$
Subfield: Plane Geometry


Question: $在平面四边形ABCD中，对角线AC与BD交于点E,\angle ABD=45^\circ ,AE=EC,DE=2BE,AB=6，AD=3\sqrt{2}．$![Image](https://cdntk.53zaixian.com/tq/1623609001336000514/img/compression/M_d86890274e7fa0fecd13e7af80e6dd36.jpg)
$求\sin  \angle ADC的值．$

Solution:
$由\angle ADB=90^\circ ,AE=\sqrt{26},DE=2\sqrt{2},AD=3\sqrt{2}，得\\sin  \angle EAD=\frac{DE}{AE}=\frac{2\sqrt{2}}{\sqrt{26}},\\cos  \angle EAD=\frac{AD}{AE}=\frac{3\sqrt{2}}{\sqrt{26}}．在\triangle ACD中，由余弦定理，得CD^{2}=AD^{2}+AC^{2}-2AD\cdot AC\cdot \\cos  \angle EAD=50，则CD=5\sqrt{2}．在\triangle ACD中，由正弦定理，得\frac{AC}{\\\sin   \angle ADC}=\frac{CD}{\\\sin   \angle EAD}，则\\sin  \angle ADC=\frac{2\sqrt{26} \times \frac{2\sqrt{2}}{\sqrt{26}}}{5\sqrt{2}}=\frac{4}{5}.$

Final answer: $\frac{4}{5}$
Subfield: Plane Geometry


Question: $已知椭圆C:\frac{x^2}{a^2}+\frac{y^2}{b^2}=1(a>b>0)过点A(0,1)，且离心率为\frac{\sqrt{3}}{2}.
$![Image](https://cdntk.53zaixian.com/tq/1623611251148337153/img/compression/M_dcd55c3f33058c32374df8adbf87b76e.jpg)
$求椭圆C的方程；$

Solution:
$\because 椭圆C:\frac{x^2}{a^2}+\frac{y^2}{b^2}=1(a>b>0)过点A(0,1)，即\frac{1}{b^2}=1，\therefore b=1.\because e=\frac{c}{a}=\frac{\sqrt{3}}{2}，又a^{2}-b^{2}=c^{2}，\therefore a=2，\therefore 椭圆C的方程为\frac{x^2}{4}+y^{2}=1.$

Final answer: $\frac{x^2}{4}+y^{2}=1$