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


Question: $向量a,b,c,d,e如图所示,根据图解答下列问题：$<img_2521>
$用b,c表示\overrightarrow{DB}；$

Solution:
$\overrightarrow{DB}=\overrightarrow{CB}-\overrightarrow{CD}=-\overrightarrow{BC}-\overrightarrow{CD}=-b-c.$

Final answer: $\overrightarrow{DB}=-b-c$
Subfield: Plane Geometry


Question: $向量a,b,c,d,e如图所示,根据图解答下列问题：$![Image](https://cdntk.53zaixian.com/tq/1664563249463762945/img/compression/M_9c6b1820930bf7166b7bd773053127d4.jpg)
$用a,b,e表示\overrightarrow{EC}；$

Solution:
$\overrightarrow{EC}=\overrightarrow{EA}+\overrightarrow{AB}+\overrightarrow{BC}=e+a+b.$

Final answer: $e+a+b$
Subfield: Plane Geometry


Question: $向量a,b,c,d,e如图所示,根据图解答下列问题：$<img_2521>
$用a,b,e表示\overrightarrow{EC}；$

Solution:
$\overrightarrow{EC}=\overrightarrow{EA}+\overrightarrow{AB}+\overrightarrow{BC}=e+a+b.$

Final answer: $e+a+b$
Subfield: Plane Geometry


Question: $向量a,b,c,d,e如图所示,根据图解答下列问题：$![Image](https://cdntk.53zaixian.com/tq/1664563249463762945/img/compression/M_9c6b1820930bf7166b7bd773053127d4.jpg)
$用d,c表示\overrightarrow{EC}.$

Solution:
$\overrightarrow{EC}=-\overrightarrow{CE}=-(\overrightarrow{CD}+\overrightarrow{DE})=-c-d.$

Final answer: $-d-c$
Subfield: Plane Geometry


Question: $向量a,b,c,d,e如图所示,根据图解答下列问题：$<img_2521>
$用d,c表示\overrightarrow{EC}.$

Solution:
$\overrightarrow{EC}=-\overrightarrow{CE}=-(\overrightarrow{CD}+\overrightarrow{DE})=-c-d.$

Final answer: $-d-c$
Subfield: Plane Geometry


Question: $如图,G是\triangle OAB的重心,P,Q分别是边OA,OB上的动点(不含端点),且P,G,Q三点共线,延长OG交AB于M.$![Image](https://cdntk.53zaixian.com/tq/1664563268472348673/img/compression/M_1f446a167511925435f7e1707279a926.jpg)
$设\overrightarrow{PG}=\lambda \overrightarrow{PQ}(\lambda \in R),将\overrightarrow{OG}用\overrightarrow{OP},\overrightarrow{OQ}表示；$

Solution:
$\overrightarrow{OG}＝\overrightarrow{OP}＋\overrightarrow{PG}＝\overrightarrow{OP}＋\lambda \overrightarrow{PQ}＝\overrightarrow{OP}＋\lambda (\overrightarrow{OQ}-\overrightarrow{OP})＝(1-\lambda )\overrightarrow{OP}＋\lambda \overrightarrow{OQ}.$

Final answer: $\overrightarrow{OG}=(1-\lambda)\overrightarrow{OP}+\lambda \overrightarrow{OQ}.$
Subfield: Plane Geometry


Question: $如图,G是\triangle OAB的重心,P,Q分别是边OA,OB上的动点(不含端点),且P,G,Q三点共线,延长OG交AB于M.$<img_1528>
$设\overrightarrow{PG}=\lambda \overrightarrow{PQ}(\lambda \in R),将\overrightarrow{OG}用\overrightarrow{OP},\overrightarrow{OQ}表示；$

Solution:
$\overrightarrow{OG}＝\overrightarrow{OP}＋\overrightarrow{PG}＝\overrightarrow{OP}＋\lambda \overrightarrow{PQ}＝\overrightarrow{OP}＋\lambda (\overrightarrow{OQ}-\overrightarrow{OP})＝(1-\lambda )\overrightarrow{OP}＋\lambda \overrightarrow{OQ}.$

Final answer: $\overrightarrow{OG}=(1-\lambda)\overrightarrow{OP}+\lambda \overrightarrow{OQ}.$
Subfield: Plane Geometry


Question: $如图,G是\triangle OAB的重心,P,Q分别是边OA,OB上的动点(不含端点),且P,G,Q三点共线,延长OG交AB于M.$![Image](https://cdntk.53zaixian.com/tq/1664563268472348673/img/compression/M_1f446a167511925435f7e1707279a926.jpg)
$设\overrightarrow{OP}=x\overrightarrow{OA},\overrightarrow{OQ}=y\overrightarrow{OB},证明：\frac{1}{x}+\frac{1}{y}是定值.$

Solution:
$证明:由(1)及题意得\overrightarrow{OG}＝(1-\lambda )\overrightarrow{OP}＋\lambda \overrightarrow{OQ}＝(1-\lambda )x\overrightarrow{OA}＋\lambda y\overrightarrow{OB}.\because G是\triangle OAB的重心，\therefore \overrightarrow{OG}＝\frac{2}{3}\overrightarrow{OM}＝\frac{2}{3}\times \frac{1}{2}(\overrightarrow{OA}＋\overrightarrow{OB})＝\frac{1}{3}\overrightarrow{OA}＋\frac{1}{3}\overrightarrow{OB}.\because \overrightarrow{OA}，\overrightarrow{OB}不共线，\therefore \left\{\begin{matrix}(1-\lambda )x=\frac{1}{3},\\ \lambda y=\frac{1}{3},\end{matrix}\right.\therefore \left\{\begin{matrix}\frac{1}{x}=3-3\lambda ,\\ \frac{1}{y}=3\lambda .\end{matrix}\right.\therefore \frac{1}{x}＋\frac{1}{y}＝3，为定值.$

Final answer: 证明题
Subfield: Plane Geometry


Question: $如图,G是\triangle OAB的重心,P,Q分别是边OA,OB上的动点(不含端点),且P,G,Q三点共线,延长OG交AB于M.$<img_1528>
$设\overrightarrow{OP}=x\overrightarrow{OA},\overrightarrow{OQ}=y\overrightarrow{OB},证明：\frac{1}{x}+\frac{1}{y}是定值.$

Solution:
$证明:由(1)及题意得\overrightarrow{OG}＝(1-\lambda )\overrightarrow{OP}＋\lambda \overrightarrow{OQ}＝(1-\lambda )x\overrightarrow{OA}＋\lambda y\overrightarrow{OB}.\because G是\triangle OAB的重心，\therefore \overrightarrow{OG}＝\frac{2}{3}\overrightarrow{OM}＝\frac{2}{3}\times \frac{1}{2}(\overrightarrow{OA}＋\overrightarrow{OB})＝\frac{1}{3}\overrightarrow{OA}＋\frac{1}{3}\overrightarrow{OB}.\because \overrightarrow{OA}，\overrightarrow{OB}不共线，\therefore \left\{\begin{matrix}(1-\lambda )x=\frac{1}{3},\\ \lambda y=\frac{1}{3},\end{matrix}\right.\therefore \left\{\begin{matrix}\frac{1}{x}=3-3\lambda ,\\ \frac{1}{y}=3\lambda .\end{matrix}\right.\therefore \frac{1}{x}＋\frac{1}{y}＝3，为定值.$
Subfield: Plane Geometry


Question: $如图,在\triangle ABC中,\overrightarrow{AE}=\frac{2}{3}\overrightarrow{AB},\overrightarrow{AD}=\frac{3}{4}\overrightarrow{AC},BD与CE交于点O.$![Image](https://cdntk.53zaixian.com/tq/1664563270712107009/img/compression/M_562e77a8349a652f92bdb249c78e8ab4.jpg)
$若\overrightarrow{AO}=m\overrightarrow{AB}+n\overrightarrow{AC}(m,n\in R),求mn的值；$

Solution:
$因为B，O，D三点共线，所以\overrightarrow{AO}=m\overrightarrow{AB}+(1-m)\overrightarrow{AD}，又\overrightarrow{AD}=\frac{3}{4}\overrightarrow{AC}，所以\overrightarrow{AO}=m\overrightarrow{AB}+\frac{3-3m}{4}\overrightarrow{AC}.因为C，O，E三点共线，所以\overrightarrow{AO}=n\overrightarrow{AC}+(1-n)\overrightarrow{AE}，又\overrightarrow{AE}=\frac{2}{3}\overrightarrow{AB}，所以\overrightarrow{AO}=n\overrightarrow{AC}+\frac{2-2n}{3}\overrightarrow{AB}.所以\left\{\begin{matrix}m=\frac{2-2n}{3},\\ \frac{3-3m}{4}=n,\end{matrix}\right.解得\left\{\begin{matrix}m=\frac{1}{3},\\ n=\frac{1}{2},\end{matrix}\right.所以mn=\frac{1}{3}\times \frac{1}{2}=\frac{1}{6}.$

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


Question: $如图,在\triangle ABC中,\overrightarrow{AE}=\frac{2}{3}\overrightarrow{AB},\overrightarrow{AD}=\frac{3}{4}\overrightarrow{AC},BD与CE交于点O.$<img_1786>
$若\overrightarrow{AO}=m\overrightarrow{AB}+n\overrightarrow{AC}(m,n\in R),求mn的值；$

Solution:
$因为B，O，D三点共线，所以\overrightarrow{AO}=m\overrightarrow{AB}+(1-m)\overrightarrow{AD}，又\overrightarrow{AD}=\frac{3}{4}\overrightarrow{AC}，所以\overrightarrow{AO}=m\overrightarrow{AB}+\frac{3-3m}{4}\overrightarrow{AC}.因为C，O，E三点共线，所以\overrightarrow{AO}=n\overrightarrow{AC}+(1-n)\overrightarrow{AE}，又\overrightarrow{AE}=\frac{2}{3}\overrightarrow{AB}，所以\overrightarrow{AO}=n\overrightarrow{AC}+\frac{2-2n}{3}\overrightarrow{AB}.所以\left\{\begin{matrix}m=\frac{2-2n}{3},\\ \frac{3-3m}{4}=n,\end{matrix}\right.解得\left\{\begin{matrix}m=\frac{1}{3},\\ n=\frac{1}{2},\end{matrix}\right.所以mn=\frac{1}{3}\times \frac{1}{2}=\frac{1}{6}.$

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


Question: $如图,在\triangle ABC中,\overrightarrow{AE}=\frac{2}{3}\overrightarrow{AB},\overrightarrow{AD}=\frac{3}{4}\overrightarrow{AC},BD与CE交于点O.$<img_1786>
$设\triangle ABC的面积为S,\triangle OBC的面积为S',求\frac{S'}{S}的值.$

Solution:
$延长AO，与BC交于点F.因为B，F，C三点共线，所以设\overrightarrow{AF}=t\overrightarrow{AB}+(1-t)\overrightarrow{AC},t\in R.由(1)知\overrightarrow{AO}=\frac{1}{3}\overrightarrow{AB}+\frac{1}{2}\overrightarrow{AC}，又\overrightarrow{AO}\parallel \overrightarrow{AF}，所以设\overrightarrow{AO}=\lambda \overrightarrow{AF}，\lambda \in R,所以\frac{1}{3}\overrightarrow{AB}+\frac{1}{2}\overrightarrow{AC}=\lambda [t\overrightarrow{AB}+(1-t)\overrightarrow{AC}]=\lambda t\overrightarrow{AB}+\lambda (1-t)\overrightarrow{AC}，所以\left\{\begin{matrix}\lambda t=\frac{1}{3},\\ \lambda (1-t)=\frac{1}{2}\end{matrix}\right.\Rightarrow \left\{\begin{matrix}t=\frac{2}{5},\\ \lambda =\frac{5}{6},\end{matrix}\right.所以\overrightarrow{AO}=\frac{5}{6}\overrightarrow{AF}，所以\overrightarrow{OF}=\frac{1}{6}\overrightarrow{AF}，则\frac{S'}{S}=\frac{1}{6}.$

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


Question: $如图,在\triangle ABC中,\overrightarrow{AE}=\frac{2}{3}\overrightarrow{AB},\overrightarrow{AD}=\frac{3}{4}\overrightarrow{AC},BD与CE交于点O.$![Image](https://cdntk.53zaixian.com/tq/1664563270712107009/img/compression/M_562e77a8349a652f92bdb249c78e8ab4.jpg)
$设\triangle ABC的面积为S,\triangle OBC的面积为S',求\frac{S'}{S}的值.$

Solution:
$延长AO，与BC交于点F.因为B，F，C三点共线，所以设\overrightarrow{AF}=t\overrightarrow{AB}+(1-t)\overrightarrow{AC},t\in R.由(1)知\overrightarrow{AO}=\frac{1}{3}\overrightarrow{AB}+\frac{1}{2}\overrightarrow{AC}，又\overrightarrow{AO}\parallel \overrightarrow{AF}，所以设\overrightarrow{AO}=\lambda \overrightarrow{AF}，\lambda \in R,所以\frac{1}{3}\overrightarrow{AB}+\frac{1}{2}\overrightarrow{AC}=\lambda [t\overrightarrow{AB}+(1-t)\overrightarrow{AC}]=\lambda t\overrightarrow{AB}+\lambda (1-t)\overrightarrow{AC}，所以\left\{\begin{matrix}\lambda t=\frac{1}{3},\\ \lambda (1-t)=\frac{1}{2}\end{matrix}\right.\Rightarrow \left\{\begin{matrix}t=\frac{2}{5},\\ \lambda =\frac{5}{6},\end{matrix}\right.所以\overrightarrow{AO}=\frac{5}{6}\overrightarrow{AF}，所以\overrightarrow{OF}=\frac{1}{6}\overrightarrow{AF}，则\frac{S'}{S}=\frac{1}{6}.$

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


Question: $已知P,Q分别是梯形ABCD的对角线AC与BD的中点,如图所示.$![Image](https://cdntk.53zaixian.com/tq/1664563290689576963/img/compression/M_4107adcc6ee8cc113d69620a786638f3.jpg)
$试用向量法证明PQ\parallel AB;$

Solution:
$证明：连接CQ.\because Q为BD的中点，\therefore \overrightarrow{CB}+\overrightarrow{CD}=2\overrightarrow{CQ}.\because P为AC的中点，\therefore \overrightarrow{CA}=2\overrightarrow{CP}.\therefore 2\overrightarrow{PQ}=2\overrightarrow{CQ}-2\overrightarrow{CP}=(\overrightarrow{CB}+\overrightarrow{CD})-\overrightarrow{CA}=\overrightarrow{CB}+\overrightarrow{CD}+\overrightarrow{AC}=\overrightarrow{AB}+\overrightarrow{CD}.\because \overrightarrow{CD}\parallel \overrightarrow{AB},\therefore 设\overrightarrow{CD}=\lambda \overrightarrow{AB},\lambda \in R,\therefore 2\overrightarrow{PQ}=\overrightarrow{AB}+\lambda \overrightarrow{AB},即2\overrightarrow{PQ}=(1+\lambda )\overrightarrow{AB},\therefore \overrightarrow{PQ}=\frac{1+\lambda }{2}\overrightarrow{AB}.又｜\overrightarrow{AB}｜\neq ｜\overrightarrow{CD}｜,\therefore \lambda \neq -1,\therefore \overrightarrow{PQ}\parallel \overrightarrow{AB},\therefore PQ\parallel AB.$

Final answer: 证明题
Subfield: Plane Geometry


Question: $已知P,Q分别是梯形ABCD的对角线AC与BD的中点,如图所示.$<img_2291>
$试用向量法证明PQ\parallel AB;$

Solution:
$证明：连接CQ.\because Q为BD的中点，\therefore \overrightarrow{CB}+\overrightarrow{CD}=2\overrightarrow{CQ}.\because P为AC的中点，\therefore \overrightarrow{CA}=2\overrightarrow{CP}.\therefore 2\overrightarrow{PQ}=2\overrightarrow{CQ}-2\overrightarrow{CP}=(\overrightarrow{CB}+\overrightarrow{CD})-\overrightarrow{CA}=\overrightarrow{CB}+\overrightarrow{CD}+\overrightarrow{AC}=\overrightarrow{AB}+\overrightarrow{CD}.\because \overrightarrow{CD}\parallel \overrightarrow{AB},\therefore 设\overrightarrow{CD}=\lambda \overrightarrow{AB},\lambda \in R,\therefore 2\overrightarrow{PQ}=\overrightarrow{AB}+\lambda \overrightarrow{AB},即2\overrightarrow{PQ}=(1+\lambda )\overrightarrow{AB},\therefore \overrightarrow{PQ}=\frac{1+\lambda }{2}\overrightarrow{AB}.又｜\overrightarrow{AB}｜\neq ｜\overrightarrow{CD}｜,\therefore \lambda \neq -1,\therefore \overrightarrow{PQ}\parallel \overrightarrow{AB},\therefore PQ\parallel AB.$
Subfield: Plane Geometry


Question: $已知P,Q分别是梯形ABCD的对角线AC与BD的中点,如图所示.$![Image](https://cdntk.53zaixian.com/tq/1664563290689576963/img/compression/M_4107adcc6ee8cc113d69620a786638f3.jpg)
$若AB=3CD,求PQ:AB.$

Solution:
$\because 向量\overrightarrow{AB}与\overrightarrow{CD}反向，且AB=3CD,\therefore \overrightarrow{AB}=-3\overrightarrow{CD},结合(1)可知\overrightarrow{PQ}=\frac{1-\frac{1}{3}}{2}\overrightarrow{AB}=\frac{1}{3}\overrightarrow{AB}，\therefore PQ:AB=1:3.$

Final answer: $1:3$
Subfield: Plane Geometry


Question: $已知P,Q分别是梯形ABCD的对角线AC与BD的中点,如图所示.$<img_2291>
$若AB=3CD,求PQ:AB.$

Solution:
$\because 向量\overrightarrow{AB}与\overrightarrow{CD}反向，且AB=3CD,\therefore \overrightarrow{AB}=-3\overrightarrow{CD},结合(1)可知\overrightarrow{PQ}=\frac{1-\frac{1}{3}}{2}\overrightarrow{AB}=\frac{1}{3}\overrightarrow{AB}，\therefore PQ:AB=1:3.$

Final answer: $1:3$
Subfield: Plane Geometry


Question: $如图所示,在\triangle ABO中,\overrightarrow{OC}=\frac{1}{3}\overrightarrow{OA},\overrightarrow{OD}=\frac{1}{2}\overrightarrow{OB},AD与BC相交于点M.设\overrightarrow{OA}=a,\overrightarrow{OB}=b.$![Image](https://cdntk.53zaixian.com/tq/1665654343734845441/img/compression/M_4ee2b5eb2cdd0c7da87ce5f97b347d46.jpg)
$试用向量a,b表示\overrightarrow{OM};$

Solution:
$设\overrightarrow{OM}=ma+nb(m,n\in R).因为A,D,M三点共线,所以存在实数\alpha (\alpha \neq -1)使得\overrightarrow{AM}=\alpha \overrightarrow{MD},即\overrightarrow{AO}+\overrightarrow{OM}=\alpha (\overrightarrow{MO}+\overrightarrow{OD}),于是\overrightarrow{OM}=\frac{\overrightarrow{OA}+\alpha \overrightarrow{OD}}{1+\alpha }.又\overrightarrow{OD}=\frac{1}{2}\overrightarrow{OB}，所以\overrightarrow{OM}=\frac{\overrightarrow{OA}+\frac{\alpha }{2}\overrightarrow{OB}}{1+\alpha }=\frac{1}{1+\alpha }a+\frac{\alpha }{2(1+\alpha )}b,所以\left\{\begin{matrix}m=\frac{1}{1+\alpha },\\ n=\frac{\alpha }{2(1+\alpha )},\end{matrix}\right.故m+2n=1.①因为B,C,M三点共线,所以存在实数\beta (\beta \neq -1)使得\overrightarrow{CM}=\beta \overrightarrow{MB},即\overrightarrow{CO}+\overrightarrow{OM}=\beta (\overrightarrow{MO}+\overrightarrow{OB}),于是\overrightarrow{OM}=\frac{\overrightarrow{OC}+\beta \overrightarrow{OB}}{1+\beta }.又\overrightarrow{OC}=\frac{1}{3}\overrightarrow{OA},所以\overrightarrow{OM}=\frac{\frac{1}{3}\overrightarrow{OA}+\beta \overrightarrow{OB}}{1+\beta }=\frac{1}{3(1+\beta )}a+\frac{\beta }{1+\beta }b,所以\left\{\begin{matrix}m=\frac{1}{3(1+\beta )},\\ n=\frac{\beta }{1+\beta },\end{matrix}\right.故3m+n=1.②由①②可得m=\frac{1}{5},n=\frac{2}{5},所以\overrightarrow{OM}=\frac{1}{5}a+\frac{2}{5}b.$

Final answer: $\overrightarrow{OM}=\frac{1}{5}a+\frac{2}{5}b$
Subfield: Plane Geometry


Question: $如图所示,在\triangle ABO中,\overrightarrow{OC}=\frac{1}{3}\overrightarrow{OA},\overrightarrow{OD}=\frac{1}{2}\overrightarrow{OB},AD与BC相交于点M.设\overrightarrow{OA}=a,\overrightarrow{OB}=b.$<img_1953>
$试用向量a,b表示\overrightarrow{OM};$

Solution:
$设\overrightarrow{OM}=ma+nb(m,n\in R).因为A,D,M三点共线,所以存在实数\alpha (\alpha \neq -1)使得\overrightarrow{AM}=\alpha \overrightarrow{MD},即\overrightarrow{AO}+\overrightarrow{OM}=\alpha (\overrightarrow{MO}+\overrightarrow{OD}),于是\overrightarrow{OM}=\frac{\overrightarrow{OA}+\alpha \overrightarrow{OD}}{1+\alpha }.又\overrightarrow{OD}=\frac{1}{2}\overrightarrow{OB}，所以\overrightarrow{OM}=\frac{\overrightarrow{OA}+\frac{\alpha }{2}\overrightarrow{OB}}{1+\alpha }=\frac{1}{1+\alpha }a+\frac{\alpha }{2(1+\alpha )}b,所以\left\{\begin{matrix}m=\frac{1}{1+\alpha },\\ n=\frac{\alpha }{2(1+\alpha )},\end{matrix}\right.故m+2n=1.①因为B,C,M三点共线,所以存在实数\beta (\beta \neq -1)使得\overrightarrow{CM}=\beta \overrightarrow{MB},即\overrightarrow{CO}+\overrightarrow{OM}=\beta (\overrightarrow{MO}+\overrightarrow{OB}),于是\overrightarrow{OM}=\frac{\overrightarrow{OC}+\beta \overrightarrow{OB}}{1+\beta }.又\overrightarrow{OC}=\frac{1}{3}\overrightarrow{OA},所以\overrightarrow{OM}=\frac{\frac{1}{3}\overrightarrow{OA}+\beta \overrightarrow{OB}}{1+\beta }=\frac{1}{3(1+\beta )}a+\frac{\beta }{1+\beta }b,所以\left\{\begin{matrix}m=\frac{1}{3(1+\beta )},\\ n=\frac{\beta }{1+\beta },\end{matrix}\right.故3m+n=1.②由①②可得m=\frac{1}{5},n=\frac{2}{5},所以\overrightarrow{OM}=\frac{1}{5}a+\frac{2}{5}b.$

Final answer: $\overrightarrow{OM}=\frac{1}{5}a+\frac{2}{5}b$
Subfield: Plane Geometry


Question: $如图所示,在\triangle ABO中,\overrightarrow{OC}=\frac{1}{3}\overrightarrow{OA},\overrightarrow{OD}=\frac{1}{2}\overrightarrow{OB},AD与BC相交于点M.设\overrightarrow{OA}=a,\overrightarrow{OB}=b.$<img_1953>
$在线段AC上取一点E,在线段BD上取一点F,使EF过点M,设\overrightarrow{OE}=\lambda \overrightarrow{OA},\overrightarrow{OF}=\mu \overrightarrow{OB},其中\lambda ,\mu \in R且均不为0,证明:\frac{1}{\lambda }+\frac{2}{\mu }为定值,并求出该定值.$

Solution:
$因为E,M,F三点共线,所以存在实数\eta (\eta \neq -1)使得\overrightarrow{EM}=\eta \overrightarrow{MF},即\overrightarrow{EO}+\overrightarrow{OM}=\eta (\overrightarrow{MO}+\overrightarrow{OF}),于是\overrightarrow{OM}=\frac{\overrightarrow{OE}+\eta \overrightarrow{OF}}{1+\eta }.又\overrightarrow{OE}=\lambda \overrightarrow{OA},\overrightarrow{OF}=\mu \overrightarrow{OB},所以\overrightarrow{OM}=\frac{\lambda \overrightarrow{OA}+\mu \eta \overrightarrow{OB}}{1+\eta }=\frac{\lambda }{1+\eta }a+\frac{\mu \eta }{1+\eta }b,由(1)知\overrightarrow{OM}=\frac{1}{5}a+\frac{2}{5}b,所以\frac{1}{5}a+\frac{2}{5}b=\frac{\lambda }{1+\eta }a+\frac{\mu \eta }{1+\eta }b,则\left\{\begin{matrix}\frac{\lambda }{1+\eta }=\frac{1}{5},\\ \frac{\mu \eta }{1+\eta }=\frac{2}{5},\end{matrix}\right.消去\eta ,得\frac{1}{\lambda }+\frac{2}{\mu }=5.故\frac{1}{\lambda }+\frac{2}{\mu }为定值，该定值为5.$
Subfield: Plane Geometry


Question: $如图所示,在\triangle ABO中,\overrightarrow{OC}=\frac{1}{3}\overrightarrow{OA},\overrightarrow{OD}=\frac{1}{2}\overrightarrow{OB},AD与BC相交于点M.设\overrightarrow{OA}=a,\overrightarrow{OB}=b.$![Image](https://cdntk.53zaixian.com/tq/1665654343734845441/img/compression/M_4ee2b5eb2cdd0c7da87ce5f97b347d46.jpg)
$在线段AC上取一点E,在线段BD上取一点F,使EF过点M,设\overrightarrow{OE}=\lambda \overrightarrow{OA},\overrightarrow{OF}=\mu \overrightarrow{OB},其中\lambda ,\mu \in R且均不为0,证明:\frac{1}{\lambda }+\frac{2}{\mu }为定值,并求出该定值.$

Solution:
$因为E,M,F三点共线,所以存在实数\eta (\eta \neq -1)使得\overrightarrow{EM}=\eta \overrightarrow{MF},即\overrightarrow{EO}+\overrightarrow{OM}=\eta (\overrightarrow{MO}+\overrightarrow{OF}),于是\overrightarrow{OM}=\frac{\overrightarrow{OE}+\eta \overrightarrow{OF}}{1+\eta }.又\overrightarrow{OE}=\lambda \overrightarrow{OA},\overrightarrow{OF}=\mu \overrightarrow{OB},所以\overrightarrow{OM}=\frac{\lambda \overrightarrow{OA}+\mu \eta \overrightarrow{OB}}{1+\eta }=\frac{\lambda }{1+\eta }a+\frac{\mu \eta }{1+\eta }b,由(1)知\overrightarrow{OM}=\frac{1}{5}a+\frac{2}{5}b,所以\frac{1}{5}a+\frac{2}{5}b=\frac{\lambda }{1+\eta }a+\frac{\mu \eta }{1+\eta }b,则\left\{\begin{matrix}\frac{\lambda }{1+\eta }=\frac{1}{5},\\ \frac{\mu \eta }{1+\eta }=\frac{2}{5},\end{matrix}\right.消去\eta ,得\frac{1}{\lambda }+\frac{2}{\mu }=5.故\frac{1}{\lambda }+\frac{2}{\mu }为定值，该定值为5.$

Final answer: 证明题
Subfield: Plane Geometry


Question: $如图,在矩形ABCD中,AB=2,AD=1,\overrightarrow{DM}=\frac{1}{3}\overrightarrow{DC},\overrightarrow{BN}=\frac{2}{3}\overrightarrow{BC},AC与MN相交于点E.$<img_2031>
$若\overrightarrow{MN}=\lambda \overrightarrow{AB}+\mu \overrightarrow{AD}，求\lambda 和\mu 的值；$

Solution:
$如图，以A为原点，AB所在直线为x轴，AD所在直线为y轴，建立平面直角坐标系，则A(0，0)，D(0,1),B(2,0),C(2，1)，M\left(\frac{2}{3},1\right),N\left(2,\frac{2}{3}\right).$<img_2791>$易得\overrightarrow{MN}=\left(\frac{4}{3},-\frac{1}{3}\right),\overrightarrow{AB}=(2,0),\overrightarrow{AD}=(0,1)，因为\overrightarrow{MN}=\lambda \overrightarrow{AB}+\mu \overrightarrow{AD}，即\left(\frac{4}{3},-\frac{1}{3}\right)=(2\lambda ,\mu )，所以\left\{\begin{matrix}2\lambda =\frac{4}{3},\\ \mu =-\frac{1}{3},\end{matrix}\right.所以\lambda =\frac{2}{3},\mu =-\frac{1}{3}.$

Final answer: $\lambda =\frac{2}{3}$, $\mu =-\frac{1}{3}.$
Subfield: Plane Geometry


Question: $如图,在矩形ABCD中,AB=2,AD=1,\overrightarrow{DM}=\frac{1}{3}\overrightarrow{DC},\overrightarrow{BN}=\frac{2}{3}\overrightarrow{BC},AC与MN相交于点E.$![Image](https://cdntk.53zaixian.com/tq/1664563278287020033/img/compression/M_f58b0e86164c3f7a95b1cb0e63e6fcb5.jpg)
$若\overrightarrow{MN}=\lambda \overrightarrow{AB}+\mu \overrightarrow{AD}，求\lambda 和\mu 的值；$

Solution:
$如图，以A为原点，AB所在直线为x轴，AD所在直线为y轴，建立平面直角坐标系，则A(0，0)，D(0,1),B(2,0),C(2，1)，M\left(\frac{2}{3},1\right),N\left(2,\frac{2}{3}\right).$![Image](https://cdntk.53zaixian.com/tq/1664563278287020033/img/compression/M_3a3f87b865e7b861f65fdcd529a84dc0.jpg)$易得\overrightarrow{MN}=\left(\frac{4}{3},-\frac{1}{3}\right),\overrightarrow{AB}=(2,0),\overrightarrow{AD}=(0,1)，因为\overrightarrow{MN}=\lambda \overrightarrow{AB}+\mu \overrightarrow{AD}，即\left(\frac{4}{3},-\frac{1}{3}\right)=(2\lambda ,\mu )，所以\left\{\begin{matrix}2\lambda =\frac{4}{3},\\ \mu =-\frac{1}{3},\end{matrix}\right.所以\lambda =\frac{2}{3},\mu =-\frac{1}{3}.$

Final answer: $\lambda =\frac{2}{3}$, $\mu =-\frac{1}{3}.$
Subfield: Plane Geometry


Question: $如图,在矩形ABCD中,AB=2,AD=1,\overrightarrow{DM}=\frac{1}{3}\overrightarrow{DC},\overrightarrow{BN}=\frac{2}{3}\overrightarrow{BC},AC与MN相交于点E.$<img_2031>
$用向量\overrightarrow{AM},\overrightarrow{AN}表示\overrightarrow{AE}.$

Solution:
$由题可设\overrightarrow{AE}=t\overrightarrow{AC},\overrightarrow{AC}=m\overrightarrow{AM}+n\overrightarrow{AN}，t,m,n\in R.因为\overrightarrow{AM}=\left(\frac{2}{3},1\right),\overrightarrow{AN}=\left(2,\frac{2}{3}\right)，\overrightarrow{AC}=(2,1),所以(2,1)=\left(\frac{2}{3}m+2n,m+\frac{2}{3}n\right),解得m=\frac{3}{7},n=\frac{6}{7}，即\overrightarrow{AC}=\frac{3}{7}\overrightarrow{AM}+\frac{6}{7}\overrightarrow{AN}，所以\overrightarrow{AE}=t\overrightarrow{AC}=\frac{3}{7}t\overrightarrow{AM}+\frac{6}{7}t\overrightarrow{AN}，又因为M，E，N三点共线，所以\frac{3}{7}t+\frac{6}{7}t=1,解得t=\frac{7}{9}，所以\overrightarrow{AE}=\frac{1}{3}\overrightarrow{AM}+\frac{2}{3}\overrightarrow{AN}.$

Final answer: $\overrightarrow{AE}=\frac{1}{3}\overrightarrow{AM}+\frac{2}{3}\overrightarrow{AN}$
Subfield: Plane Geometry


Question: $如图,在矩形ABCD中,AB=2,AD=1,\overrightarrow{DM}=\frac{1}{3}\overrightarrow{DC},\overrightarrow{BN}=\frac{2}{3}\overrightarrow{BC},AC与MN相交于点E.$![Image](https://cdntk.53zaixian.com/tq/1664563278287020033/img/compression/M_f58b0e86164c3f7a95b1cb0e63e6fcb5.jpg)
$用向量\overrightarrow{AM},\overrightarrow{AN}表示\overrightarrow{AE}.$

Solution:
$由题可设\overrightarrow{AE}=t\overrightarrow{AC},\overrightarrow{AC}=m\overrightarrow{AM}+n\overrightarrow{AN}，t,m,n\in R.因为\overrightarrow{AM}=\left(\frac{2}{3},1\right),\overrightarrow{AN}=\left(2,\frac{2}{3}\right)，\overrightarrow{AC}=(2,1),所以(2,1)=\left(\frac{2}{3}m+2n,m+\frac{2}{3}n\right),解得m=\frac{3}{7},n=\frac{6}{7}，即\overrightarrow{AC}=\frac{3}{7}\overrightarrow{AM}+\frac{6}{7}\overrightarrow{AN}，所以\overrightarrow{AE}=t\overrightarrow{AC}=\frac{3}{7}t\overrightarrow{AM}+\frac{6}{7}t\overrightarrow{AN}，又因为M，E，N三点共线，所以\frac{3}{7}t+\frac{6}{7}t=1,解得t=\frac{7}{9}，所以\overrightarrow{AE}=\frac{1}{3}\overrightarrow{AM}+\frac{2}{3}\overrightarrow{AN}.$

Final answer: $\overrightarrow{AE}=\frac{1}{3}\overrightarrow{AM}+\frac{2}{3}\overrightarrow{AN}$
Subfield: Plane Geometry


Question: $如图,在直角梯形OABC中,OA\parallel CB,OA\perp OC,OA=2BC=2OC,M为线段AB上靠近点B的三等分点,OM交AC于D,P为线段BC上的一个动点(包含端点).$![Image](https://cdntk.53zaixian.com/tq/1664563281000734721/img/compression/M_ef42a5a75e6bf6aa7df15634ab2a1863.jpg)
$用\overrightarrow{OA}和\overrightarrow{OC}表示\overrightarrow{OM}；$

Solution:
$因为M为线段AB上靠近点B的三等分点，所以\overrightarrow{MA}=\frac{2}{3}\overrightarrow{BA}=\frac{2}{3}(\overrightarrow{OA}-\overrightarrow{OB})，又CB\parallel OA，且OA=2BC，故\overrightarrow{CB}=\frac{1}{2}\overrightarrow{OA}，则\overrightarrow{OM}=\overrightarrow{OA}-\overrightarrow{MA}=\overrightarrow{OA}-\frac{2}{3}(\overrightarrow{OA}-\overrightarrow{OB})=\frac{1}{3}\overrightarrow{OA}+\frac{2}{3}\overrightarrow{OB}=\frac{1}{3}\overrightarrow{OA}+\frac{2}{3}(\overrightarrow{OC}+\overrightarrow{CB})=\frac{1}{3}\overrightarrow{OA}+\frac{2}{3}\left(\overrightarrow{OC}+\frac{1}{2}\overrightarrow{OA}\right)=\frac{2}{3}\overrightarrow{OA}+\frac{2}{3}\overrightarrow{OC},即\overrightarrow{OM}=\frac{2}{3}\overrightarrow{OA}+\frac{2}{3}\overrightarrow{OC}.$

Final answer: $\overrightarrow{OM}=\frac{2}{3}\overrightarrow{OA}+\frac{2}{3}\overrightarrow{OC}$
Subfield: Plane Geometry


Question: $如图,在直角梯形OABC中,OA\parallel CB,OA\perp OC,OA=2BC=2OC,M为线段AB上靠近点B的三等分点,OM交AC于D,P为线段BC上的一个动点(包含端点).$<img_2354>
$用\overrightarrow{OA}和\overrightarrow{OC}表示\overrightarrow{OM}；$

Solution:
$因为M为线段AB上靠近点B的三等分点，所以\overrightarrow{MA}=\frac{2}{3}\overrightarrow{BA}=\frac{2}{3}(\overrightarrow{OA}-\overrightarrow{OB})，又CB\parallel OA，且OA=2BC，故\overrightarrow{CB}=\frac{1}{2}\overrightarrow{OA}，则\overrightarrow{OM}=\overrightarrow{OA}-\overrightarrow{MA}=\overrightarrow{OA}-\frac{2}{3}(\overrightarrow{OA}-\overrightarrow{OB})=\frac{1}{3}\overrightarrow{OA}+\frac{2}{3}\overrightarrow{OB}=\frac{1}{3}\overrightarrow{OA}+\frac{2}{3}(\overrightarrow{OC}+\overrightarrow{CB})=\frac{1}{3}\overrightarrow{OA}+\frac{2}{3}\left(\overrightarrow{OC}+\frac{1}{2}\overrightarrow{OA}\right)=\frac{2}{3}\overrightarrow{OA}+\frac{2}{3}\overrightarrow{OC},即\overrightarrow{OM}=\frac{2}{3}\overrightarrow{OA}+\frac{2}{3}\overrightarrow{OC}.$

Final answer: $\overrightarrow{OM}=\frac{2}{3}\overrightarrow{OA}+\frac{2}{3}\overrightarrow{OC}$