di-zhang-fdu/OlympiadExtended huggingface dataset:
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>
$设\overrightarrow{OB}=\lambda \overrightarrow{CA}+\mu \overrightarrow{OP},求\lambda ,\mu 的取值范围.$

Solution:
$以O为坐标原点，建立如图所示的平面直角坐标系.设OA=2，则A(2,0),C(0,1),B(1,1),O(0,0)，因为点P在线段BC上运动(包含端点)，所以可设其坐标为(m,1),0\leq m\leq 1，则\overrightarrow{OB}=(1,1),\overrightarrow{CA}=(2,-1),\overrightarrow{OP}=(m,1)，由\overrightarrow{OB}=\lambda \overrightarrow{CA}+\mu \overrightarrow{OP}可得1=2\lambda +\mu m,1=-\lambda +\mu ，则\mu =\frac{3}{m+2},\lambda =\mu -1，因为m\in [0,1]，所以m+2\in [2,3]，故\mu \in \left[1,\frac{3}{2}\right],\lambda \in \left[0,\frac{1}{2}\right].$<img_2570>



Final answer: $ \left[0,\frac{1}{2}\right]$, $ \left[1,\frac{3}{2}\right]$
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)
$设\overrightarrow{OB}=\lambda \overrightarrow{CA}+\mu \overrightarrow{OP},求\lambda ,\mu 的取值范围.$

Solution:
$以O为坐标原点，建立如图所示的平面直角坐标系.设OA=2，则A(2,0),C(0,1),B(1,1),O(0,0)，因为点P在线段BC上运动(包含端点)，所以可设其坐标为(m,1),0\leq m\leq 1，则\overrightarrow{OB}=(1,1),\overrightarrow{CA}=(2,-1),\overrightarrow{OP}=(m,1)，由\overrightarrow{OB}=\lambda \overrightarrow{CA}+\mu \overrightarrow{OP}可得1=2\lambda +\mu m,1=-\lambda +\mu ，则\mu =\frac{3}{m+2},\lambda =\mu -1，因为m\in [0,1]，所以m+2\in [2,3]，故\mu \in \left[1,\frac{3}{2}\right],\lambda \in \left[0,\frac{1}{2}\right].$![Image](https://cdntk.53zaixian.com/tq/1664563281218838530/img/compression/M_9b7f19047c8f179d5a2057c4e368d5d1.jpg)

Final answer: $ \left[0,\frac{1}{2}\right]$, $ \left[1,\frac{3}{2}\right]$
Subfield: Plane Geometry


Question: $如图，在\triangle OAB中,点D在线段OB上,且 OD =2DB,延长BA到点C,使AC=BA.设\overrightarrow{OA}=a,\overrightarrow{OB}=b.$![Image](https://cdntk.53zaixian.com/tq/1664565246345543681/img/compression/M_e9b9917228d40eade3c1d69b02c5089d.jpg)
$用a,b表示向量\overrightarrow{OC},\overrightarrow{DC};$

Solution:
$\because AC=BA,\therefore A为BC的中点,\therefore \overrightarrow{OA}=\frac{1}{2}(\overrightarrow{OB}+\overrightarrow{OC}),\therefore \overrightarrow{OC}=2\overrightarrow{OA}-\overrightarrow{OB}=2a-b.\because OD=2DB,\therefore \overrightarrow{DC}=\overrightarrow{OC}-\overrightarrow{OD}=\overrightarrow{OC}-\frac{2}{3}\overrightarrow{OB}=2a-\frac{5}{3}b.$

Final answer: $\overrightarrow{OC}=2a-b$，$\overrightarrow{DC}=2a-\frac{5}{3}b.$
Subfield: Plane Geometry


Question: $如图，在\triangle OAB中,点D在线段OB上,且 OD =2DB,延长BA到点C,使AC=BA.设\overrightarrow{OA}=a,\overrightarrow{OB}=b.$<img_2183>
$用a,b表示向量\overrightarrow{OC},\overrightarrow{DC};$

Solution:
$\because AC=BA,\therefore A为BC的中点,\therefore \overrightarrow{OA}=\frac{1}{2}(\overrightarrow{OB}+\overrightarrow{OC}),\therefore \overrightarrow{OC}=2\overrightarrow{OA}-\overrightarrow{OB}=2a-b.\because OD=2DB,\therefore \overrightarrow{DC}=\overrightarrow{OC}-\overrightarrow{OD}=\overrightarrow{OC}-\frac{2}{3}\overrightarrow{OB}=2a-\frac{5}{3}b.$

Final answer: $\overrightarrow{OC}=2a-b$，$\overrightarrow{DC}=2a-\frac{5}{3}b.$
Subfield: Plane Geometry


Question: $如图，在\triangle OAB中,点D在线段OB上,且 OD =2DB,延长BA到点C,使AC=BA.设\overrightarrow{OA}=a,\overrightarrow{OB}=b.$![Image](https://cdntk.53zaixian.com/tq/1664565246345543681/img/compression/M_e9b9917228d40eade3c1d69b02c5089d.jpg)
$若向量\overrightarrow{OC}与\overrightarrow{OA}+k\overrightarrow{DC}共线,求实数k的值.$

Solution:
$由(1)得\overrightarrow{OA}+k\overrightarrow{DC}=(2k+1)a-\frac{5}{3}kb.\because \overrightarrow{OC}与\overrightarrow{OA}+k\overrightarrow{DC}共线,\therefore 设\overrightarrow{OC}=\lambda (\overrightarrow{OA}+k\overrightarrow{DC}),\lambda \in R,即2a-b=\lambda (2k+1)a-\frac{5}{3}\lambda kb，\therefore \left\{\begin{matrix}2=\lambda (2k+1),\\ -1=- \frac{5}{3}\lambda k,\end{matrix}\right.解得k=\frac{3}{4}.$

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


Question: $如图，在\triangle OAB中,点D在线段OB上,且 OD =2DB,延长BA到点C,使AC=BA.设\overrightarrow{OA}=a,\overrightarrow{OB}=b.$<img_2183>
$若向量\overrightarrow{OC}与\overrightarrow{OA}+k\overrightarrow{DC}共线,求实数k的值.$

Solution:
$由(1)得\overrightarrow{OA}+k\overrightarrow{DC}=(2k+1)a-\frac{5}{3}kb.\because \overrightarrow{OC}与\overrightarrow{OA}+k\overrightarrow{DC}共线,\therefore 设\overrightarrow{OC}=\lambda (\overrightarrow{OA}+k\overrightarrow{DC}),\lambda \in R,即2a-b=\lambda (2k+1)a-\frac{5}{3}\lambda kb，\therefore \left\{\begin{matrix}2=\lambda (2k+1),\\ -1=- \frac{5}{3}\lambda k,\end{matrix}\right.解得k=\frac{3}{4}.$

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


Question: $如图,在平行四边形ABCD中,\overrightarrow{BM}=\frac{1}{2}\overrightarrow{MC},N为线段CD的中点,E为线段MN上的点，且\overrightarrow{ME}=2\overrightarrow{EN}.$<img_2834>
$若\overrightarrow{AE}=\lambda \overrightarrow{AB}+\mu \overrightarrow{AD}(\lambda ,\mu \in R),求\lambda \mu 的值;$

Solution:
$根据题意可得\overrightarrow{AE}=\overrightarrow{AB}+\overrightarrow{BM}+\overrightarrow{ME}=\overrightarrow{AB}+\frac{1}{3}\overrightarrow{BC}+\frac{2}{3}\overrightarrow{MN}=\overrightarrow{AB}+\frac{1}{3}\overrightarrow{AD}+\frac{2}{3}(\overrightarrow{MC}+\overrightarrow{CN})=\overrightarrow{AB}+\frac{1}{3}\overrightarrow{AD}+\frac{2}{3}\times \left(\frac{2}{3}\overrightarrow{AD}-\frac{1}{2}\overrightarrow{AB}\right)=\frac{2}{3}\overrightarrow{AB}+\frac{7}{9}\overrightarrow{AD}.又\overrightarrow{AE}=\lambda \overrightarrow{AB}+\mu \overrightarrow{AD},所以\lambda =\frac{2}{3},\mu =\frac{7}{9},所以\lambda \mu =\frac{14}{27}.$

Final answer: $\frac{14}{27}$
Subfield: Plane Geometry


Question: $如图,在平行四边形ABCD中,\overrightarrow{BM}=\frac{1}{2}\overrightarrow{MC},N为线段CD的中点,E为线段MN上的点，且\overrightarrow{ME}=2\overrightarrow{EN}.$![Image](https://cdntk.53zaixian.com/tq/1664565246718836739/img/compression/M_f7a258377dff41875de81f6ac9c5c522.jpg)
$若\overrightarrow{AE}=\lambda \overrightarrow{AB}+\mu \overrightarrow{AD}(\lambda ,\mu \in R),求\lambda \mu 的值;$

Solution:
$根据题意可得\overrightarrow{AE}=\overrightarrow{AB}+\overrightarrow{BM}+\overrightarrow{ME}=\overrightarrow{AB}+\frac{1}{3}\overrightarrow{BC}+\frac{2}{3}\overrightarrow{MN}=\overrightarrow{AB}+\frac{1}{3}\overrightarrow{AD}+\frac{2}{3}(\overrightarrow{MC}+\overrightarrow{CN})=\overrightarrow{AB}+\frac{1}{3}\overrightarrow{AD}+\frac{2}{3}\times \left(\frac{2}{3}\overrightarrow{AD}-\frac{1}{2}\overrightarrow{AB}\right)=\frac{2}{3}\overrightarrow{AB}+\frac{7}{9}\overrightarrow{AD}.又\overrightarrow{AE}=\lambda \overrightarrow{AB}+\mu \overrightarrow{AD},所以\lambda =\frac{2}{3},\mu =\frac{7}{9},所以\lambda \mu =\frac{14}{27}.$

Final answer: $\frac{14}{27}$
Subfield: Plane Geometry


Question: $如图,在平行四边形ABCD中,\overrightarrow{BM}=\frac{1}{2}\overrightarrow{MC},N为线段CD的中点,E为线段MN上的点，且\overrightarrow{ME}=2\overrightarrow{EN}.$![Image](https://cdntk.53zaixian.com/tq/1664565246718836739/img/compression/M_f7a258377dff41875de81f6ac9c5c522.jpg)
$延长MN,AD，二者交于点P,点F在线段 NP上(包含端点),若\overrightarrow{AF}=t\overrightarrow{AM}+(1-t)\overrightarrow{AN},求实数t的取值范围.$

Solution:
$如图所示,连接AM,AN.由题意可得\overrightarrow{NP}=\overrightarrow{MN}.$![Image](https://cdntk.53zaixian.com/tq/1664565247276679169/img/compression/M_d554897a68ca309c59cd81ff0ec116ed.jpg)$因为点F在线段NP上(包含端点),所以可设\overrightarrow{NF}=k\overrightarrow{NP}=k\overrightarrow{MN}(0\leq k\leq 1),所以\overrightarrow{AF}=\overrightarrow{AM}+\overrightarrow{MF}=\overrightarrow{AM}+(1+k)\overrightarrow{MN}=\overrightarrow{AM}+(1+k)(\overrightarrow{AN}-\overrightarrow{AM})=-k\overrightarrow{AM}+(1+k)\overrightarrow{AN}.又\overrightarrow{AF}=t\overrightarrow{AM}+(1-t)\overrightarrow{AN},所以t=-k\in [-1,0].故实数t的取值范围是[-1,0].$

Final answer: $[-1,0]$
Subfield: Plane Geometry


Question: $如图,在平行四边形ABCD中,\overrightarrow{BM}=\frac{1}{2}\overrightarrow{MC},N为线段CD的中点,E为线段MN上的点，且\overrightarrow{ME}=2\overrightarrow{EN}.$<img_2834>
$延长MN,AD，二者交于点P,点F在线段 NP上(包含端点),若\overrightarrow{AF}=t\overrightarrow{AM}+(1-t)\overrightarrow{AN},求实数t的取值范围.$

Solution:
$如图所示,连接AM,AN.由题意可得\overrightarrow{NP}=\overrightarrow{MN}.$<img_3006>$因为点F在线段NP上(包含端点),所以可设\overrightarrow{NF}=k\overrightarrow{NP}=k\overrightarrow{MN}(0\leq k\leq 1),所以\overrightarrow{AF}=\overrightarrow{AM}+\overrightarrow{MF}=\overrightarrow{AM}+(1+k)\overrightarrow{MN}=\overrightarrow{AM}+(1+k)(\overrightarrow{AN}-\overrightarrow{AM})=-k\overrightarrow{AM}+(1+k)\overrightarrow{AN}.又\overrightarrow{AF}=t\overrightarrow{AM}+(1-t)\overrightarrow{AN},所以t=-k\in [-1,0].故实数t的取值范围是[-1,0].$

Final answer: $[-1,0]$
Subfield: Plane Geometry


Question: $如图所示，在\triangle ABC中，\overrightarrow{AB}=a，\overrightarrow{AC}=b，D为AB的中点，E为CD上一点，且DC=4EC，AE的延长线与BC交于点F.$![Image](https://cdntk.53zaixian.com/tq/1668099013714575361/img/compression/M_da248de3e6363785b94c7955712fc847.jpg)
$请用向量a，b表示\overrightarrow{AE}；$

Solution:
$因为\overrightarrow{DC}=4\overrightarrow{EC},所以\overrightarrow{AC}-\overrightarrow{AD}=4(\overrightarrow{AC}-\overrightarrow{AE}),所以\overrightarrow{AE}=\frac{3}{4}\overrightarrow{AC}+\frac{1}{4}\overrightarrow{AD}.(3分)因为D为AB的中点,所以\overrightarrow{AD}=\frac{1}{2}\overrightarrow{AB}=\frac{1}{2}a,所以\overrightarrow{AE}=\frac{1}{8}a+\frac{3}{4}b.(6分)$

Final answer: $\overrightarrow{AE}=\frac{1}{8}a+\frac{3}{4}b$
Subfield: Plane Geometry


Question: $如图所示，在\triangle ABC中，\overrightarrow{AB}=a，\overrightarrow{AC}=b，D为AB的中点，E为CD上一点，且DC=4EC，AE的延长线与BC交于点F.$<img_1652>
$请用向量a，b表示\overrightarrow{AE}；$

Solution:
$因为\overrightarrow{DC}=4\overrightarrow{EC},所以\overrightarrow{AC}-\overrightarrow{AD}=4(\overrightarrow{AC}-\overrightarrow{AE}),所以\overrightarrow{AE}=\frac{3}{4}\overrightarrow{AC}+\frac{1}{4}\overrightarrow{AD}.(3分)因为D为AB的中点,所以\overrightarrow{AD}=\frac{1}{2}\overrightarrow{AB}=\frac{1}{2}a,所以\overrightarrow{AE}=\frac{1}{8}a+\frac{3}{4}b.(6分)$

Final answer: $\overrightarrow{AE}=\frac{1}{8}a+\frac{3}{4}b$
Subfield: Plane Geometry


Question: $如图所示，在\triangle ABC中，\overrightarrow{AB}=a，\overrightarrow{AC}=b，D为AB的中点，E为CD上一点，且DC=4EC，AE的延长线与BC交于点F.$![Image](https://cdntk.53zaixian.com/tq/1668099013714575361/img/compression/M_da248de3e6363785b94c7955712fc847.jpg)
$请用向量a，b表示\overrightarrow{AF}，并求出\frac{AE}{EF}和\frac{BF}{FC}的值.$

Solution:
$因为B,F,C三点共线,所以可设\overrightarrow{BF}=t\overrightarrow{BC},t\in R,所以\overrightarrow{AF}-\overrightarrow{AB}=t(\overrightarrow{AC}-\overrightarrow{AB}),即\overrightarrow{AF}=(1-t)\overrightarrow{AB}+t\overrightarrow{AC},即\overrightarrow{AF}=tb+(1-t)a,(8分)因为A,F,E三点共线,所以存在实数\lambda ,使得\overrightarrow{AF}=\lambda \overrightarrow{AE}=\frac{\lambda }{8}a+\frac{3\lambda }{4}b,因为a,b不共线,所以\left\{\begin{matrix}\frac{\lambda }{8}=1-t,\\ \frac{3\lambda }{4}=t,\end{matrix}\right.所以\left\{\begin{matrix}\lambda =\frac{8}{7},\\ t=\frac{6}{7},\end{matrix}\right.所以\overrightarrow{AF}=\frac{8}{7}\overrightarrow{AE},\overrightarrow{BF}=\frac{6}{7}\overrightarrow{BC},(10分)所以\overrightarrow{AF}=\frac{1}{7}a+\frac{6}{7}b,\frac{AE}{EF}的值为7,\frac{BF}{FC}的值为6.(12分)$

Final answer: 7, $6$
Subfield: Plane Geometry


Question: $如图所示，在\triangle ABC中，\overrightarrow{AB}=a，\overrightarrow{AC}=b，D为AB的中点，E为CD上一点，且DC=4EC，AE的延长线与BC交于点F.$<img_1652>
$请用向量a，b表示\overrightarrow{AF}，并求出\frac{AE}{EF}和\frac{BF}{FC}的值.$

Solution:
$因为B,F,C三点共线,所以可设\overrightarrow{BF}=t\overrightarrow{BC},t\in R,所以\overrightarrow{AF}-\overrightarrow{AB}=t(\overrightarrow{AC}-\overrightarrow{AB}),即\overrightarrow{AF}=(1-t)\overrightarrow{AB}+t\overrightarrow{AC},即\overrightarrow{AF}=tb+(1-t)a,(8分)因为A,F,E三点共线,所以存在实数\lambda ,使得\overrightarrow{AF}=\lambda \overrightarrow{AE}=\frac{\lambda }{8}a+\frac{3\lambda }{4}b,因为a,b不共线,所以\left\{\begin{matrix}\frac{\lambda }{8}=1-t,\\ \frac{3\lambda }{4}=t,\end{matrix}\right.所以\left\{\begin{matrix}\lambda =\frac{8}{7},\\ t=\frac{6}{7},\end{matrix}\right.所以\overrightarrow{AF}=\frac{8}{7}\overrightarrow{AE},\overrightarrow{BF}=\frac{6}{7}\overrightarrow{BC},(10分)所以\overrightarrow{AF}=\frac{1}{7}a+\frac{6}{7}b,\frac{AE}{EF}的值为7,\frac{BF}{FC}的值为6.(12分)$

Final answer: 7, $6$
Subfield: Plane Geometry


Question: $如图，已知河水自西向东流，流速为|v_{0}|=1 m/s，设某人在静水中游泳的速度为v_{1}，在水中的实际速度为v_{2}.$<img_2683>
$若此人朝正南方向游去，且|v_{1}|=\sqrt{3} m/s，求他实际前进方向与水流方向的夹角\alpha 和v_{2}的大小；$

Solution:
$设\overrightarrow{OA}=v_{0},\overrightarrow{OB}=v_{1},\overrightarrow{OC}=v_{2}，由题意可知v_{2}=v_{0}+v_{1}，|\overrightarrow{OA}|=1，易得四边形OACB为平行四边形.
(1分)
当此人朝正南方向游去时，四边形OACB为矩形，且|\overrightarrow{OB}|=AC=\sqrt{3}，如图①所示，
$<img_2629>$
图①
则在Rt\triangle OAC中，|v_{2}|=OC=\sqrt{OA^2+AC^2}=2，
(3分)
所以\\tan   \alpha =\\tan  \angle AOC=\frac{\sqrt{3}}{1}=\sqrt{3}，又0^\circ <\alpha <90^\circ ,所以\alpha =60^\circ .
(5分)
故此人实际前进方向与水流方向的夹角\alpha 为60^\circ ，v_{2}的大小为2 m/s.
(6分)$

Final answer: $60^\circ$, $2$ ,  m/s
Subfield: Plane Geometry


Question: $如图，已知河水自西向东流，流速为|v_{0}|=1 m/s，设某人在静水中游泳的速度为v_{1}，在水中的实际速度为v_{2}.$![Image](https://cdntk.53zaixian.com/tq/1668099014045925378/img/compression/M_f85b340b5f03a15d7a2d1a411bd612ea.jpg)
$若此人朝正南方向游去，且|v_{1}|=\sqrt{3} m/s，求他实际前进方向与水流方向的夹角\alpha 和v_{2}的大小；$

Solution:
$设\overrightarrow{OA}=v_{0},\overrightarrow{OB}=v_{1},\overrightarrow{OC}=v_{2}，由题意可知v_{2}=v_{0}+v_{1}，|\overrightarrow{OA}|=1，易得四边形OACB为平行四边形.
(1分)
当此人朝正南方向游去时，四边形OACB为矩形，且|\overrightarrow{OB}|=AC=\sqrt{3}，如图①所示，
$![Image](https://cdntk.53zaixian.com/tq/1668099014045925378/img/compression/M_54d38970bc6141775aaf7066c4d20e25.jpg)$
图①
则在Rt\triangle OAC中，|v_{2}|=OC=\sqrt{OA^2+AC^2}=2，
(3分)
所以\\tan   \alpha =\\tan  \angle AOC=\frac{\sqrt{3}}{1}=\sqrt{3}，又0^\circ <\alpha <90^\circ ,所以\alpha =60^\circ .
(5分)
故此人实际前进方向与水流方向的夹角\alpha 为60^\circ ，v_{2}的大小为2 m/s.
(6分)$

Final answer: $60^\circ$, $2$ ,  m/s
Subfield: Plane Geometry


Question: $如图，已知河水自西向东流，流速为|v_{0}|=1 m/s，设某人在静水中游泳的速度为v_{1}，在水中的实际速度为v_{2}.$![Image](https://cdntk.53zaixian.com/tq/1668099014045925378/img/compression/M_f85b340b5f03a15d7a2d1a411bd612ea.jpg)
$若此人实际的前进方向与水流的方向垂直，且|v_{2}|=\sqrt{3} m/s，求他游泳的方向与水流方向的夹角\beta 和v_{1}的大小.$

Solution:
$由题意知\angle OCB=90^\circ ，且|v_{2}|=|\overrightarrow{OC}|=\sqrt{3}，BC=|\overrightarrow{OA}|=1，如图②所示，
(7分)
$![Image](https://cdntk.53zaixian.com/tq/1668099014045925378/img/compression/M_c40ab71fd705e5aca3c454ace7ededd6.jpg)$
图②
则在Rt\triangle OCB中，|v_{1}|=OB=\sqrt{OC^2+BC^2}=2，
(9分)
\\tan  \angle BOC=\frac{1}{\sqrt{3}}=\frac{\sqrt{3}}{3}，又0^\circ <\angle BOC<90^\circ ，所以\angle BOC=30^\circ ，则\beta =90^\circ +30^\circ =120^\circ .
(11分)
故此人游泳的方向与水流方向的夹角\beta 为120^\circ ，v_{1}的大小为2 m/s.
(12分)$

Final answer: $120^\circ$ , $2$ ， m/s
Subfield: Plane Geometry


Question: $如图，已知河水自西向东流，流速为|v_{0}|=1 m/s，设某人在静水中游泳的速度为v_{1}，在水中的实际速度为v_{2}.$<img_2683>
$若此人实际的前进方向与水流的方向垂直，且|v_{2}|=\sqrt{3} m/s，求他游泳的方向与水流方向的夹角\beta 和v_{1}的大小.$

Solution:
$由题意知\angle OCB=90^\circ ，且|v_{2}|=|\overrightarrow{OC}|=\sqrt{3}，BC=|\overrightarrow{OA}|=1，如图②所示，
(7分)
$<img_3112>$
图②
则在Rt\triangle OCB中，|v_{1}|=OB=\sqrt{OC^2+BC^2}=2，
(9分)
\\tan  \angle BOC=\frac{1}{\sqrt{3}}=\frac{\sqrt{3}}{3}，又0^\circ <\angle BOC<90^\circ ，所以\angle BOC=30^\circ ，则\beta =90^\circ +30^\circ =120^\circ .
(11分)
故此人游泳的方向与水流方向的夹角\beta 为120^\circ ，v_{1}的大小为2 m/s.
(12分)$

Final answer: $120^\circ$ , $2$ , m/s
Subfield: Plane Geometry


Question: $如图，点D，E，F分别在\triangle ABC的边AB，BC，AC上做匀速运动，当t=0时，点D，E，F分别从点A，B，C出发，以各自的定速度向点B，C，A前进，当t=1时，点D，E，F分别到达点B，C，A.$![Image](https://cdntk.53zaixian.com/tq/1668099014800900098/img/compression/M_8bc370ff7012eeffdd9eee21bcfc7893.jpg)
$证明：在运动过程中(t\in [0,1])，\triangle DEF的重心保持不变；$

Solution:
$证明：设A(x_{A},y_{A})，B(x_{B},y_{B}),C(x_{C},y_{C}),\triangle DEF的重心为O(x_{0},y_{0})，由题意可得，在某一时刻t，\overrightarrow{AD}=\frac{t}{1-t}\overrightarrow{DB},\overrightarrow{BE}=\frac{t}{1-t}\overrightarrow{EC},\overrightarrow{CF}=\frac{t}{1-t}\overrightarrow{FA},(2分)则D(tx_{B}+(1-t)x_{A}，ty_{B}+(1-t)y_{A})，E(tx_{C}+(1-t)x_{B}，ty_{C}+(1-t)y_{B}),F(tx_{A}+(1-t)x_{C},ty_{A}+(1-t)y_{C}).(4分)由三角形的重心坐标公式，得x_{0}=\frac{x_D+x_E+x_F}{3},y_{0}=\frac{y_D+y_E+y_F}{3}，把点D，E，F的坐标代入x_{0}，y_{0}中，求得\triangle DEF的重心坐标为\left(\frac{x_A+x_B+x_C}{3},\frac{y_A+y_B+y_C}{3}\right)，与t无关，即在运动过程中(t\in [0,1])，\triangle DEF的重心保持不变.(6分)$

Final answer: 证明题
Subfield: Plane Geometry


Question: $如图，点D，E，F分别在\triangle ABC的边AB，BC，AC上做匀速运动，当t=0时，点D，E，F分别从点A，B，C出发，以各自的定速度向点B，C，A前进，当t=1时，点D，E，F分别到达点B，C，A.$<img_1960>
$证明：在运动过程中(t\in [0,1])，\triangle DEF的重心保持不变；$

Solution:
$证明：设A(x_{A},y_{A})，B(x_{B},y_{B}),C(x_{C},y_{C}),\triangle DEF的重心为O(x_{0},y_{0})，由题意可得，在某一时刻t，\overrightarrow{AD}=\frac{t}{1-t}\overrightarrow{DB},\overrightarrow{BE}=\frac{t}{1-t}\overrightarrow{EC},\overrightarrow{CF}=\frac{t}{1-t}\overrightarrow{FA},(2分)则D(tx_{B}+(1-t)x_{A}，ty_{B}+(1-t)y_{A})，E(tx_{C}+(1-t)x_{B}，ty_{C}+(1-t)y_{B}),F(tx_{A}+(1-t)x_{C},ty_{A}+(1-t)y_{C}).(4分)由三角形的重心坐标公式，得x_{0}=\frac{x_D+x_E+x_F}{3},y_{0}=\frac{y_D+y_E+y_F}{3}，把点D，E，F的坐标代入x_{0}，y_{0}中，求得\triangle DEF的重心坐标为\left(\frac{x_A+x_B+x_C}{3},\frac{y_A+y_B+y_C}{3}\right)，与t无关，即在运动过程中(t\in [0,1])，\triangle DEF的重心保持不变.(6分)$
Subfield: Plane Geometry


Question: $如图，点D，E，F分别在\triangle ABC的边AB，BC，AC上做匀速运动，当t=0时，点D，E，F分别从点A，B，C出发，以各自的定速度向点B，C，A前进，当t=1时，点D，E，F分别到达点B，C，A.$![Image](https://cdntk.53zaixian.com/tq/1668099014800900098/img/compression/M_8bc370ff7012eeffdd9eee21bcfc7893.jpg)
$若\triangle ABC的面积为S，求\triangle DEF的面积的最小值(t\in [0,1]).$

Solution:
$\because \frac{AD}{AB}=t,\frac{AF}{AC}=1-t,\therefore S_{\triangle DFA}:S=(AD\cdot AF):(AB\cdot AC)=t(1-t),即S_{\triangle DFA}=t(1-t)S，(8分)同理，S_{\triangle EFC}=S_{\triangle DEB}=t(1-t)S，\therefore S_{\triangle DEF}=S-(S_{\triangle DFA}+S_{\triangle DEB}+S_{\triangle EFC})=(3t^{2}-3t+1)S=\left[3\left(t-\frac{1}{2}\right)^2+\frac{1}{4}\right]S,t\in [0,1],(10分)\therefore 当t=\frac{1}{2}时，\triangle DEF的面积取得最小值,为\frac{1}{4}S.(12分)$

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


Question: $如图，点D，E，F分别在\triangle ABC的边AB，BC，AC上做匀速运动，当t=0时，点D，E，F分别从点A，B，C出发，以各自的定速度向点B，C，A前进，当t=1时，点D，E，F分别到达点B，C，A.$<img_1960>
$若\triangle ABC的面积为S，求\triangle DEF的面积的最小值(t\in [0,1]).$

Solution:
$\because \frac{AD}{AB}=t,\frac{AF}{AC}=1-t,\therefore S_{\triangle DFA}:S=(AD\cdot AF):(AB\cdot AC)=t(1-t),即S_{\triangle DFA}=t(1-t)S，(8分)同理，S_{\triangle EFC}=S_{\triangle DEB}=t(1-t)S，\therefore S_{\triangle DEF}=S-(S_{\triangle DFA}+S_{\triangle DEB}+S_{\triangle EFC})=(3t^{2}-3t+1)S=\left[3\left(t-\frac{1}{2}\right)^2+\frac{1}{4}\right]S,t\in [0,1],(10分)\therefore 当t=\frac{1}{2}时，\triangle DEF的面积取得最小值,为\frac{1}{4}S.(12分)$

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


Question: $如图，某人从A点出发，向西走了200 m后到达B点，然后改变方向，沿北偏西一定角度的某方向行走了200\sqrt{2} m到达C点，最后又改变方向，向东走了200 m到达D点，发现D点在B点的正北方.以A为原点建立平面直角坐标系，图中1个单位长度表示100 m.$![Image](https://cdntk.53zaixian.com/tq/1704016028642189315/img/compression/M_fb1d2885b4d318a74b2b4f3506a91fcf.jpg)
$求向量\overrightarrow{DA}的模.$

Solution:
$如图，作出向量\overrightarrow{DA}，由题意可知，CD\parallel AB且CD=AB，所以四边形ABCD是平行四边形，则|\overrightarrow{DA}|=|\overrightarrow{BC}|=200\sqrt{2} m.$![Image](https://cdntk.53zaixian.com/tq/1704016029170671618/img/compression/M_8dbe35a27201fb151485f2ad37c5a048.jpg)

Final answer: $200\sqrt 2$ m
Subfield: Plane Geometry


Question: $如图，某人从A点出发，向西走了200 m后到达B点，然后改变方向，沿北偏西一定角度的某方向行走了200\sqrt{2} m到达C点，最后又改变方向，向东走了200 m到达D点，发现D点在B点的正北方.以A为原点建立平面直角坐标系，图中1个单位长度表示100 m.$<img_3104>
$求向量\overrightarrow{DA}的模.$

Solution:
$如图，作出向量\overrightarrow{DA}，由题意可知，CD\parallel AB且CD=AB，所以四边形ABCD是平行四边形，则|\overrightarrow{DA}|=|\overrightarrow{BC}|=200\sqrt{2} m.$<img_1549>

Final answer: $200\sqrt 2$ m
Subfield: Plane Geometry


Question: $如图，\triangle ABC和\triangle A'B'C'是在各边的三等分点处相交的两个全等的正三角形，设\triangle ABC的边长为a，分别写出图中给出的长度为\frac{a}{3}的所有向量中，满足下列条件的向量.$![Image](https://cdntk.53zaixian.com/tq/1704016030575763459/img/compression/M_e1cfbe8d642cb0b620c704a921d825d9.jpg)
$与向量\overrightarrow{GH}相等的向量；$

Solution:
$与向量\overrightarrow{GH}相等的向量，即与向量\overrightarrow{GH}大小相等，方向相同的向量，有\overrightarrow{HC}，\overrightarrow{LB'}.$

Final answer: $\overrightarrow{HC}, \overrightarrow{LB'}$
Subfield: Plane Geometry


Question: $如图，\triangle ABC和\triangle A'B'C'是在各边的三等分点处相交的两个全等的正三角形，设\triangle ABC的边长为a，分别写出图中给出的长度为\frac{a}{3}的所有向量中，与向量\overrightarrow{GH}相等的向量.$<img_2141>

Solution:
$与向量\overrightarrow{GH}相等的向量，即与向量\overrightarrow{GH}大小相等，方向相同的向量，有\overrightarrow{HC}，\overrightarrow{LB'}.$



Final answer: $\overrightarrow{HC}, \overrightarrow{LB'}$
Subfield: Plane Geometry


Question: $如图，\triangle ABC和\triangle A'B'C'是在各边的三等分点处相交的两个全等的正三角形，设\triangle ABC的边长为a，分别写出图中给出的长度为\frac{a}{3}的所有向量中，满足下列条件的向量.$![Image](https://cdntk.53zaixian.com/tq/1704016030575763459/img/compression/M_e1cfbe8d642cb0b620c704a921d825d9.jpg)
$与向量\overrightarrow{GH}共线的向量；$

Solution:
$与向量\overrightarrow{GH}共线的向量，即与向量\overrightarrow{GH}方向相同或相反的向量，有\overrightarrow{HC}，\overrightarrow{LB'}，\overrightarrow{GB}，\overrightarrow{LE}，\overrightarrow{EC'}.$

Final answer: $\overrightarrow{HC}$, $\overrightarrow{LB'}$, $\overrightarrow{GB}$, $\overrightarrow{LE}$, $\overrightarrow{EC'}$.
Subfield: Plane Geometry


Question: $如图，\triangle ABC和\triangle A'B'C'是在各边的三等分点处相交的两个全等的正三角形，设\triangle ABC的边长为a，分别写出图中给出的长度为\frac{a}{3}的所有向量中，与向量\overrightarrow{GH}共线的向量.$<img_2141>

Solution:
$与向量\overrightarrow{GH}共线的向量，即与向量\overrightarrow{GH}方向相同或相反的向量，有\overrightarrow{HC}，\overrightarrow{LB'}，\overrightarrow{GB}，\overrightarrow{LE}，\overrightarrow{EC'}.$



Final answer: $\overrightarrow{HC}$, $\overrightarrow{LB'}$, $\overrightarrow{GB}$, $\overrightarrow{LE}$, $\overrightarrow{EC'}$.
Subfield: Plane Geometry


Question: $如图，\triangle ABC和\triangle A'B'C'是在各边的三等分点处相交的两个全等的正三角形，设\triangle ABC的边长为a，分别写出图中给出的长度为\frac{a}{3}的所有向量中，向量\overrightarrow{EA}的相反向量.$<img_2141>
$.$

Solution:
$向量\overrightarrow{EA}的相反向量，即与\overrightarrow{EA}大小相等，且方向相反的向量，有\overrightarrow{EF}，\overrightarrow{FB}，\overrightarrow{HA'}.$



Final answer: $\overrightarrow{EF}, \overrightarrow{FB}, \overrightarrow{HA'}$
Subfield: Plane Geometry


Question: $如图，\triangle ABC和\triangle A'B'C'是在各边的三等分点处相交的两个全等的正三角形，设\triangle ABC的边长为a，分别写出图中给出的长度为\frac{a}{3}的所有向量中，满足下列条件的向量.$![Image](https://cdntk.53zaixian.com/tq/1704016030575763459/img/compression/M_e1cfbe8d642cb0b620c704a921d825d9.jpg)
$向量\overrightarrow{EA}的相反向量.$

Solution:
$向量\overrightarrow{EA}的相反向量，即与\overrightarrow{EA}大小相等，且方向相反的向量，有\overrightarrow{EF}，\overrightarrow{FB}，\overrightarrow{HA'}.$

Final answer: $\overrightarrow{EF}, \overrightarrow{FB}, \overrightarrow{HA'}$
Subfield: Plane Geometry


Question: $如图，已知以O为圆心，1为半径的圆上有8个等分点A，B，C，D，E，F，G，H，以图中标出的9个点为起点或终点作向量.$![Image](https://cdntk.53zaixian.com/tq/1704016031234269185/img/compression/M_2d462b6754c8434e02afbc3c4a9348f9.jpg)
$向量\overrightarrow{OD}与\overrightarrow{OE}的夹角是多少？$

Solution:
$因为以O为圆心，1为半径的圆上的8个等分点分别为A，B，C，D，E，F，G，H，所以弧DE所对的圆心角是45^\circ ，即\angle DOE=45^\circ ，所以\overrightarrow{OD}与\overrightarrow{OE}的夹角为45^\circ .$

Final answer: $45^\circ$
Subfield: Plane Geometry


Question: $如图，已知以O为圆心，1为半径的圆上有8个等分点A，B，C，D，E，F，G，H，以图中标出的9个点为起点或终点作向量.$<img_1863>
$向量\overrightarrow{OD}与\overrightarrow{OE}的夹角是多少？$

Solution:
$因为以O为圆心，1为半径的圆上的8个等分点分别为A，B，C，D，E，F，G，H，所以弧DE所对的圆心角是45^\circ ，即\angle DOE=45^\circ ，所以\overrightarrow{OD}与\overrightarrow{OE}的夹角为45^\circ .$

Final answer: $45^\circ$
Subfield: Plane Geometry


Question: $如图，已知以O为圆心，1为半径的圆上有8个等分点A，B，C，D，E，F，G，H，以图中标出的9个点为起点或终点作向量.$![Image](https://cdntk.53zaixian.com/tq/1704016031234269185/img/compression/M_2d462b6754c8434e02afbc3c4a9348f9.jpg)
$与向量\overrightarrow{OD}垂直的向量有哪些？$

Solution:
$如图，BF是圆O的直径，OD\perp BF，CE\parallel BF\parallel AG，$![Image](https://cdntk.53zaixian.com/tq/1704016031527870466/img/compression/M_90bddbed0a733f07caca6d9a8e0644eb.jpg)$所以与向量\overrightarrow{OD}垂直的向量有\overrightarrow{OB},\overrightarrow{BO},\overrightarrow{OF},\overrightarrow{FO},\overrightarrow{BF},\overrightarrow{FB},\overrightarrow{CE},\overrightarrow{EC},\overrightarrow{AG},\overrightarrow{GA}.$

Final answer: $\overrightarrow{OB},\overrightarrow{BO},\overrightarrow{OF},\overrightarrow{FO},\overrightarrow{BF},\overrightarrow{FB},\overrightarrow{CE},\overrightarrow{EC},\overrightarrow{AG},\overrightarrow{GA}$
Subfield: Plane Geometry


Question: $如图，已知以O为圆心，1为半径的圆上有8个等分点A，B，C，D，E，F，G，H，以图中标出的9个点为起点或终点作向量.$<img_1863>
$与向量\overrightarrow{OD}垂直的向量有哪些？$

Solution:
$如图，BF是圆O的直径，OD\perp BF，CE\parallel BF\parallel AG，$<img_3118>$所以与向量\overrightarrow{OD}垂直的向量有\overrightarrow{OB},\overrightarrow{BO},\overrightarrow{OF},\overrightarrow{FO},\overrightarrow{BF},\overrightarrow{FB},\overrightarrow{CE},\overrightarrow{EC},\overrightarrow{AG},\overrightarrow{GA}.$



Final answer: $\overrightarrow{OB},\overrightarrow{BO},\overrightarrow{OF},\overrightarrow{FO},\overrightarrow{BF},\overrightarrow{FB},\overrightarrow{CE},\overrightarrow{EC},\overrightarrow{AG},\overrightarrow{GA}$