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


Question: $如图，已知\triangle ABC是边长为2的正三角形，点P_{1}，P_{2}，P_{3}是BC边的四等分点.$![Image](https://cdntk.53zaixian.com/tq/1706233178586976257/img/compression/M_43a2bc1b0a974064e60864ff8b19174a.jpg)
$若P为线段AP_{3}上的动点，求\overrightarrow{PA}\cdot \overrightarrow{PC}的最小值，并指出当\overrightarrow{PA}\cdot \overrightarrow{PC}取得最小值时点P的位置.$

Solution:
$记\overrightarrow{AB}=a，\overrightarrow{AC}=b，由\overrightarrow{BP_3}=\frac{3}{4}\overrightarrow{BC}得\overrightarrow{AP_3}=\overrightarrow{AB}+\overrightarrow{BP_3}=\overrightarrow{AB}+\frac{3}{4}\overrightarrow{BC}=\overrightarrow{AB}+\frac{3}{4}(\overrightarrow{AC}-\overrightarrow{AB})=\frac{1}{4}\overrightarrow{AB}+\frac{3}{4}\overrightarrow{AC}=\frac{1}{4}a+\frac{3}{4}b.设\overrightarrow{AP}=t\overrightarrow{AP_3}，0\leq t\leq 1，则\overrightarrow{PA}=-\frac{t}{4}a-\frac{3t}{4}b，\overrightarrow{PC}=\overrightarrow{PA}+\overrightarrow{AC}=-\frac{t}{4}a+\left(1-\frac{3t}{4}\right)b,则\overrightarrow{PA}\cdot \overrightarrow{PC}=\left(-\frac{t}{4}\boldsymbol{a}-\frac{3t}{4}\boldsymbol{b}\right)\cdot \left[-\frac{t}{4}\boldsymbol{a}+\left(1-\frac{3t}{4}\right)\boldsymbol{b}\right]=\frac{t}{16}[ta^{2}+(6t-4)a\cdot b+3(3t-4)b^{2}]=\frac{1}{4}(13t^{2}-14t)=\frac{13}{4}\left(t-\frac{7}{13}\right)^2-\frac{49}{52}，t\in [0,1].所以当t=\frac{7}{13}，即AP=\frac{7}{13}AP_{3}时，\overrightarrow{PA}\cdot \overrightarrow{PC}取得最小值，为-\frac{49}{52}.$

Final answer: $\frac{-49}{52}$
Subfield: Plane Geometry


Question: $已知\triangle ABO中，延长BA到C，使AC=BA,D是OB上靠近点B的一个三等分点，DC与OA交于点E，设\overrightarrow{OA}=a,\overrightarrow{OB}=b.$<img_3133>
$用a,b表示向量\overrightarrow{OC},\overrightarrow{DC}；$

Solution:
$由题意知A是BC的中点，则\overrightarrow{OB}+\overrightarrow{OC}=2\overrightarrow{OA}，所以\overrightarrow{OC}=2\overrightarrow{OA}-\overrightarrow{OB}=2a-b，易知\overrightarrow{OD}=\frac{2}{3}\overrightarrow{OB},则\overrightarrow{DC}=\overrightarrow{OC}-\overrightarrow{OD}=\overrightarrow{OC}-\frac{2}{3}\overrightarrow{OB}=2a-b-\frac{2}{3}b=2a-\frac{5}{3}b.$

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


Question: $已知\triangle ABO中，延长BA到C，使AC=BA,D是OB上靠近点B的一个三等分点，DC与OA交于点E，设\overrightarrow{OA}=a,\overrightarrow{OB}=b.$![Image](https://cdntk.53zaixian.com/tq/1704016048301527041/img/compression/M_e494fb23d6abb3ccd1f093f38bf4efba.jpg)
$用a,b表示向量\overrightarrow{OC},\overrightarrow{DC}；$

Solution:
$由题意知A是BC的中点，则\overrightarrow{OB}+\overrightarrow{OC}=2\overrightarrow{OA}，所以\overrightarrow{OC}=2\overrightarrow{OA}-\overrightarrow{OB}=2a-b，易知\overrightarrow{OD}=\frac{2}{3}\overrightarrow{OB},则\overrightarrow{DC}=\overrightarrow{OC}-\overrightarrow{OD}=\overrightarrow{OC}-\frac{2}{3}\overrightarrow{OB}=2a-b-\frac{2}{3}b=2a-\frac{5}{3}b.$

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


Question: $已知\triangle ABO中，延长BA到C，使AC=BA,D是OB上靠近点B的一个三等分点，DC与OA交于点E，设\overrightarrow{OA}=a,\overrightarrow{OB}=b.$<img_3133>
$若\overrightarrow{OE}=\lambda \overrightarrow{OA}，求实数\lambda 的值.$

Solution:
$因为\overrightarrow{EC}=\overrightarrow{OC}-\overrightarrow{OE}=2a-b-\lambda a=(2-\lambda )a-b，\overrightarrow{DC}=2a-\frac{5}{3}b,且\overrightarrow{EC}\parallel \overrightarrow{DC}，所以\frac{2-\lambda }{2}=\frac{-1}{-\frac{5}{3}}=\frac{3}{5}，解得\lambda =\frac{4}{5}.$

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


Question: $已知\triangle ABO中，延长BA到C，使AC=BA,D是OB上靠近点B的一个三等分点，DC与OA交于点E，设\overrightarrow{OA}=a,\overrightarrow{OB}=b.$![Image](https://cdntk.53zaixian.com/tq/1704016048301527041/img/compression/M_e494fb23d6abb3ccd1f093f38bf4efba.jpg)
$若\overrightarrow{OE}=\lambda \overrightarrow{OA}，求实数\lambda 的值.$

Solution:
$因为\overrightarrow{EC}=\overrightarrow{OC}-\overrightarrow{OE}=2a-b-\lambda a=(2-\lambda )a-b，\overrightarrow{DC}=2a-\frac{5}{3}b,且\overrightarrow{EC}\parallel \overrightarrow{DC}，所以\frac{2-\lambda }{2}=\frac{-1}{-\frac{5}{3}}=\frac{3}{5}，解得\lambda =\frac{4}{5}.$

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


Question: $如图，已知正方形ABCD的边长为2，过中心O的直线与两边AB，CD分别交于点M，N.$![Image](https://cdntk.53zaixian.com/tq/1706233211844243458/img/compression/M_9d95b2f9a7b913e892f75db825a3bd5a.jpg)
$若Q是BC的中点，求\overrightarrow{QM}\cdot \overrightarrow{QN}的取值范围；$

Solution:
$连接QO，由题意知O为MN的中点，\therefore \overrightarrow{ON}=-\overrightarrow{OM}，\therefore \overrightarrow{QM}\cdot \overrightarrow{QN}=(\overrightarrow{QO}+\overrightarrow{OM})\cdot (\overrightarrow{QO}+\overrightarrow{ON})=\overrightarrow{QO}^{2}-\overrightarrow{OM}^{2}.\because Q是BC的中点，\therefore |\overrightarrow{QO}|=1，1\leq |\overrightarrow{OM}|\leq \sqrt{2}，\therefore -1\leq \overrightarrow{QO}^{2}-\overrightarrow{OM}^{2}\leq 0，即\overrightarrow{QM}\cdot \overrightarrow{QN}的取值范围为[-1,0].$

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


Question: $如图，已知正方形ABCD的边长为2，过中心O的直线与两边AB，CD分别交于点M，N.$<img_2445>
$若Q是BC的中点，求\overrightarrow{QM}\cdot \overrightarrow{QN}的取值范围；$

Solution:
$连接QO，由题意知O为MN的中点，\therefore \overrightarrow{ON}=-\overrightarrow{OM}，\therefore \overrightarrow{QM}\cdot \overrightarrow{QN}=(\overrightarrow{QO}+\overrightarrow{OM})\cdot (\overrightarrow{QO}+\overrightarrow{ON})=\overrightarrow{QO}^{2}-\overrightarrow{OM}^{2}.\because Q是BC的中点，\therefore |\overrightarrow{QO}|=1，1\leq |\overrightarrow{OM}|\leq \sqrt{2}，\therefore -1\leq \overrightarrow{QO}^{2}-\overrightarrow{OM}^{2}\leq 0，即\overrightarrow{QM}\cdot \overrightarrow{QN}的取值范围为[-1,0].$

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


Question: $如图，已知正方形ABCD的边长为2，过中心O的直线与两边AB，CD分别交于点M，N.$<img_2445>
$若P是正方形ABCD所在平面内一点，且满足2\overrightarrow{OP}=\lambda \overrightarrow{OB}+(1-\lambda )\overrightarrow{OC}，\lambda \in R，求\overrightarrow{PM}\cdot \overrightarrow{PN}的最小值.$

Solution:
$令\overrightarrow{OT}=2\overrightarrow{OP}，则 \overrightarrow{OT}=2\overrightarrow{OP}=\lambda \overrightarrow{OB}+(1-\lambda )\overrightarrow{OC}，\therefore \overrightarrow{OT}=\lambda \overrightarrow{OB}+\overrightarrow{OC}-\lambda \overrightarrow{OC}，即\overrightarrow{OT}-\overrightarrow{OC}=\lambda \overrightarrow{OB}-\lambda \overrightarrow{OC}，\therefore \overrightarrow{CT}=\lambda \overrightarrow{CB}，\therefore 点T 在直线BC上.\because O为MN的中点，\therefore \overrightarrow{PM}\cdot \overrightarrow{PN}=(\overrightarrow{PO}+\overrightarrow{OM})\cdot (\overrightarrow{PO}+\overrightarrow{ON})=\overrightarrow{PO}^{2}-\overrightarrow{OM}^{2}，|\overrightarrow{OT}|\geq 1，从而|\overrightarrow{OP}|\geq \frac{1}{2}，\because 1\leq |\overrightarrow{OM}|\leq \sqrt{2}，\therefore \overrightarrow{PM}\cdot \overrightarrow{PN}=\overrightarrow{PO}^{2}-\overrightarrow{OM}^{2}\geq \frac{1}{4}-2=-\frac{7}{4}，即\overrightarrow{PM}\cdot \overrightarrow{PN}的最小值为-\frac{7}{4}.$

Final answer: $-\frac{7}{4}$
Subfield: Plane Geometry


Question: $如图，已知正方形ABCD的边长为2，过中心O的直线与两边AB，CD分别交于点M，N.$![Image](https://cdntk.53zaixian.com/tq/1706233211844243458/img/compression/M_9d95b2f9a7b913e892f75db825a3bd5a.jpg)
$若P是正方形ABCD所在平面内一点，且满足2\overrightarrow{OP}=\lambda \overrightarrow{OB}+(1-\lambda )\overrightarrow{OC}，\lambda \in R，求\overrightarrow{PM}\cdot \overrightarrow{PN}的最小值.$

Solution:
$令\overrightarrow{OT}=2\overrightarrow{OP}，则 \overrightarrow{OT}=2\overrightarrow{OP}=\lambda \overrightarrow{OB}+(1-\lambda )\overrightarrow{OC}，\therefore \overrightarrow{OT}=\lambda \overrightarrow{OB}+\overrightarrow{OC}-\lambda \overrightarrow{OC}，即\overrightarrow{OT}-\overrightarrow{OC}=\lambda \overrightarrow{OB}-\lambda \overrightarrow{OC}，\therefore \overrightarrow{CT}=\lambda \overrightarrow{CB}，\therefore 点T 在直线BC上.\because O为MN的中点，\therefore \overrightarrow{PM}\cdot \overrightarrow{PN}=(\overrightarrow{PO}+\overrightarrow{OM})\cdot (\overrightarrow{PO}+\overrightarrow{ON})=\overrightarrow{PO}^{2}-\overrightarrow{OM}^{2}，|\overrightarrow{OT}|\geq 1，从而|\overrightarrow{OP}|\geq \frac{1}{2}，\because 1\leq |\overrightarrow{OM}|\leq \sqrt{2}，\therefore \overrightarrow{PM}\cdot \overrightarrow{PN}=\overrightarrow{PO}^{2}-\overrightarrow{OM}^{2}\geq \frac{1}{4}-2=-\frac{7}{4}，即\overrightarrow{PM}\cdot \overrightarrow{PN}的最小值为-\frac{7}{4}.$

Final answer: $-\frac{7}{4}$
Subfield: Plane Geometry


Question: $如图，在平行四边形ABCD中，E是AB的中点，F，G分别是AD，BC的三等分点，且AF=\frac{1}{3}AD,BG=\frac{1}{3}BC.设\overrightarrow{AB}=a，\overrightarrow{AD}=b.$![Image](https://cdntk.53zaixian.com/tq/1706233215644282881/img/compression/M_25fe453c99d135642ea82397f7ad104d.jpg)
$用a，b表示\overrightarrow{EF}，\overrightarrow{EG}；$

Solution:
$\overrightarrow{EF}=\overrightarrow{AF}-\overrightarrow{AE}=\frac{1}{3}\overrightarrow{AD}-\frac{1}{2}\overrightarrow{AB}=\frac{1}{3}b-\frac{1}{2}a，\overrightarrow{EG}=\overrightarrow{EB}+\overrightarrow{BG}=\frac{1}{2}\overrightarrow{AB}+\overrightarrow{AF}=\frac{1}{2}\overrightarrow{AB}+\frac{1}{3}\overrightarrow{AD}=\frac{1}{2}a+\frac{1}{3}b.$

Final answer: $\overrightarrow{EF}=\frac{1}{3}b-\frac{1}{2}a,\overrightarrow{EG}=\frac{1}{2}a+\frac{1}{3}b.$
Subfield: Plane Geometry


Question: $如图，在平行四边形ABCD中，E是AB的中点，F，G分别是AD，BC的三等分点，且AF=\frac{1}{3}AD,BG=\frac{1}{3}BC.设\overrightarrow{AB}=a，\overrightarrow{AD}=b.$<img_2701>
$用a，b表示\overrightarrow{EF}，\overrightarrow{EG}；$

Solution:
$\overrightarrow{EF}=\overrightarrow{AF}-\overrightarrow{AE}=\frac{1}{3}\overrightarrow{AD}-\frac{1}{2}\overrightarrow{AB}=\frac{1}{3}b-\frac{1}{2}a，\overrightarrow{EG}=\overrightarrow{EB}+\overrightarrow{BG}=\frac{1}{2}\overrightarrow{AB}+\overrightarrow{AF}=\frac{1}{2}\overrightarrow{AB}+\frac{1}{3}\overrightarrow{AD}=\frac{1}{2}a+\frac{1}{3}b.$

Final answer: $\overrightarrow{EF}=\frac{1}{3}b-\frac{1}{2}a,\overrightarrow{EG}=\frac{1}{2}a+\frac{1}{3}b.$
Subfield: Plane Geometry


Question: $如图，在平行四边形ABCD中，E是AB的中点，F，G分别是AD，BC的三等分点，且AF=\frac{1}{3}AD,BG=\frac{1}{3}BC.设\overrightarrow{AB}=a，\overrightarrow{AD}=b.$![Image](https://cdntk.53zaixian.com/tq/1706233215644282881/img/compression/M_25fe453c99d135642ea82397f7ad104d.jpg)
$如果|b|=\frac{3}{2}|a|，那么EF，EG有什么位置关系?用向量法证明你的结论.$

Solution:
$EF\perp EG.证明如下：\because \overrightarrow{EF}=\frac{1}{3}b-\frac{1}{2}a，\overrightarrow{EG}=\frac{1}{3}b+\frac{1}{2}a，|b|=\frac{3}{2}|a|，\therefore \overrightarrow{EF}\cdot \overrightarrow{EG}=\left(\frac{1}{3}\boldsymbol{b}-\frac{1}{2}\boldsymbol{a}\right)\cdot \left(\frac{1}{3}\boldsymbol{b}+\frac{1}{2}\boldsymbol{a}\right)=\frac{1}{9}b^{2}-\frac{1}{4}a^{2}=\frac{1}{9}\times \frac{9}{4}a^{2}-\frac{1}{4}a^{2}=0.\therefore \overrightarrow{EF}\perp \overrightarrow{EG}，\therefore EF\perp EG.$

Final answer: 证明题
Subfield: Plane Geometry


Question: $如图，在平行四边形ABCD中，E是AB的中点，F，G分别是AD，BC的三等分点，且AF=\frac{1}{3}AD,BG=\frac{1}{3}BC.设\overrightarrow{AB}=a，\overrightarrow{AD}=b.$<img_2701>
$如果|b|=\frac{3}{2}|a|，那么EF，EG有什么位置关系?用向量法证明你的结论.$

Solution:
$EF\perp EG.证明如下：\because \overrightarrow{EF}=\frac{1}{3}b-\frac{1}{2}a，\overrightarrow{EG}=\frac{1}{3}b+\frac{1}{2}a，|b|=\frac{3}{2}|a|，\therefore \overrightarrow{EF}\cdot \overrightarrow{EG}=\left(\frac{1}{3}\boldsymbol{b}-\frac{1}{2}\boldsymbol{a}\right)\cdot \left(\frac{1}{3}\boldsymbol{b}+\frac{1}{2}\boldsymbol{a}\right)=\frac{1}{9}b^{2}-\frac{1}{4}a^{2}=\frac{1}{9}\times \frac{9}{4}a^{2}-\frac{1}{4}a^{2}=0.\therefore \overrightarrow{EF}\perp \overrightarrow{EG}，\therefore EF\perp EG.$
Subfield: Plane Geometry


Question: $如图，在\triangle ABC中，已知\angle BAC=120^\circ ，AB=2，AC=4，点D在BC上，且BD=2DC，E是AC的中点，AD，BE相交于点O.$<img_1486>
$求线段AD，BE的长；$

Solution:
$易得\overrightarrow{BE}=\overrightarrow{AE}-\overrightarrow{AB}=\frac{1}{2}\overrightarrow{AC}-\overrightarrow{AB}，\therefore |\overrightarrow{BE}|^{2}=\overrightarrow{BE}^{2}=\left(\frac{1}{2}\overrightarrow{AC}-\overrightarrow{AB}\right)^{2}=\frac{1}{4}\overrightarrow{AC}^{2}-\overrightarrow{AC}\cdot \overrightarrow{AB}+\overrightarrow{AB}^{2}=\frac{1}{4}|\overrightarrow{AC}|^{2}-|\overrightarrow{AC}|\cdot |\overrightarrow{AB}|\\cos  \angle BAC+|\overrightarrow{AB}|^{2}=12，\therefore |\overrightarrow{BE}|=2\sqrt{3}，即BE=2\sqrt{3}.易得\overrightarrow{AD}=\overrightarrow{AB}+\overrightarrow{BD}=\overrightarrow{AB}+\frac{2}{3}\overrightarrow{BC}=\overrightarrow{AB}+\frac{2}{3}(\overrightarrow{AC}-\overrightarrow{AB})=\frac{2}{3}\overrightarrow{AC}+\frac{1}{3}\overrightarrow{AB}，\therefore |\overrightarrow{AD}|^{2}=\overrightarrow{AD}^{2}=\left(\frac{2}{3}\overrightarrow{AC}+\frac{1}{3}\overrightarrow{AB}\right)^{2}=\frac{4}{9}\overrightarrow{AC}^{2}+\frac{4}{9}\overrightarrow{AC}\cdot \overrightarrow{AB}+\frac{1}{9}\overrightarrow{AB}^{2}=\frac{4}{9}|\overrightarrow{AC}|^{2}+\frac{4}{9}|\overrightarrow{AC}|\cdot |\overrightarrow{AB}|\\cos  \angle BAC+\frac{1}{9}|\overrightarrow{AB}|^{2}=\frac{52}{9},\therefore |\overrightarrow{AD}|=\frac{2\sqrt{13}}{3},即AD=\frac{2\sqrt{13}}{3}.$

Final answer: $2\sqrt 3$, $\frac{2\sqrt 13}{3}$
Subfield: Plane Geometry


Question: $如图，在\triangle ABC中，已知\angle BAC=120^\circ ，AB=2，AC=4，点D在BC上，且BD=2DC，E是AC的中点，AD，BE相交于点O.$![Image](https://cdntk.53zaixian.com/tq/1706233216252456962/img/compression/M_4fa609f693dc95a029395fd2d1c518d0.jpg)
$求线段AD，BE的长；$

Solution:
$易得\overrightarrow{BE}=\overrightarrow{AE}-\overrightarrow{AB}=\frac{1}{2}\overrightarrow{AC}-\overrightarrow{AB}，\therefore |\overrightarrow{BE}|^{2}=\overrightarrow{BE}^{2}=\left(\frac{1}{2}\overrightarrow{AC}-\overrightarrow{AB}\right)^{2}=\frac{1}{4}\overrightarrow{AC}^{2}-\overrightarrow{AC}\cdot \overrightarrow{AB}+\overrightarrow{AB}^{2}=\frac{1}{4}|\overrightarrow{AC}|^{2}-|\overrightarrow{AC}|\cdot |\overrightarrow{AB}|\\cos  \angle BAC+|\overrightarrow{AB}|^{2}=12，\therefore |\overrightarrow{BE}|=2\sqrt{3}，即BE=2\sqrt{3}.易得\overrightarrow{AD}=\overrightarrow{AB}+\overrightarrow{BD}=\overrightarrow{AB}+\frac{2}{3}\overrightarrow{BC}=\overrightarrow{AB}+\frac{2}{3}(\overrightarrow{AC}-\overrightarrow{AB})=\frac{2}{3}\overrightarrow{AC}+\frac{1}{3}\overrightarrow{AB}，\therefore |\overrightarrow{AD}|^{2}=\overrightarrow{AD}^{2}=\left(\frac{2}{3}\overrightarrow{AC}+\frac{1}{3}\overrightarrow{AB}\right)^{2}=\frac{4}{9}\overrightarrow{AC}^{2}+\frac{4}{9}\overrightarrow{AC}\cdot \overrightarrow{AB}+\frac{1}{9}\overrightarrow{AB}^{2}=\frac{4}{9}|\overrightarrow{AC}|^{2}+\frac{4}{9}|\overrightarrow{AC}|\cdot |\overrightarrow{AB}|\\cos  \angle BAC+\frac{1}{9}|\overrightarrow{AB}|^{2}=\frac{52}{9},\therefore |\overrightarrow{AD}|=\frac{2\sqrt{13}}{3},即AD=\frac{2\sqrt{13}}{3}.$

Final answer: $2\sqrt 3$, $\frac{2\sqrt 13}{3}$
Subfield: Plane Geometry


Question: $如图，在\triangle ABC中，已知\angle BAC=120^\circ ，AB=2，AC=4，点D在BC上，且BD=2DC，E是AC的中点，AD，BE相交于点O.$<img_1486>
$求\angle EOD的余弦值.$

Solution:
$由(1)知\overrightarrow{BE}=\frac{1}{2}\overrightarrow{AC}-\overrightarrow{AB}，\overrightarrow{AD}=\frac{2}{3}\overrightarrow{AC}+\frac{1}{3}\overrightarrow{AB}，\therefore \overrightarrow{AD}\cdot \overrightarrow{BE}=\left(\frac{2}{3}\overrightarrow{AC}+\frac{1}{3}\overrightarrow{AB}\right)\cdot \left(\frac{1}{2}\overrightarrow{AC}-\overrightarrow{AB}\right)=\frac{1}{3}\overrightarrow{AC}^{2}-\frac{1}{2}\overrightarrow{AC}\cdot \overrightarrow{AB}-\frac{1}{3}\overrightarrow{AB}^{2}=\frac{1}{3}\times 4^{2}-\frac{1}{2}\times (-4)-\frac{1}{3}\times 2^{2}=6.易知\angle EOD为\overrightarrow{AD}与\overrightarrow{BE}的夹角，\therefore \\cos  \angle EOD=\frac{\overrightarrow{AD}\cdot \overrightarrow{BE}}{|\overrightarrow{AD}||\overrightarrow{BE}|}=\frac{6}{\frac{2\sqrt{13}}{3}\times 2\sqrt{3}}=\frac{3\sqrt{39}}{26}.$

Final answer: $\frac{3\sqrt{39}}{26}$
Subfield: Plane Geometry


Question: $如图，在\triangle ABC中，已知\angle BAC=120^\circ ，AB=2，AC=4，点D在BC上，且BD=2DC，E是AC的中点，AD，BE相交于点O.$![Image](https://cdntk.53zaixian.com/tq/1706233216252456962/img/compression/M_4fa609f693dc95a029395fd2d1c518d0.jpg)
$求\angle EOD的余弦值.$

Solution:
$由(1)知\overrightarrow{BE}=\frac{1}{2}\overrightarrow{AC}-\overrightarrow{AB}，\overrightarrow{AD}=\frac{2}{3}\overrightarrow{AC}+\frac{1}{3}\overrightarrow{AB}，\therefore \overrightarrow{AD}\cdot \overrightarrow{BE}=\left(\frac{2}{3}\overrightarrow{AC}+\frac{1}{3}\overrightarrow{AB}\right)\cdot \left(\frac{1}{2}\overrightarrow{AC}-\overrightarrow{AB}\right)=\frac{1}{3}\overrightarrow{AC}^{2}-\frac{1}{2}\overrightarrow{AC}\cdot \overrightarrow{AB}-\frac{1}{3}\overrightarrow{AB}^{2}=\frac{1}{3}\times 4^{2}-\frac{1}{2}\times (-4)-\frac{1}{3}\times 2^{2}=6.易知\angle EOD为\overrightarrow{AD}与\overrightarrow{BE}的夹角，\therefore \\cos  \angle EOD=\frac{\overrightarrow{AD}\cdot \overrightarrow{BE}}{|\overrightarrow{AD}||\overrightarrow{BE}|}=\frac{6}{\frac{2\sqrt{13}}{3}\times 2\sqrt{3}}=\frac{3\sqrt{39}}{26}.$

Final answer: $\frac{3\sqrt{39}}{26}$
Subfield: Plane Geometry


Question: $在矩形ABCD中，E是AC的中点，点F在边CD上.$<img_1583>
$若F是边CD上靠近C的三等分点，试用\overrightarrow{AB}，\overrightarrow{AD}表示\overrightarrow{EF}；$

Solution:
$\overrightarrow{EF}=\overrightarrow{EA}+\overrightarrow{AD}+\overrightarrow{DF}=-\frac{1}{2}\overrightarrow{AC}+\overrightarrow{AD}+\frac{2}{3}\overrightarrow{DC}=-\frac{1}{2}(\overrightarrow{AB}+\overrightarrow{AD})+\overrightarrow{AD}+\frac{2}{3}\overrightarrow{AB}=\frac{1}{6}\overrightarrow{AB}+\frac{1}{2}\overrightarrow{AD}.$

Final answer: $\frac{1}{6}\overrightarrow{AB}+\frac{1}{2}\overrightarrow{AD}$
Subfield: Plane Geometry


Question: $在矩形ABCD中，E是AC的中点，点F在边CD上.$![Image](https://cdntk.53zaixian.com/tq/1706233228340187138/img/compression/M_1a6e46e4c30bc444e716934b1deaaaa8.jpg)
$若F是边CD上靠近C的三等分点，试用\overrightarrow{AB}，\overrightarrow{AD}表示\overrightarrow{EF}；$

Solution:
$\overrightarrow{EF}=\overrightarrow{EA}+\overrightarrow{AD}+\overrightarrow{DF}=-\frac{1}{2}\overrightarrow{AC}+\overrightarrow{AD}+\frac{2}{3}\overrightarrow{DC}=-\frac{1}{2}(\overrightarrow{AB}+\overrightarrow{AD})+\overrightarrow{AD}+\frac{2}{3}\overrightarrow{AB}=\frac{1}{6}\overrightarrow{AB}+\frac{1}{2}\overrightarrow{AD}.$

Final answer: $\frac{1}{6}\overrightarrow{AB}+\frac{1}{2}\overrightarrow{AD}$
Subfield: Plane Geometry


Question: $在矩形ABCD中，E是AC的中点，点F在边CD上.$![Image](https://cdntk.53zaixian.com/tq/1706233228340187138/img/compression/M_1a6e46e4c30bc444e716934b1deaaaa8.jpg)
$若\overrightarrow{BM}=\lambda \overrightarrow{BC}，F是边CD上靠近C的四等分点，且\overrightarrow{AM}\parallel \overrightarrow{EF}，求\lambda 的值.$

Solution:
$根据图形特点建立平面直角坐标系，将形转化为数，通过向量的坐标运算进行求解，充分体现了数形结合的思想.以A为坐标原点，建立如图所示的平面直角坐标系,$![Image](https://cdntk.53zaixian.com/tq/1706233228340187138/img/compression/M_683e3c74cf908adfe0e620645dfbfca1.jpg)$设AB=a,AD=b,a>0,b>0，则A(0,0),B(a,0),C(a,b),E\left(\frac{a}{2},\frac{b}{2}\right),F\left(\frac{3}{4}a,b\right),故\overrightarrow{AM}=\overrightarrow{AB}+\overrightarrow{BM}=\overrightarrow{AB}+\lambda \overrightarrow{BC}=(a,0)+\lambda (0,b)=(a,\lambda b),\overrightarrow{EF}=\left(\frac{a}{4},\frac{b}{2}\right).\because \overrightarrow{AM}\parallel \overrightarrow{EF},\therefore \frac{a}{4}\cdot \lambda b=a\cdot \frac{b}{2},\therefore \lambda =2.$

Final answer: $2$
Subfield: Plane Geometry


Question: $在矩形ABCD中，E是AC的中点，点F在边CD上.$<img_1583>
$若\overrightarrow{BM}=\lambda \overrightarrow{BC}，F是边CD上靠近C的四等分点，且\overrightarrow{AM}\parallel \overrightarrow{EF}，求\lambda 的值.$

Solution:
$根据图形特点建立平面直角坐标系，将形转化为数，通过向量的坐标运算进行求解，充分体现了数形结合的思想.以A为坐标原点，建立如图所示的平面直角坐标系,$<img_1680>$设AB=a,AD=b,a>0,b>0，则A(0,0),B(a,0),C(a,b),E\left(\frac{a}{2},\frac{b}{2}\right),F\left(\frac{3}{4}a,b\right),故\overrightarrow{AM}=\overrightarrow{AB}+\overrightarrow{BM}=\overrightarrow{AB}+\lambda \overrightarrow{BC}=(a,0)+\lambda (0,b)=(a,\lambda b),\overrightarrow{EF}=\left(\frac{a}{4},\frac{b}{2}\right).\because \overrightarrow{AM}\parallel \overrightarrow{EF},\therefore \frac{a}{4}\cdot \lambda b=a\cdot \frac{b}{2},\therefore \lambda =2.$

Final answer: $2$
Subfield: Plane Geometry


Question: $如图，在直角三角形ABC中，\angle BAC=90^\circ ,CB=2CA=2，D,E分别是线段AB,BC上的点，满足\overrightarrow{AD}=\lambda \overrightarrow{AB},\overrightarrow{BE}=\lambda \overrightarrow{BC},\lambda \in (0,1).$<img_3091>
$求\overrightarrow{AE}\cdot \overrightarrow{BC}的取值范围；$

Solution:
$\because 在直角三角形ABC中，\angle BAC=90^\circ ,CB=2CA=2，\therefore \angle B=30^\circ ,BA=\sqrt{3}，\therefore \overrightarrow{BA}\cdot \overrightarrow{BC}=\sqrt{3}\times 2\times \\cos   30^\circ =3，\therefore \overrightarrow{AE}\cdot \overrightarrow{BC}=(\overrightarrow{AB}+\overrightarrow{BE})\cdot \overrightarrow{BC}=(\overrightarrow{AB}+\lambda \overrightarrow{BC})\cdot \overrightarrow{BC}=\overrightarrow{AB}\cdot \overrightarrow{BC}+\lambda \overrightarrow{BC}^{2}=-\overrightarrow{BA}\cdot \overrightarrow{BC}+\lambda \overrightarrow{BC}^{2}=-3+4\lambda ，\because \lambda \in (0,1)，\therefore \overrightarrow{AE}\cdot \overrightarrow{BC}\in (-3,1).$

Final answer: $(-3, 1)$
Subfield: Plane Geometry


Question: $如图，在直角三角形ABC中，\angle BAC=90^\circ ,CB=2CA=2，D,E分别是线段AB,BC上的点，满足\overrightarrow{AD}=\lambda \overrightarrow{AB},\overrightarrow{BE}=\lambda \overrightarrow{BC},\lambda \in (0,1).$![Image](https://cdntk.53zaixian.com/tq/1706233229904662531/img/compression/M_48f1c5a0cf4b3a0ae75db4705baa531d.jpg)
$求\overrightarrow{AE}\cdot \overrightarrow{BC}的取值范围；$

Solution:
$\because 在直角三角形ABC中，\angle BAC=90^\circ ,CB=2CA=2，\therefore \angle B=30^\circ ,BA=\sqrt{3}，\therefore \overrightarrow{BA}\cdot \overrightarrow{BC}=\sqrt{3}\times 2\times \\cos   30^\circ =3，\therefore \overrightarrow{AE}\cdot \overrightarrow{BC}=(\overrightarrow{AB}+\overrightarrow{BE})\cdot \overrightarrow{BC}=(\overrightarrow{AB}+\lambda \overrightarrow{BC})\cdot \overrightarrow{BC}=\overrightarrow{AB}\cdot \overrightarrow{BC}+\lambda \overrightarrow{BC}^{2}=-\overrightarrow{BA}\cdot \overrightarrow{BC}+\lambda \overrightarrow{BC}^{2}=-3+4\lambda ，\because \lambda \in (0,1)，\therefore \overrightarrow{AE}\cdot \overrightarrow{BC}\in (-3,1).$

Final answer: $(-3, 1)$
Subfield: Plane Geometry


Question: $如图，在直角三角形ABC中，\angle BAC=90^\circ ,CB=2CA=2，D,E分别是线段AB,BC上的点，满足\overrightarrow{AD}=\lambda \overrightarrow{AB},\overrightarrow{BE}=\lambda \overrightarrow{BC},\lambda \in (0,1).$![Image](https://cdntk.53zaixian.com/tq/1706233229904662531/img/compression/M_48f1c5a0cf4b3a0ae75db4705baa531d.jpg)
$是否存在实数\lambda ，使得\overrightarrow{AE}\perp \overrightarrow{CD}？若存在，求出\lambda 的值；若不存在，请说明理由.$

Solution:
$存在.\overrightarrow{AE}\cdot \overrightarrow{CD}=(\overrightarrow{AB}+\overrightarrow{BE})\cdot (\overrightarrow{AD}-\overrightarrow{AC})=(\overrightarrow{AB}+\lambda \overrightarrow{BC})\cdot (\lambda \overrightarrow{AB}-\overrightarrow{AC})=\lambda \overrightarrow{AB}^{2}-\overrightarrow{AB}\cdot \overrightarrow{AC}+\lambda ^{2}\overrightarrow{BC}\cdot \overrightarrow{AB}-\lambda \overrightarrow{BC}\cdot \overrightarrow{AC}=3\lambda -0+\lambda ^{2}\times 2\times \sqrt{3}\times \\cos   150^\circ -\lambda \times 2\times 1\times \\cos   60^\circ =2\lambda -3\lambda ^{2},令2\lambda -3\lambda ^{2}=0，得\lambda =\frac{2}{3}或\lambda =0(舍去).由互相垂直的两个向量的数量积为0得到关于\lambda 的一元二次方程，体现了方程思想.\therefore 存在实数\lambda =\frac{2}{3}，使得\overrightarrow{AE}\perp \overrightarrow{CD}.$

Final answer: $\frac{2}{3}$
Subfield: Plane Geometry


Question: $如图，在直角三角形ABC中，\angle BAC=90^\circ ,CB=2CA=2，D,E分别是线段AB,BC上的点，满足\overrightarrow{AD}=\lambda \overrightarrow{AB},\overrightarrow{BE}=\lambda \overrightarrow{BC},\lambda \in (0,1).$<img_3091>
$是否存在实数\lambda ，使得\overrightarrow{AE}\perp \overrightarrow{CD}？若存在，求出\lambda 的值；若不存在，请说明理由.$

Solution:
$存在.\overrightarrow{AE}\cdot \overrightarrow{CD}=(\overrightarrow{AB}+\overrightarrow{BE})\cdot (\overrightarrow{AD}-\overrightarrow{AC})=(\overrightarrow{AB}+\lambda \overrightarrow{BC})\cdot (\lambda \overrightarrow{AB}-\overrightarrow{AC})=\lambda \overrightarrow{AB}^{2}-\overrightarrow{AB}\cdot \overrightarrow{AC}+\lambda ^{2}\overrightarrow{BC}\cdot \overrightarrow{AB}-\lambda \overrightarrow{BC}\cdot \overrightarrow{AC}=3\lambda -0+\lambda ^{2}\times 2\times \sqrt{3}\times \\cos   150^\circ -\lambda \times 2\times 1\times \\cos   60^\circ =2\lambda -3\lambda ^{2},令2\lambda -3\lambda ^{2}=0，得\lambda =\frac{2}{3}或\lambda =0(舍去).由互相垂直的两个向量的数量积为0得到关于\lambda 的一元二次方程，体现了方程思想.\therefore 存在实数\lambda =\frac{2}{3}，使得\overrightarrow{AE}\perp \overrightarrow{CD}.$

Final answer: $\frac{2}{3}$
Subfield: Plane Geometry


Question: $如图，在矩形ABCD中，E是BC的中点，点F在边CD上.$![Image](https://cdntk.53zaixian.com/tq/1706233230424756226/img/compression/M_0c6055fbc13e380012f2a952b9776aff.jpg)
$若AB=BC=2，F是边CD上靠近C的三等分点，求\overrightarrow{AE}\cdot \overrightarrow{EF}的值；$

Solution:
$根据题目特征建立平面直角坐标系，将几何问题转化为向量的坐标运算问题求解.以A为坐标原点,AB所在直线为x轴，AD所在直线为y轴建立平面直角坐标系，如图所示.$![Image](https://cdntk.53zaixian.com/tq/1706233230424756226/img/compression/M_e1cf3cf6f54c14e349b37e84a3713dbb.jpg)$因为E是BC的中点，所以E(2,1)，又A(0，0)，所以\overrightarrow{AE}=(2,1).因为F是CD上靠近C的三等分点，所以F\left(\frac{4}{3},2\right),所以\overrightarrow{EF}=\left(-\frac{2}{3},1\right)，所以\overrightarrow{AE}\cdot \overrightarrow{EF}=2\times \left(-\frac{2}{3}\right)+1\times 1=-\frac{1}{3}.$

Final answer: $-\frac{1}{3}$