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


Question: $如图，平行四边形ABCD中，AB=4，AD=2，\angle BAD=60^\circ ，点E，F分别为AD，DC边的中点，BE与AF相交于点O，记 \overrightarrow{AB}=a,\overrightarrow{AD}=b.$![Image](https://cdntk.53zaixian.com/tq/1704021149165953025/img/compression/M_cbf9bdcd3957f543bc0fbeeeed7590ab.jpg)
$(2)若\overrightarrow{AO}=\lambda \overrightarrow{AF}，求实数\lambda 的值.$

Solution:
$因为\overrightarrow{EB}=\overrightarrow{AB}-\overrightarrow{AE}=a-\frac{1}{2}b,\overrightarrow{EO}与\overrightarrow{EB}共线，所以设\overrightarrow{EO}=\mu \overrightarrow{EB}=\mu \left(\right.a-\frac{1}{2}b\left)\right.,\mu \in R，则\overrightarrow{AO}=\overrightarrow{AE}+\overrightarrow{EO}=\frac{1}{2}b+\mu \left(\right.a-\frac{1}{2}b\left)\right.,又\overrightarrow{AF}=\overrightarrow{AD}+\overrightarrow{DF}=\overrightarrow{AD}+\frac{1}{2}\overrightarrow{AB}=\frac{1}{2}a+b,且\overrightarrow{AO}=\lambda \overrightarrow{AF},所以\frac{1}{2}b+\mu \left(\right.a-\frac{1}{2}b\left)\right.=\lambda \left(\right.\frac{1}{2}a+b\left)\right.,即\frac{1}{2}\lambda a+\lambda b=\mu a+\frac{1}{2}(1-\mu )b,则\left\{\begin{matrix}\frac{1}{2}\lambda =\mu ,\\ \lambda =\frac{1}{2}(1-\mu ),\end{matrix}\right.解得\left\{\begin{matrix}\lambda =\frac{2}{5},\\ \mu =\frac{1}{5},\end{matrix}\right.所以\lambda =\frac{2}{5}.$

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


Question: $如图，平行四边形ABCD中，AB=4，AD=2，\angle BAD=60^\circ ，点E，F分别为AD，DC边的中点，BE与AF相交于点O，记 \overrightarrow{AB}=a,\overrightarrow{AD}=b.$<img_1579>
$(2)若\overrightarrow{AO}=\lambda \overrightarrow{AF}，求实数\lambda 的值.$

Solution:
$因为\overrightarrow{EB}=\overrightarrow{AB}-\overrightarrow{AE}=a-\frac{1}{2}b,\overrightarrow{EO}与\overrightarrow{EB}共线，所以设\overrightarrow{EO}=\mu \overrightarrow{EB}=\mu \left(\right.a-\frac{1}{2}b\left)\right.,\mu \in R，则\overrightarrow{AO}=\overrightarrow{AE}+\overrightarrow{EO}=\frac{1}{2}b+\mu \left(\right.a-\frac{1}{2}b\left)\right.,又\overrightarrow{AF}=\overrightarrow{AD}+\overrightarrow{DF}=\overrightarrow{AD}+\frac{1}{2}\overrightarrow{AB}=\frac{1}{2}a+b,且\overrightarrow{AO}=\lambda \overrightarrow{AF},所以\frac{1}{2}b+\mu \left(\right.a-\frac{1}{2}b\left)\right.=\lambda \left(\right.\frac{1}{2}a+b\left)\right.,即\frac{1}{2}\lambda a+\lambda b=\mu a+\frac{1}{2}(1-\mu )b,则\left\{\begin{matrix}\frac{1}{2}\lambda =\mu ,\\ \lambda =\frac{1}{2}(1-\mu ),\end{matrix}\right.解得\left\{\begin{matrix}\lambda =\frac{2}{5},\\ \mu =\frac{1}{5},\end{matrix}\right.所以\lambda =\frac{2}{5}.$

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


Question: $如图，在长方形ABCD中，E为边DC的中点，F为边BC上一点，且\frac{CF}{CB}=\frac{3}{4}.设\overrightarrow{AB}=a，\overrightarrow{AD}=b.$![Image](https://cdntk.53zaixian.com/tq/1704021149715406850/img/compression/M_aee74f0ee110da93d895def27b21c199.jpg)
$(1)试用基｛a,b｝表示\overrightarrow{AE}，\overrightarrow{EF}；$

Solution:
$\overrightarrow{AE}=\overrightarrow{AD}+\overrightarrow{DE}=\overrightarrow{AD}+\frac{1}{2}\overrightarrow{AB}=b+\frac{1}{2}a.\overrightarrow{EF}=\overrightarrow{AF}－\overrightarrow{AE}=\overrightarrow{AB}+\overrightarrow{BF}－\overrightarrow{AD}－\frac{1}{2}\overrightarrow{AB}=\overrightarrow{AB}+\frac{1}{4}\overrightarrow{AD}－\overrightarrow{AD}－\frac{1}{2}\overrightarrow{AB}=\frac{1}{2}\overrightarrow{AB}－\frac{3}{4}\overrightarrow{AD}=\frac{1}{2}a－\frac{3}{4}b.$

Final answer: $\overrightarrow{AE}=b+\frac{1}{2}a$,
$\overrightarrow{EF}=\frac{1}{2}a-\frac{3}{4}b$
Subfield: Plane Geometry


Question: $如图，在长方形ABCD中，E为边DC的中点，F为边BC上一点，且\frac{CF}{CB}=\frac{3}{4}.设\overrightarrow{AB}=a，\overrightarrow{AD}=b.$<img_2800>
$(1)试用基｛a,b｝表示\overrightarrow{AE}$

Solution:
$\overrightarrow{AE}=\overrightarrow{AD}+\overrightarrow{DE}=\overrightarrow{AD}+\frac{1}{2}\overrightarrow{AB}=b+\frac{1}{2}a.\overrightarrow{EF}=\overrightarrow{AF}－\overrightarrow{AE}=\overrightarrow{AB}+\overrightarrow{BF}－\overrightarrow{AD}－\frac{1}{2}\overrightarrow{AB}=\overrightarrow{AB}+\frac{1}{4}\overrightarrow{AD}－\overrightarrow{AD}－\frac{1}{2}\overrightarrow{AB}=\frac{1}{2}\overrightarrow{AB}－\frac{3}{4}\overrightarrow{AD}=\frac{1}{2}a－\frac{3}{4}b.$



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


Question: $如图，在长方形ABCD中，E为边DC的中点，F为边BC上一点，且\frac{CF}{CB}=\frac{3}{4}.设\overrightarrow{AB}=a，\overrightarrow{AD}=b.$![Image](https://cdntk.53zaixian.com/tq/1704021149715406850/img/compression/M_aee74f0ee110da93d895def27b21c199.jpg)
$(2)若\overrightarrow{AG}=\frac{3}{4}a+\frac{5}{8}b,求证：E，G，F三点共线.$

Solution:
$证明：若E，G，F三点共线，则存在实数\lambda ，使得\overrightarrow{EG}=\lambda \overrightarrow{EF}，即\overrightarrow{AG}－\overrightarrow{AE}=\lambda \overrightarrow{EF}，因为\overrightarrow{AG}－\overrightarrow{AE}=\frac{3}{4}a+\frac{5}{8}b－b－\frac{1}{2}a=\frac{1}{4}a－\frac{3}{8}b，\overrightarrow{EF}=\frac{1}{2}a－\frac{3}{4}b，所以\frac{1}{4}a－\frac{3}{8}b=\lambda \left(\frac{1}{2}\boldsymbol{a}-\frac{3}{4}{\boldsymbol{b}}\right)，所以\lambda =\frac{1}{2}，所以E，G，F三点共线.$

Final answer: 证明题
Subfield: Plane Geometry


Question: $如图，在长方形ABCD中，E为边DC的中点，F为边BC上一点，且\frac{CF}{CB}=\frac{3}{4}.设\overrightarrow{AB}=a，\overrightarrow{AD}=b.$<img_2800>
$(2)若\overrightarrow{AG}=\frac{3}{4}a+\frac{5}{8}b,求证：E，G，F三点共线.$

Solution:
$证明：若E，G，F三点共线，则存在实数\lambda ，使得\overrightarrow{EG}=\lambda \overrightarrow{EF}，即\overrightarrow{AG}－\overrightarrow{AE}=\lambda \overrightarrow{EF}，因为\overrightarrow{AG}－\overrightarrow{AE}=\frac{3}{4}a+\frac{5}{8}b－b－\frac{1}{2}a=\frac{1}{4}a－\frac{3}{8}b，\overrightarrow{EF}=\frac{1}{2}a－\frac{3}{4}b，所以\frac{1}{4}a－\frac{3}{8}b=\lambda \left(\frac{1}{2}\boldsymbol{a}-\frac{3}{4}{\boldsymbol{b}}\right)，所以\lambda =\frac{1}{2}，所以E，G，F三点共线.$
Subfield: Plane Geometry


Question: $如图所示，在\triangle ABC中，D是边BC的中点，E是线段AD上靠近点A的三等分点.过点E的直线与边AB,AC分别交于点P,Q.设\overrightarrow{PB}=\lambda \overrightarrow{AP},\overrightarrow{QC}=\mu \overrightarrow{AQ}，其中\lambda ,\mu \geq 0.$<img_2205>
$(1)试用\overrightarrow{AD}与\overrightarrow{BC}分别表示\overrightarrow{AB}$

Solution:
$因为D是边BC的中点，所以\overrightarrow{AB}=\overrightarrow{AD}+\overrightarrow{DB}=\overrightarrow{AD}+\frac{1}{2}\overrightarrow{CB}=\overrightarrow{AD}-\frac{1}{2}\overrightarrow{BC}，\overrightarrow{AC}=\overrightarrow{AD}+\overrightarrow{DC}=\overrightarrow{AD}+\frac{1}{2}\overrightarrow{BC}.$



Final answer: $\overrightarrow{AB}=\overrightarrow{AD}-\frac{1}{2}\overrightarrow{BC}$
Subfield: Plane Geometry


Question: $如图所示，在\triangle ABC中，D是边BC的中点，E是线段AD上靠近点A的三等分点.过点E的直线与边AB,AC分别交于点P,Q.设\overrightarrow{PB}=\lambda \overrightarrow{AP},\overrightarrow{QC}=\mu \overrightarrow{AQ}，其中\lambda ,\mu \geq 0.$![Image](https://cdntk.53zaixian.com/tq/1704021153037295618/img/compression/M_3861179f1d656510c0af5d1af897c957.jpg)
$(1)试用\overrightarrow{AD}与\overrightarrow{BC}分别表示\overrightarrow{AB}，\overrightarrow{AC}；$

Solution:
$因为D是边BC的中点，所以\overrightarrow{AB}=\overrightarrow{AD}+\overrightarrow{DB}=\overrightarrow{AD}+\frac{1}{2}\overrightarrow{CB}=\overrightarrow{AD}-\frac{1}{2}\overrightarrow{BC}，\overrightarrow{AC}=\overrightarrow{AD}+\overrightarrow{DC}=\overrightarrow{AD}+\frac{1}{2}\overrightarrow{BC}.$

Final answer: $\overrightarrow{AB}=\overrightarrow{AD}-\frac{1}{2}\overrightarrow{BC}，\overrightarrow{AC}=\overrightarrow{AD}+\frac{1}{2}\overrightarrow{BC}$
Subfield: Plane Geometry


Question: $如图所示，在\triangle ABC中，D是边BC的中点，E是线段AD上靠近点A的三等分点.过点E的直线与边AB,AC分别交于点P,Q.设\overrightarrow{PB}=\lambda \overrightarrow{AP},\overrightarrow{QC}=\mu \overrightarrow{AQ}，其中\lambda ,\mu \geq 0.$<img_2205>
$(2)求证：\lambda +\mu 为定值，并求此定值.$

Solution:
$因为\overrightarrow{PB}=\lambda \overrightarrow{AP},\overrightarrow{QC}=\mu \overrightarrow{AQ}，所以\overrightarrow{AC}=(1+\mu )\overrightarrow{AQ}，\overrightarrow{AB}=(1+\lambda )\overrightarrow{AP}，因为D是边BC的中点，所以\overrightarrow{AD}=\frac{1}{2}(\overrightarrow{AB}+\overrightarrow{AC})，所以\overrightarrow{AE}=\frac{1}{3}\overrightarrow{AD}=\frac{1}{6}(\overrightarrow{AB}+\overrightarrow{AC})=\frac{\lambda +1}{6}\overrightarrow{AP}+\frac{\mu +1}{6}\overrightarrow{AQ}，因为P，E，Q三点共线，所以\frac{\lambda +1}{6}+\frac{\mu +1}{6}=1，解得\lambda +\mu =4，所以\lambda +\mu 为定值4.$

Final answer: $4$
Subfield: Plane Geometry


Question: $如图所示，在\triangle ABC中，D是边BC的中点，E是线段AD上靠近点A的三等分点.过点E的直线与边AB,AC分别交于点P,Q.设\overrightarrow{PB}=\lambda \overrightarrow{AP},\overrightarrow{QC}=\mu \overrightarrow{AQ}，其中\lambda ,\mu \geq 0.$![Image](https://cdntk.53zaixian.com/tq/1704021153037295618/img/compression/M_3861179f1d656510c0af5d1af897c957.jpg)
$(2)求证：\lambda +\mu 为定值，并求此定值.$

Solution:
$因为\overrightarrow{PB}=\lambda \overrightarrow{AP},\overrightarrow{QC}=\mu \overrightarrow{AQ}，所以\overrightarrow{AC}=(1+\mu )\overrightarrow{AQ}，\overrightarrow{AB}=(1+\lambda )\overrightarrow{AP}，因为D是边BC的中点，所以\overrightarrow{AD}=\frac{1}{2}(\overrightarrow{AB}+\overrightarrow{AC})，所以\overrightarrow{AE}=\frac{1}{3}\overrightarrow{AD}=\frac{1}{6}(\overrightarrow{AB}+\overrightarrow{AC})=\frac{\lambda +1}{6}\overrightarrow{AP}+\frac{\mu +1}{6}\overrightarrow{AQ}，因为P，E，Q三点共线，所以\frac{\lambda +1}{6}+\frac{\mu +1}{6}=1，解得\lambda +\mu =4，所以\lambda +\mu 为定值4.$

Final answer: $4$
Subfield: Plane Geometry


Question: $如图，在\triangle OAB中，P为线段AB上的一个动点(不含端点)，且满足\overrightarrow{AP}=\lambda \overrightarrow{PB}.$![Image](https://cdntk.53zaixian.com/tq/1704021174252720131/img/compression/M_34ab6683f8144dbbd897a6691633bf92.jpg)
$(1)若\lambda =\frac{1}{3}，用向量\overrightarrow{OA},\overrightarrow{OB}表示\overrightarrow{OP}；$

Solution:
$因为\overrightarrow{AP}=\lambda \overrightarrow{PB}，所以\overrightarrow{AP}=\frac{\lambda }{\lambda +1}\overrightarrow{AB}，所以\overrightarrow{OP}=\overrightarrow{OA}+\overrightarrow{AP}=\overrightarrow{OA}+\frac{\lambda }{\lambda +1}(\overrightarrow{OB}-\overrightarrow{OA})=\frac{1}{\lambda +1}\overrightarrow{OA}+\frac{\lambda }{\lambda +1}\overrightarrow{OB}，当\lambda =\frac{1}{3}时，\overrightarrow{OP}=\frac{3}{4}\overrightarrow{OA}+\frac{1}{4}\overrightarrow{OB}.$

Final answer: $当\lambda =1/3时，\overrightarrow{OP}=\frac{3}{4}\overrightarrow{OA}+\frac{1}{4}\overrightarrow{OB}.$
Subfield: Plane Geometry


Question: $如图，在\triangle OAB中，P为线段AB上的一个动点(不含端点)，且满足\overrightarrow{AP}=\lambda \overrightarrow{PB}.$<img_2685>
$(1)若\lambda =\frac{1}{3}，用向量\overrightarrow{OA},\overrightarrow{OB}表示\overrightarrow{OP}；$

Solution:
$因为\overrightarrow{AP}=\lambda \overrightarrow{PB}，所以\overrightarrow{AP}=\frac{\lambda }{\lambda +1}\overrightarrow{AB}，所以\overrightarrow{OP}=\overrightarrow{OA}+\overrightarrow{AP}=\overrightarrow{OA}+\frac{\lambda }{\lambda +1}(\overrightarrow{OB}-\overrightarrow{OA})=\frac{1}{\lambda +1}\overrightarrow{OA}+\frac{\lambda }{\lambda +1}\overrightarrow{OB}，当\lambda =\frac{1}{3}时，\overrightarrow{OP}=\frac{3}{4}\overrightarrow{OA}+\frac{1}{4}\overrightarrow{OB}.$



Final answer: $\overrightarrow{OP}=\frac{3}{4}\overrightarrow{OA}+\frac{1}{4}\overrightarrow{OB}$
Subfield: Plane Geometry


Question: $如图，在\triangle OAB中，P为线段AB上的一个动点(不含端点)，且满足\overrightarrow{AP}=\lambda \overrightarrow{PB}.$<img_2685>
$(2)若|\overrightarrow{OA}|=6,|\overrightarrow{OB}|=2，且\angle AOB=120^\circ ，求\overrightarrow{OP}\cdot \overrightarrow{AB}的取值范围.$

Solution:
$由(1)可知\overrightarrow{OP}=\frac{1}{\lambda +1}\overrightarrow{OA}+\frac{\lambda }{\lambda +1}\overrightarrow{OB}，所以\overrightarrow{OP}\cdot \overrightarrow{AB}=\left(\frac{1}{\lambda +1}\overrightarrow{OA}+\frac{\lambda }{\lambda +1}\overrightarrow{OB}\right)\cdot (\overrightarrow{OB}-\overrightarrow{OA})=-\frac{1}{\lambda +1}|\overrightarrow{OA}|^{2}+\frac{1-\lambda }{\lambda +1}\overrightarrow{OA}\cdot \overrightarrow{OB}+\frac{\lambda }{\lambda +1}|\overrightarrow{OB}|^{2}.因为|\overrightarrow{OA}|=6,|\overrightarrow{OB}|=2，\angle AOB=120^\circ ，所以\overrightarrow{OP}\cdot \overrightarrow{AB}=-\frac{36}{\lambda +1}+\frac{1-\lambda }{\lambda +1}\times 6\times 2\times \left(-\frac{1}{2}\right)+\frac{4\lambda }{\lambda +1}=10-\frac{52}{\lambda +1}，由题知0<\lambda <1，所以-52<-\frac{52}{\lambda +1}<-26，所以-42<\overrightarrow{OP}\cdot \overrightarrow{AB}<-16，即\overrightarrow{OP}\cdot \overrightarrow{AB}的取值范围为(-42,-16).$

Final answer: $(-42,-16)$
Subfield: Plane Geometry


Question: $如图，在\triangle OAB中，P为线段AB上的一个动点(不含端点)，且满足\overrightarrow{AP}=\lambda \overrightarrow{PB}.$![Image](https://cdntk.53zaixian.com/tq/1704021174252720131/img/compression/M_34ab6683f8144dbbd897a6691633bf92.jpg)
$(2)若|\overrightarrow{OA}|=6,|\overrightarrow{OB}|=2，且\angle AOB=120^\circ ，求\overrightarrow{OP}\cdot \overrightarrow{AB}的取值范围.$

Solution:
$由(1)可知\overrightarrow{OP}=\frac{1}{\lambda +1}\overrightarrow{OA}+\frac{\lambda }{\lambda +1}\overrightarrow{OB}，所以\overrightarrow{OP}\cdot \overrightarrow{AB}=\left(\frac{1}{\lambda +1}\overrightarrow{OA}+\frac{\lambda }{\lambda +1}\overrightarrow{OB}\right)\cdot (\overrightarrow{OB}-\overrightarrow{OA})=-\frac{1}{\lambda +1}|\overrightarrow{OA}|^{2}+\frac{1-\lambda }{\lambda +1}\overrightarrow{OA}\cdot \overrightarrow{OB}+\frac{\lambda }{\lambda +1}|\overrightarrow{OB}|^{2}.因为|\overrightarrow{OA}|=6,|\overrightarrow{OB}|=2，\angle AOB=120^\circ ，所以\overrightarrow{OP}\cdot \overrightarrow{AB}=-\frac{36}{\lambda +1}+\frac{1-\lambda }{\lambda +1}\times 6\times 2\times \left(-\frac{1}{2}\right)+\frac{4\lambda }{\lambda +1}=10-\frac{52}{\lambda +1}，由题知0<\lambda <1，所以-52<-\frac{52}{\lambda +1}<-26，所以-42<\overrightarrow{OP}\cdot \overrightarrow{AB}<-16，即\overrightarrow{OP}\cdot \overrightarrow{AB}的取值范围为(-42,-16).$

Final answer: $(-42,-16)$
Subfield: Plane Geometry


Question: $如图，在同一平面内，一个质点O受三个力F_{1}，F_{2}，F_{3}的作用保持平衡，其中F_{3}与F_{2}的夹角为\alpha ，F_{3}与F_{1}的夹角为\beta .$![Image](https://cdntk.53zaixian.com/tq/1704021226285117442/img/compression/M_808fc72ccbbfa1cdd12b3f80a144704f.jpg)
$(1)若\alpha =120^\circ ，\beta =150^\circ ，|F_{3}|=10，求力F_{1}，F_{2}的大小；$

Solution:
$因为质点在F_{1}，F_{2}，F_{3}的作用下保持平衡，所以F_{1}+F_{2}+F_{3}=0，所以F_{3}=-(F_{1}+F_{2})，又\alpha =120^\circ ，\beta =150^\circ ,所以F_{1}与F_{2}的夹角为90^\circ ，所以F_{1}\cdot F_{2}=0，则|F_{3}|^{2}=[-(F_{1}+F_{2})]^{2}=F^2_1+2F_{1}\cdot F_{2}+F^2_2=F^2_1+F^2_2，将|F_{3}|=10代入可得F^2_1+F^2_2=100. 如图.$![Image](https://cdntk.53zaixian.com/tq/1704021226578718723/img/compression/M_7540528ba0dcb4e0ccef83fbd7a1dab5.jpg)$易得\angle 1=30^\circ ， 所以|F_{1}|=|F_{3}|\times \\cos   30^\circ =10\times \frac{\sqrt{3}}{2}=5\sqrt{3}， |F_{2}|=|F_{3}|\times \\sin   30^\circ =10\times \frac{1}{2}=5.$

Final answer: $|F_{1}|=5\sqrt{3}, |F_{2}|=5$
Subfield: Plane Geometry


Question: $如图，在同一平面内，一个质点O受三个力F_{1}，F_{2}，F_{3}的作用保持平衡，其中F_{3}与F_{2}的夹角为\alpha ，F_{3}与F_{1}的夹角为\beta .$<img_2166>
$(1)若\alpha =120^\circ ，\beta =150^\circ ，|F_{3}|=10，求力F_{1}，F_{2}的大小；$

Solution:
$因为质点在F_{1}，F_{2}，F_{3}的作用下保持平衡，所以F_{1}+F_{2}+F_{3}=0，所以F_{3}=-(F_{1}+F_{2})，又\alpha =120^\circ ，\beta =150^\circ ,所以F_{1}与F_{2}的夹角为90^\circ ，所以F_{1}\cdot F_{2}=0，则|F_{3}|^{2}=[-(F_{1}+F_{2})]^{2}=F^2_1+2F_{1}\cdot F_{2}+F^2_2=F^2_1+F^2_2，将|F_{3}|=10代入可得F^2_1+F^2_2=100. 如图.$<img_1818>$易得\angle 1=30^\circ ， 所以|F_{1}|=|F_{3}|\times \\cos   30^\circ =10\times \frac{\sqrt{3}}{2}=5\sqrt{3}， |F_{2}|=|F_{3}|\times \\sin   30^\circ =10\times \frac{1}{2}=5.$



Final answer: $|F_{1}|=5\sqrt{3}, |F_{2}|=5$
Subfield: Plane Geometry


Question: $如图，在同一平面内，一个质点O受三个力F_{1}，F_{2}，F_{3}的作用保持平衡，其中F_{3}与F_{2}的夹角为\alpha ，F_{3}与F_{1}的夹角为\beta .$<img_2166>
$(2)若|F_{1}|:|F_{2}|:|F_{3}|=1:\sqrt{2}:\sqrt{3}，求\alpha 与\beta 的余弦值.$

Solution:
$因为|F_{1}|:|F_{2}|:|F_{3}|=1:\sqrt{2}:\sqrt{3}，且质点处于平衡状态， 所以以|F_{1}|，|F_{2}|，|F_{3}|为边长的三角形为直角三角形，如图所示，$<img_1557>$则\\cos  \angle 3=\frac{1}{\sqrt{3}}=\frac{\sqrt{3}}{3}，\\cos  \angle 2=\frac{\sqrt{2}}{\sqrt{3}}=\frac{\sqrt{6}}{3},所以\\cos   \beta =\\cos  (\pi -\angle 3)=-\\cos  \angle 3=-\frac{\sqrt{3}}{3}，\\cos   \alpha =\\cos  (\pi -\angle 2)=-\\cos  \angle 2=-\frac{\sqrt{6}}{3}.$

Final answer: $\\cos   \beta =-\frac{\sqrt{3}}{3}, \\cos   \alpha =-\frac{\sqrt{6}}{3}$
Subfield: Plane Geometry


Question: $如图，在同一平面内，一个质点O受三个力F_{1}，F_{2}，F_{3}的作用保持平衡，其中F_{3}与F_{2}的夹角为\alpha ，F_{3}与F_{1}的夹角为\beta .$![Image](https://cdntk.53zaixian.com/tq/1704021226285117442/img/compression/M_808fc72ccbbfa1cdd12b3f80a144704f.jpg)
$(2)若|F_{1}|:|F_{2}|:|F_{3}|=1:\sqrt{2}:\sqrt{3}，求\alpha 与\beta 的余弦值.$

Solution:
$因为|F_{1}|:|F_{2}|:|F_{3}|=1:\sqrt{2}:\sqrt{3}，且质点处于平衡状态， 所以以|F_{1}|，|F_{2}|，|F_{3}|为边长的三角形为直角三角形，如图所示，$![Image](https://cdntk.53zaixian.com/tq/1704021227157532674/img/compression/M_fcba7b5ef83fc6055a6792f4c19139be.jpg)$则\\cos  \angle 3=\frac{1}{\sqrt{3}}=\frac{\sqrt{3}}{3}，\\cos  \angle 2=\frac{\sqrt{2}}{\sqrt{3}}=\frac{\sqrt{6}}{3},所以\\cos   \beta =\\cos  (\pi -\angle 3)=-\\cos  \angle 3=-\frac{\sqrt{3}}{3}，\\cos   \alpha =\\cos  (\pi -\angle 2)=-\\cos  \angle 2=-\frac{\sqrt{6}}{3}.$

Final answer: $\\cos   \beta =-\frac{\sqrt{3}}{3}, \\cos   \alpha =-\frac{\sqrt{6}}{3}$
Subfield: Plane Geometry


Question: $某市一棚户区改造用地平面示意图如图所示.该区域是半径为R的圆形，圆的内接四边形ABCD内是原棚户区建筑用地，测量可知AB=AD=4 km,BC=6 km,CD=2 km.$![Image](https://cdntk.53zaixian.com/tq/1704021214786473987/img/compression/M_0ff1836e290a6d90e34e7742500016bc.jpg)
$(1)求原棚户区建筑用地ABCD的对角线AC的长度；$

Solution:
$由题意得\angle ABC+\angle ADC=180^\circ ，\therefore \\cos  \angle ABC+\\cos  \angle ADC=0,由余弦定理，得AC^{2}=4^{2}+6^{2}-2\times 4\times 6\times \\cos  \angle ABC=4^{2}+2^{2}-2\times 4\times 2\times \\cos  \angle ADC，\therefore \\cos  \angle ABC=\frac{1}{2}.\because 0^\circ ＜\angle ABC＜180^\circ ,\therefore \angle ABC=60^\circ ,\therefore AC^{2}=28，\therefore AC=2\sqrt{7} km.故原棚户区建筑用地ABCD的对角线AC的长度为2\sqrt{7} km.$

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


Question: $某市一棚户区改造用地平面示意图如图所示.该区域是半径为R的圆形，圆的内接四边形ABCD内是原棚户区建筑用地，测量可知AB=AD=4 km,BC=6 km,CD=2 km.$<img_2017>
$(1)求原棚户区建筑用地ABCD的对角线AC的长度；$

Solution:
$由题意得\angle ABC+\angle ADC=180^\circ ，\therefore \\cos  \angle ABC+\\cos  \angle ADC=0,由余弦定理，得AC^{2}=4^{2}+6^{2}-2\times 4\times 6\times \\cos  \angle ABC=4^{2}+2^{2}-2\times 4\times 2\times \\cos  \angle ADC，\therefore \\cos  \angle ABC=\frac{1}{2}.\because 0^\circ ＜\angle ABC＜180^\circ ,\therefore \angle ABC=60^\circ ,\therefore AC^{2}=28，\therefore AC=2\sqrt{7} km.故原棚户区建筑用地ABCD的对角线AC的长度为2\sqrt{7} km.$

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


Question: $某市一棚户区改造用地平面示意图如图所示.该区域是半径为R的圆形，圆的内接四边形ABCD内是原棚户区建筑用地，测量可知AB=AD=4 km,BC=6 km,CD=2 km.$![Image](https://cdntk.53zaixian.com/tq/1704021214786473987/img/compression/M_0ff1836e290a6d90e34e7742500016bc.jpg)
$(2)请计算原棚户区建筑用地ABCD的面积.$

Solution:
$由(1)知\angle ABC=60^\circ ，\angle ADC=120^\circ 故S_{四边形ABCD}=S_{\triangle ABC}+S_{\triangle ADC}=\frac{1}{2}\times 4\times 6\times \\sin   60^\circ +\frac{1}{2}\times 2\times 4\times \\sin   120^\circ =8\sqrt{3}(km^{2}).即原棚户区建筑用地ABCD的面积为8\sqrt{3} km^{2}.$

Final answer: $8\sqrt 3$ km^2
Subfield: Plane Geometry


Question: $某市一棚户区改造用地平面示意图如图所示.该区域是半径为R的圆形，圆的内接四边形ABCD内是原棚户区建筑用地，测量可知AB=AD=4 km,BC=6 km,CD=2 km.$<img_2017>
$(2)请计算原棚户区建筑用地ABCD的面积.$

Solution:
$由(1)知\angle ABC=60^\circ ，\angle ADC=120^\circ 故S_{四边形ABCD}=S_{\triangle ABC}+S_{\triangle ADC}=\frac{1}{2}\times 4\times 6\times \\sin   60^\circ +\frac{1}{2}\times 2\times 4\times \\sin   120^\circ =8\sqrt{3}(km^{2}).即原棚户区建筑用地ABCD的面积为8\sqrt{3} km^{2}.$

Final answer: $8\sqrt 3$ km^2
Subfield: Plane Geometry


Question: $湖面上甲、乙、丙三艘船沿着同一条直线航行，某一时刻，甲船在最前面的A处，乙船在中间B处，丙船在最后面的C处，且BC:AB=3:1.一架无人机在空中的P处对它们进行数据测量，在同一时刻测得\angle APB=30^\circ ，\angle BPC=90^\circ .(船只与无人机的大小及其他因素忽略不计)$<img_2958>
$(1)求此时无人机到甲、丙两船的距离之比的值；$

Solution:
$在\triangle APB中，由正弦定理得，\frac{AP}{\\\sin   \angle ABP}=\frac{AB}{\\\sin   \angle APB}=\frac{AB}{\frac{1}{2}},在\triangle BPC中，由正弦定理得,\frac{CP}{\\\sin   \angle CBP}=\frac{BC}{\\\sin   \angle CPB}=\frac{BC}{1},又BC=3AB，\\sin  \angle ABP=\\sin  \angle CBP，所以\frac{AP}{CP}=\frac{2}{3}，即此时无人机到甲、丙两船的距离之比为2:3.$

Final answer: $2/3$
Subfield: Plane Geometry


Question: $湖面上甲、乙、丙三艘船沿着同一条直线航行，某一时刻，甲船在最前面的A处，乙船在中间B处，丙船在最后面的C处，且BC:AB=3:1.一架无人机在空中的P处对它们进行数据测量，在同一时刻测得\angle APB=30^\circ ，\angle BPC=90^\circ .(船只与无人机的大小及其他因素忽略不计)$![Image](https://cdntk.53zaixian.com/tq/1704021215247847426/img/compression/M_b3c4a49038b45e1d82ef544edcfc518b.jpg)
$(1)求此时无人机到甲、丙两船的距离之比的值；$

Solution:
$在\triangle APB中，由正弦定理得，\frac{AP}{\\\sin   \angle ABP}=\frac{AB}{\\\sin   \angle APB}=\frac{AB}{\frac{1}{2}},在\triangle BPC中，由正弦定理得,\frac{CP}{\\\sin   \angle CBP}=\frac{BC}{\\\sin   \angle CPB}=\frac{BC}{1},又BC=3AB，\\sin  \angle ABP=\\sin  \angle CBP，所以\frac{AP}{CP}=\frac{2}{3}，即此时无人机到甲、丙两船的距离之比为2:3.$

Final answer: $2/3$
Subfield: Plane Geometry


Question: $湖面上甲、乙、丙三艘船沿着同一条直线航行，某一时刻，甲船在最前面的A处，乙船在中间B处，丙船在最后面的C处，且BC:AB=3:1.一架无人机在空中的P处对它们进行数据测量，在同一时刻测得\angle APB=30^\circ ，\angle BPC=90^\circ .(船只与无人机的大小及其他因素忽略不计)$![Image](https://cdntk.53zaixian.com/tq/1704021215247847426/img/compression/M_b3c4a49038b45e1d82ef544edcfc518b.jpg)
$(2)若此时甲、乙两船相距100米，求无人机到丙船的距离.(精确到1米，参考数据：\sqrt{19}\approx 4.359)$

Solution:
$由BC:AB=3:1,AB=100米,得AC=400米，设AP=2x米，则CP=3x米，在\triangle APC中，\angle APC=\angle APB+\angle BPC=120^\circ ,由余弦定理得160 000=(2x)^{2}+(3x)^{2}-2\cdot (2x)\cdot (3x)\cdot \\cos   120^\circ ，所以x=\frac{400\sqrt{19}}{19}，故无人机到丙船的距离CP=3x=\frac{1 200\sqrt{19}}{19}\approx 275(米).$

Final answer: $275$ (米)
Subfield: Plane Geometry


Question: $湖面上甲、乙、丙三艘船沿着同一条直线航行，某一时刻，甲船在最前面的A处，乙船在中间B处，丙船在最后面的C处，且BC:AB=3:1.一架无人机在空中的P处对它们进行数据测量，在同一时刻测得\angle APB=30^\circ ，\angle BPC=90^\circ .(船只与无人机的大小及其他因素忽略不计)$<img_2958>
$(2)若此时甲、乙两船相距100米，求无人机到丙船的距离.(精确到1米，参考数据：\sqrt{19}\approx 4.359)$

Solution:
$由BC:AB=3:1,AB=100米,得AC=400米，设AP=2x米，则CP=3x米，在\triangle APC中，\angle APC=\angle APB+\angle BPC=120^\circ ,由余弦定理得160 000=(2x)^{2}+(3x)^{2}-2\cdot (2x)\cdot (3x)\cdot \\cos   120^\circ ，所以x=\frac{400\sqrt{19}}{19}，故无人机到丙船的距离CP=3x=\frac{1 200\sqrt{19}}{19}\approx 275(米).$

Final answer: $275$ (米)
Subfield: Plane Geometry


Question: $在三角形ABC中，AB=2，AC=1，\angle ACB=\frac{\pi }{2}，D是线段BC上一点，且\overrightarrow{BD}=\frac{1}{2}\overrightarrow{DC}，F为线段AB上一点．$![Image](https://cdntk.53zaixian.com/tq/1706223416172019714/img/compression/M_2e6053602823f0f6b4175de4f3e60a41.jpg)
$(1)设\overrightarrow{AB}=a，\overrightarrow{AC}=b，\overrightarrow{AD}=xa+yb，求x－y；$

Solution:
$解法一：因为\overrightarrow{AD}=\overrightarrow{AC}+\frac{2}{3}\overrightarrow{CB}=\overrightarrow{AC}+\frac{2}{3}(\overrightarrow{AB}-\overrightarrow{AC})=\frac{2}{3}\overrightarrow{AB}+\frac{1}{3}\overrightarrow{AC}=\frac{2}{3}a+\frac{1}{3}b=xa+yb，所以x=\frac{2}{3}，y=\frac{1}{3}，所以x-y=\frac{1}{3}．解法二：以点C为坐标原点，\overrightarrow{CB},\overrightarrow{CA}的方向分别为x轴，y轴的正方向建立平面直角坐标系(图略)，则C(0,0)，A(0,1)，B(\sqrt{3},0)，D\left(\frac{2\sqrt{3}}{3},0\right)，所以\overrightarrow{AD}=\left(\frac{2\sqrt{3}}{3},-1\right)，\overrightarrow{AB}=a=(\sqrt{3},-1)，\overrightarrow{AC}=b=(0,-1)，\overrightarrow{AD}=xa+yb=x(\sqrt{3},-1)+y(0,-1)=(\sqrt{3}x,-x-y)，所以\left\{\begin{matrix}\frac{2\sqrt[　]{3}}{3}=\sqrt[　]{3}x,\\ -1=-x-y,\end{matrix}\right.解得\left\{\begin{matrix}x=\frac{2}{3},\\ y=\frac{1}{3},\end{matrix}\right.所以x-y=\frac{1}{3}．$

Final answer: $x-y=\frac{1}{3}$
Subfield: Plane Geometry


Question: $在三角形ABC中，AB=2，AC=1，\angle ACB=\frac{\pi }{2}，D是线段BC上一点，且\overrightarrow{BD}=\frac{1}{2}\overrightarrow{DC}，F为线段AB上一点．$<img_2978>
$(1)设\overrightarrow{AB}=a，\overrightarrow{AC}=b，\overrightarrow{AD}=xa+yb，求x－y；$

Solution:
$解法一：因为\overrightarrow{AD}=\overrightarrow{AC}+\frac{2}{3}\overrightarrow{CB}=\overrightarrow{AC}+\frac{2}{3}(\overrightarrow{AB}-\overrightarrow{AC})=\frac{2}{3}\overrightarrow{AB}+\frac{1}{3}\overrightarrow{AC}=\frac{2}{3}a+\frac{1}{3}b=xa+yb，所以x=\frac{2}{3}，y=\frac{1}{3}，所以x-y=\frac{1}{3}．解法二：以点C为坐标原点，\overrightarrow{CB},\overrightarrow{CA}的方向分别为x轴，y轴的正方向建立平面直角坐标系(图略)，则C(0,0)，A(0,1)，B(\sqrt{3},0)，D\left(\frac{2\sqrt{3}}{3},0\right)，所以\overrightarrow{AD}=\left(\frac{2\sqrt{3}}{3},-1\right)，\overrightarrow{AB}=a=(\sqrt{3},-1)，\overrightarrow{AC}=b=(0,-1)，\overrightarrow{AD}=xa+yb=x(\sqrt{3},-1)+y(0,-1)=(\sqrt{3}x,-x-y)，所以\left\{\begin{matrix}\frac{2\sqrt[　]{3}}{3}=\sqrt[　]{3}x,\\ -1=-x-y,\end{matrix}\right.解得\left\{\begin{matrix}x=\frac{2}{3},\\ y=\frac{1}{3},\end{matrix}\right.所以x-y=\frac{1}{3}．$

Final answer: $x-y=\frac{1}{3}$