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


Question: $如图，在平行四边形ABCD中，\angle DAB=60^\circ ，AB=2AD=2，平面AED\perp 平面ABCD，三角形AED为等边三角形，EF\parallel AB，EF=1，M，N分别为AD，AB的中点.$![Image](https://cdntk.53zaixian.com/tq/1706899161349849089/img/compression/M_ba052f49dd967ff0bdfceab6bd252357.jpg)
$求证：平面EMN\parallel 平面BDF；$

Solution:
$证明：因为M，N分别为AD，AB的中点，所以MN为\triangle ABD的中位线，所以MN\parallel BD，又MN\not\subset 平面BDF,BD\subset 平面BDF，所以MN\parallel 平面BDF.因为EF\parallel AB，EF=1，AB=2NB=2，所以NB=1,所以EF\parallel NB，EF=NB，所以四边形FENB是平行四边形，所以EN\parallel BF，又EN\not\subset 平面BDF,BF\subset 平面BDF,所以EN\parallel 平面BDF，又因为EN\cap MN=N，EN,MN\subset 平面EMN,所以平面EMN\parallel 平面BDF.$

Final answer: 证明题
Subfield: Plane Geometry


Question: $如图，在平行四边形ABCD中，\angle DAB=60^\circ ，AB=2AD=2，平面AED\perp 平面ABCD，三角形AED为等边三角形，EF\parallel AB，EF=1，M，N分别为AD，AB的中点.$![Image](https://cdntk.53zaixian.com/tq/1706899161349849089/img/compression/M_ba052f49dd967ff0bdfceab6bd252357.jpg)
$求证：平面BDF\perp 平面ABCD；$

Solution:
$证明：因为三角形AED为等边三角形，M为AD的中点，所以EM\perp AD,因为平面AED\perp 平面ABCD,平面AED\cap 平面ABCD=AD,EM\subset 平面AED,所以EM\perp 平面ABCD，又EM\subset 平面EMN，所以平面EMN\perp 平面ABCD.由(1)知，平面EMN\parallel 平面BDF,所以平面BDF\perp 平面ABCD.$

Final answer: 证明题
Subfield: Plane Geometry


Question: $如图，在平行四边形ABCD中，\angle DAB=60^\circ ，AB=2AD=2，平面AED\perp 平面ABCD，三角形AED为等边三角形，EF\parallel AB，EF=1，M，N分别为AD，AB的中点.$<img_1733>
$求证：平面BDF\perp 平面ABCD；$

Solution:
$证明：因为三角形AED为等边三角形，M为AD的中点，所以EM\perp AD,因为平面AED\perp 平面ABCD,平面AED\cap 平面ABCD=AD,EM\subset 平面AED,所以EM\perp 平面ABCD，又EM\subset 平面EMN，所以平面EMN\perp 平面ABCD.由(1)知，平面EMN\parallel 平面BDF,所以平面BDF\perp 平面ABCD.$
Subfield: Plane Geometry


Question: $如图，在平行四边形ABCD中，\angle DAB=60^\circ ，AB=2AD=2，平面AED\perp 平面ABCD，三角形AED为等边三角形，EF\parallel AB，EF=1，M，N分别为AD，AB的中点.$![Image](https://cdntk.53zaixian.com/tq/1706899161349849089/img/compression/M_ba052f49dd967ff0bdfceab6bd252357.jpg)
$求直线FC与平面BDF所成角的正切值.$

Solution:
$已知平行四边形ABCD中，\angle DAB=60^\circ ，AB=2AD=2，所以AD\perp BD,所以BC\perp BD，又平面BDF\perp 平面ABCD,平面BDF\cap 平面ABCD=BD,BC\subset 平面ABCD,所以BC\perp 平面BDF，所以\angle CFB即为直线FC与平面BDF所成的角.在Rt\triangle EMN中，EM=\frac{\sqrt{3}}{2},MN=\frac{1}{2}BD=\frac{\sqrt{3}}{2}，所以BF=EN=\frac{\sqrt{6}}{2}，所以\\tan  \angle CFB=\frac{BC}{BF}=\frac{1}{\frac{\sqrt{6}}{2}}=\frac{\sqrt{6}}{3},所以直线FC与平面BDF所成角的正切值为\frac{\sqrt{6}}{3}.$

Final answer: $\frac{\sqrt{6}}{3}$
Subfield: Plane Geometry


Question: $如图，在平行四边形ABCD中，\angle DAB=60^\circ ，AB=2AD=2，平面AED\perp 平面ABCD，三角形AED为等边三角形，EF\parallel AB，EF=1，M，N分别为AD，AB的中点.$<img_1733>
$求直线FC与平面BDF所成角的正切值.$

Solution:
$已知平行四边形ABCD中，\angle DAB=60^\circ ，AB=2AD=2，所以AD\perp BD,所以BC\perp BD，又平面BDF\perp 平面ABCD,平面BDF\cap 平面ABCD=BD,BC\subset 平面ABCD,所以BC\perp 平面BDF，所以\angle CFB即为直线FC与平面BDF所成的角.在Rt\triangle EMN中，EM=\frac{\sqrt{3}}{2},MN=\frac{1}{2}BD=\frac{\sqrt{3}}{2}，所以BF=EN=\frac{\sqrt{6}}{2}，所以\\tan  \angle CFB=\frac{BC}{BF}=\frac{1}{\frac{\sqrt{6}}{2}}=\frac{\sqrt{6}}{3},所以直线FC与平面BDF所成角的正切值为\frac{\sqrt{6}}{3}.$

Final answer: $\frac{\sqrt{6}}{3}$
Subfield: Plane Geometry


Question: $如图,在\triangle ABC中,\overrightarrow{AB}\cdot \overrightarrow{AC}=0,|\overrightarrow{AB}|=8,|\overrightarrow{AC}|=6,l为线段BC的垂直平分线,l与BC交于点D,E为l上异于点D的任意一点.$![Image](https://cdntk.53zaixian.com/tq/1704026520310460417/img/compression/M_31e615371e92bfff238cdc2c383117ed.jpg)
$求\overrightarrow{AD}\cdot \overrightarrow{CB}的值;$

Solution:
$如图，以D为坐标原点,BC所在直线为x轴,直线l为y轴建立平面直角坐标系,(2分)因为\overrightarrow{AB}\cdot \overrightarrow{AC}=0,所以\overrightarrow{AB}\perp \overrightarrow{AC},又|\overrightarrow{AB}|=8,|\overrightarrow{AC}|=6，所以|\overrightarrow{CB}|=10,易得D(0,0),B(-5,0),C(5,0),A\left(\frac{7}{5},\frac{24}{5}\right),(4分)所以\overrightarrow{AD}=\left(-\frac{7}{5},-\frac{24}{5}\right),\overrightarrow{CB}=(-10,0),所以\overrightarrow{AD}\cdot \overrightarrow{CB}=-\frac{7}{5}\times (-10)+\left(-\frac{24}{5}\right)\times 0=14.(6分)$![Image](https://cdntk.53zaixian.com/tq/1704026520599867393/img/compression/M_14d699b89fb05b0b8f5d4b833edf0ebf.jpg)

Final answer: $14$
Subfield: Plane Geometry


Question: $如图,在\triangle ABC中,\overrightarrow{AB}\cdot \overrightarrow{AC}=0,|\overrightarrow{AB}|=8,|\overrightarrow{AC}|=6,l为线段BC的垂直平分线,l与BC交于点D,E为l上异于点D的任意一点.$<img_3072>
$求\overrightarrow{AD}\cdot \overrightarrow{CB}的值;$

Solution:
$如图，以D为坐标原点,BC所在直线为x轴,直线l为y轴建立平面直角坐标系,(2分)因为\overrightarrow{AB}\cdot \overrightarrow{AC}=0,所以\overrightarrow{AB}\perp \overrightarrow{AC},又|\overrightarrow{AB}|=8,|\overrightarrow{AC}|=6，所以|\overrightarrow{CB}|=10,易得D(0,0),B(-5,0),C(5,0),A\left(\frac{7}{5},\frac{24}{5}\right),(4分)所以\overrightarrow{AD}=\left(-\frac{7}{5},-\frac{24}{5}\right),\overrightarrow{CB}=(-10,0),所以\overrightarrow{AD}\cdot \overrightarrow{CB}=-\frac{7}{5}\times (-10)+\left(-\frac{24}{5}\right)\times 0=14.(6分)$<img_1815>

Final answer: $14$
Subfield: Plane Geometry


Question: $如图所示，在▱ABCD中，\overrightarrow{AB}=a，\overrightarrow{AD}=b，BM=\frac{2}{3}BC，AN=\frac{1}{4}AB.$<img_1808>
$试用向量a，b分别表示\overrightarrow{DN}，\overrightarrow{AM};$

Solution:
$\because AN=\frac{1}{4}AB,\therefore \overrightarrow{AN}=\frac{1}{4}\overrightarrow{AB}=\frac{1}{4}a,(1分)\therefore \overrightarrow{DN}=\overrightarrow{AN}-\overrightarrow{AD}=\frac{1}{4}a-b.(2分)\because BM=\frac{2}{3}BC,\therefore \overrightarrow{BM}=\frac{2}{3}\overrightarrow{BC}=\frac{2}{3}\overrightarrow{AD}=\frac{2}{3}b,(4分)\therefore \overrightarrow{AM}=\overrightarrow{AB}+\overrightarrow{BM}=a+\frac{2}{3}b.(6分)$

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


Question: $如图所示，在▱ABCD中，\overrightarrow{AB}=a，\overrightarrow{AD}=b，BM=\frac{2}{3}BC，AN=\frac{1}{4}AB.$![Image](https://cdntk.53zaixian.com/tq/1704026521103183873/img/compression/M_4a9bc1d59492f9e465f5677d03fe9479.jpg)
$试用向量a，b分别表示\overrightarrow{DN}，\overrightarrow{AM};$

Solution:
$\because AN=\frac{1}{4}AB,\therefore \overrightarrow{AN}=\frac{1}{4}\overrightarrow{AB}=\frac{1}{4}a,(1分)\therefore \overrightarrow{DN}=\overrightarrow{AN}-\overrightarrow{AD}=\frac{1}{4}a-b.(2分)\because BM=\frac{2}{3}BC,\therefore \overrightarrow{BM}=\frac{2}{3}\overrightarrow{BC}=\frac{2}{3}\overrightarrow{AD}=\frac{2}{3}b,(4分)\therefore \overrightarrow{AM}=\overrightarrow{AB}+\overrightarrow{BM}=a+\frac{2}{3}b.(6分)$

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


Question: $如图所示，在▱ABCD中，\overrightarrow{AB}=a，\overrightarrow{AD}=b，BM=\frac{2}{3}BC，AN=\frac{1}{4}AB.$![Image](https://cdntk.53zaixian.com/tq/1704026521103183873/img/compression/M_4a9bc1d59492f9e465f5677d03fe9479.jpg)
$若AM交DN于点O，求\frac{AO}{OM}的值.$

Solution:
$\because A,O,M三点共线，\therefore \overrightarrow{AO}\parallel \overrightarrow{AM}.(7分)设\overrightarrow{AO}=\lambda \overrightarrow{AM}(0<\lambda <1)，则\overrightarrow{DO}=\overrightarrow{AO}－\overrightarrow{AD}=\lambda \overrightarrow{AM}－\overrightarrow{AD}=\lambda \left(\boldsymbol{a}+\frac{2}{3}\boldsymbol{b}\right)－b=\lambda a+\left(\frac{2}{3}\lambda -1\right)b.(8分)\because D，O，N三点共线,\therefore \overrightarrow{DO}\parallel \overrightarrow{DN},设\overrightarrow{DO}=\mu \overrightarrow{DN}(0<\mu <1)，则\lambda a+\left(\frac{2}{3}\lambda -1\right)b=\mu \left(\frac{1}{4}\boldsymbol{a}-\boldsymbol{b}\right).(9分)\because 向量a，b不共线，\therefore \left\{\begin{matrix}\lambda =\frac{1}{4}\mu ,\\ \frac{2}{3}\lambda -1=-\mu ,\end{matrix}\right.解得\left\{\begin{matrix}\lambda =\frac{3}{14},\\ \mu =\frac{6}{7},\end{matrix}\right.(10分)\therefore \overrightarrow{AO}=\frac{3}{14}\overrightarrow{AM}，\overrightarrow{OM}=\frac{11}{14}\overrightarrow{AM}，\therefore \frac{AO}{OM}=\frac{3}{11}.(12分)$

Final answer: $\frac{3}{11}$
Subfield: Plane Geometry


Question: $如图所示，在▱ABCD中，\overrightarrow{AB}=a，\overrightarrow{AD}=b，BM=\frac{2}{3}BC，AN=\frac{1}{4}AB.$<img_1808>
$若AM交DN于点O，求\frac{AO}{OM}的值.$

Solution:
$\because A,O,M三点共线，\therefore \overrightarrow{AO}\parallel \overrightarrow{AM}.(7分)设\overrightarrow{AO}=\lambda \overrightarrow{AM}(0<\lambda <1)，则\overrightarrow{DO}=\overrightarrow{AO}－\overrightarrow{AD}=\lambda \overrightarrow{AM}－\overrightarrow{AD}=\lambda \left(\boldsymbol{a}+\frac{2}{3}\boldsymbol{b}\right)－b=\lambda a+\left(\frac{2}{3}\lambda -1\right)b.(8分)\because D，O，N三点共线,\therefore \overrightarrow{DO}\parallel \overrightarrow{DN},设\overrightarrow{DO}=\mu \overrightarrow{DN}(0<\mu <1)，则\lambda a+\left(\frac{2}{3}\lambda -1\right)b=\mu \left(\frac{1}{4}\boldsymbol{a}-\boldsymbol{b}\right).(9分)\because 向量a，b不共线，\therefore \left\{\begin{matrix}\lambda =\frac{1}{4}\mu ,\\ \frac{2}{3}\lambda -1=-\mu ,\end{matrix}\right.解得\left\{\begin{matrix}\lambda =\frac{3}{14},\\ \mu =\frac{6}{7},\end{matrix}\right.(10分)\therefore \overrightarrow{AO}=\frac{3}{14}\overrightarrow{AM}，\overrightarrow{OM}=\frac{11}{14}\overrightarrow{AM}，\therefore \frac{AO}{OM}=\frac{3}{11}.(12分)$

Final answer: $\frac{3}{11}$
Subfield: Plane Geometry


Question: $已知C，D是两个小区的所在地，C，D到一条公路AB的垂直距离分别为CA=1 km，DB=2 km，A，B两地之间的距离为4 km.$
$如图1所示，某移动公司将在A，B之间找一点M，在M处建造一个信号塔，使得M对C，D的张角与M对C，A的张角相等，试确定点M与点A之间的距离；
$![Image](https://cdntk.53zaixian.com/tq/1704026528550211587/img/editor/compression/6b00ddf2-56f0-4d21-b482-9d75121b554b.png)

Solution:
$设MA=m km,\angle CMA=\theta ,则MB=(4-m)km,\angle CMD=\theta ,\angle BMD=\pi -2\theta .依题意得\\tan   \theta =\frac{1}{m},\\tan   2\theta =-\frac{2}{4-m},(2分)由\\tan   2\theta =\frac{2\\\tan    \theta }{1-\\\tan   ^2\theta }得-\frac{2}{4-m}=\frac{\frac{2}{m}}{1-\left(\frac{1}{m}\right)^2}，解得m=\frac{1}{4},故点M与点A之间的距离为\frac{1}{4} km.(4分)$

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


Question: $已知C，D是两个小区的所在地，C，D到一条公路AB的垂直距离分别为CA=1 km，DB=2 km，A，B两地之间的距离为4 km.$
$如图1所示，某移动公司将在A，B之间找一点M，在M处建造一个信号塔，使得M对C，D的张角与M对C，A的张角相等，试确定点M与点A之间的距离；
$<img_3007>

Solution:
$设MA=m km,\angle CMA=\theta ,则MB=(4-m)km,\angle CMD=\theta ,\angle BMD=\pi -2\theta .依题意得\\tan   \theta =\frac{1}{m},\\tan   2\theta =-\frac{2}{4-m},(2分)由\\tan   2\theta =\frac{2\\\tan    \theta }{1-\\\tan   ^2\theta }得-\frac{2}{4-m}=\frac{\frac{2}{m}}{1-\left(\frac{1}{m}\right)^2}，解得m=\frac{1}{4},故点M与点A之间的距离为\frac{1}{4} km.(4分)$

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


Question: $已知C，D是两个小区的所在地，C，D到一条公路AB的垂直距离分别为CA=1 km，DB=2 km，A，B两地之间的距离为4 km.$
$如图2所示，某公交公司将在A，B之间找一点N，在N处建造一个公交站台，使得N对C，D两个小区的视角\angle CND最大，试确定点N与点A之间的距离.
$![Image](https://cdntk.53zaixian.com/tq/1704026528579571714/img/editor/compression/62ae9386-0831-47be-bd4c-fada36a62ecb.png)

Solution:
$设\angle CNA=\alpha ,\angle DNB=\beta ,则\angle CND=\pi -(\alpha +\beta ).设AN=x km，则NB=(4-x)km,所以\\tan   \alpha =\frac{1}{x},\\tan   \beta =\frac{2}{4-x} ，\\tan  \angle CND=\\tan  [\pi -(\alpha +\beta )]=-\\tan  (\alpha +\beta )=\frac{x+4}{x^2-4x+2},(6分)记f(x)=\frac{x+4}{x^2-4x+2}(0<x<4,且x\neq 2\pm \sqrt{2}),对x的范围进行分类讨论：①当x接近2\pm \sqrt{2}这两个值时，f(x)趋近于正无穷，此时\angle CND趋近于90^\circ ；②当0<x<2-\sqrt{2}或2+\sqrt{2}<x<4时，\angle CND为锐角;(8分)③当2-\sqrt{2}<x<2+\sqrt{2}时，\angle CND为钝角,令x+4=t,则6-\sqrt{2}<t<6+\sqrt{2},则y=\frac{t}{t^2-12t+34}=\frac{1}{t+\frac{34}{t}-12}\leq \frac{1}{2\sqrt{34}-12}=-\frac{6+\sqrt{34}}{4},当且仅当t=\sqrt{34}时，等号成立，此时y_{max}=-\frac{6+\sqrt{34}}{4},\angle CND最大，x=t-4=\sqrt{34}-4.故当\angle CND最大时，点N与点A之间的距离为(\sqrt{34}-4)km.(12分)$

Final answer: $\sqrt{34}-4$ km
Subfield: Plane Geometry


Question: $已知C，D是两个小区的所在地，C，D到一条公路AB的垂直距离分别为CA=1 km，DB=2 km，A，B两地之间的距离为4 km.$
$如图2所示，某公交公司将在A，B之间找一点N，在N处建造一个公交站台，使得N对C，D两个小区的视角\angle CND最大，试确定点N与点A之间的距离.
$<img_2040>

Solution:
$设\angle CNA=\alpha ,\angle DNB=\beta ,则\angle CND=\pi -(\alpha +\beta ).设AN=x km，则NB=(4-x)km,所以\\tan   \alpha =\frac{1}{x},\\tan   \beta =\frac{2}{4-x} ，\\tan  \angle CND=\\tan  [\pi -(\alpha +\beta )]=-\\tan  (\alpha +\beta )=\frac{x+4}{x^2-4x+2},(6分)记f(x)=\frac{x+4}{x^2-4x+2}(0<x<4,且x\neq 2\pm \sqrt{2}),对x的范围进行分类讨论：①当x接近2\pm \sqrt{2}这两个值时，f(x)趋近于正无穷，此时\angle CND趋近于90^\circ ；②当0<x<2-\sqrt{2}或2+\sqrt{2}<x<4时，\angle CND为锐角;(8分)③当2-\sqrt{2}<x<2+\sqrt{2}时，\angle CND为钝角,令x+4=t,则6-\sqrt{2}<t<6+\sqrt{2},则y=\frac{t}{t^2-12t+34}=\frac{1}{t+\frac{34}{t}-12}\leq \frac{1}{2\sqrt{34}-12}=-\frac{6+\sqrt{34}}{4},当且仅当t=\sqrt{34}时，等号成立，此时y_{max}=-\frac{6+\sqrt{34}}{4},\angle CND最大，x=t-4=\sqrt{34}-4.故当\angle CND最大时，点N与点A之间的距离为(\sqrt{34}-4)km.(12分)$

Final answer: $\sqrt{34}-4$ km
Subfield: Plane Geometry


Question: $如图，在\triangle OAB中，已知P为线段AB上一点，\overrightarrow{OP}=x\overrightarrow{OA}+y\overrightarrow{OB}.$![Image](https://cdntk.53zaixian.com/tq/1704026574263930882/img/compression/M_bbaa268d5e53d0118f51de6c30d4ee45.jpg)
$若\overrightarrow{BP}=\overrightarrow{PA}，求实数x，y的值；$

Solution:
$\because \overrightarrow{BP}=\overrightarrow{PA},\therefore \overrightarrow{BO}+\overrightarrow{OP}=\overrightarrow{PO}+\overrightarrow{OA}，即2\overrightarrow{OP}=\overrightarrow{OB}+\overrightarrow{OA}，(2分)\therefore \overrightarrow{OP}=\frac{1}{2}\overrightarrow{OA}+\frac{1}{2}\overrightarrow{OB}，即x=\frac{1}{2},y=\frac{1}{2}.(4分)$

Final answer: $x=\frac{1}{2}$,$ y=\frac{1}{2}$
Subfield: Plane Geometry


Question: $如图，在\triangle OAB中，已知P为线段AB上一点，\overrightarrow{OP}=x\overrightarrow{OA}+y\overrightarrow{OB}.$<img_3076>
$若\overrightarrow{BP}=\overrightarrow{PA}，求实数x，y的值；$

Solution:
$\because \overrightarrow{BP}=\overrightarrow{PA},\therefore \overrightarrow{BO}+\overrightarrow{OP}=\overrightarrow{PO}+\overrightarrow{OA}，即2\overrightarrow{OP}=\overrightarrow{OB}+\overrightarrow{OA}，(2分)\therefore \overrightarrow{OP}=\frac{1}{2}\overrightarrow{OA}+\frac{1}{2}\overrightarrow{OB}，即x=\frac{1}{2},y=\frac{1}{2}.(4分)$

Final answer: $x=\frac{1}{2}$,$ y=\frac{1}{2}$
Subfield: Plane Geometry


Question: $如图，在\triangle OAB中，已知P为线段AB上一点，\overrightarrow{OP}=x\overrightarrow{OA}+y\overrightarrow{OB}.$<img_3076>
$若\overrightarrow{BP}=3\overrightarrow{PA}，|\overrightarrow{OA}|=4，|\overrightarrow{OB}|=2，且\overrightarrow{OA}与\overrightarrow{OB}的夹角为60^\circ ，求\overrightarrow{OP}\cdot \overrightarrow{AB}的值.$

Solution:
$\because \overrightarrow{BP}=3\overrightarrow{PA}，\therefore \overrightarrow{BO}+\overrightarrow{OP}=3\overrightarrow{PO}+3\overrightarrow{OA}，即4\overrightarrow{OP}=\overrightarrow{OB}+3\overrightarrow{OA}，(6分)\therefore \overrightarrow{OP}=\frac{3}{4}\overrightarrow{OA}+\frac{1}{4}\overrightarrow{OB}，(8分)\therefore \overrightarrow{OP}\cdot \overrightarrow{AB}=\left(\frac{3}{4}\overrightarrow{OA}+\frac{1}{4}\overrightarrow{OB}\right)\cdot (\overrightarrow{OB}-\overrightarrow{OA})=\frac{1}{4}\overrightarrow{OB}^{2}-\frac{3}{4}\overrightarrow{OA}^{2}+\frac{1}{2}\overrightarrow{OA}\cdot \overrightarrow{OB}=\frac{1}{4}\times 2^{2}-\frac{3}{4}\times 4^{2}+\frac{1}{2}\times 4\times 2\times \frac{1}{2}=-9.(12分)$

Final answer: $-9$
Subfield: Plane Geometry


Question: $如图，在\triangle OAB中，已知P为线段AB上一点，\overrightarrow{OP}=x\overrightarrow{OA}+y\overrightarrow{OB}.$![Image](https://cdntk.53zaixian.com/tq/1704026574263930882/img/compression/M_bbaa268d5e53d0118f51de6c30d4ee45.jpg)
$若\overrightarrow{BP}=3\overrightarrow{PA}，|\overrightarrow{OA}|=4，|\overrightarrow{OB}|=2，且\overrightarrow{OA}与\overrightarrow{OB}的夹角为60^\circ ，求\overrightarrow{OP}\cdot \overrightarrow{AB}的值.$

Solution:
$\because \overrightarrow{BP}=3\overrightarrow{PA}，\therefore \overrightarrow{BO}+\overrightarrow{OP}=3\overrightarrow{PO}+3\overrightarrow{OA}，即4\overrightarrow{OP}=\overrightarrow{OB}+3\overrightarrow{OA}，(6分)\therefore \overrightarrow{OP}=\frac{3}{4}\overrightarrow{OA}+\frac{1}{4}\overrightarrow{OB}，(8分)\therefore \overrightarrow{OP}\cdot \overrightarrow{AB}=\left(\frac{3}{4}\overrightarrow{OA}+\frac{1}{4}\overrightarrow{OB}\right)\cdot (\overrightarrow{OB}-\overrightarrow{OA})=\frac{1}{4}\overrightarrow{OB}^{2}-\frac{3}{4}\overrightarrow{OA}^{2}+\frac{1}{2}\overrightarrow{OA}\cdot \overrightarrow{OB}=\frac{1}{4}\times 2^{2}-\frac{3}{4}\times 4^{2}+\frac{1}{2}\times 4\times 2\times \frac{1}{2}=-9.(12分)$

Final answer: $-9$
Subfield: Plane Geometry


Question: $如图，在四边形OABC中，\overrightarrow{OA}=2\overrightarrow{CB},\overrightarrow{BM}=\frac{1}{3}\overrightarrow{BA},\overrightarrow{CP}=x\overrightarrow{CB}(0\leq x\leq 1),\overrightarrow{BA}\cdot \overrightarrow{BC}+\overrightarrow{BO}^{2}=(\overrightarrow{BA}+\overrightarrow{BC})\cdot \overrightarrow{BO}.$![Image](https://cdntk.53zaixian.com/tq/1702594545426038785/img/compression/M_eb13417db39f8f16a44d5d8fd2bffcab.jpg)
$证明:OA\perp OC；$

Solution:
$证明：\because \overrightarrow{BA}\cdot \overrightarrow{BC}+\overrightarrow{BO}^{2}=(\overrightarrow{BA}+\overrightarrow{BC})\cdot \overrightarrow{BO}，\therefore \overrightarrow{BA}\cdot \overrightarrow{BC}-\overrightarrow{BC}\cdot \overrightarrow{BO}+\overrightarrow{BO}^{2}-\overrightarrow{BA}\cdot \overrightarrow{BO}=0，即\overrightarrow{BC}\cdot (\overrightarrow{BA}-\overrightarrow{BO})-\overrightarrow{BO}\cdot (\overrightarrow{BA}-\overrightarrow{BO})=0，\therefore \overrightarrow{BC}\cdot \overrightarrow{OA}-\overrightarrow{BO}\cdot \overrightarrow{OA}=0，即(\overrightarrow{BC}-\overrightarrow{BO})\cdot \overrightarrow{OA}=0，即\overrightarrow{OC}\cdot \overrightarrow{OA}=0，\therefore OA\perp OC.$

Final answer: 证明题
Subfield: Plane Geometry


Question: $如图，在四边形OABC中，\overrightarrow{OA}=2\overrightarrow{CB},\overrightarrow{BM}=\frac{1}{3}\overrightarrow{BA},\overrightarrow{CP}=x\overrightarrow{CB}(0\leq x\leq 1),\overrightarrow{BA}\cdot \overrightarrow{BC}+\overrightarrow{BO}^{2}=(\overrightarrow{BA}+\overrightarrow{BC})\cdot \overrightarrow{BO}.$<img_1836>
$证明:OA\perp OC；$

Solution:
$证明：\because \overrightarrow{BA}\cdot \overrightarrow{BC}+\overrightarrow{BO}^{2}=(\overrightarrow{BA}+\overrightarrow{BC})\cdot \overrightarrow{BO}，\therefore \overrightarrow{BA}\cdot \overrightarrow{BC}-\overrightarrow{BC}\cdot \overrightarrow{BO}+\overrightarrow{BO}^{2}-\overrightarrow{BA}\cdot \overrightarrow{BO}=0，即\overrightarrow{BC}\cdot (\overrightarrow{BA}-\overrightarrow{BO})-\overrightarrow{BO}\cdot (\overrightarrow{BA}-\overrightarrow{BO})=0，\therefore \overrightarrow{BC}\cdot \overrightarrow{OA}-\overrightarrow{BO}\cdot \overrightarrow{OA}=0，即(\overrightarrow{BC}-\overrightarrow{BO})\cdot \overrightarrow{OA}=0，即\overrightarrow{OC}\cdot \overrightarrow{OA}=0，\therefore OA\perp OC.$
Subfield: Plane Geometry


Question: $如图，在四边形OABC中，\overrightarrow{OA}=2\overrightarrow{CB},\overrightarrow{BM}=\frac{1}{3}\overrightarrow{BA},\overrightarrow{CP}=x\overrightarrow{CB}(0\leq x\leq 1),\overrightarrow{BA}\cdot \overrightarrow{BC}+\overrightarrow{BO}^{2}=(\overrightarrow{BA}+\overrightarrow{BC})\cdot \overrightarrow{BO}.$<img_1836>
$设\overrightarrow{OM}=\lambda \overrightarrow{CA}+\mu \overrightarrow{OP}，求\lambda \mu 的最大值，并求出\lambda \mu 取得最大值时x的值.$

Solution:
$由题意得\overrightarrow{OM}=\overrightarrow{OA}+\overrightarrow{AM}=\overrightarrow{OA}+\frac{2}{3}\overrightarrow{AB}=\overrightarrow{OA}+\frac{2}{3}(\overrightarrow{OB}-\overrightarrow{OA})=\frac{2}{3}\overrightarrow{OB}+\frac{1}{3}\overrightarrow{OA}=\frac{2}{3}(\overrightarrow{OC}+\overrightarrow{CB})+\frac{1}{3}\overrightarrow{OA}=\frac{2}{3}\overrightarrow{OC}+\frac{2}{3}\times \frac{1}{2}\overrightarrow{OA}+\frac{1}{3}\overrightarrow{OA}=\frac{2}{3}\overrightarrow{OA}+\frac{2}{3}\overrightarrow{OC}.又\overrightarrow{OM}=\lambda \overrightarrow{CA}+\mu \overrightarrow{OP}=\lambda (\overrightarrow{OA}-\overrightarrow{OC})+\mu (\overrightarrow{OC}+\overrightarrow{CP})=\lambda (\overrightarrow{OA}-\overrightarrow{OC})+\mu \left(\overrightarrow{OC}+\frac{x}{2}\overrightarrow{OA}\right)=\left(\lambda +\frac{\mu x}{2}\right)\overrightarrow{OA}+(\mu -\lambda )\overrightarrow{OC}，\therefore \left\{\begin{matrix}\lambda +\frac{\mu x}{2}=\frac{2}{3},\\ \mu -\lambda =\frac{2}{3},\end{matrix}\right.即\left\{\begin{matrix}\mu =\frac{8}{3(x+2)},\\ \lambda =\mu -\frac{2}{3},\end{matrix}\right.\therefore \lambda \mu =\mu \left(\mu -\frac{2}{3}\right)=\left(\mu -\frac{1}{3}\right)^{2}-\frac{1}{9}.\because 0\leq x\leq 1,\therefore \frac{8}{9}\leq \mu \leq \frac{4}{3}，\therefore 当\mu =\frac{4}{3}时，\lambda \mu 取得最大值\frac{8}{9}，此时x=0.$

Final answer: $\frac{8}{9}$,$0$
Subfield: Plane Geometry


Question: $如图，在四边形OABC中，\overrightarrow{OA}=2\overrightarrow{CB},\overrightarrow{BM}=\frac{1}{3}\overrightarrow{BA},\overrightarrow{CP}=x\overrightarrow{CB}(0\leq x\leq 1),\overrightarrow{BA}\cdot \overrightarrow{BC}+\overrightarrow{BO}^{2}=(\overrightarrow{BA}+\overrightarrow{BC})\cdot \overrightarrow{BO}.$![Image](https://cdntk.53zaixian.com/tq/1702594545426038785/img/compression/M_eb13417db39f8f16a44d5d8fd2bffcab.jpg)
$设\overrightarrow{OM}=\lambda \overrightarrow{CA}+\mu \overrightarrow{OP}，求\lambda \mu 的最大值，并求出\lambda \mu 取得最大值时x的值.$

Solution:
$由题意得\overrightarrow{OM}=\overrightarrow{OA}+\overrightarrow{AM}=\overrightarrow{OA}+\frac{2}{3}\overrightarrow{AB}=\overrightarrow{OA}+\frac{2}{3}(\overrightarrow{OB}-\overrightarrow{OA})=\frac{2}{3}\overrightarrow{OB}+\frac{1}{3}\overrightarrow{OA}=\frac{2}{3}(\overrightarrow{OC}+\overrightarrow{CB})+\frac{1}{3}\overrightarrow{OA}=\frac{2}{3}\overrightarrow{OC}+\frac{2}{3}\times \frac{1}{2}\overrightarrow{OA}+\frac{1}{3}\overrightarrow{OA}=\frac{2}{3}\overrightarrow{OA}+\frac{2}{3}\overrightarrow{OC}.又\overrightarrow{OM}=\lambda \overrightarrow{CA}+\mu \overrightarrow{OP}=\lambda (\overrightarrow{OA}-\overrightarrow{OC})+\mu (\overrightarrow{OC}+\overrightarrow{CP})=\lambda (\overrightarrow{OA}-\overrightarrow{OC})+\mu \left(\overrightarrow{OC}+\frac{x}{2}\overrightarrow{OA}\right)=\left(\lambda +\frac{\mu x}{2}\right)\overrightarrow{OA}+(\mu -\lambda )\overrightarrow{OC}，\therefore \left\{\begin{matrix}\lambda +\frac{\mu x}{2}=\frac{2}{3},\\ \mu -\lambda =\frac{2}{3},\end{matrix}\right.即\left\{\begin{matrix}\mu =\frac{8}{3(x+2)},\\ \lambda =\mu -\frac{2}{3},\end{matrix}\right.\therefore \lambda \mu =\mu \left(\mu -\frac{2}{3}\right)=\left(\mu -\frac{1}{3}\right)^{2}-\frac{1}{9}.\because 0\leq x\leq 1,\therefore \frac{8}{9}\leq \mu \leq \frac{4}{3}，\therefore 当\mu =\frac{4}{3}时，\lambda \mu 取得最大值\frac{8}{9}，此时x=0.$

Final answer: $\frac{8}{9}$,$0$
Subfield: Plane Geometry


Question: $如图，一个直角走廊的宽分别为a，b，一铁棒与廊壁成\theta 角，该铁棒欲通过该直角走廊.$<img_1963>
$求铁棒的长度L(用含\theta 的表达式表示)；$

Solution:
$作出示意图如图所示，$<img_2941>$易知\angle ACD=\angle CBE=\theta ，AD=b,BE=a.在\triangle ACD中，AC=\frac{AD}{\\\sin    \theta }=\frac{b}{\\\sin    \theta }，在\triangle BCE中，BC=\frac{BE}{\\\cos    \theta }=\frac{a}{\\\cos    \theta }，所以L=AC+BC=\frac{b}{\\\sin    \theta }+\frac{a}{\\\cos    \theta },\theta \in \left(0,\frac{\pi }{2}\right).$

Final answer: $L = b/\sin (\theta ) + a/\cos (\theta )$
Subfield: Plane Geometry


Question: $如图，一个直角走廊的宽分别为a，b，一铁棒与廊壁成\theta 角，该铁棒欲通过该直角走廊.$![Image](https://cdntk.53zaixian.com/tq/1702594597028560898/img/compression/M_d90bc66c4f5c5d43e4f6559cb7e80196.jpg)
$求铁棒的长度L(用含\theta 的表达式表示)；$

Solution:
$作出示意图如图所示，$![Image](https://cdntk.53zaixian.com/tq/1702594597401853954/img/compression/M_7dee053b23a14f61342b5edbd53d7346.jpg)$易知\angle ACD=\angle CBE=\theta ，AD=b,BE=a.在\triangle ACD中，AC=\frac{AD}{\\\sin    \theta }=\frac{b}{\\\sin    \theta }，在\triangle BCE中，BC=\frac{BE}{\\\cos    \theta }=\frac{a}{\\\cos    \theta }，所以L=AC+BC=\frac{b}{\\\sin    \theta }+\frac{a}{\\\cos    \theta },\theta \in \left(0,\frac{\pi }{2}\right).$

Final answer: $L = b/\sin (\theta ) + a/\cos (\theta )$
Subfield: Plane Geometry


Question: $如图，一个直角走廊的宽分别为a，b，一铁棒与廊壁成\theta 角，该铁棒欲通过该直角走廊.$<img_1963>
$当a=b=2时，求能够通过这个直角走廊的铁棒的长度的最大值.$

Solution:
$当a=b=2时，L=\frac{b}{\\\sin    \theta }+\frac{a}{\\\cos    \theta }=\frac{2}{\\\sin    \theta }+\frac{2}{\\\cos    \theta }=2\left(\frac{\\\sin    \theta +\\\cos    \theta }{\\\sin    \theta \\\cos    \theta }\right)=2\sqrt{\frac{{(\\\mathrm{\\sin  }    \theta +\\\mathrm{\\cos   }   \theta )}^2}{{\\\mathrm{\\sin  }    }^2\theta \\\mathrm{\\cos  }   ^2\theta }}=4\sqrt{\frac{1+\\\mathrm{\\sin  }    2\theta }{{\\\mathrm{\\sin  }    }^22\theta }}=4\sqrt{\frac{1}{{\\\mathrm{\\sin  }    }^22\theta }+\frac{1}{\\\mathrm{\\sin  }    2\theta }}.令t=\frac{1}{\\\sin    2\theta }，因为\theta \in \left(0,\frac{\pi }{2}\right)，所以2\theta \in (0,\pi )，所以\\sin   2\theta \in (0,1]，所以t\geq 1.所以L=4\sqrt{\frac{1}{{\\\mathrm{\\sin  }    }^22\theta }+ \frac{1}{\\\mathrm{\\sin  }    2\theta }}=4\sqrt{t^2+t}=4\sqrt{{\left(t+ \frac{1}{2}\right)}^2- \frac{1}{4}},t\geq 1，易知当t=1，即\theta =\frac{\pi }{4}时，L取得最小值，为4\sqrt{2}，所以能够通过这个直角走廊的铁棒的长度的最大值为4\sqrt{2}.$

Final answer: $4\sqrt 2$