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


Question: $如图，在平面四边形ABCD中，BC\parallel AD，AB=BC=2，AD=4，\angle BAD=120^\circ ，E，F分别是AD，DC的中点，G为线段BC上一点，且\overrightarrow{BG}=\lambda \overrightarrow{BC}.设\overrightarrow{AB}=a，\overrightarrow{AD}=b.$<img_2450>
$若\lambda \in (0,1)，求\overrightarrow{AF}\cdot \overrightarrow{EG}的取值范围.$

Solution:
$\overrightarrow{EG}=\overrightarrow{EA}+\overrightarrow{AB}+\overrightarrow{BG}=-\frac{1}{2}b+a+\lambda \overrightarrow{BC}=-\frac{1}{2}b+a+\frac{1}{2}\lambda b=a+\frac{1}{2}(\lambda -1)b.(6分)因为\angle BAD=120^\circ ，|a|=2，|b|=4，所以a\cdot b=|a|\cdot |b|\\cos   120^\circ =2\times 4\times \left(-\frac{1}{2}\right)=-4，(7分)所以\overrightarrow{AF}\cdot \overrightarrow{EG}=\left(\frac{1}{2}\boldsymbol{a}+\frac{3}{4}\boldsymbol{b}\right)\cdot \left[\boldsymbol{a}+\frac{1}{2}(\lambda -1)\boldsymbol{b}\right]=\frac{1}{2}a^{2}+\left(\frac{1}{4}\lambda +\frac{1}{2}\right)a\cdot b+\frac{3}{8}(\lambda -1)b^{2}=\frac{1}{2}\times 2^{2}+\left(\frac{1}{4}\lambda +\frac{1}{2}\right)\times (-4)+\frac{3}{8}(\lambda -1)\times 4^{2}=5\lambda -6，(10分)因为0<\lambda <1，所以-6<5\lambda -6<-1，所以-6<\overrightarrow{AF}\cdot \overrightarrow{EG}<-1，故\overrightarrow{AF}\cdot \overrightarrow{EG}的取值范围为(-6，-1).(12分)$

Final answer: $(-6, -1)$
Subfield: Plane Geometry


Question: $如图，在平面四边形ABCD中，BC\parallel AD，AB=BC=2，AD=4，\angle BAD=120^\circ ，E，F分别是AD，DC的中点，G为线段BC上一点，且\overrightarrow{BG}=\lambda \overrightarrow{BC}.设\overrightarrow{AB}=a，\overrightarrow{AD}=b.$![Image](https://cdntk.53zaixian.com/tq/1703969347091439619/img/compression/M_2718d250de363a341692b50678229215.jpg)
$若\lambda \in (0,1)，求\overrightarrow{AF}\cdot \overrightarrow{EG}的取值范围.$

Solution:
$\overrightarrow{EG}=\overrightarrow{EA}+\overrightarrow{AB}+\overrightarrow{BG}=-\frac{1}{2}b+a+\lambda \overrightarrow{BC}=-\frac{1}{2}b+a+\frac{1}{2}\lambda b=a+\frac{1}{2}(\lambda -1)b.(6分)因为\angle BAD=120^\circ ，|a|=2，|b|=4，所以a\cdot b=|a|\cdot |b|\\cos   120^\circ =2\times 4\times \left(-\frac{1}{2}\right)=-4，(7分)所以\overrightarrow{AF}\cdot \overrightarrow{EG}=\left(\frac{1}{2}\boldsymbol{a}+\frac{3}{4}\boldsymbol{b}\right)\cdot \left[\boldsymbol{a}+\frac{1}{2}(\lambda -1)\boldsymbol{b}\right]=\frac{1}{2}a^{2}+\left(\frac{1}{4}\lambda +\frac{1}{2}\right)a\cdot b+\frac{3}{8}(\lambda -1)b^{2}=\frac{1}{2}\times 2^{2}+\left(\frac{1}{4}\lambda +\frac{1}{2}\right)\times (-4)+\frac{3}{8}(\lambda -1)\times 4^{2}=5\lambda -6，(10分)因为0<\lambda <1，所以-6<5\lambda -6<-1，所以-6<\overrightarrow{AF}\cdot \overrightarrow{EG}<-1，故\overrightarrow{AF}\cdot \overrightarrow{EG}的取值范围为(-6，-1).(12分)$

Final answer: $(-6, -1)$
Subfield: Plane Geometry


Question: $如图，正方形ABCD的边长为6，E是AB的中点，F是BC边上靠近点B的三等分点，AF与DE交于点M.$<img_1868>
$求\angle EMF的余弦值；$

Solution:
$如图所示，以点A为原点建立平面直角坐标系，$<img_2787>$则D(0,6),E(3,0),A(0,0),F(6,2)，\therefore \overrightarrow{DE}=(3,-6)，\overrightarrow{AF}=(6,2)，(2分)\because \angle EMF是\overrightarrow{DE}与\overrightarrow{AF}的夹角，\therefore \\cos  \angle EMF=\\cos  \langle \overrightarrow{DE},\overrightarrow{AF}\rangle =\frac{18-12}{\sqrt{9+36}\times \sqrt{36+4}}=\frac{\sqrt{2}}{10}，\therefore \angle EMF的余弦值为\frac{\sqrt{2}}{10}.(4分)$

Final answer: $\frac{\sqrt{2}}{10}$
Subfield: Plane Geometry


Question: $如图，正方形ABCD的边长为6，E是AB的中点，F是BC边上靠近点B的三等分点，AF与DE交于点M.$![Image](https://cdntk.53zaixian.com/tq/1703969347443761153/img/compression/M_0b3fa33efb16f3fb3c827cd3a85fc125.jpg)
$求\angle EMF的余弦值；$

Solution:
$如图所示，以点A为原点建立平面直角坐标系，$![Image](https://cdntk.53zaixian.com/tq/1703969347695419393/img/compression/M_133eb130fc1a89171068752c2cf4993e.jpg)$则D(0,6),E(3,0),A(0,0),F(6,2)，\therefore \overrightarrow{DE}=(3,-6)，\overrightarrow{AF}=(6,2)，(2分)\because \angle EMF是\overrightarrow{DE}与\overrightarrow{AF}的夹角，\therefore \\cos  \angle EMF=\\cos  \langle \overrightarrow{DE},\overrightarrow{AF}\rangle =\frac{18-12}{\sqrt{9+36}\times \sqrt{36+4}}=\frac{\sqrt{2}}{10}，\therefore \angle EMF的余弦值为\frac{\sqrt{2}}{10}.(4分)$

Final answer: $\frac{\sqrt{2}}{10}$
Subfield: Plane Geometry


Question: $如图，正方形ABCD的边长为6，E是AB的中点，F是BC边上靠近点B的三等分点，AF与DE交于点M.$<img_1868>
$设\overrightarrow{AM}=\lambda \overrightarrow{AF}，求\lambda 的值及点M的坐标；$

Solution:
$\because \overrightarrow{AM}=\lambda \overrightarrow{AF}，\therefore \overrightarrow{AM}=(6\lambda ,2\lambda )，则M(6\lambda ,2\lambda ),由D,M,E三点共线，可设\overrightarrow{DM}=t\overrightarrow{DE},0即(6\lambda ,2\lambda -6)=t(3,-6)，则\left\{\begin{matrix}6\lambda =3t,\\ 2\lambda -6=-6t,\end{matrix}\right.解得\lambda =\frac{3}{7}，故M\left(\frac{18}{7},\frac{6}{7}\right).(6分)$

Final answer: $\frac{3}{7}$,$ (\frac{18}{7}, \frac{6}{7})$
Subfield: Plane Geometry


Question: $如图，正方形ABCD的边长为6，E是AB的中点，F是BC边上靠近点B的三等分点，AF与DE交于点M.$![Image](https://cdntk.53zaixian.com/tq/1703969347443761153/img/compression/M_0b3fa33efb16f3fb3c827cd3a85fc125.jpg)
$设\overrightarrow{AM}=\lambda \overrightarrow{AF}，求\lambda 的值及点M的坐标；$

Solution:
$\because \overrightarrow{AM}=\lambda \overrightarrow{AF}，\therefore \overrightarrow{AM}=(6\lambda ,2\lambda )，则M(6\lambda ,2\lambda ),由D,M,E三点共线，可设\overrightarrow{DM}=t\overrightarrow{DE},0即(6\lambda ,2\lambda -6)=t(3,-6)，则\left\{\begin{matrix}6\lambda =3t,\\ 2\lambda -6=-6t,\end{matrix}\right.解得\lambda =\frac{3}{7}，故M\left(\frac{18}{7},\frac{6}{7}\right).(6分)$

Final answer: $\frac{3}{7}$,$ (\frac{18}{7}, \frac{6}{7})$
Subfield: Plane Geometry


Question: $如图，正方形ABCD的边长为6，E是AB的中点，F是BC边上靠近点B的三等分点，AF与DE交于点M.$![Image](https://cdntk.53zaixian.com/tq/1703969347443761153/img/compression/M_0b3fa33efb16f3fb3c827cd3a85fc125.jpg)
$若点P自点A按逆时针方向沿正方形的边运动一周回到点A，在这个过程中，是否存在这样的点P，使得EF\perp MP？若存在，求出MP的长度；若不存在，请说明理由.$

Solution:
$存在.由(1)得\overrightarrow{EF}=(3,2)，假设存在点P，使得EF\perp MP，则\overrightarrow{EF}\cdot \overrightarrow{MP}=0.①当点P在AB上时，设P(x,0)(0\leq x\leq 6)，由(2)知M\left(\frac{18}{7},\frac{6}{7}\right)，则\overrightarrow{MP}=\left(x-\frac{18}{7},-\frac{6}{7}\right)，则3x-\frac{54}{7}-\frac{12}{7}=0，\therefore x=\frac{22}{7}，满足题意，\therefore P\left(\frac{22}{7},0\right)，故MP=\sqrt{\left(\frac{22}{7}-\frac{18}{7}\right)^2+\left(-\frac{6}{7}\right)^2}=\frac{2\sqrt{13}}{7}；②当点P在BC上时，设P(6,y),0\leq y\leq 6，由(2)知M\left(\frac{18}{7},\frac{6}{7}\right)，则\overrightarrow{MP}=\left(\frac{24}{7},y-\frac{6}{7}\right)，则\frac{72}{7}+2y-\frac{12}{7}=0,\therefore y=-\frac{30}{7}(舍去)；(9分)③当点P在CD上时，设P(x,6),0\leq x\leq 6，由(2)知M\left(\frac{18}{7},\frac{6}{7}\right)，则\overrightarrow{MP}=\left(x-\frac{18}{7},\frac{36}{7}\right)，则3x-\frac{54}{7}+\frac{72}{7}=0，\therefore x=-\frac{6}{7}(舍去)；④当点P在DA上时，设P(0,y),0\leq y\leq 6，由(2)知M\left(\frac{18}{7},\frac{6}{7}\right)，则\overrightarrow{MP}=\left(-\frac{18}{7},y-\frac{6}{7}\right)，则-\frac{54}{7}+2y-\frac{12}{7}=0,\therefore y=\frac{33}{7}，满足题意，\therefore P\left(0,\frac{33}{7}\right),故MP=\sqrt{\left(-\frac{18}{7}\right)^2+\left(\frac{33}{7}-\frac{6}{7}\right)^2}=\frac{9\sqrt{13}}{7}.综上，存在符合题意的点P\left(\frac{22}{7},0\right)或P\left(0,\frac{33}{7}\right)，使得EF\perp MP，此时MP的长度为\frac{2\sqrt{13}}{7}或\frac{9\sqrt{13}}{7}.(12分)$

Final answer: $存在符合题意的点P$, $\frac{2\sqrt 13}{7}, \frac{9\sqrt 13}{7}$
Subfield: Plane Geometry


Question: $如图，正方形ABCD的边长为6，E是AB的中点，F是BC边上靠近点B的三等分点，AF与DE交于点M.$<img_1868>
$若点P自点A按逆时针方向沿正方形的边运动一周回到点A，在这个过程中，存在点P，使得EF\perp MP, 请求出MP的所有可能长度.$

Solution:
$存在.由(1)得\overrightarrow{EF}=(3,2)，假设存在点P，使得EF\perp MP，则\overrightarrow{EF}\cdot \overrightarrow{MP}=0.①当点P在AB上时，设P(x,0)(0\leq x\leq 6)，由(2)知M\left(\frac{18}{7},\frac{6}{7}\right)，则\overrightarrow{MP}=\left(x-\frac{18}{7},-\frac{6}{7}\right)，则3x-\frac{54}{7}-\frac{12}{7}=0，\therefore x=\frac{22}{7}，满足题意，\therefore P\left(\frac{22}{7},0\right)，故MP=\sqrt{\left(\frac{22}{7}-\frac{18}{7}\right)^2+\left(-\frac{6}{7}\right)^2}=\frac{2\sqrt{13}}{7}；②当点P在BC上时，设P(6,y),0\leq y\leq 6，由(2)知M\left(\frac{18}{7},\frac{6}{7}\right)，则\overrightarrow{MP}=\left(\frac{24}{7},y-\frac{6}{7}\right)，则\frac{72}{7}+2y-\frac{12}{7}=0,\therefore y=-\frac{30}{7}(舍去)；(9分)③当点P在CD上时，设P(x,6),0\leq x\leq 6，由(2)知M\left(\frac{18}{7},\frac{6}{7}\right)，则\overrightarrow{MP}=\left(x-\frac{18}{7},\frac{36}{7}\right)，则3x-\frac{54}{7}+\frac{72}{7}=0，\therefore x=-\frac{6}{7}(舍去)；④当点P在DA上时，设P(0,y),0\leq y\leq 6，由(2)知M\left(\frac{18}{7},\frac{6}{7}\right)，则\overrightarrow{MP}=\left(-\frac{18}{7},y-\frac{6}{7}\right)，则-\frac{54}{7}+2y-\frac{12}{7}=0,\therefore y=\frac{33}{7}，满足题意，\therefore P\left(0,\frac{33}{7}\right),故MP=\sqrt{\left(-\frac{18}{7}\right)^2+\left(\frac{33}{7}-\frac{6}{7}\right)^2}=\frac{9\sqrt{13}}{7}.综上，存在符合题意的点P\left(\frac{22}{7},0\right)或P\left(0,\frac{33}{7}\right)，使得EF\perp MP，此时MP的长度为\frac{2\sqrt{13}}{7}或\frac{9\sqrt{13}}{7}.(12分)$



Final answer: $\frac{2\sqrt 13}{7}, \frac{9\sqrt 13}{7}$
Subfield: Plane Geometry


Question: $如图，有一四边形休闲区域ABCD，其四周是步道，中间是花卉种植区域，为减少拥堵，中间穿插了新能源环保电动车道AC,\angle D=2\angle B，且AD=1,CD=3,\\cos   B=\frac{\sqrt{3}}{3}.$![Image](https://cdntk.53zaixian.com/tq/1703968916042432514/img/compression/M_b8f2f920ea603d0dca858a74d3bacdbc.jpg)$
注：如果选择条件①和条件②分别解答，按第一个解答计分.$
$求新能源环保电动车道AC的长；$

Solution:
$\because \\cos   B=\frac{\sqrt{3}}{3}，\angle D=2\angle B，\therefore \\cos   D=\\cos   2B=2\\cos  ^{2}B-1=-\frac{1}{3}.(2分)在\triangle ADC中，AD=1，CD=3，由余弦定理得AC^{2}=AD^{2}+CD^{2}-2AD\cdot CD\\cos   D=1+9-6\times \left(-\frac{1}{3}\right)=12，(4分)\therefore AC=2\sqrt{3}．(5分)$

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


Question: $如图，有一四边形休闲区域ABCD，其四周是步道，中间是花卉种植区域，为减少拥堵，中间穿插了新能源环保电动车道AC,\angle D=2\angle B，且AD=1,CD=3,\\cos   B=\frac{\sqrt{3}}{3}.$<img_2194>$
注：如果选择条件①和条件②分别解答，按第一个解答计分.$
$求新能源环保电动车道AC的长；$

Solution:
$\because \\cos   B=\frac{\sqrt{3}}{3}，\angle D=2\angle B，\therefore \\cos   D=\\cos   2B=2\\cos  ^{2}B-1=-\frac{1}{3}.(2分)在\triangle ADC中，AD=1，CD=3，由余弦定理得AC^{2}=AD^{2}+CD^{2}-2AD\cdot CD\\cos   D=1+9-6\times \left(-\frac{1}{3}\right)=12，(4分)\therefore AC=2\sqrt{3}．(5分)$

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


Question: $如图，有一四边形休闲区域ABCD，其四周是步道，中间是花卉种植区域，为减少拥堵，中间穿插了新能源环保电动车道AC,\angle D=2\angle B，且AD=1,CD=3,\\cos   B=\frac{\sqrt{3}}{3}.$![Image](https://cdntk.53zaixian.com/tq/1703968916042432514/img/compression/M_b8f2f920ea603d0dca858a74d3bacdbc.jpg)$
注：如果选择条件①和条件②分别解答，按第一个解答计分.$
$若　　　，求花卉种植区域的总面积(道路宽度不计).从①\angle ACB=\frac{\pi }{3}，②BC=\sqrt{6}这两个条件中任选一个，补充在上面问题中并作答.$

Solution:
$若选①：\because \\cos   B=\frac{\sqrt{3}}{3}，\therefore \\sin   B=\frac{\sqrt{6}}{3}.在\triangle ABC中，由正弦定理得\frac{AB}{\\\sin   \angle ACB}=\frac{AC}{\\\sin    B}，即\frac{AB}{\frac{\sqrt{3}}{2}}=\frac{2\sqrt{3}}{\frac{\sqrt{6}}{3}}，解得AB=\frac{3\sqrt{6}}{2}，(6分)\because \\sin  \angle BAC=\\sin  (\angle B+\angle ACB)=\\sin   B\\cos  \angle ACB+\\cos   B\\sin  \angle ACB=\frac{\sqrt{6}}{3}\times \frac{1}{2}+\frac{\sqrt{3}}{3}\times \frac{\sqrt{3}}{2}=\frac{3+\sqrt{6}}{6}，\therefore S_{\triangle ABC}=\frac{1}{2}AB\cdot AC\cdot \\sin  \angle BAC=\frac{1}{2}\times \frac{3\sqrt{6}}{2}\times 2\sqrt{3}\times \frac{3+\sqrt{6}}{6}=\frac{6\sqrt{3}+9\sqrt{2}}{4}.(8分)\because \\cos   D=-\frac{1}{3}，\therefore \\sin   D=\sqrt{1-\frac{1}{9}}=\frac{2\sqrt{2}}{3}，故S_{\triangle ADC}=\frac{1}{2}AD\cdot CD\cdot \\sin   D=\frac{1}{2}\times 1\times 3\times \frac{2\sqrt{2}}{3}=\sqrt{2}，(9分)\therefore 花卉种植区域的总面积为\frac{6\sqrt{3}+9\sqrt{2}}{4}+\sqrt{2}=\frac{6\sqrt{3}+13\sqrt{2}}{4}．(10分)若选②：在\triangle ABC中，由余弦定理得\\cos   B=\frac{AB^2+6-12}{2\sqrt{6}\times AB}=\frac{\sqrt{3}}{3}，解得AB=3\sqrt{2}或AB=-\sqrt{2}(舍去).(6分)\because \\cos   B=\frac{\sqrt{3}}{3}，\therefore \\sin   B=\frac{\sqrt{6}}{3}，\therefore S_{\triangle ABC}=\frac{1}{2}AB\cdot BC\cdot \\sin   B=\frac{1}{2}\times 3\sqrt{2}\times \sqrt{6}\times \frac{\sqrt{6}}{3}=3\sqrt{2}.(8分)\because \\cos   D=-\frac{1}{3}，\therefore \\sin   D=\sqrt{1-\frac{1}{9}}=\frac{2\sqrt{2}}{3}，故S_{\triangle ADC}=\frac{1}{2}AD\cdot CD\cdot \\sin   D=\frac{1}{2}\times 1\times 3\times \frac{2\sqrt{2}}{3}=\sqrt{2}，(9分)\therefore 花卉种植区域的总面积为3\sqrt{2}+\sqrt{2}=4\sqrt{2}．(10分)$

Final answer: $问题不完整，没有给出选项(①和②)。请给出选项后再进行提问。$
Subfield: Plane Geometry


Question: $如图，有一四边形休闲区域ABCD，其四周是步道，中间是花卉种植区域，为减少拥堵，中间穿插了新能源环保电动车道AC,\angle D=2\angle B，且AD=1,CD=3,\\cos   B=\frac{\sqrt{3}}{3}.$<img_2194>$
注：如果选择条件①和条件②分别解答，按第一个解答计分.$
$若\angle ACB=\frac{\pi }{3}，求花卉种植区域的总面积(道路宽度不计).$

Solution:
\because \\cos   B=\frac{\sqrt{3}}{3}，\therefore \\sin   B=\frac{\sqrt{6}}{3}.在\triangle ABC中，由正弦定理得\frac{AB}{\\\sin   \angle ACB}=\frac{AC}{\\\sin    B}，即\frac{AB}{\frac{\sqrt{3}}{2}}=\frac{2\sqrt{3}}{\frac{\sqrt{6}}{3}}，解得AB=\frac{3\sqrt{6}}{2}，(6分)\because \\sin  \angle BAC=\\sin  (\angle B+\angle ACB)=\\sin   B\\cos  \angle ACB+\\cos   B\\sin  \angle ACB=\frac{\sqrt{6}}{3}\times \frac{1}{2}+\frac{\sqrt{3}}{3}\times \frac{\sqrt{3}}{2}=\frac{3+\sqrt{6}}{6}，\therefore S_{\triangle ABC}=\frac{1}{2}AB\cdot AC\cdot \\sin  \angle BAC=\frac{1}{2}\times \frac{3\sqrt{6}}{2}\times 2\sqrt{3}\times \frac{3+\sqrt{6}}{6}=\frac{6\sqrt{3}+9\sqrt{2}}{4}.(8分)\because \\cos   D=-\frac{1}{3}，\therefore \\sin   D=\sqrt{1-\frac{1}{9}}=\frac{2\sqrt{2}}{3}，故S_{\triangle ADC}=\frac{1}{2}AD\cdot CD\cdot \\sin   D=\frac{1}{2}\times 1\times 3\times \frac{2\sqrt{2}}{3}=\sqrt{2}，(9分)\therefore 花卉种植区域的总面积为\frac{6\sqrt{3}+9\sqrt{2}}{4}+\sqrt{2}=\frac{6\sqrt{3}+13\sqrt{2}}{4}．



Final answer: \frac{6\sqrt{3}+13\sqrt{2}}{4}
Subfield: Plane Geometry


Question: $某数学课外活动小组为了测量一个摩天轮的最高点P距地面的高度，选取了与点P在地面上的射影A在同一水平面内的两个测量基点B，C(如图所示)，现测得\angle ABC=\angle ACB=\angle ACP=30^\circ ，B，C两点间的距离是390米.$![Image](https://cdntk.53zaixian.com/tq/1703968916453474306/img/compression/M_e6e85f03fee6ea27acb37a1fc920dc2c.jpg)
$求最高点P距地面的高度PA；$

Solution:
$由题意得\angle BAC=120^\circ .(1分)在\triangle ABC中，由正弦定理，得\frac{AC}{\\\sin   \angle ABC}=\frac{BC}{\\\sin   \angle BAC}，所以AC=\frac{390}{\frac{\sqrt{3}}{2}}\times \frac{1}{2}=\frac{390}{\sqrt{3}}=130\sqrt{3}(米).(3分)因为PA\perp AC，\angle PCA=30^\circ ，所以\frac{PA}{AC}=\\tan   30^\circ =\frac{\sqrt{3}}{3}，所以PA=\frac{\sqrt{3}}{3}AC=\frac{\sqrt{3}}{3}\times 130\sqrt{3}=130(米)，故最高点P距地面的高度PA为130米.(6分)$

Final answer: $130$ 米
Subfield: Plane Geometry


Question: $某数学课外活动小组为了测量一个摩天轮的最高点P距地面的高度，选取了与点P在地面上的射影A在同一水平面内的两个测量基点B，C(如图所示)，现测得\angle ABC=\angle ACB=\angle ACP=30^\circ ，B，C两点间的距离是390米.$<img_2079>
$求最高点P距地面的高度PA；$

Solution:
$由题意得\angle BAC=120^\circ .(1分)在\triangle ABC中，由正弦定理，得\frac{AC}{\\\sin   \angle ABC}=\frac{BC}{\\\sin   \angle BAC}，所以AC=\frac{390}{\frac{\sqrt{3}}{2}}\times \frac{1}{2}=\frac{390}{\sqrt{3}}=130\sqrt{3}(米).(3分)因为PA\perp AC，\angle PCA=30^\circ ，所以\frac{PA}{AC}=\\tan   30^\circ =\frac{\sqrt{3}}{3}，所以PA=\frac{\sqrt{3}}{3}AC=\frac{\sqrt{3}}{3}\times 130\sqrt{3}=130(米)，故最高点P距地面的高度PA为130米.(6分)$

Final answer: $130$ 米
Subfield: Plane Geometry


Question: $某数学课外活动小组为了测量一个摩天轮的最高点P距地面的高度，选取了与点P在地面上的射影A在同一水平面内的两个测量基点B，C(如图所示)，现测得\angle ABC=\angle ACB=\angle ACP=30^\circ ，B，C两点间的距离是390米.$![Image](https://cdntk.53zaixian.com/tq/1703968916453474306/img/compression/M_e6e85f03fee6ea27acb37a1fc920dc2c.jpg)
$若摩天轮最低点Q距地面的高度QA=20米，开启后按逆时针方向匀速旋转，转动一周需要20分钟.从游客进入摩天轮位于最低点Q处的轿厢开始计时，转动t分钟后距地面的高度为h米.若在摩天轮所在的平面内，以PQ的中点为坐标原点，PQ所在直线为y轴建立平面直角坐标系，求在转动一周的过程中，h(单位：米)关于t(单位：分钟)的函数解析式.$

Solution:
$如图所示，O为PQ的中点，$![Image](https://cdntk.53zaixian.com/tq/1703968916705132545/img/compression/M_eefe13aaaf0457ab424491d05af0518a.jpg)$因为PA=130米，QA=20米，所以摩天轮的半径r=\frac{130-20}{2}=55(米)，所以OA=75米.(8分)由题意知，摩天轮每20分钟转动一周，所以每分钟转过的角度为\frac{2\pi }{20}=\frac{\pi }{10}.从Q点开始计时，转动t分钟后轿厢运动到M点，则\angle QOM=\frac{\pi }{10}t，所以\angle xOM=\frac{\pi }{10}t-\frac{\pi }{2}.(10分)设M(x_{0},y_{0})，由三角函数的定义得y_{0}=55\\sin  \left(\frac{\pi }{10}t-\frac{\pi }{2}\right)=-55\\cos  \frac{\pi }{10}t.又因为h=y_{0}+OA，所以游客距地面的高度h关于t的函数解析式为h=75-55\\cos  \frac{\pi }{10}t(0\leq t\leq 20).(12分)$

Final answer: $h=75-55\cos (\pi /10t) (0\leq t\leq 20)$
Subfield: Plane Geometry


Question: $如图，在四边形ABCD中，已知BC=1，AC^{2}=AB^{2}+AB+1.$![Image](https://cdntk.53zaixian.com/tq/1703968917158117378/img/compression/M_b957359f79d3074a6eb3620445ccdf16.jpg)
$若\triangle ABC的面积为\sqrt{3}，求\triangle ABC的周长；$

Solution:
$在\triangle ABC中，由余弦定理得AC^{2}=AB^{2}+BC^{2}-2AB\cdot BC\\cos  \angle ABC.\because BC=1，AC^{2}=AB^{2}+AB+1，\therefore 1-2AB\\cos  \angle ABC=AB+1，\therefore \\cos  \angle ABC=-\frac{1}{2}，故\angle ABC=120^\circ .(2分)又\triangle ABC的面积为\sqrt{3}，\therefore \frac{1}{2}AB\cdot BC\\sin   120^\circ =\sqrt{3}，解得AB=4，由余弦定理得AC^{2}=AB^{2}+BC^{2}-2AB\cdot BC\\cos  \angle ABC=21，解得AC=\sqrt{21}(负值舍去)，(4分)\therefore \triangle ABC的周长为AB+BC+AC=5+\sqrt{21}.(6分)$

Final answer: $5 +\sqrt 21$
Subfield: Plane Geometry


Question: $如图，在四边形ABCD中，已知BC=1，AC^{2}=AB^{2}+AB+1.$<img_2948>
$若\triangle ABC的面积为\sqrt{3}，求\triangle ABC的周长；$

Solution:
$在\triangle ABC中，由余弦定理得AC^{2}=AB^{2}+BC^{2}-2AB\cdot BC\\cos  \angle ABC.\because BC=1，AC^{2}=AB^{2}+AB+1，\therefore 1-2AB\\cos  \angle ABC=AB+1，\therefore \\cos  \angle ABC=-\frac{1}{2}，故\angle ABC=120^\circ .(2分)又\triangle ABC的面积为\sqrt{3}，\therefore \frac{1}{2}AB\cdot BC\\sin   120^\circ =\sqrt{3}，解得AB=4，由余弦定理得AC^{2}=AB^{2}+BC^{2}-2AB\cdot BC\\cos  \angle ABC=21，解得AC=\sqrt{21}(负值舍去)，(4分)\therefore \triangle ABC的周长为AB+BC+AC=5+\sqrt{21}.(6分)$

Final answer: $5 +\sqrt 21$
Subfield: Plane Geometry


Question: $如图，在四边形ABCD中，已知BC=1，AC^{2}=AB^{2}+AB+1.$<img_2948>
$若AB=3，\angle ADB=60^\circ ，\angle BCD=120^\circ ，求\angle BDC的大小.$

Solution:
$由(1)可知\angle ABC=120^\circ ，又\angle ADB=60^\circ ，\angle BCD=120^\circ ，四边形的内角和为360^\circ ，\therefore \angle BAD+\angle BDC=60^\circ .设\angle BDC=\theta (0^\circ <\theta <60^\circ )，则\angle BAD=60^\circ -\theta .在\triangle ABD中，由正弦定理得\frac{AB}{\\\sin   \angle ADB}=\frac{BD}{\\\sin   \angle BAD}，即\frac{3}{\frac{\sqrt{3}}{2}}=\frac{BD}{\\\sin   (60^{\circ}-\theta )}，(7分)在\triangle CBD中，由正弦定理得\frac{BC}{\\\sin   \angle BDC}=\frac{BD}{\\\sin   \angle BCD}，即\frac{1}{\\\sin    \theta }=\frac{BD}{\frac{\sqrt{3}}{2}}，(8分)消去BD，得\\sin  (60^\circ -\theta )\\sin   \theta =\frac{1}{4}，即\left(\frac{\sqrt{3}}{2}\\\cos    \theta -\frac{1}{2}\\\sin    \theta \right)\\sin   \theta =\frac{1}{4}，\therefore \frac{\sqrt{3}}{4}\\sin   2\theta +\frac{1}{4}\\cos   2\theta =\frac{1}{2}，即\frac{1}{2}\\sin  (2\theta +30^\circ )=\frac{1}{2}，故\\sin  (2\theta +30^\circ )=1.(10分)\because 0^\circ <\theta <60^\circ ，\therefore 30^\circ <2\theta +30^\circ <150^\circ ，则2\theta +30^\circ =90^\circ ，解得\theta =30^\circ ，则\angle BDC=30^\circ .(12分)$

Final answer: $30^\circ$
Subfield: Plane Geometry


Question: $如图，在四边形ABCD中，已知BC=1，AC^{2}=AB^{2}+AB+1.$![Image](https://cdntk.53zaixian.com/tq/1703968917158117378/img/compression/M_b957359f79d3074a6eb3620445ccdf16.jpg)
$若AB=3，\angle ADB=60^\circ ，\angle BCD=120^\circ ，求\angle BDC的大小.$

Solution:
$由(1)可知\angle ABC=120^\circ ，又\angle ADB=60^\circ ，\angle BCD=120^\circ ，四边形的内角和为360^\circ ，\therefore \angle BAD+\angle BDC=60^\circ .设\angle BDC=\theta (0^\circ <\theta <60^\circ )，则\angle BAD=60^\circ -\theta .在\triangle ABD中，由正弦定理得\frac{AB}{\\\sin   \angle ADB}=\frac{BD}{\\\sin   \angle BAD}，即\frac{3}{\frac{\sqrt{3}}{2}}=\frac{BD}{\\\sin   (60^{\circ}-\theta )}，(7分)在\triangle CBD中，由正弦定理得\frac{BC}{\\\sin   \angle BDC}=\frac{BD}{\\\sin   \angle BCD}，即\frac{1}{\\\sin    \theta }=\frac{BD}{\frac{\sqrt{3}}{2}}，(8分)消去BD，得\\sin  (60^\circ -\theta )\\sin   \theta =\frac{1}{4}，即\left(\frac{\sqrt{3}}{2}\\\cos    \theta -\frac{1}{2}\\\sin    \theta \right)\\sin   \theta =\frac{1}{4}，\therefore \frac{\sqrt{3}}{4}\\sin   2\theta +\frac{1}{4}\\cos   2\theta =\frac{1}{2}，即\frac{1}{2}\\sin  (2\theta +30^\circ )=\frac{1}{2}，故\\sin  (2\theta +30^\circ )=1.(10分)\because 0^\circ <\theta <60^\circ ，\therefore 30^\circ <2\theta +30^\circ <150^\circ ，则2\theta +30^\circ =90^\circ ，解得\theta =30^\circ ，则\angle BDC=30^\circ .(12分)$

Final answer: $30^\circ$
Subfield: Plane Geometry


Question: $已知\triangle ABC的内角A，B，C的对边分别为a，b，c，c=1且2c\\sin   A\\cos   B=a\\sin   A－b\\sin   B+\frac{1}{4}b\\sin   C.
$<img_2839>
$求b；$

Solution:
$\because 2c\\sin   A\\cos   B=a\\sin   A-b\\sin   B+\frac{1}{4}b\\sin   C，\therefore 2ca\\cos   B=a^{2}-b^{2}+\frac{1}{4}bc.(2分)\therefore 由余弦定理得2ca\cdot \frac{c^2+a^2-b^2}{2ac}=a^{2}-b^{2}+\frac{1}{4}bc，整理得b=4c.又c=1，\therefore b=4.(3分)$

Final answer: $4$
Subfield: Plane Geometry


Question: $已知\triangle ABC的内角A，B，C的对边分别为a，b，c，c=1且2c\\sin   A\\cos   B=a\\sin   A－b\\sin   B+\frac{1}{4}b\\sin   C.
$![Image](https://cdntk.53zaixian.com/tq/1703968917644656642/img/compression/M_1ace49af54bd63b7ae303577fa477a5c.jpg)
$求b；$

Solution:
$\because 2c\\sin   A\\cos   B=a\\sin   A-b\\sin   B+\frac{1}{4}b\\sin   C，\therefore 2ca\\cos   B=a^{2}-b^{2}+\frac{1}{4}bc.(2分)\therefore 由余弦定理得2ca\cdot \frac{c^2+a^2-b^2}{2ac}=a^{2}-b^{2}+\frac{1}{4}bc，整理得b=4c.又c=1，\therefore b=4.(3分)$

Final answer: $4$
Subfield: Plane Geometry


Question: $已知\triangle ABC的内角A，B，C的对边分别为a，b，c，c=1且2c\\sin   A\\cos   B=a\\sin   A－b\\sin   B+\frac{1}{4}b\\sin   C.
$![Image](https://cdntk.53zaixian.com/tq/1703968917644656642/img/compression/M_1ace49af54bd63b7ae303577fa477a5c.jpg)
$若AD为BC边上的中线，\\cos  \angle BAD=\frac{\sqrt{21}}{7}，求\angle BAC的余弦值；$

Solution:
$\because D为BC的中点，\therefore \overrightarrow{AD}=\frac{1}{2}(\overrightarrow{AB}+\overrightarrow{AC}).设\angle BAC=\theta ，则|\overrightarrow{AD}|=\frac{1}{2}\sqrt{{\overrightarrow{AB}}^2+{\overrightarrow{AC}}^2+2\overrightarrow{AB}\cdot \overrightarrow{AC}}=\frac{1}{2}\sqrt{c^2+b^2+2bc\\\cos    \theta }=\frac{\sqrt{17+8\\\cos    \theta }}{2}，(4分)又\overrightarrow{AB}\cdot \overrightarrow{AD}=\overrightarrow{AB}\cdot \frac{1}{2}(\overrightarrow{AB}+\overrightarrow{AC})=\frac{1}{2}(\overrightarrow{AB}^{2}+\overrightarrow{AB}\cdot \overrightarrow{AC})=\frac{c^2+cb\\\cos    \theta }{2}=\frac{1+4\\\cos    \theta }{2}，\therefore \\cos  \angle BAD=\frac{\overrightarrow{AB}\cdot \overrightarrow{AD}}{|\overrightarrow{AB}||\overrightarrow{AD}|}=\frac{1+4\\\cos    \theta }{\sqrt{17+8\\\cos    \theta }}=\frac{\sqrt{21}}{7}，即28\\cos  ^{2}\theta +8\\cos   \theta -11=0，解得\\cos   \theta =\frac{1}{2}或\\cos   \theta =-\frac{11}{14}，(5分)\because \angle BAD为锐角，\therefore \overrightarrow{AB}\cdot \overrightarrow{AD}>0，\therefore 1+4\\cos   \theta >0，\therefore \\cos   \theta =\frac{1}{2}，\therefore \angle BAC的余弦值为\frac{1}{2}.(6分)$

Final answer: $\frac{1}{2}$
Subfield: Plane Geometry


Question: $已知\triangle ABC的内角A，B，C的对边分别为a，b，c，c=1且2c\\sin   A\\cos   B=a\\sin   A－b\\sin   B+\frac{1}{4}b\\sin   C.
$<img_2839>
$若AD为BC边上的中线，\\cos  \angle BAD=\frac{\sqrt{21}}{7}，求\angle BAC的余弦值；$

Solution:
$\because D为BC的中点，\therefore \overrightarrow{AD}=\frac{1}{2}(\overrightarrow{AB}+\overrightarrow{AC}).设\angle BAC=\theta ，则|\overrightarrow{AD}|=\frac{1}{2}\sqrt{{\overrightarrow{AB}}^2+{\overrightarrow{AC}}^2+2\overrightarrow{AB}\cdot \overrightarrow{AC}}=\frac{1}{2}\sqrt{c^2+b^2+2bc\\\cos    \theta }=\frac{\sqrt{17+8\\\cos    \theta }}{2}，(4分)又\overrightarrow{AB}\cdot \overrightarrow{AD}=\overrightarrow{AB}\cdot \frac{1}{2}(\overrightarrow{AB}+\overrightarrow{AC})=\frac{1}{2}(\overrightarrow{AB}^{2}+\overrightarrow{AB}\cdot \overrightarrow{AC})=\frac{c^2+cb\\\cos    \theta }{2}=\frac{1+4\\\cos    \theta }{2}，\therefore \\cos  \angle BAD=\frac{\overrightarrow{AB}\cdot \overrightarrow{AD}}{|\overrightarrow{AB}||\overrightarrow{AD}|}=\frac{1+4\\\cos    \theta }{\sqrt{17+8\\\cos    \theta }}=\frac{\sqrt{21}}{7}，即28\\cos  ^{2}\theta +8\\cos   \theta -11=0，解得\\cos   \theta =\frac{1}{2}或\\cos   \theta =-\frac{11}{14}，(5分)\because \angle BAD为锐角，\therefore \overrightarrow{AB}\cdot \overrightarrow{AD}>0，\therefore 1+4\\cos   \theta >0，\therefore \\cos   \theta =\frac{1}{2}，\therefore \angle BAC的余弦值为\frac{1}{2}.(6分)$

Final answer: $\frac{1}{2}$