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


Question: $如图，扇形OAB的半径为1，圆心角为\frac{\pi }{3}，过扇形弧上一点C分别向OA，OB作垂线，垂足分别为D，E，得到\triangle CDE，当点C(与A，B不重合)在扇形弧上从A到B运动时.$![Image](https://cdntk.53zaixian.com/tq/1696417094169976834/img/compression/M_03c6348f8e7a7c30b2d2524467468d9d.jpg)
$\triangle CDE的面积是如何变化的？$

Solution:
$如图，过点E作EF\perp OD，垂足为F，连接OC，$![Image](https://cdntk.53zaixian.com/tq/1696417094522298370/img/compression/M_3a540ce1defd99d15909b16654e6c0f7.jpg)$记\angle COD=\alpha \left(0<\alpha <\frac{\pi }{3}\right)，则\angle COE=\frac{\pi }{3}-\alpha .在\triangle OCD中，OD=\\cos   \alpha ，CD=\\sin   \alpha ，则S_{\triangle OCD}=\frac{1}{2}\\sin   \alpha \\cos   \alpha =\frac{1}{4}\\sin   2\alpha ，在\triangle OCE中，OE=\\cos  \left(\frac{\pi }{3}-\alpha \right)，CE=\\sin  \left(\frac{\pi }{3}-\alpha \right)，则S_{\triangle OCE}=\frac{1}{2}\\sin  \left(\frac{\pi }{3}-\alpha \right)\\cos  \left(\frac{\pi }{3}-\alpha \right)=\frac{1}{4}\\sin  \left(\frac{2\pi }{3}-2\alpha \right)，在\triangle ODE中，EF=OE\times \\sin   \frac{\pi }{3}=\frac{\sqrt{3}}{2}\\cos  \left(\frac{\pi }{3}-\alpha \right)，则S_{\triangle ODE}=\frac{1}{2}\times OD\cdot EF=\frac{\sqrt{3}}{4}\\cos   \alpha \\cos  \left(\frac{\pi }{3}-\alpha \right)，(4分)所以S_{\triangle CDE}=S_{\triangle OCD}+S_{\triangle OCE}-S_{\triangle ODE}=\frac{1}{4}\\sin   2\alpha +\frac{1}{4}\\sin  \left(\frac{2\pi }{3}-2\alpha \right)-\frac{\sqrt{3}}{4}\\cos   \alpha \\cos  \left(\frac{\pi }{3}-\alpha \right)=\frac{1}{4}\\sin   2\alpha +\frac{1}{4}\times \left(\frac{\sqrt{3}}{2}\\\cos    2\alpha +\frac{1}{2}\\\sin    2\alpha \right)-\frac{\sqrt{3}}{4}\\cos   \alpha \left(\right.\frac{1}{2}\\cos   \alpha +\frac{\sqrt{3}}{2}\\sin   \alpha \left)\right.=\frac{3}{8}\\sin   2\alpha +\frac{\sqrt{3}}{8}\\cos   2\alpha -\frac{\sqrt{3}}{16}\\cos   2\alpha -\frac{3}{16}\\sin   2\alpha -\frac{\sqrt{3}}{16}=\frac{3}{16}\\sin   2\alpha +\frac{\sqrt{3}}{16}\\cos   2\alpha -\frac{\sqrt{3}}{16}=\frac{\sqrt{3}}{8}\\sin  \left(2\alpha +\frac{\pi }{6}\right)-\frac{\sqrt{3}}{16}.(7分)由\alpha \in \left(0,\frac{\pi }{3}\right)，得2\alpha +\frac{\pi }{6}\in \left(\frac{\pi }{6},\frac{5\pi }{6}\right)，令2\alpha +\frac{\pi }{6}=\frac{\pi }{2}，得\alpha =\frac{\pi }{6}，所以点C在扇形弧上从点A运动到弧AB的中点的过程中，\triangle CDE的面积越来越大，点C在扇形弧上从弧AB的中点运动到点B的过程中，\triangle CDE的面积越来越小.(8分)$

Final answer: $点C在扇形弧上从点A运动到弧AB的中点的过程中，\triangle CDE的面积越来越大；
点C在扇形弧上从弧AB的中点运动到点B的过程中，\triangle CDE的面积越来越小。$
Subfield: Plane Geometry


Question: $如图，扇形OAB的半径为1，圆心角为\frac{\pi }{3}，过扇形弧上一点C分别向OA，OB作垂线，垂足分别为D，E，得到\triangle CDE，当点C(与A，B不重合)在扇形弧上从A到B运动时.$![Image](https://cdntk.53zaixian.com/tq/1696417094169976834/img/compression/M_03c6348f8e7a7c30b2d2524467468d9d.jpg)
$求\triangle CDE面积的最大值.$

Solution:
$由(1)知，当点C位于弧AB的中点，即\alpha =\frac{\pi }{6}时，\triangle CDE的面积有最大值，为\frac{\sqrt{3}}{8}\\sin   \frac{\pi }{2}-\frac{\sqrt{3}}{16}=\frac{\sqrt{3}}{16}.(10分)$

Final answer: $\frac{\sqrt{3}}{16}$
Subfield: Plane Geometry


Question: $如图，扇形OAB的半径为1，圆心角为\frac{\pi }{3}，过扇形弧上一点C分别向OA，OB作垂线，垂足分别为D，E，得到\triangle CDE，当点C(与A，B不重合)在扇形弧上从A到B运动时.$<img_2264>
$求\triangle CDE面积的最大值.$

Solution:
$由(1)知，当点C位于弧AB的中点，即\alpha =\frac{\pi }{6}时，\triangle CDE的面积有最大值，为\frac{\sqrt{3}}{8}\\sin   \frac{\pi }{2}-\frac{\sqrt{3}}{16}=\frac{\sqrt{3}}{16}.(10分)$

Final answer: $\frac{\sqrt{3}}{16}$
Subfield: Plane Geometry


Question: $如图，有一条宽为60 m的笔直的河道(假设河道足够长)，规划在河道内围出一块直角三角形区域(图中\triangle ABC)养殖观赏鱼，AB\perp AC，顶点A到河岸的距离AE=h_{1} m，AD=h_{2} m，C，B两点分别在两岸l_{1}，l_{2}上，设\angle ABD=\alpha .$<img_1651>
$若\alpha =30^\circ ，求养殖区域面积的最大值；$

Solution:
$当\alpha =30^\circ 时，AB=\frac{h_2}{\\\sin    \alpha }=2h_{2}，AC=\frac{h_1}{\\\cos    \alpha }=\frac{2}{\sqrt{3}}h_{1}，(1分)所以S_{\triangle ABC}=\frac{1}{2}AB\cdot AC=\frac{2}{\sqrt{3}}h_{1}h_{2}，易知h_{1}>0，h_{2}>0，因为h_{1}+h_{2}=60\geq 2\sqrt{h_1h_2}(当且仅当h_{1}=h_{2}时，等号成立)，所以h_{1}h_{2}\leq 900，所以S_{\triangle ABC}=\frac{2}{\sqrt{3}}h_{1}h_{2}\leq 600\sqrt{3}，所以养殖区域面积的最大值为600\sqrt{3} m^{2}.(5分)$

Final answer: $600\sqrt 3 $ $m^2$
Subfield: Plane Geometry


Question: $如图，有一条宽为60 m的笔直的河道(假设河道足够长)，规划在河道内围出一块直角三角形区域(图中\triangle ABC)养殖观赏鱼，AB\perp AC，顶点A到河岸的距离AE=h_{1} m，AD=h_{2} m，C，B两点分别在两岸l_{1}，l_{2}上，设\angle ABD=\alpha .$![Image](https://cdntk.53zaixian.com/tq/1696417113300750337/img/compression/M_57888487c13383f2604d6d7dd5d31142.jpg)
$若\alpha =30^\circ ，求养殖区域面积的最大值；$

Solution:
$当\alpha =30^\circ 时，AB=\frac{h_2}{\\\sin    \alpha }=2h_{2}，AC=\frac{h_1}{\\\cos    \alpha }=\frac{2}{\sqrt{3}}h_{1}，(1分)所以S_{\triangle ABC}=\frac{1}{2}AB\cdot AC=\frac{2}{\sqrt{3}}h_{1}h_{2}，易知h_{1}>0，h_{2}>0，因为h_{1}+h_{2}=60\geq 2\sqrt{h_1h_2}(当且仅当h_{1}=h_{2}时，等号成立)，所以h_{1}h_{2}\leq 900，所以S_{\triangle ABC}=\frac{2}{\sqrt{3}}h_{1}h_{2}\leq 600\sqrt{3}，所以养殖区域面积的最大值为600\sqrt{3} m^{2}.(5分)$

Final answer: $600\sqrt 3 $ $m^2$
Subfield: Plane Geometry


Question: $如图，有一条宽为60 m的笔直的河道(假设河道足够长)，规划在河道内围出一块直角三角形区域(图中\triangle ABC)养殖观赏鱼，AB\perp AC，顶点A到河岸的距离AE=h_{1} m，AD=h_{2} m，C，B两点分别在两岸l_{1}，l_{2}上，设\angle ABD=\alpha .$![Image](https://cdntk.53zaixian.com/tq/1696417113300750337/img/compression/M_57888487c13383f2604d6d7dd5d31142.jpg)
$现拟沿着养殖区域\triangle ABC三边搭建观赏长廊(宽度忽略不计)，若h_{1}=30，求观赏长廊总长f(\alpha )(单位：m)的最小值.$

Solution:
$由题意得AB=\frac{30}{\\\sin    \alpha }，AC=\frac{30}{\\\cos    \alpha }，\alpha \in \left(0,\frac{\pi }{2}\right)，所以BC=\sqrt{\left(\frac{30}{\\\sin     \alpha }\right)^2+\left(\frac{30}{\\\cos     \alpha }\right)^2}=30\sqrt{\frac{1}{\\\sin    ^2\alpha }+\frac{1}{\\\cos    ^2\alpha }}=\frac{30}{\\\sin    \alpha \\\cos    \alpha }，所以y=f(\alpha )=30\left(\frac{1}{\\\sin    \alpha }+\frac{1}{\\\cos    \alpha }+\frac{1}{\\\sin    \alpha \\\cos    \alpha }\right)=30\left(\frac{\\\sin    \alpha +\\\cos    \alpha +1}{\\\sin    \alpha \\\cos    \alpha }\right)，　
(8分)
设t=\\sin   \alpha +\\cos   \alpha ，则t=\\sin   \alpha +\\cos   \alpha =\sqrt{2}\\sin  \left(\alpha +\frac{\pi }{4}\right)\in (1，\sqrt{2}]，所以\\sin   \alpha \\cos   \alpha =\frac{t^2-1}{2}.所以y=30\cdot \frac{t+1}{\frac{t^2-1}{2}}=\frac{60}{t-1}，易知当t=\sqrt{2}时，y取得最小值，所以f(\alpha )_{min}=\frac{60}{\sqrt{2}-1}=60\sqrt{2}+60，所以观赏长廊总长的最小值为(60\sqrt{2}+60)m.
(12分)$

Final answer: $(60\sqrt 2+60)$ m
Subfield: Plane Geometry


Question: $如图，有一条宽为60 m的笔直的河道(假设河道足够长)，规划在河道内围出一块直角三角形区域(图中\triangle ABC)养殖观赏鱼，AB\perp AC，顶点A到河岸的距离AE=h_{1} m，AD=h_{2} m，C，B两点分别在两岸l_{1}，l_{2}上，设\angle ABD=\alpha .$<img_1651>
$现拟沿着养殖区域\triangle ABC三边搭建观赏长廊(宽度忽略不计)，若h_{1}=30，求观赏长廊总长f(\alpha )(单位：m)的最小值.$

Solution:
$由题意得AB=\frac{30}{\\\sin    \alpha }，AC=\frac{30}{\\\cos    \alpha }，\alpha \in \left(0,\frac{\pi }{2}\right)，所以BC=\sqrt{\left(\frac{30}{\\\sin     \alpha }\right)^2+\left(\frac{30}{\\\cos     \alpha }\right)^2}=30\sqrt{\frac{1}{\\\sin    ^2\alpha }+\frac{1}{\\\cos    ^2\alpha }}=\frac{30}{\\\sin    \alpha \\\cos    \alpha }，所以y=f(\alpha )=30\left(\frac{1}{\\\sin    \alpha }+\frac{1}{\\\cos    \alpha }+\frac{1}{\\\sin    \alpha \\\cos    \alpha }\right)=30\left(\frac{\\\sin    \alpha +\\\cos    \alpha +1}{\\\sin    \alpha \\\cos    \alpha }\right)，　
(8分)
设t=\\sin   \alpha +\\cos   \alpha ，则t=\\sin   \alpha +\\cos   \alpha =\sqrt{2}\\sin  \left(\alpha +\frac{\pi }{4}\right)\in (1，\sqrt{2}]，所以\\sin   \alpha \\cos   \alpha =\frac{t^2-1}{2}.所以y=30\cdot \frac{t+1}{\frac{t^2-1}{2}}=\frac{60}{t-1}，易知当t=\sqrt{2}时，y取得最小值，所以f(\alpha )_{min}=\frac{60}{\sqrt{2}-1}=60\sqrt{2}+60，所以观赏长廊总长的最小值为(60\sqrt{2}+60)m.
(12分)$

Final answer: $(60\sqrt 2+60)$ m
Subfield: Plane Geometry


Question: $小明将上周每天骑车上学路上的情况用图象表示：$![Image](https://cdntk.53zaixian.com/tq/1696448812469903362/img/compression/M_d92098a1f3c3f7ec50008f5dbd35fac6.jpg)$　$![Image](https://cdntk.53zaixian.com/tq/1696448812469903362/img/compression/M_e995d4295926f7f2bd9ea5e2cb59079b.jpg)$　$![Image](https://cdntk.53zaixian.com/tq/1696448812469903362/img/compression/M_4177638eb8e9554d2c7936aec033aed5.jpg)![Image](https://cdntk.53zaixian.com/tq/1696448812469903362/img/compression/M_7e0c38e65895975571842cfa31e4a78d.jpg)$　$![Image](https://cdntk.53zaixian.com/tq/1696448812469903362/img/compression/M_168658ce52d15472f45152124e42dbba.jpg)$很遗憾图象的先后次序不小心被打乱了.还好小明同时用文字进行了记录：周一：匀速骑车前进；周二：匀速骑车前进，中间遇到红灯停了一次；周三：骑车出门晚了，越骑越快；周四：骑车出门后一会儿想起忘带东西又加速回去拿；周五：\ldots \ldots $
$本周小明打算跑步上学，多消耗点热量.已知单位时间消耗的热量y(卡/小时)与跑步的平均速度v(千米/小时)满足y=-\frac{50}{3}v^{2}+350v-\frac{3 200}{3}，小明家到学校的距离是1.5千米，假设小明上学路上不停顿，则他从家跑步到学校最多可以消耗多少热量？$

Solution:
$设消耗的热量为S卡，求出S=-25\left(v+\frac{64}{v}\right)+525，再利用基本不等式求解即可.$

Final answer: $525$ 卡
Subfield: Plane Geometry


Question: $小明将上周每天骑车上学路上的情况用图象表示：$<img_3071>$　$<img_2748>$　$<img_2511><img_2503>$　$<img_2904>$很遗憾图象的先后次序不小心被打乱了.还好小明同时用文字进行了记录：周一：匀速骑车前进；周二：匀速骑车前进，中间遇到红灯停了一次；周三：骑车出门晚了，越骑越快；周四：骑车出门后一会儿想起忘带东西又加速回去拿；周五：\ldots \ldots $
$本周小明打算跑步上学，多消耗点热量.已知单位时间消耗的热量y(卡/小时)与跑步的平均速度v(千米/小时)满足y=-\frac{50}{3}v^{2}+350v-\frac{3 200}{3}，小明家到学校的距离是1.5千米，假设小明上学路上不停顿，则他从家跑步到学校最多可以消耗多少热量？$

Solution:
$设消耗的热量为S卡，求出S=-25\left(v+\frac{64}{v}\right)+525，再利用基本不等式求解即可.$

Final answer: $525$ 卡
Subfield: Plane Geometry


Question: $如图，边长为1的正三角形ABC的中心为O，过点O的直线与边AB、AC分别交于点M、N(点M、N可以和端点重合).$<img_2289>
$用\overrightarrow{AB}、\overrightarrow{AC}表示\overrightarrow{AO}；$

Solution:
$连接AO并延长交BC于D，则D为BC的中点，故\overrightarrow{AD}=\frac{1}{2}(\overrightarrow{AB}+\overrightarrow{AC})，则\overrightarrow{AO}=\frac{2}{3}\overrightarrow{AD}=\frac{1}{3}(\overrightarrow{AB}+\overrightarrow{AC})=\frac{1}{3}\overrightarrow{AB}+\frac{1}{3}\overrightarrow{AC}.(4分)$

Final answer: $\overrightarrow{AO}=\frac{1}{3}\overrightarrow{AB}+\frac{1}{3}\overrightarrow{AC}$
Subfield: Plane Geometry


Question: $如图，边长为1的正三角形ABC的中心为O，过点O的直线与边AB、AC分别交于点M、N(点M、N可以和端点重合).$![Image](https://cdntk.53zaixian.com/tq/1699313601134796802/img/compression/M_e42665ca5fa6432d0c57686639fbab1a.jpg)
$用\overrightarrow{AB}、\overrightarrow{AC}表示\overrightarrow{AO}；$

Solution:
$连接AO并延长交BC于D，则D为BC的中点，故\overrightarrow{AD}=\frac{1}{2}(\overrightarrow{AB}+\overrightarrow{AC})，则\overrightarrow{AO}=\frac{2}{3}\overrightarrow{AD}=\frac{1}{3}(\overrightarrow{AB}+\overrightarrow{AC})=\frac{1}{3}\overrightarrow{AB}+\frac{1}{3}\overrightarrow{AC}.(4分)$

Final answer: $\overrightarrow{AO}=\frac{1}{3}\overrightarrow{AB}+\frac{1}{3}\overrightarrow{AC}$
Subfield: Plane Geometry


Question: $如图，边长为1的正三角形ABC的中心为O，过点O的直线与边AB、AC分别交于点M、N(点M、N可以和端点重合).$![Image](https://cdntk.53zaixian.com/tq/1699313601134796802/img/compression/M_e42665ca5fa6432d0c57686639fbab1a.jpg)
$设\overrightarrow{AM}=\lambda \overrightarrow{AB}，\overrightarrow{AN}=\mu \overrightarrow{AC}，求\frac{1}{\lambda }+\frac{1}{\mu }的值.$

Solution:
$由于M、O、N三点共线，故设\overrightarrow{MO}=k\overrightarrow{MN}，k\in (0，1)，故\overrightarrow{MO}=\overrightarrow{AO}-\overrightarrow{AM}=k(\overrightarrow{AN}-\overrightarrow{AM})，则\overrightarrow{AO}=(1-k)\overrightarrow{AM}+k\overrightarrow{AN}，(5分)故\overrightarrow{AO}=(1-k)\lambda \overrightarrow{AB}+k\mu \overrightarrow{AC}，即\frac{2}{3}\overrightarrow{AD}=(1-k)\lambda \overrightarrow{AB}+k\mu \overrightarrow{AC}，所以\overrightarrow{AD}=\frac{3}{2}(1-k)\lambda \overrightarrow{AB}+\frac{3}{2}k\mu \overrightarrow{AC}，(6分)故\frac{1}{2}(\overrightarrow{AB}+\overrightarrow{AC})=\frac{3}{2}(1-k)\lambda \overrightarrow{AB}+\frac{3}{2}k\mu \overrightarrow{AC}，则\frac{3}{2}(1-k)\lambda =\frac{1}{2}且\frac{3}{2}k\mu =\frac{1}{2}，(8分)则\frac{1}{\lambda }+\frac{1}{\mu }=3(1-k)+3k=3.(10分)$

Final answer: $3$
Subfield: Plane Geometry


Question: $如图，边长为1的正三角形ABC的中心为O，过点O的直线与边AB、AC分别交于点M、N(点M、N可以和端点重合).$<img_2289>
$设\overrightarrow{AM}=\lambda \overrightarrow{AB}，\overrightarrow{AN}=\mu \overrightarrow{AC}，求\frac{1}{\lambda }+\frac{1}{\mu }的值.$

Solution:
$由于M、O、N三点共线，故设\overrightarrow{MO}=k\overrightarrow{MN}，k\in (0，1)，故\overrightarrow{MO}=\overrightarrow{AO}-\overrightarrow{AM}=k(\overrightarrow{AN}-\overrightarrow{AM})，则\overrightarrow{AO}=(1-k)\overrightarrow{AM}+k\overrightarrow{AN}，(5分)故\overrightarrow{AO}=(1-k)\lambda \overrightarrow{AB}+k\mu \overrightarrow{AC}，即\frac{2}{3}\overrightarrow{AD}=(1-k)\lambda \overrightarrow{AB}+k\mu \overrightarrow{AC}，所以\overrightarrow{AD}=\frac{3}{2}(1-k)\lambda \overrightarrow{AB}+\frac{3}{2}k\mu \overrightarrow{AC}，(6分)故\frac{1}{2}(\overrightarrow{AB}+\overrightarrow{AC})=\frac{3}{2}(1-k)\lambda \overrightarrow{AB}+\frac{3}{2}k\mu \overrightarrow{AC}，则\frac{3}{2}(1-k)\lambda =\frac{1}{2}且\frac{3}{2}k\mu =\frac{1}{2}，(8分)则\frac{1}{\lambda }+\frac{1}{\mu }=3(1-k)+3k=3.(10分)$

Final answer: $3$
Subfield: Plane Geometry


Question: $如图，在梯形ABCD中，\overrightarrow{AB}=2\overrightarrow{DC}，E，F是DC的两个三等分点，G，H是AB的两个三等分点，线段BC上一动点P满足\overrightarrow{BP}=\lambda \overrightarrow{BC}(0\leq \lambda \leq 1)，AP分别交EG，FH于M，N两点，记\overrightarrow{AB}=a，\overrightarrow{AD}=b.$![Image](https://cdntk.53zaixian.com/tq/1699313607094902787/img/compression/M_2b114f1efa05b55532fd676fa6f16e29.jpg)
$当\lambda =\frac{1}{2}时，用a，b表示\overrightarrow{AP}；$

Solution:
$当\lambda =\frac{1}{2}时，\overrightarrow{BP}=\frac{1}{2}\overrightarrow{BC}，则\overrightarrow{AP}=\overrightarrow{AB}+\overrightarrow{BP}=\overrightarrow{AB}+\frac{1}{2}\overrightarrow{BC}=\overrightarrow{AB}+\frac{1}{2}(\overrightarrow{BA}+\overrightarrow{AD}+\overrightarrow{DC})=\overrightarrow{AB}+\frac{1}{2}\left(-\overrightarrow{AB}+\overrightarrow{AD}+\frac{1}{2}\overrightarrow{AB}\right)=\frac{3}{4}\overrightarrow{AB}+\frac{1}{2}\overrightarrow{AD}=\frac{3}{4}a+\frac{1}{2}b，所以\overrightarrow{AP}=\frac{3}{4}a+\frac{1}{2}b.(4分)$

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


Question: $如图，在梯形ABCD中，\overrightarrow{AB}=2\overrightarrow{DC}，E，F是DC的两个三等分点，G，H是AB的两个三等分点，线段BC上一动点P满足\overrightarrow{BP}=\lambda \overrightarrow{BC}(0\leq \lambda \leq 1)，AP分别交EG，FH于M，N两点，记\overrightarrow{AB}=a，\overrightarrow{AD}=b.$<img_2051>
$当\lambda =\frac{1}{2}时，用a，b表示\overrightarrow{AP}；$

Solution:
$当\lambda =\frac{1}{2}时，\overrightarrow{BP}=\frac{1}{2}\overrightarrow{BC}，则\overrightarrow{AP}=\overrightarrow{AB}+\overrightarrow{BP}=\overrightarrow{AB}+\frac{1}{2}\overrightarrow{BC}=\overrightarrow{AB}+\frac{1}{2}(\overrightarrow{BA}+\overrightarrow{AD}+\overrightarrow{DC})=\overrightarrow{AB}+\frac{1}{2}\left(-\overrightarrow{AB}+\overrightarrow{AD}+\frac{1}{2}\overrightarrow{AB}\right)=\frac{3}{4}\overrightarrow{AB}+\frac{1}{2}\overrightarrow{AD}=\frac{3}{4}a+\frac{1}{2}b，所以\overrightarrow{AP}=\frac{3}{4}a+\frac{1}{2}b.(4分)$

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


Question: $如图，在梯形ABCD中，\overrightarrow{AB}=2\overrightarrow{DC}，E，F是DC的两个三等分点，G，H是AB的两个三等分点，线段BC上一动点P满足\overrightarrow{BP}=\lambda \overrightarrow{BC}(0\leq \lambda \leq 1)，AP分别交EG，FH于M，N两点，记\overrightarrow{AB}=a，\overrightarrow{AD}=b.$<img_2051>
$若\overrightarrow{MN}=\mu \overrightarrow{AP}，试写出\lambda 和\mu 的关系，并求出\mu 的取值范围.$

Solution:
$连接AE，AF，则\overrightarrow{AE}=\overrightarrow{AD}+\frac{1}{3}\overrightarrow{DC}=b+\frac{1}{3}\times \frac{1}{2}a=\frac{1}{6}a+b，\overrightarrow{AF}=\overrightarrow{AD}+\frac{2}{3}\overrightarrow{DC}=b+\frac{2}{3}\times \frac{1}{2}a=\frac{1}{3}a+b，(6分)$<img_2174>$因为E，M，G三点共线，F，N，H三点共线，所以设\overrightarrow{AM}=x\overrightarrow{AE}+(1-x)\overrightarrow{AG}=xb+\left(\frac{1}{3}-\frac{1}{6}x\right)a，\overrightarrow{AN}=y\overrightarrow{AF}+(1-y)\overrightarrow{AH}=yb+\left(\frac{2}{3}-\frac{1}{3}y\right)a，(8分)又因为\overrightarrow{AN}\parallel \overrightarrow{AM}，所以x\left(\frac{2}{3}-\frac{1}{3}y\right)=y\left(\frac{1}{3}-\frac{1}{6}x\right)，得\frac{xy}{2}=2x-y，因为\overrightarrow{BP}=\lambda \overrightarrow{BC}，所以\overrightarrow{AP}-\overrightarrow{AB}=\lambda (\overrightarrow{AC}-\overrightarrow{AB})，所以\overrightarrow{AP}=\lambda \overrightarrow{AC}+(1-\lambda )\overrightarrow{AB}=\lambda b+\left(1-\frac{\lambda }{2}\right)a，\overrightarrow{MN}=\overrightarrow{AN}-\overrightarrow{AM}=(y-x)b+\left(\frac{1}{3}-\frac{1}{3}y+\frac{1}{6}x\right)a，因为\overrightarrow{MN}=\mu \overrightarrow{AP}，所以\left\{\begin{matrix}y-x=\lambda \mu ,\\ \frac{1}{3}-\frac{1}{3}y+\frac{1}{6}x=\mu -\frac{\lambda \mu }{2},\end{matrix}\right.即\left\{\begin{matrix}y=2\lambda \mu -6\mu +2,\\ x=\lambda \mu -6\mu +2,\end{matrix}\right.代入\frac{xy}{2}=2x-y得(\lambda -3)(\lambda -6)\mu ^{2}+(3\lambda -6)\mu =0，因为\mu \neq 0，所以\mu =\frac{-3(\lambda -2)}{(\lambda -3)(\lambda -6)}=\frac{-3}{\lambda -2+\frac{4}{\lambda -2}-5}，(10分)令t=\lambda -2，则t\in [-2，-1]，易知y=t+\frac{4}{t}在[-2，-1]上单调递减，所以y=t+\frac{4}{t}\in [-5，-4]，所以\mu \in \left[\frac{3}{10},\frac{1}{3}\right]，所以\mu 的取值范围为\left[\frac{3}{10},\frac{1}{3}\right].(12分)$



Final answer: $\mu = \frac{-3(\lambda -2)}{(\lambda -3)(\lambda -6)}$, $\left[\frac{3}{10},\frac{1}{3}\right]$
Subfield: Plane Geometry


Question: $如图，在梯形ABCD中，\overrightarrow{AB}=2\overrightarrow{DC}，E，F是DC的两个三等分点，G，H是AB的两个三等分点，线段BC上一动点P满足\overrightarrow{BP}=\lambda \overrightarrow{BC}(0\leq \lambda \leq 1)，AP分别交EG，FH于M，N两点，记\overrightarrow{AB}=a，\overrightarrow{AD}=b.$![Image](https://cdntk.53zaixian.com/tq/1699313607094902787/img/compression/M_2b114f1efa05b55532fd676fa6f16e29.jpg)
$若\overrightarrow{MN}=\mu \overrightarrow{AP}，试写出\lambda 和\mu 的关系，并求出\mu 的取值范围.$

Solution:
$连接AE，AF，则\overrightarrow{AE}=\overrightarrow{AD}+\frac{1}{3}\overrightarrow{DC}=b+\frac{1}{3}\times \frac{1}{2}a=\frac{1}{6}a+b，\overrightarrow{AF}=\overrightarrow{AD}+\frac{2}{3}\overrightarrow{DC}=b+\frac{2}{3}\times \frac{1}{2}a=\frac{1}{3}a+b，(6分)$![Image](https://cdntk.53zaixian.com/tq/1699313607510138882/img/compression/M_98d1db3a91c2d0ab5f948898cd10bb0c.jpg)$因为E，M，G三点共线，F，N，H三点共线，所以设\overrightarrow{AM}=x\overrightarrow{AE}+(1-x)\overrightarrow{AG}=xb+\left(\frac{1}{3}-\frac{1}{6}x\right)a，\overrightarrow{AN}=y\overrightarrow{AF}+(1-y)\overrightarrow{AH}=yb+\left(\frac{2}{3}-\frac{1}{3}y\right)a，(8分)又因为\overrightarrow{AN}\parallel \overrightarrow{AM}，所以x\left(\frac{2}{3}-\frac{1}{3}y\right)=y\left(\frac{1}{3}-\frac{1}{6}x\right)，得\frac{xy}{2}=2x-y，因为\overrightarrow{BP}=\lambda \overrightarrow{BC}，所以\overrightarrow{AP}-\overrightarrow{AB}=\lambda (\overrightarrow{AC}-\overrightarrow{AB})，所以\overrightarrow{AP}=\lambda \overrightarrow{AC}+(1-\lambda )\overrightarrow{AB}=\lambda b+\left(1-\frac{\lambda }{2}\right)a，\overrightarrow{MN}=\overrightarrow{AN}-\overrightarrow{AM}=(y-x)b+\left(\frac{1}{3}-\frac{1}{3}y+\frac{1}{6}x\right)a，因为\overrightarrow{MN}=\mu \overrightarrow{AP}，所以\left\{\begin{matrix}y-x=\lambda \mu ,\\ \frac{1}{3}-\frac{1}{3}y+\frac{1}{6}x=\mu -\frac{\lambda \mu }{2},\end{matrix}\right.即\left\{\begin{matrix}y=2\lambda \mu -6\mu +2,\\ x=\lambda \mu -6\mu +2,\end{matrix}\right.代入\frac{xy}{2}=2x-y得(\lambda -3)(\lambda -6)\mu ^{2}+(3\lambda -6)\mu =0，因为\mu \neq 0，所以\mu =\frac{-3(\lambda -2)}{(\lambda -3)(\lambda -6)}=\frac{-3}{\lambda -2+\frac{4}{\lambda -2}-5}，(10分)令t=\lambda -2，则t\in [-2，-1]，易知y=t+\frac{4}{t}在[-2，-1]上单调递减，所以y=t+\frac{4}{t}\in [-5，-4]，所以\mu \in \left[\frac{3}{10},\frac{1}{3}\right]，所以\mu 的取值范围为\left[\frac{3}{10},\frac{1}{3}\right].(12分)$

Final answer: $\mu = \frac{-3(\lambda -2)}{(\lambda -3)(\lambda -6)}$, $\left[\frac{3}{10},\frac{1}{3}\right]$
Subfield: Plane Geometry


Question: $在平面直角坐标系中，已知射线y=\sqrt{3}x(x\geq 0)与射线y=-\sqrt{3}x(x\geq 0)，过点M(1,0)作直线l分别交两射线于点A、B(不同于原点O).
$<img_1647>
$当|OA|+|OB|取得最小值时，求直线l的方程；$

Solution:
$设A(a，\sqrt{3}a)，B(b，-\sqrt{3}b)(a，b>0)，利用A，B，M三点共线可得a，b的关系，计算出|OA|+|OB|后由基本不等式求得最小值，从而得直线l的方程；$

Final answer: $x-2y=0$
Subfield: Plane Geometry


Question: $在平面直角坐标系中，已知射线y=\sqrt{3}x(x\geq 0)与射线y=-\sqrt{3}x(x\geq 0)，过点M(1,0)作直线l分别交两射线于点A、B(不同于原点O).
$![Image](https://cdntk.53zaixian.com/tq/1699313607908597762/img/compression/M_f30319b58100a20b939d53b579378367.jpg)
$当|OA|+|OB|取得最小值时，求直线l的方程；$

Solution:
$设A(a，\sqrt{3}a)，B(b，-\sqrt{3}b)(a，b>0)，利用A，B，M三点共线可得a，b的关系，计算出|OA|+|OB|后由基本不等式求得最小值，从而得直线l的方程；$

Final answer: $x-2y=0$
Subfield: Plane Geometry


Question: $如图所示，AD是\triangle ABC的一条中线，点O满足\overrightarrow{AO}=2\overrightarrow{OD}，过点O的直线分别与射线AB、射线AC交于M、N两点.$![Image](https://cdntk.53zaixian.com/tq/1699313608613240834/img/compression/M_3b651de02c4cf74051fcdb80f31c939f.jpg)
$求证：\overrightarrow{AD}=\frac{1}{2}\overrightarrow{AB}+\frac{1}{2}\overrightarrow{AC}；$

Solution:
$以\overrightarrow{AB}，\overrightarrow{AC}为基底，根据线性运算求证即可；$

Final answer: 证明题
Subfield: Plane Geometry


Question: $如图所示，AD是\triangle ABC的一条中线，点O满足\overrightarrow{AO}=2\overrightarrow{OD}，过点O的直线分别与射线AB、射线AC交于M、N两点.$<img_1893>
$求证：\overrightarrow{AD}=\frac{1}{2}\overrightarrow{AB}+\frac{1}{2}\overrightarrow{AC}；$

Solution:
$以\overrightarrow{AB}，\overrightarrow{AC}为基底，根据线性运算求证即可；$
Subfield: Plane Geometry


Question: $如图所示，AD是\triangle ABC的一条中线，点O满足\overrightarrow{AO}=2\overrightarrow{OD}，过点O的直线分别与射线AB、射线AC交于M、N两点.$![Image](https://cdntk.53zaixian.com/tq/1699313608613240834/img/compression/M_3b651de02c4cf74051fcdb80f31c939f.jpg)
$设\overrightarrow{AM}=m\overrightarrow{AB}，\overrightarrow{AN}=n\overrightarrow{AC}，m>0，n>0，求\frac{1}{m}+\frac{1}{n}的值.$

Solution:
$由M，O，N三点共线可设\overrightarrow{MO}=\lambda \overrightarrow{ON}，根据向量线性运算求出m，n，即可求解.$

Final answer: $\frac{1}{m}+\frac{1}{n}=\frac{1}{\lambda}$
Subfield: Plane Geometry


Question: $如图所示，AD是\triangle ABC的一条中线，点O满足\overrightarrow{AO}=2\overrightarrow{OD}，过点O的直线分别与射线AB、射线AC交于M、N两点.$<img_1893>
$设\overrightarrow{AM}=m\overrightarrow{AB}，\overrightarrow{AN}=n\overrightarrow{AC}，m>0，n>0，求\frac{1}{m}+\frac{1}{n}的值.$

Solution:
$由M，O，N三点共线可设\overrightarrow{MO}=\lambda \overrightarrow{ON}，根据向量线性运算求出m，n，即可求解.$

Final answer: $\frac{1}{m}+\frac{1}{n}=\frac{1}{\lambda}$
Subfield: Plane Geometry


Question: $如图，E，F分别是矩形ABCD的边CD和BC的中点，AC与EF交于点N.$<img_2077>
$若\overrightarrow{CN}=\lambda \overrightarrow{CE}+\mu \overrightarrow{CF}，求\lambda +\mu 的值；$

Solution:
$设\overrightarrow{EN}=t\overrightarrow{EF}(0则\overrightarrow{CN}=\overrightarrow{CE}+\overrightarrow{EN}=\overrightarrow{CE}+t\overrightarrow{EF}=\overrightarrow{CE}+t(\overrightarrow{CF}-\overrightarrow{CE})=(1-t)\overrightarrow{CE}+t\overrightarrow{CF}，又\overrightarrow{CN}=\lambda \overrightarrow{CE}+\mu \overrightarrow{CF}，所以\left\{\begin{matrix}\lambda =1-t,\\ \mu =t,\end{matrix}\right.所以\lambda +\mu =1.(6分)$

Final answer: $1$
Subfield: Plane Geometry


Question: $如图，E，F分别是矩形ABCD的边CD和BC的中点，AC与EF交于点N.$![Image](https://cdntk.53zaixian.com/tq/1699316733586825219/img/compression/M_94fd75e4b7bf5c29dd394c3f158ea494.jpg)
$若\overrightarrow{CN}=\lambda \overrightarrow{CE}+\mu \overrightarrow{CF}，求\lambda +\mu 的值；$

Solution:
$设\overrightarrow{EN}=t\overrightarrow{EF}(0则\overrightarrow{CN}=\overrightarrow{CE}+\overrightarrow{EN}=\overrightarrow{CE}+t\overrightarrow{EF}=\overrightarrow{CE}+t(\overrightarrow{CF}-\overrightarrow{CE})=(1-t)\overrightarrow{CE}+t\overrightarrow{CF}，又\overrightarrow{CN}=\lambda \overrightarrow{CE}+\mu \overrightarrow{CF}，所以\left\{\begin{matrix}\lambda =1-t,\\ \mu =t,\end{matrix}\right.所以\lambda +\mu =1.(6分)$

Final answer: $1$
Subfield: Plane Geometry


Question: $如图，E，F分别是矩形ABCD的边CD和BC的中点，AC与EF交于点N.$<img_2077>
$设\overrightarrow{AE}=a，\overrightarrow{AF}=b，试用a，b表示\overrightarrow{AC}.$

Solution:
$取AC的中点为O，连接OE，OF，如图所示，$<img_1633>$\overrightarrow{AC}=2\overrightarrow{OC}=2(\overrightarrow{OE}+\overrightarrow{OF})=2\left(\overrightarrow{AE}-\frac{1}{2}\overrightarrow{AC}+\overrightarrow{AF}-\frac{1}{2}\overrightarrow{AC}\right)，所以3\overrightarrow{AC}=2\overrightarrow{AE}+2\overrightarrow{AF}，所以\overrightarrow{AC}=\frac{2}{3}a+\frac{2}{3}b.(12分)$

Final answer: $\overrightarrow{AC}=\frac{2}{3}a+\frac{2}{3}b$
Subfield: Plane Geometry


Question: $如图，E，F分别是矩形ABCD的边CD和BC的中点，AC与EF交于点N.$![Image](https://cdntk.53zaixian.com/tq/1699316733586825219/img/compression/M_94fd75e4b7bf5c29dd394c3f158ea494.jpg)
$设\overrightarrow{AE}=a，\overrightarrow{AF}=b，试用a，b表示\overrightarrow{AC}.$

Solution:
$取AC的中点为O，连接OE，OF，如图所示，$![Image](https://cdntk.53zaixian.com/tq/1699316733934952449/img/compression/M_d384f90d140412b2d3a81a037b54df41.jpg)$\overrightarrow{AC}=2\overrightarrow{OC}=2(\overrightarrow{OE}+\overrightarrow{OF})=2\left(\overrightarrow{AE}-\frac{1}{2}\overrightarrow{AC}+\overrightarrow{AF}-\frac{1}{2}\overrightarrow{AC}\right)，所以3\overrightarrow{AC}=2\overrightarrow{AE}+2\overrightarrow{AF}，所以\overrightarrow{AC}=\frac{2}{3}a+\frac{2}{3}b.(12分)$

Final answer: $\overrightarrow{AC}=\frac{2}{3}a+\frac{2}{3}b$
Subfield: Plane Geometry


Question: $已知O，A，B是平面上不共线的三点，直线AB上有一点C，满足2\overrightarrow{AC}+\overrightarrow{CB}=0.
$<img_2252>
$用\overrightarrow{OA}，\overrightarrow{OB}表示\overrightarrow{OC}；$

Solution:
$\because 2\overrightarrow{AC}+\overrightarrow{CB}=0，\therefore \overrightarrow{BC}=2\overrightarrow{AC}，\therefore A为线段BC的中点，(2分)\therefore \overrightarrow{OB}+\overrightarrow{OC}=2\overrightarrow{OA}，\therefore \overrightarrow{OC}=2\overrightarrow{OA}-\overrightarrow{OB}.(6分)$

Final answer: $\overrightarrow{OC}=2\overrightarrow{OA}-\overrightarrow{OB}$