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


Question: $已知函数f(x)=2\\sin  (\omega x+\phi )\left(\right.\omega >0,-\frac{\pi }{2}<\phi <\frac{\pi }{2}\left)\right.的部分图象如图所示.$<img_1778>
$求\omega ,\phi 的值；$

Solution:
$设函数f(x)的最小正周期为T，由题图可知，\frac{3}{4}T=\frac{\pi }{3}-\left(-\frac{5\pi }{12}\right)=\frac{3\pi }{4}，所以T=\pi .又T=\frac{2\pi }{|\omega |}，\omega >0，所以\omega =\frac{2\pi }{T}=2.又f\left(\frac{\pi }{3}\right)=2，所以\\sin  \left(\frac{2\pi }{3}+\varphi \right)=1.因为-\frac{\pi }{2}<\phi <\frac{\pi }{2}，所以\frac{\pi }{6}<\frac{2\pi }{3}+\phi <\frac{7\pi }{6}，所以\frac{2\pi }{3}+\phi =\frac{\pi }{2}，所以\phi =-\frac{\pi }{6}.$

Final answer: $\omega=2$, $\phi=-\frac{\pi}{6}$
Subfield: Plane Geometry


Question: $已知函数f(x)=2\\sin  (\omega x+\phi )\left(\right.\omega >0,-\frac{\pi }{2}<\phi <\frac{\pi }{2}\left)\right.的部分图象如图所示.$![Image](https://cdntk.53zaixian.com/tq/1704021062174584834/img/compression/M_f1c346006c0521a0d25e4644f82615d4.jpg)
$求\omega ,\phi 的值；$

Solution:
$设函数f(x)的最小正周期为T，由题图可知，\frac{3}{4}T=\frac{\pi }{3}-\left(-\frac{5\pi }{12}\right)=\frac{3\pi }{4}，所以T=\pi .又T=\frac{2\pi }{|\omega |}，\omega >0，所以\omega =\frac{2\pi }{T}=2.又f\left(\frac{\pi }{3}\right)=2，所以\\sin  \left(\frac{2\pi }{3}+\varphi \right)=1.因为-\frac{\pi }{2}<\phi <\frac{\pi }{2}，所以\frac{\pi }{6}<\frac{2\pi }{3}+\phi <\frac{7\pi }{6}，所以\frac{2\pi }{3}+\phi =\frac{\pi }{2}，所以\phi =-\frac{\pi }{6}.$

Final answer: $\omega=2$, $\phi=-\frac{\pi}{6}$
Subfield: Plane Geometry


Question: $已知函数f(x)=2\\sin  (\omega x+\phi )\left(\right.\omega >0,-\frac{\pi }{2}<\phi <\frac{\pi }{2}\left)\right.的部分图象如图所示.$<img_1778>
$求函数f(x)在[0,\pi ]上的单调递减区间；$

Solution:
$由(1)知，f(x)=2\\sin  \left(2x-\frac{\pi }{6}\right).当x\in [0,\pi ]时，2x-\frac{\pi }{6}\in \left[-\frac{\pi }{6},\frac{11\pi }{6}\right]，结合正弦函数的性质可知，当2x-\frac{\pi }{6}\in \left[-\frac{\pi }{6},\frac{\pi }{2}\right]，即x\in \left[0,\frac{\pi }{3}\right]时，f(x)单调递增；当2x-\frac{\pi }{6}\in \left(\frac{\pi }{2},\frac{3\pi }{2}\right]，即x\in \left(\frac{\pi }{3},\frac{5\pi }{6}\right]时，f(x)单调递减；当2x-\frac{\pi }{6}\in \left(\frac{3\pi }{2},\frac{11\pi }{6}\right]，即x\in \left(\frac{5\pi }{6},\pi \right]时，f(x)单调递增.所以函数f(x)在[0，\pi ]上的单调递增区间为\left[0,\frac{\pi }{3}\right]和\left(\frac{5\pi }{6},\pi \right]，单调递减区间为\left(\frac{\pi }{3},\frac{5\pi }{6}\right].$



Final answer: (\pi /3,5\pi /6]
Subfield: Plane Geometry


Question: $已知函数f(x)=2\\sin  (\omega x+\phi )\left(\right.\omega >0,-\frac{\pi }{2}<\phi <\frac{\pi }{2}\left)\right.的部分图象如图所示.$![Image](https://cdntk.53zaixian.com/tq/1704021062174584834/img/compression/M_f1c346006c0521a0d25e4644f82615d4.jpg)
$求函数f(x)在[0,\pi ]上的单调区间；$

Solution:
$由(1)知，f(x)=2\\sin  \left(2x-\frac{\pi }{6}\right).当x\in [0,\pi ]时，2x-\frac{\pi }{6}\in \left[-\frac{\pi }{6},\frac{11\pi }{6}\right]，结合正弦函数的性质可知，当2x-\frac{\pi }{6}\in \left[-\frac{\pi }{6},\frac{\pi }{2}\right]，即x\in \left[0,\frac{\pi }{3}\right]时，f(x)单调递增；当2x-\frac{\pi }{6}\in \left(\frac{\pi }{2},\frac{3\pi }{2}\right]，即x\in \left(\frac{\pi }{3},\frac{5\pi }{6}\right]时，f(x)单调递减；当2x-\frac{\pi }{6}\in \left(\frac{3\pi }{2},\frac{11\pi }{6}\right]，即x\in \left(\frac{5\pi }{6},\pi \right]时，f(x)单调递增.所以函数f(x)在[0，\pi ]上的单调递增区间为\left[0,\frac{\pi }{3}\right]和\left(\frac{5\pi }{6},\pi \right]，单调递减区间为\left(\frac{\pi }{3},\frac{5\pi }{6}\right].$

Final answer: $函数f(x)在[0，\pi ]上的单调递增区间为[0,\pi /3]和(5\pi /6,\pi ]，单调递减区间为(\pi /3,5\pi /6]。$
Subfield: Plane Geometry


Question: $已知函数f(x)=2\\sin  (\omega x+\phi )\left(\right.\omega >0,-\frac{\pi }{2}<\phi <\frac{\pi }{2}\left)\right.的部分图象如图所示.$![Image](https://cdntk.53zaixian.com/tq/1704021062174584834/img/compression/M_f1c346006c0521a0d25e4644f82615d4.jpg)
$若对任意x_{1},x_{2}\in [0，\pi ]都有|f(x_{1})-f(x_{2})|<m，求实数m的取值范围.$

Solution:
$结合(2)易知，函数f(x)在[0,\pi ]上的最大值为f \left(\frac{\pi }{3}\right)=2，最小值为f \left(\frac{5\pi }{6}\right)=-2，所以对任意x_{1},x_{2}\in [0，\pi ]，都有|f(x_{1})-f(x_{2})|\leq |2-(-2)|=4，因为对任意x_{1},x_{2}\in [0，\pi ]，都有|f(x_{1})-f(x_{2})|所以m>4，即m的取值范围是{m|m>4}.$

Final answer: $(4, +\infin)$
Subfield: Plane Geometry


Question: $已知函数f(x)=2\\sin  (\omega x+\phi )\left(\right.\omega >0,-\frac{\pi }{2}<\phi <\frac{\pi }{2}\left)\right.的部分图象如图所示.$<img_1778>
$若对任意x_{1},x_{2}\in [0，\pi ]都有|f(x_{1})-f(x_{2})|<m，求实数m的取值范围.$

Solution:
$结合(2)易知，函数f(x)在[0,\pi ]上的最大值为f \left(\frac{\pi }{3}\right)=2，最小值为f \left(\frac{5\pi }{6}\right)=-2，所以对任意x_{1},x_{2}\in [0，\pi ]，都有|f(x_{1})-f(x_{2})|\leq |2-(-2)|=4，因为对任意x_{1},x_{2}\in [0，\pi ]，都有|f(x_{1})-f(x_{2})|所以m>4，即m的取值范围是{m|m>4}.$

Final answer: $(4, +\infin)$
Subfield: Plane Geometry


Question: $如图，一只蚂蚁绕一个竖直放置的圆逆时针匀速爬行，已知圆的半径为1米，圆心O距离地面的高度为1.5米，蚂蚁爬行一圈需要4分钟，且蚂蚁的起始位置在最低点P_{0}处.$![Image](https://cdntk.53zaixian.com/tq/1704021090910879747/img/compression/M_c917d598abe97d599dd11c0fc3bf7410.jpg)
$试写出蚂蚁距离地面的高度h(米)关于爬行时间t(分钟)的函数关系式；$

Solution:
$如图所示，设t分钟时蚂蚁爬到A点，连接OA，过点A作AB\perp OP_{0},垂足为B.因为蚂蚁爬行一圈需要4分钟，所以t分钟时蚂蚁所转过的圆心角为\angle BOA=\frac{2\pi }{4}t=\frac{\pi }{2}t，在Rt\triangle OBA中，OB=\\cos  \frac{\pi }{2}t(米)，所以h=1.5-\\cos  \frac{\pi }{2}t.$![Image](https://cdntk.53zaixian.com/tq/1704021091267395586/img/compression/M_4df8dd96ebb1e86b2dcdc54ce8859ccf.jpg)

Final answer: $h=1.5 - \cos (\pi /2)t$
Subfield: Plane Geometry


Question: $如图，一只蚂蚁绕一个竖直放置的圆逆时针匀速爬行，已知圆的半径为1米，圆心O距离地面的高度为1.5米，蚂蚁爬行一圈需要4分钟，且蚂蚁的起始位置在最低点P_{0}处.$<img_2365>
$试写出蚂蚁距离地面的高度h(米)关于爬行时间t(分钟)的函数关系式；$

Solution:
$如图所示，设t分钟时蚂蚁爬到A点，连接OA，过点A作AB\perp OP_{0},垂足为B.因为蚂蚁爬行一圈需要4分钟，所以t分钟时蚂蚁所转过的圆心角为\angle BOA=\frac{2\pi }{4}t=\frac{\pi }{2}t，在Rt\triangle OBA中，OB=\\cos  \frac{\pi }{2}t(米)，所以h=1.5-\\cos  \frac{\pi }{2}t.$<img_2256>

Final answer: $h=1.5 - \cos (\pi /2)t$
Subfield: Plane Geometry


Question: $如图，一只蚂蚁绕一个竖直放置的圆逆时针匀速爬行，已知圆的半径为1米，圆心O距离地面的高度为1.5米，蚂蚁爬行一圈需要4分钟，且蚂蚁的起始位置在最低点P_{0}处.$![Image](https://cdntk.53zaixian.com/tq/1704021090910879747/img/compression/M_c917d598abe97d599dd11c0fc3bf7410.jpg)
$在蚂蚁绕圆爬行一圈的时间内，有多长时间蚂蚁距离地面超过1米？$

Solution:
$令h=1.5-\\cos  \frac{\pi }{2}t>1，得\\cos  \frac{\pi }{2}t<\frac{1}{2}，即\frac{\pi }{3}<\frac{\pi }{2}t<\frac{5\pi }{3}，所以\frac{2}{3}<t<\frac{10}{3}，则蚂蚁距离地面超过1米的时长为\frac{10}{3}-\frac{2}{3}=\frac{8}{3}(分钟).$

Final answer: $\frac{8}{3}$ 分钟
Subfield: Plane Geometry


Question: $如图，一只蚂蚁绕一个竖直放置的圆逆时针匀速爬行，已知圆的半径为1米，圆心O距离地面的高度为1.5米，蚂蚁爬行一圈需要4分钟，且蚂蚁的起始位置在最低点P_{0}处.$<img_2365>
$在蚂蚁绕圆爬行一圈的时间内，有多长时间蚂蚁距离地面超过1米？$

Solution:
$令h=1.5-\\cos  \frac{\pi }{2}t>1，得\\cos  \frac{\pi }{2}t<\frac{1}{2}，即\frac{\pi }{3}<\frac{\pi }{2}t<\frac{5\pi }{3}，所以\frac{2}{3}<t<\frac{10}{3}，则蚂蚁距离地面超过1米的时长为\frac{10}{3}-\frac{2}{3}=\frac{8}{3}(分钟).$

Final answer: $\frac{8}{3}$ 分钟
Subfield: Plane Geometry


Question: $如图，O是正方形ABCD对角线的交点，四边形OAED和四边形OCFB都是正方形，在向量\overrightarrow{AO},\overrightarrow{BO}，\overrightarrow{CO},\overrightarrow{DO},\overrightarrow{BF},\overrightarrow{CF},\overrightarrow{AE},\overrightarrow{DE}中：$<img_2763>
$分别写出与\overrightarrow{AO},\overrightarrow{BO}相等的向量；$

Solution:
$\overrightarrow{BF}=\overrightarrow{AO}，\overrightarrow{AE}=\overrightarrow{BO}.$

Final answer: $\overrightarrow{BF}, \overrightarrow{AE}$
Subfield: Plane Geometry


Question: $如图，O是正方形ABCD对角线的交点，四边形OAED和四边形OCFB都是正方形，在向量\overrightarrow{AO},\overrightarrow{BO}，\overrightarrow{CO},\overrightarrow{DO},\overrightarrow{BF},\overrightarrow{CF},\overrightarrow{AE},\overrightarrow{DE}中：$![Image](https://cdntk.53zaixian.com/tq/1704021122700529665/img/compression/M_6957742921058517a7dd9a972c71eb51.jpg)
$分别写出与\overrightarrow{AO},\overrightarrow{BO}相等的向量；$

Solution:
$\overrightarrow{BF}=\overrightarrow{AO}，\overrightarrow{AE}=\overrightarrow{BO}.$

Final answer: $\overrightarrow{BF}, \overrightarrow{AE}$
Subfield: Plane Geometry


Question: $如图，O是正方形ABCD对角线的交点，四边形OAED和四边形OCFB都是正方形，在向量\overrightarrow{AO},\overrightarrow{BO}，\overrightarrow{CO},\overrightarrow{DO},\overrightarrow{BF},\overrightarrow{CF},\overrightarrow{AE},\overrightarrow{DE}中：$![Image](https://cdntk.53zaixian.com/tq/1704021122700529665/img/compression/M_6957742921058517a7dd9a972c71eb51.jpg)
$写出与\overrightarrow{AO}共线的向量；$

Solution:
$与\overrightarrow{AO}共线的向量有\overrightarrow{BF},\overrightarrow{CO},\overrightarrow{DE}.$

Final answer: $\overrightarrow{BF},\overrightarrow{CO},\overrightarrow{DE}$
Subfield: Plane Geometry


Question: $如图，O是正方形ABCD对角线的交点，四边形OAED和四边形OCFB都是正方形，在向量\overrightarrow{AO},\overrightarrow{BO}，\overrightarrow{CO},\overrightarrow{DO},\overrightarrow{BF},\overrightarrow{CF},\overrightarrow{AE},\overrightarrow{DE}中：$<img_2763>
$写出与\overrightarrow{AO}共线的向量；$

Solution:
$与\overrightarrow{AO}共线的向量有\overrightarrow{BF},\overrightarrow{CO},\overrightarrow{DE}.$

Final answer: $\overrightarrow{BF},\overrightarrow{CO},\overrightarrow{DE}$
Subfield: Plane Geometry


Question: $如图，O是正方形ABCD对角线的交点，四边形OAED和四边形OCFB都是正方形，在向量\overrightarrow{AO},\overrightarrow{BO}，\overrightarrow{CO},\overrightarrow{DO},\overrightarrow{BF},\overrightarrow{CF},\overrightarrow{AE},\overrightarrow{DE}中：$![Image](https://cdntk.53zaixian.com/tq/1704021122700529665/img/compression/M_6957742921058517a7dd9a972c71eb51.jpg)
$写出与\overrightarrow{AO}长度相等的向量；$

Solution:
$与\overrightarrow{AO}长度相等的向量有\overrightarrow{CO},\overrightarrow{DO},\overrightarrow{BO},\overrightarrow{BF},\overrightarrow{CF},\overrightarrow{AE},\overrightarrow{DE}.$

Final answer: $ \overrightarrow{CO},\overrightarrow{DO},\overrightarrow{BO},\overrightarrow{BF},\overrightarrow{CF},\overrightarrow{AE},\overrightarrow{DE} $
Subfield: Plane Geometry


Question: $如图，O是正方形ABCD对角线的交点，四边形OAED和四边形OCFB都是正方形，在向量\overrightarrow{AO},\overrightarrow{BO}，\overrightarrow{CO},\overrightarrow{DO},\overrightarrow{BF},\overrightarrow{CF},\overrightarrow{AE},\overrightarrow{DE}中：$<img_2763>
$写出与\overrightarrow{AO}长度相等的向量；$

Solution:
$与\overrightarrow{AO}长度相等的向量有\overrightarrow{CO},\overrightarrow{DO},\overrightarrow{BO},\overrightarrow{BF},\overrightarrow{CF},\overrightarrow{AE},\overrightarrow{DE}.$

Final answer: $ \overrightarrow{CO},\overrightarrow{DO},\overrightarrow{BO},\overrightarrow{BF},\overrightarrow{CF},\overrightarrow{AE},\overrightarrow{DE} $
Subfield: Plane Geometry


Question: $如图，O是正方形ABCD对角线的交点，四边形OAED和四边形OCFB都是正方形，在向量\overrightarrow{AO},\overrightarrow{BO}，\overrightarrow{CO},\overrightarrow{DO},\overrightarrow{BF},\overrightarrow{CF},\overrightarrow{AE},\overrightarrow{DE}中：$<img_2763>
$(4)求向量\overrightarrow{AE}与\overrightarrow{DC}的夹角的大小.$

Solution:
$因为\overrightarrow{DC}=\overrightarrow{AB}，所以向量\overrightarrow{AE}与\overrightarrow{DC}的夹角即向量\overrightarrow{AE}与\overrightarrow{AB}的夹角，为135^\circ .$

Final answer: $135^\circ $
Subfield: Plane Geometry


Question: $如图，O是正方形ABCD对角线的交点，四边形OAED和四边形OCFB都是正方形，在向量\overrightarrow{AO},\overrightarrow{BO}，\overrightarrow{CO},\overrightarrow{DO},\overrightarrow{BF},\overrightarrow{CF},\overrightarrow{AE},\overrightarrow{DE}中：$![Image](https://cdntk.53zaixian.com/tq/1704021122700529665/img/compression/M_6957742921058517a7dd9a972c71eb51.jpg)
$(4)求向量\overrightarrow{AE}与\overrightarrow{DC}的夹角的大小.$

Solution:
$因为\overrightarrow{DC}=\overrightarrow{AB}，所以向量\overrightarrow{AE}与\overrightarrow{DC}的夹角即向量\overrightarrow{AE}与\overrightarrow{AB}的夹角，为135^\circ .$

Final answer: $135^\circ $
Subfield: Plane Geometry


Question: $如图，半圆的直径AB=6，C是半圆上的一点，D，E分别是AB，BC上的点，且AD=1，BE=4，DE=3.$![Image](https://cdntk.53zaixian.com/tq/1704021124030124033/img/compression/M_02fd71df14af58f613e7a6f739049338.jpg)
$求证：\overrightarrow{AC}\parallel \overrightarrow{DE}；$

Solution:
$证明：因为AB=6，AD=1，所以BD=5，又DE=3，BE=4，所以DE^{2}+BE^{2}=BD^{2}，所以\triangle DBE是直角三角形，\angle DEB=90^\circ .因为AB为直径，C为半圆上一点，所以\angle ACB=90^\circ .所以AC\parallel DE，故\overrightarrow{AC}\parallel \overrightarrow{DE}.$

Final answer: 证明题
Subfield: Plane Geometry


Question: $如图，半圆的直径AB=6，C是半圆上的一点，D，E分别是AB，BC上的点，且AD=1，BE=4，DE=3.$<img_2407>
$求证：\overrightarrow{AC}\parallel \overrightarrow{DE}；$

Solution:
$证明：因为AB=6，AD=1，所以BD=5，又DE=3，BE=4，所以DE^{2}+BE^{2}=BD^{2}，所以\triangle DBE是直角三角形，\angle DEB=90^\circ .因为AB为直径，C为半圆上一点，所以\angle ACB=90^\circ .所以AC\parallel DE，故\overrightarrow{AC}\parallel \overrightarrow{DE}.$
Subfield: Plane Geometry


Question: $如图，半圆的直径AB=6，C是半圆上的一点，D，E分别是AB，BC上的点，且AD=1，BE=4，DE=3.$<img_2407>
$求|\overrightarrow{AC}|；$

Solution:
$因为AC\parallel DE，所以\triangle ABC∽\triangle DBE，所以\frac{AC}{DE}=\frac{AB}{BD}，即\frac{AC}{3}=\frac{6}{5}，解得AC=\frac{18}{5}，即|\overrightarrow{AC}|=\frac{18}{5}.$

Final answer: $\frac{18}{5}$
Subfield: Plane Geometry


Question: $如图，半圆的直径AB=6，C是半圆上的一点，D，E分别是AB，BC上的点，且AD=1，BE=4，DE=3.$![Image](https://cdntk.53zaixian.com/tq/1704021124030124033/img/compression/M_02fd71df14af58f613e7a6f739049338.jpg)
$求|\overrightarrow{AC}|；$

Solution:
$因为AC\parallel DE，所以\triangle ABC∽\triangle DBE，所以\frac{AC}{DE}=\frac{AB}{BD}，即\frac{AC}{3}=\frac{6}{5}，解得AC=\frac{18}{5}，即|\overrightarrow{AC}|=\frac{18}{5}.$

Final answer: $\frac{18}{5}$
Subfield: Plane Geometry


Question: $如图，半圆的直径AB=6，C是半圆上的一点，D，E分别是AB，BC上的点，且AD=1，BE=4，DE=3.$![Image](https://cdntk.53zaixian.com/tq/1704021124030124033/img/compression/M_02fd71df14af58f613e7a6f739049338.jpg)
$求向量\overrightarrow{DE}与向量\overrightarrow{AB}的夹角的余弦值.$

Solution:
$向量\overrightarrow{DE}与向量\overrightarrow{AB}的夹角即向量\overrightarrow{DE}与向量\overrightarrow{DB}的夹角\angle EDB，而在Rt\triangle DBE中，\\cos  \angle EDB=\frac{DE}{DB}=\frac{3}{5}，所以向量\overrightarrow{DE}与向量\overrightarrow{AB}的夹角的余弦值为\frac{3}{5}.$

Final answer: $\frac{3}{5}$
Subfield: Plane Geometry


Question: $如图，半圆的直径AB=6，C是半圆上的一点，D，E分别是AB，BC上的点，且AD=1，BE=4，DE=3.$<img_2407>
$求向量\overrightarrow{DE}与向量\overrightarrow{AB}的夹角的余弦值.$

Solution:
$向量\overrightarrow{DE}与向量\overrightarrow{AB}的夹角即向量\overrightarrow{DE}与向量\overrightarrow{DB}的夹角\angle EDB，而在Rt\triangle DBE中，\\cos  \angle EDB=\frac{DE}{DB}=\frac{3}{5}，所以向量\overrightarrow{DE}与向量\overrightarrow{AB}的夹角的余弦值为\frac{3}{5}.$

Final answer: $\frac{3}{5}$
Subfield: Plane Geometry


Question: $如图，在▱ABCD中，\overrightarrow{AB}=a,\overrightarrow{AD}=b.$<img_3109>
$用a,b表示\overrightarrow{AC},\overrightarrow{DB}；$

Solution:
$\overrightarrow{AC}=\overrightarrow{AB}+\overrightarrow{AD}=a+b,\overrightarrow{DB}=\overrightarrow{AB}-\overrightarrow{AD}=a-b.$

Final answer: $\overrightarrow{AC}=a+b$,$\overrightarrow{DB}=a-b$
Subfield: Plane Geometry


Question: $如图，在▱ABCD中，\overrightarrow{AB}=a,\overrightarrow{AD}=b.$![Image](https://cdntk.53zaixian.com/tq/1704021131592454146/img/compression/M_b0d7d6a545cb21f3fbc1084dcdd73e37.jpg)
$用a,b表示\overrightarrow{AC},\overrightarrow{DB}；$

Solution:
$\overrightarrow{AC}=\overrightarrow{AB}+\overrightarrow{AD}=a+b,\overrightarrow{DB}=\overrightarrow{AB}-\overrightarrow{AD}=a-b.$

Final answer: $\overrightarrow{AC}=a+b$,$\overrightarrow{DB}=a-b$
Subfield: Plane Geometry


Question: $如图，在▱ABCD中，\overrightarrow{AB}=a,\overrightarrow{AD}=b.$![Image](https://cdntk.53zaixian.com/tq/1704021131592454146/img/compression/M_b0d7d6a545cb21f3fbc1084dcdd73e37.jpg)
$(2)当a,b满足什么条件时，a+b与a-b所在的直线互相垂直？$

Solution:
$由(1)知a+b=\overrightarrow{AC},a-b=\overrightarrow{DB}.当a+b，a-b所在的直线互相垂直时，AC\perp BD.又\because 四边形ABCD为平行四边形，\therefore 四边形ABCD为菱形，即a,b应满足|a|=|b|.$

Final answer: $a,b应满足|a|=|b|$
Subfield: Plane Geometry


Question: $如图，在▱ABCD中，\overrightarrow{AB}=a,\overrightarrow{AD}=b.$![Image](https://cdntk.53zaixian.com/tq/1704021131592454146/img/compression/M_b0d7d6a545cb21f3fbc1084dcdd73e37.jpg)
$(3)当a,b满足什么条件时，|a+b|=|a-b|?$

Solution:
$|a+b|=|a-b|，即|\overrightarrow{AC}|=|\overrightarrow{BD}|,\therefore 平行四边形ABCD为矩形，\therefore 当a与b所在的直线互相垂直时，满足|a+b|=|a-b|.$

Final answer: $当a与b所在的直线互相垂直时。$
Subfield: Plane Geometry


Question: $如图，在▱ABCD中，\overrightarrow{AB}=a,\overrightarrow{AD}=b.$![Image](https://cdntk.53zaixian.com/tq/1704021131592454146/img/compression/M_b0d7d6a545cb21f3fbc1084dcdd73e37.jpg)
$(4)a+b与a-b有可能为相等向量吗？为什么？$

Solution:
$不可能,▱ABCD的两条对角线不可能平行，因此a+b与a-b不可能为共线向量，\therefore a+b与a-b不可能为相等向量.$

Final answer: $不可能, a+b与a-b不可能为相等向量.$
Subfield: Plane Geometry


Question: $如图，在▱ABCD中，\overrightarrow{AB}=a,\overrightarrow{AD}=b.$<img_3109>
$(4)a+b与a-b有可能为相等向量吗？为什么？$

Solution:
$不可能,▱ABCD的两条对角线不可能平行，因此a+b与a-b不可能为共线向量，\therefore a+b与a-b不可能为相等向量.$



Final answer: $不可能, a+b与a-b不可能为相等向量.$
Subfield: Plane Geometry


Question: $如图，在\triangle ABC中，\overrightarrow{AM}=\frac{1}{3}\overrightarrow{AB},\overrightarrow{BN}=\frac{1}{2}\overrightarrow{BC}.设\overrightarrow{AB}=a,\overrightarrow{AC}=b.$![Image](https://cdntk.53zaixian.com/tq/1704021139754569729/img/compression/M_20dd8e6d703efaebe980d509b148a041.jpg)
$(1)用a,b表示\overrightarrow{BC},\overrightarrow{MN}；$

Solution:
$由题图知，\overrightarrow{BC}=\overrightarrow{AC}-\overrightarrow{AB}=b-a，\overrightarrow{MN}=\overrightarrow{BN}-\overrightarrow{BM}=\frac{1}{2}\overrightarrow{BC}+\frac{2}{3}\overrightarrow{AB}=\frac{1}{2}(b-a)+\frac{2}{3}a=\frac{1}{2}b+\frac{1}{6}a.$

Final answer: $\overrightarrow{BC}=b-a,\overrightarrow{MN}=\frac{1}{2}b+\frac{1}{6}a$
Subfield: Plane Geometry


Question: $如图，在\triangle ABC中，\overrightarrow{AM}=\frac{1}{3}\overrightarrow{AB},\overrightarrow{BN}=\frac{1}{2}\overrightarrow{BC}.设\overrightarrow{AB}=a,\overrightarrow{AC}=b.$<img_1984>
$(1)用a,b表示\overrightarrow{BC},\overrightarrow{MN}；$

Solution:
$由题图知，\overrightarrow{BC}=\overrightarrow{AC}-\overrightarrow{AB}=b-a，\overrightarrow{MN}=\overrightarrow{BN}-\overrightarrow{BM}=\frac{1}{2}\overrightarrow{BC}+\frac{2}{3}\overrightarrow{AB}=\frac{1}{2}(b-a)+\frac{2}{3}a=\frac{1}{2}b+\frac{1}{6}a.$

Final answer: $\overrightarrow{BC}=b-a,\overrightarrow{MN}=\frac{1}{2}b+\frac{1}{6}a$
Subfield: Plane Geometry


Question: $如图，在\triangle ABC中，\overrightarrow{AM}=\frac{1}{3}\overrightarrow{AB},\overrightarrow{BN}=\frac{1}{2}\overrightarrow{BC}.设\overrightarrow{AB}=a,\overrightarrow{AC}=b.$<img_1984>
$(2)若P为\triangle ABC内部一点，且\overrightarrow{AP}=\frac{5}{12}a+\frac{1}{4}b.求证：M,P,N三点共线.$

Solution:
$证明：连接AN，如图.易得\overrightarrow{AM}+\overrightarrow{AN}=\frac{1}{3}\overrightarrow{AB}+\overrightarrow{AC}+\overrightarrow{CN}=\frac{1}{3}\overrightarrow{AB}+\overrightarrow{AC}-\frac{1}{2}\overrightarrow{BC}=\frac{1}{3}a+b-\frac{1}{2}(b-a)=\frac{5}{6}a+\frac{1}{2}b，又\overrightarrow{AP}=\frac{5}{12}a+\frac{1}{4}b，所以\overrightarrow{AP}=\frac{1}{2}\overrightarrow{AM}+\frac{1}{2}\overrightarrow{AN}，故M,P,N三点共线.$<img_2707>
Subfield: Plane Geometry


Question: $如图，在\triangle ABC中，\overrightarrow{AM}=\frac{1}{3}\overrightarrow{AB},\overrightarrow{BN}=\frac{1}{2}\overrightarrow{BC}.设\overrightarrow{AB}=a,\overrightarrow{AC}=b.$![Image](https://cdntk.53zaixian.com/tq/1704021139754569729/img/compression/M_20dd8e6d703efaebe980d509b148a041.jpg)
$(2)若P为\triangle ABC内部一点，且\overrightarrow{AP}=\frac{5}{12}a+\frac{1}{4}b.求证：M,P,N三点共线.$

Solution:
$证明：连接AN，如图.易得\overrightarrow{AM}+\overrightarrow{AN}=\frac{1}{3}\overrightarrow{AB}+\overrightarrow{AC}+\overrightarrow{CN}=\frac{1}{3}\overrightarrow{AB}+\overrightarrow{AC}-\frac{1}{2}\overrightarrow{BC}=\frac{1}{3}a+b-\frac{1}{2}(b-a)=\frac{5}{6}a+\frac{1}{2}b，又\overrightarrow{AP}=\frac{5}{12}a+\frac{1}{4}b，所以\overrightarrow{AP}=\frac{1}{2}\overrightarrow{AM}+\frac{1}{2}\overrightarrow{AN}，故M,P,N三点共线.$![Image](https://cdntk.53zaixian.com/tq/1704021140194971649/img/compression/M_106c3558716b24cf81d05e3bf0b8ee2f.jpg)

Final answer: 证明题
Subfield: Plane Geometry


Question: $如图，在\triangle ABC中，点D在线段BC上,满足\overrightarrow{BD}=2\overrightarrow{DC}，M是AD的中点，过M的直线分别交AB，AC于P，Q两点，记 \overrightarrow{AP}=\lambda \overrightarrow{AB}，\overrightarrow{AQ}=\mu \overrightarrow{AC}，且\lambda >0，\mu >0．$![Image](https://cdntk.53zaixian.com/tq/1704021142078214146/img/compression/M_99391aeddb177d7f778c63ba0841a5ba.jpg)
$(1)试用\overrightarrow{AB},\overrightarrow{AC}分别表示\overrightarrow{AD}与\overrightarrow{AM}；$

Solution:
$由\overrightarrow{BD}=2\overrightarrow{DC}得\overrightarrow{AD}－\overrightarrow{AB}=2\overrightarrow{AC}－2\overrightarrow{AD}，则\overrightarrow{AD}=\frac{1}{3}\overrightarrow{AB}+\frac{2}{3}\overrightarrow{AC}.因为M是AD的中点，所以\overrightarrow{AM}=\frac{1}{2}\overrightarrow{AD}，所以\overrightarrow{AM}=\frac{1}{6}\overrightarrow{AB}+\frac{1}{3}\overrightarrow{AC}.$

Final answer: $\overrightarrow{AD}=\frac{1}{3}\overrightarrow{AB}+\frac{2}{3}\overrightarrow{AC}$，
$\overrightarrow{AM}=\frac{1}{6}\overrightarrow{AB}+\frac{1}{3}\overrightarrow{AC}$
Subfield: Plane Geometry


Question: $如图，在\triangle ABC中，点D在线段BC上,满足\overrightarrow{BD}=2\overrightarrow{DC}，M是AD的中点，过M的直线分别交AB，AC于P，Q两点，记 \overrightarrow{AP}=\lambda \overrightarrow{AB}，\overrightarrow{AQ}=\mu \overrightarrow{AC}，且\lambda >0，\mu >0．$<img_2486>
$试用\overrightarrow{AB},\overrightarrow{AC}表示\overrightarrow{AD}；$

Solution:
$由\overrightarrow{BD}=2\overrightarrow{DC}得\overrightarrow{AD}－\overrightarrow{AB}=2\overrightarrow{AC}－2\overrightarrow{AD}，则\overrightarrow{AD}=\frac{1}{3}\overrightarrow{AB}+\frac{2}{3}\overrightarrow{AC}.因为M是AD的中点，所以\overrightarrow{AM}=\frac{1}{2}\overrightarrow{AD}，所以\overrightarrow{AM}=\frac{1}{6}\overrightarrow{AB}+\frac{1}{3}\overrightarrow{AC}.$



Final answer: $\overrightarrow{AD}=\frac{1}{3}\overrightarrow{AB}+\frac{2}{3}\overrightarrow{AC}$
Subfield: Plane Geometry


Question: $如图，在\triangle ABC中，点D在线段BC上,满足\overrightarrow{BD}=2\overrightarrow{DC}，M是AD的中点，过M的直线分别交AB，AC于P，Q两点，记 \overrightarrow{AP}=\lambda \overrightarrow{AB}，\overrightarrow{AQ}=\mu \overrightarrow{AC}，且\lambda >0，\mu >0．$![Image](https://cdntk.53zaixian.com/tq/1704021142078214146/img/compression/M_99391aeddb177d7f778c63ba0841a5ba.jpg)
$(2)求\lambda +2\mu 的最小值.$

Solution:
$由\overrightarrow{AP}=\lambda \overrightarrow{AB},\overrightarrow{AQ}=\mu \overrightarrow{AC}，且\lambda >0，\mu >0得\overrightarrow{AB}=\frac{1}{\lambda }\overrightarrow{AP},\overrightarrow{AC}=\frac{1}{\mu }\overrightarrow{AQ}，所以\overrightarrow{AM}=\frac{1}{6\lambda }\overrightarrow{AP}+\frac{1}{3\mu }\overrightarrow{AQ}.又因为P，M，Q三点共线，所以\frac{1}{6\lambda }+\frac{1}{3\mu }=1,所以\lambda +2\mu =(\lambda +2\mu )\cdot \left(\frac{1}{6\lambda }+\frac{1}{3\mu }\right)=\frac{1}{6}+\frac{2}{3}+\frac{\mu }{3\lambda }+\frac{\lambda }{3\mu }\geq \frac{5}{6}+2\sqrt{\frac{\mu }{3\lambda }\cdot \frac{\lambda }{3\mu }}=\frac{5}{6}+\frac{2}{3}=\frac{3}{2}，当且仅当\frac{\mu }{3\lambda }=\frac{\lambda }{3\mu }，即\lambda =\mu =\frac{1}{2}时取等号，所以\lambda +2\mu 的最小值为\frac{3}{2}.$

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


Question: $如图，在\triangle ABC中，点D在线段BC上,满足\overrightarrow{BD}=2\overrightarrow{DC}，M是AD的中点，过M的直线分别交AB，AC于P，Q两点，记 \overrightarrow{AP}=\lambda \overrightarrow{AB}，\overrightarrow{AQ}=\mu \overrightarrow{AC}，且\lambda >0，\mu >0．$<img_2486>
$(2)求\lambda +2\mu 的最小值.$

Solution:
$由\overrightarrow{AP}=\lambda \overrightarrow{AB},\overrightarrow{AQ}=\mu \overrightarrow{AC}，且\lambda >0，\mu >0得\overrightarrow{AB}=\frac{1}{\lambda }\overrightarrow{AP},\overrightarrow{AC}=\frac{1}{\mu }\overrightarrow{AQ}，所以\overrightarrow{AM}=\frac{1}{6\lambda }\overrightarrow{AP}+\frac{1}{3\mu }\overrightarrow{AQ}.又因为P，M，Q三点共线，所以\frac{1}{6\lambda }+\frac{1}{3\mu }=1,所以\lambda +2\mu =(\lambda +2\mu )\cdot \left(\frac{1}{6\lambda }+\frac{1}{3\mu }\right)=\frac{1}{6}+\frac{2}{3}+\frac{\mu }{3\lambda }+\frac{\lambda }{3\mu }\geq \frac{5}{6}+2\sqrt{\frac{\mu }{3\lambda }\cdot \frac{\lambda }{3\mu }}=\frac{5}{6}+\frac{2}{3}=\frac{3}{2}，当且仅当\frac{\mu }{3\lambda }=\frac{\lambda }{3\mu }，即\lambda =\mu =\frac{1}{2}时取等号，所以\lambda +2\mu 的最小值为\frac{3}{2}.$

Final answer: $\frac{3}{2}$
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>
$(1)以{a,b}为基，写出向量\overrightarrow{BE}的分解式；$

Solution:
$\overrightarrow{BE}=\overrightarrow{BA}+\overrightarrow{AE}=-\overrightarrow{AB}+\frac{1}{2}\overrightarrow{AD}=-a+\frac{1}{2}b.$



Final answer: $-a+\frac{1}{2}b$
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)
$(1)以{a,b}为基，写出向量\overrightarrow{BE}的分解式；$

Solution:
$\overrightarrow{BE}=\overrightarrow{BA}+\overrightarrow{AE}=-\overrightarrow{AB}+\frac{1}{2}\overrightarrow{AD}=-a+\frac{1}{2}b.$

Final answer: $-\overrightarrow{AB}+\frac{1}{2}\overrightarrow{AD}=-a+\frac{1}{2}b$