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


Question: $已知抛物线C:y^{2}=2px(p>0)过点M(1，-2\sqrt{2})，直线l经过抛物线的焦点F与抛物线交于A，B两点.$<img_2277>
$若直线OA，OB的斜率分别为k_{1}，k_{2}，且k_{1}+k_{2}=2，求直线l的方程.$

Solution:
$由题意知直线l的斜率存在且不为0，设直线l的方程为y=k(x-2)，A(x_{1},y_{1})，B(x_{2},y_{2})，联立\left\{\begin{matrix}y^2=8x,\\  y=k(x-2),\end{matrix}\right.可得k^{2}x^{2}-(4k^{2}+8)x+4k^{2}=0，显然\Delta >0，从而x_{1}+x_{2}=\frac{4k^2+8}{k^2}，x_{1}x_{2}=4.\therefore k_{1}+k_{2}=\frac{y_1}{x_1}+\frac{y_2}{x_2}=\frac{k(x_1-2)}{x_1}+\frac{k(x_2-2)}{x_2}=2k-\left(\frac{2k}{x_1}+\frac{2k}{x_2}\right)=2k-\frac{2k(x_1+x_2)}{x_1x_2}=2k-2k\frac{\frac{4k^2+8}{k^2}}{4}=-\frac{4}{k}=2，\therefore k=-2.\therefore 直线l的方程为2x+y-4=0.$

Final answer: $2x+y-4=0$
Subfield: Plane Geometry


Question: $已知F为抛物线T:x^{2}=2py(p>0)的焦点，点P在抛物线T上，O为坐标原点，\triangle OPF的外接圆与抛物线T的准线相切，且该圆的周长为3\pi .$
$求抛物线T的方程；$

Solution:
$易得F\left(0,\frac{p}{2}\right)，所以\triangle OPF的外接圆圆心在直线y=\frac{p}{4}上，又外接圆与准线y=-\frac{p}{2}相切，所以外接圆的半径为\frac{p}{4}+\frac{p}{2}=\frac{3}{4}p，所以周长为2\pi \cdot \frac{3}{4}p=3\pi ，所以p=2，故抛物线T的方程为x^{2}=4y.$

Final answer: $x^{2}=4y$
Subfield: Plane Geometry


Question: $已知F为抛物线T:x^{2}=2py(p>0)的焦点，点P在抛物线T上，O为坐标原点，\triangle OPF的外接圆与抛物线T的准线相切，且该圆的周长为3\pi .$
$如图，设点A，B，C都在抛物线T上，若\triangle ABC是以AC为斜边的等腰直角三角形，求\overrightarrow{AB}\cdot \overrightarrow{AC}的最小值.$![Image](https://cdntk.53zaixian.com/tq/1623894924585533441/img/compression/M_8a260b9a388dbdbed79ced5306d72afa.jpg)

Solution:
$设点A\left(x_1,\frac{x^2_1}{4}\right)，B\left(x_2,\frac{x^2_2}{4}\right)，C\left(x_3,\frac{x^2_3}{4}\right)(x_{1}> x_{2}> x_{3}    )，直线AB的斜率为k(k>0)，因为AB\perp BC，所以直线BC的斜率为-\frac{1}{k}.因为|AB|=|BC|，所以\sqrt{1+k^2}\cdot |x_{1}-x_{2}   |=\sqrt{1+\frac{1}{k^2}}\cdot |x_{2}- x_{3}  |，得x_{2}- x_{3}  =k(x_{1}-x_{2}   )，①因为k=\frac{\frac{x^2_1}{4}-\frac{x^2_2}{4}}{x_1-x_2}=\frac{x_1+x_2}{4}，所以x_{1} +  x_{2} =4k，即x_{1} =4k-x_{2}，②因为-\frac{1}{k}=\frac{\frac{x^2_2}{4}-\frac{x^2_3}{4}}{x_2-x_3}=\frac{x_2+x_3}{4}，所以x_{2} +  x_{3} =-\frac{4}{k}，即x_{3} =-x_{2} -\frac{4}{k}，③将②③代入①，得2x_{2} +\frac{4}{k}=k(4k-2x_{2} )，即(k+1)x_{2} =2k^{2}-\frac{2}{k}=\frac{2(k^3-1)}{k}，则x_{2}=\frac{2(k^3-1)}{k(k+1)}，所以\overrightarrow{AB}\cdot \overrightarrow{AC}=|\overrightarrow{AB}|\cdot |\overrightarrow{AC}|\\cos   45^\circ =|\overrightarrow{AB}|^{2}=(x_{1}-x_{2}   )^{2}(1+k^{2})=(4k-2x_{2} )^{2}(1+k^{2})=\left[4k-\frac{4(k^3-1)}{k(k+1)}\right]^2(1+k^{2})=\frac{16(k^2+1)^3}{k^2(k+1)^2}因为k^{2}+1\geq 2k，所以(k^{2}+1)^{2}\geq 4k^{2}，又k^{2}+1\geq \frac{(k+1)^2}{2}，所以(k^{2}+1)^{3}\geq 2k^{2}(k+1)^{2}，从而\frac{{(k^2+1)}^3}{k^2(k+1)^2}\geq 2，当且仅当k=1时取等号，则\overrightarrow{AB}\cdot \overrightarrow{AC}\geq 32，所以\overrightarrow{AB}\cdot \overrightarrow{AC}的最小值为32.$

Final answer: $32$
Subfield: Plane Geometry


Question: $已知F为抛物线T:x^{2}=2py(p>0)的焦点，点P在抛物线T上，O为坐标原点，\triangle OPF的外接圆与抛物线T的准线相切，且该圆的周长为3\pi .$
$如图，设点A，B，C都在抛物线T上，若\triangle ABC是以AC为斜边的等腰直角三角形，求\overrightarrow{AB}\cdot \overrightarrow{AC}的最小值.$<img_1745>

Solution:
$设点A\left(x_1,\frac{x^2_1}{4}\right)，B\left(x_2,\frac{x^2_2}{4}\right)，C\left(x_3,\frac{x^2_3}{4}\right)(x_{1}> x_{2}> x_{3}    )，直线AB的斜率为k(k>0)，因为AB\perp BC，所以直线BC的斜率为-\frac{1}{k}.因为|AB|=|BC|，所以\sqrt{1+k^2}\cdot |x_{1}-x_{2}   |=\sqrt{1+\frac{1}{k^2}}\cdot |x_{2}- x_{3}  |，得x_{2}- x_{3}  =k(x_{1}-x_{2}   )，①因为k=\frac{\frac{x^2_1}{4}-\frac{x^2_2}{4}}{x_1-x_2}=\frac{x_1+x_2}{4}，所以x_{1} +  x_{2} =4k，即x_{1} =4k-x_{2}，②因为-\frac{1}{k}=\frac{\frac{x^2_2}{4}-\frac{x^2_3}{4}}{x_2-x_3}=\frac{x_2+x_3}{4}，所以x_{2} +  x_{3} =-\frac{4}{k}，即x_{3} =-x_{2} -\frac{4}{k}，③将②③代入①，得2x_{2} +\frac{4}{k}=k(4k-2x_{2} )，即(k+1)x_{2} =2k^{2}-\frac{2}{k}=\frac{2(k^3-1)}{k}，则x_{2}=\frac{2(k^3-1)}{k(k+1)}，所以\overrightarrow{AB}\cdot \overrightarrow{AC}=|\overrightarrow{AB}|\cdot |\overrightarrow{AC}|\\cos   45^\circ =|\overrightarrow{AB}|^{2}=(x_{1}-x_{2}   )^{2}(1+k^{2})=(4k-2x_{2} )^{2}(1+k^{2})=\left[4k-\frac{4(k^3-1)}{k(k+1)}\right]^2(1+k^{2})=\frac{16(k^2+1)^3}{k^2(k+1)^2}因为k^{2}+1\geq 2k，所以(k^{2}+1)^{2}\geq 4k^{2}，又k^{2}+1\geq \frac{(k+1)^2}{2}，所以(k^{2}+1)^{3}\geq 2k^{2}(k+1)^{2}，从而\frac{{(k^2+1)}^3}{k^2(k+1)^2}\geq 2，当且仅当k=1时取等号，则\overrightarrow{AB}\cdot \overrightarrow{AC}\geq 32，所以\overrightarrow{AB}\cdot \overrightarrow{AC}的最小值为32.$

Final answer: $32$
Subfield: Plane Geometry


Question: $如图，A为椭圆C:\frac{x^2}{8}+\frac{y^2}{4}=1的左顶点，过原点且异于x轴的直线与椭圆C交于M,N两点，直线AM,AN与圆O:x^{2}+y^{2}=8的另一交点分别为P,Q.$![Image](https://cdntk.53zaixian.com/tq/1623894946241208321/img/compression/M_8f15e62c45f2d703b7c1a5002e8f65b6.jpg)
$设直线AM,AN的斜率分别为k_{1},k_{2}，证明：k_{1}\cdot k_{2}为定值；$

Solution:
$证明：因为A为椭圆C:\frac{x^2}{8}+\frac{y^2}{4}=1的左顶点，故A(-2\sqrt{2},0).设M(x_{0},y_{0})，则N(-x_{0},-y_{0})，故直线AM的斜率k_{1}=\frac{y_0}{x_0+2\sqrt{2}}，直线AN的斜率k_{2}=\frac{-y_0}{-x_0+2\sqrt{2}}=\frac{y_0}{x_0-2\sqrt{2}}，故k_{1}\cdot k_{2}=\frac{y^2_0}{(x_0+2\sqrt{2})(x_0-2\sqrt{2})}=\frac{y^2_0}{x^2_0-8}.因为M(x_{0},y_{0})是椭圆C:\frac{x^2}{8}+\frac{y^2}{4}=1上的点，所以\frac{x^2_0}{8}+\frac{y^2_0}{4}=1，即y^2_0=-\frac{1}{2}(x^2_0-8)，故k_{1}\cdot k_{2}=-\frac{1}{2}.$

Final answer: 证明题
Subfield: Plane Geometry


Question: $如图，A为椭圆C:\frac{x^2}{8}+\frac{y^2}{4}=1的左顶点，过原点且异于x轴的直线与椭圆C交于M,N两点，直线AM,AN与圆O:x^{2}+y^{2}=8的另一交点分别为P,Q.$<img_2462>
$设直线AM,AN的斜率分别为k_{1},k_{2}，证明：k_{1}\cdot k_{2}为定值；$

Solution:
$证明：因为A为椭圆C:\frac{x^2}{8}+\frac{y^2}{4}=1的左顶点，故A(-2\sqrt{2},0).设M(x_{0},y_{0})，则N(-x_{0},-y_{0})，故直线AM的斜率k_{1}=\frac{y_0}{x_0+2\sqrt{2}}，直线AN的斜率k_{2}=\frac{-y_0}{-x_0+2\sqrt{2}}=\frac{y_0}{x_0-2\sqrt{2}}，故k_{1}\cdot k_{2}=\frac{y^2_0}{(x_0+2\sqrt{2})(x_0-2\sqrt{2})}=\frac{y^2_0}{x^2_0-8}.因为M(x_{0},y_{0})是椭圆C:\frac{x^2}{8}+\frac{y^2}{4}=1上的点，所以\frac{x^2_0}{8}+\frac{y^2_0}{4}=1，即y^2_0=-\frac{1}{2}(x^2_0-8)，故k_{1}\cdot k_{2}=-\frac{1}{2}.$
Subfield: Plane Geometry


Question: $如图，A为椭圆C:\frac{x^2}{8}+\frac{y^2}{4}=1的左顶点，过原点且异于x轴的直线与椭圆C交于M,N两点，直线AM,AN与圆O:x^{2}+y^{2}=8的另一交点分别为P,Q.$![Image](https://cdntk.53zaixian.com/tq/1623894946241208321/img/compression/M_8f15e62c45f2d703b7c1a5002e8f65b6.jpg)
$设\triangle AMN与\triangle APQ的面积分别为S_{1},S_{2}，求\frac{S_1}{S_2}的最大值.$

Solution:
$设P(x_{1},y_{1})，Q(x_{2},y_{2})，直线AM的方程为x=m_{1}y-2\sqrt{2}(m_{1}\neq 0).联立\left\{\begin{matrix}\frac{x^2}{8}+\frac{y^2}{4}=1,\\ x=m_1y-2\sqrt{2},\end{matrix}\right.得(m^2_1+2)y^{2}-4\sqrt{2}m_{1}y=0.则y=0或y=\frac{4\sqrt{2}m_1}{m^2_1+2}，故y_{0}=\frac{4\sqrt{2}m_1}{m^2_1+2}.将x=m_{1}y-2\sqrt{2}代入x^{2}+y^{2}=8，得(m^2_1+1)y^{2}-4\sqrt{2}m_{1}y=0，则y=0或y=\frac{4\sqrt{2}m_1}{m^2_1+1}，故y_{1}=\frac{4\sqrt{2}m_1}{m^2_1+1}.所以\frac{|AM|}{|AP|}=\left|\frac{y_0}{y_1}\right|=\frac{m^2_1+1}{m^2_1+2}.设直线AN的方程为x=m_{2}y-2\sqrt{2}(m_{2}\neq 0)，同理可得\frac{|AN|}{|AQ|}=\frac{m^2_2+1}{m^2_2+2}.又k_{1}\cdot k_{2}=-\frac{1}{2}，所以\frac{1}{m_1m_2}=-\frac{1}{2}，即m_{1}m_{2}=-2.故\frac{|AN|}{|AQ|}=\frac{m^2_2+1}{m^2_2+2}=\frac{m^2_1+4}{2m^2_1+4}.所以\frac{S_1}{S_2}=\frac{|AM| \cdot |AN|}{|AP| \cdot |AQ|}=\frac{m^2_1+1}{m^2_1+2}\cdot \frac{m^2_1+4}{2m^2_1+4}=\frac{(m^2_1+1)(m^2_1+4)}{2(m^2_1+2)^2}.令m^2_1+2=t，则\frac{S_1}{S_2}=\frac{(t-1)(t+2)}{2t^2}=\frac{1}{2}\left(-\frac{2}{t^2}+\frac{1}{t}+1\right)=-\left(\frac{1}{t}-\frac{1}{4}\right)^2+\frac{9}{16}.当\frac{1}{t}=\frac{1}{4}，即t=4，即m^2_1+2=4，即m^2_1=2时，\frac{S_1}{S_2}取得最大值\frac{9}{16}.$

Final answer: $\frac{9}{16}$
Subfield: Plane Geometry


Question: $如图，A为椭圆C:\frac{x^2}{8}+\frac{y^2}{4}=1的左顶点，过原点且异于x轴的直线与椭圆C交于M,N两点，直线AM,AN与圆O:x^{2}+y^{2}=8的另一交点分别为P,Q.$<img_2462>
$设\triangle AMN与\triangle APQ的面积分别为S_{1},S_{2}，求\frac{S_1}{S_2}的最大值.$

Solution:
$设P(x_{1},y_{1})，Q(x_{2},y_{2})，直线AM的方程为x=m_{1}y-2\sqrt{2}(m_{1}\neq 0).联立\left\{\begin{matrix}\frac{x^2}{8}+\frac{y^2}{4}=1,\\ x=m_1y-2\sqrt{2},\end{matrix}\right.得(m^2_1+2)y^{2}-4\sqrt{2}m_{1}y=0.则y=0或y=\frac{4\sqrt{2}m_1}{m^2_1+2}，故y_{0}=\frac{4\sqrt{2}m_1}{m^2_1+2}.将x=m_{1}y-2\sqrt{2}代入x^{2}+y^{2}=8，得(m^2_1+1)y^{2}-4\sqrt{2}m_{1}y=0，则y=0或y=\frac{4\sqrt{2}m_1}{m^2_1+1}，故y_{1}=\frac{4\sqrt{2}m_1}{m^2_1+1}.所以\frac{|AM|}{|AP|}=\left|\frac{y_0}{y_1}\right|=\frac{m^2_1+1}{m^2_1+2}.设直线AN的方程为x=m_{2}y-2\sqrt{2}(m_{2}\neq 0)，同理可得\frac{|AN|}{|AQ|}=\frac{m^2_2+1}{m^2_2+2}.又k_{1}\cdot k_{2}=-\frac{1}{2}，所以\frac{1}{m_1m_2}=-\frac{1}{2}，即m_{1}m_{2}=-2.故\frac{|AN|}{|AQ|}=\frac{m^2_2+1}{m^2_2+2}=\frac{m^2_1+4}{2m^2_1+4}.所以\frac{S_1}{S_2}=\frac{|AM| \cdot |AN|}{|AP| \cdot |AQ|}=\frac{m^2_1+1}{m^2_1+2}\cdot \frac{m^2_1+4}{2m^2_1+4}=\frac{(m^2_1+1)(m^2_1+4)}{2(m^2_1+2)^2}.令m^2_1+2=t，则\frac{S_1}{S_2}=\frac{(t-1)(t+2)}{2t^2}=\frac{1}{2}\left(-\frac{2}{t^2}+\frac{1}{t}+1\right)=-\left(\frac{1}{t}-\frac{1}{4}\right)^2+\frac{9}{16}.当\frac{1}{t}=\frac{1}{4}，即t=4，即m^2_1+2=4，即m^2_1=2时，\frac{S_1}{S_2}取得最大值\frac{9}{16}.$

Final answer: $\frac{9}{16}$
Subfield: Plane Geometry


Question: $已知抛物线C_{1}：y^{2}=4x与椭圆C_{2}：\frac{x^2}{a^2}+\frac{y^2}{b^2}=1(a>b>0)有公共的焦点，C_{2}的左、右焦点分别为F_{1}，F_{2}，该椭圆的离心率为\frac{1}{2}.$
$求椭圆C_{2}的方程；$

Solution:
$由题意可得，抛物线的焦点为(1,0)，所以椭圆的半焦距c=1，又椭圆的离心率e=\frac{c}{a}=\frac{1}{2}，所以a=2，则b^{2}=a^{2}-c^{2}=4-1=3，即b=\sqrt{3}，所以椭圆C_{2}的方程为\frac{x^2}{4}+\frac{y^2}{3}=1.$

Final answer: $\frac{x^2}{4}+\frac{y^2}{3}=1$
Subfield: Plane Geometry


Question: $已知抛物线C_{1}：y^{2}=4x与椭圆C_{2}：\frac{x^2}{a^2}+\frac{y^2}{b^2}=1(a>b>0)有公共的焦点，C_{2}的左、右焦点分别为F_{1}，F_{2}，该椭圆的离心率为\frac{1}{2}.$
$如图，若直线l与x轴、椭圆C_{2}顺次交于P，Q，R(点P在椭圆左顶点的左侧)，且\angle PF_{1}Q与\angle PF_{1}R互为补角，求\triangle F_{1}QR面积S的最大值.$![Image](https://cdntk.53zaixian.com/tq/1623894967469084673/img/compression/M_8663348533850efe435f2172cd3d4b8f.jpg)

Solution:
$设Q(x_{1},y_{1})，R(x_{2},y_{2})，F_{1}(-1,0)，因为\angle PF_{1}Q与\angle PF_{1}R互补，所以k_{QF_{1}}+k_{RF_{1}}=0，所以\frac{y_1}{x_1+1}+\frac{y_2}{x_2+1}=0，化简整理得x_{1}y_{2}+y_{2}+x_{2}y_{1}+y_{1}=0①.设直线PQ的方程为x=my+n(m\neq 0)，联立直线与椭圆方程得\left\{\begin{matrix}x=my+n,\\ \frac{x^2}{4}+\frac{y^2}{3}=1,\end{matrix}\right.化简整理可得(3m^{2}+4)y^{2}+6mny+3n^{2}-12=0，\Delta =36m^{2}n^{2}-4(3m^{2}+4)(3n^{2}-12)>0，可得n^{2}<3m^{2}+4②，由韦达定理，可得y_{1}+y_{2}=-\frac{6mn}{3m^2+4}，y_{1}y_{2}=\frac{3n^2-12}{3m^2+4}③.将x_{1}=my_{1}+n，x_{2}=my_{2}+n代入①，可得2my_{1}y_{2}+(n+1)(y_{1}+y_{2})=0④，将③代入④，可得\frac{6m(n^2-4)}{3m^2+4}=\frac{6mn(n+1)}{3m^2+4}，解得n=-4，所以直线PQ的方程为x=my-4.由②可得，3m^{2}+4>16，即m^{2}>4.点F_{1}(-1,0)到直线PQ的距离d=\frac{|-1 \times 1-0+4|}{\sqrt{1+m^2}}=\frac{3}{\sqrt{1+m^2}}，所以S=\frac{1}{2}|QR|\cdot d=\frac{1}{2}\sqrt{1+m^2}\cdot \sqrt{(y_1+y_2)^2-4y_1y_2}\cdot \frac{3}{\sqrt{1+m^2}}=18\sqrt{\frac{m^2-4}{(3m^2+4)^2}}.令\sqrt{m^2-4}=t，t>0，则S=\frac{18t}{3t^2+16}=\frac{18}{3t+\frac{16}{t}}\leq \frac{18}{2\sqrt{3 \times 16}}=\frac{3\sqrt{3}}{4}，当且仅当3t=\frac{16}{t}，m=\pm \frac{2\sqrt{21}}{3}时，等号成立，所以\triangle F_{1}QR面积S的最大值为\frac{3\sqrt{3}}{4}.$

Final answer: $\frac{3\sqrt 3}{4}$
Subfield: Plane Geometry


Question: $已知抛物线C_{1}：y^{2}=4x与椭圆C_{2}：\frac{x^2}{a^2}+\frac{y^2}{b^2}=1(a>b>0)有公共的焦点，C_{2}的左、右焦点分别为F_{1}，F_{2}，该椭圆的离心率为\frac{1}{2}.$
$如图，若直线l与x轴、椭圆C_{2}顺次交于P，Q，R(点P在椭圆左顶点的左侧)，且\angle PF_{1}Q与\angle PF_{1}R互为补角，求\triangle F_{1}QR面积S的最大值.$<img_2857>

Solution:
$设Q(x_{1},y_{1})，R(x_{2},y_{2})，F_{1}(-1,0)，因为\angle PF_{1}Q与\angle PF_{1}R互补，所以k_{QF_{1}}+k_{RF_{1}}=0，所以\frac{y_1}{x_1+1}+\frac{y_2}{x_2+1}=0，化简整理得x_{1}y_{2}+y_{2}+x_{2}y_{1}+y_{1}=0①.设直线PQ的方程为x=my+n(m\neq 0)，联立直线与椭圆方程得\left\{\begin{matrix}x=my+n,\\ \frac{x^2}{4}+\frac{y^2}{3}=1,\end{matrix}\right.化简整理可得(3m^{2}+4)y^{2}+6mny+3n^{2}-12=0，\Delta =36m^{2}n^{2}-4(3m^{2}+4)(3n^{2}-12)>0，可得n^{2}<3m^{2}+4②，由韦达定理，可得y_{1}+y_{2}=-\frac{6mn}{3m^2+4}，y_{1}y_{2}=\frac{3n^2-12}{3m^2+4}③.将x_{1}=my_{1}+n，x_{2}=my_{2}+n代入①，可得2my_{1}y_{2}+(n+1)(y_{1}+y_{2})=0④，将③代入④，可得\frac{6m(n^2-4)}{3m^2+4}=\frac{6mn(n+1)}{3m^2+4}，解得n=-4，所以直线PQ的方程为x=my-4.由②可得，3m^{2}+4>16，即m^{2}>4.点F_{1}(-1,0)到直线PQ的距离d=\frac{|-1 \times 1-0+4|}{\sqrt{1+m^2}}=\frac{3}{\sqrt{1+m^2}}，所以S=\frac{1}{2}|QR|\cdot d=\frac{1}{2}\sqrt{1+m^2}\cdot \sqrt{(y_1+y_2)^2-4y_1y_2}\cdot \frac{3}{\sqrt{1+m^2}}=18\sqrt{\frac{m^2-4}{(3m^2+4)^2}}.令\sqrt{m^2-4}=t，t>0，则S=\frac{18t}{3t^2+16}=\frac{18}{3t+\frac{16}{t}}\leq \frac{18}{2\sqrt{3 \times 16}}=\frac{3\sqrt{3}}{4}，当且仅当3t=\frac{16}{t}，m=\pm \frac{2\sqrt{21}}{3}时，等号成立，所以\triangle F_{1}QR面积S的最大值为\frac{3\sqrt{3}}{4}.$

Final answer: $\frac{3\sqrt 3}{4}$
Subfield: Plane Geometry


Question: $如图所示，在\triangle OAB中，C为AB的中点，点D在边OB上，且\overrightarrow{OD}=2\overrightarrow{DB}，OC，AD相交于点E，设\overrightarrow{OC}=a，\overrightarrow{OB}=b.$<img_2925>
$用a，b表示向量\overrightarrow{OA}，\overrightarrow{DA}；$

Solution:
$\because C为AB的中点，\therefore \overrightarrow{OA}+\overrightarrow{OB}=2\overrightarrow{OC}，\therefore \overrightarrow{OA}=2\overrightarrow{OC}-\overrightarrow{OB}=2a-b.\because \overrightarrow{OD}=2\overrightarrow{DB}，\therefore \overrightarrow{OD}=\frac{2}{3}\overrightarrow{OB}=\frac{2}{3}b，\therefore \overrightarrow{DA}=\overrightarrow{OA}-\overrightarrow{OD}=2a-b-\frac{2}{3}b=2a-\frac{5}{3}b.$

Final answer: $\overrightarrow{OA}=2a-b$, $\overrightarrow{DA}=2a-\frac{5}{3}b$
Subfield: Plane Geometry


Question: $如图所示，在\triangle OAB中，C为AB的中点，点D在边OB上，且\overrightarrow{OD}=2\overrightarrow{DB}，OC，AD相交于点E，设\overrightarrow{OC}=a，\overrightarrow{OB}=b.$![Image](https://cdntk.53zaixian.com/tq/1703664185721778178/img/compression/M_84f868e8d0bfdee4f3e191900e0c9f3b.jpg)
$用a，b表示向量\overrightarrow{OA}，\overrightarrow{DA}；$

Solution:
$\because C为AB的中点，\therefore \overrightarrow{OA}+\overrightarrow{OB}=2\overrightarrow{OC}，\therefore \overrightarrow{OA}=2\overrightarrow{OC}-\overrightarrow{OB}=2a-b.\because \overrightarrow{OD}=2\overrightarrow{DB}，\therefore \overrightarrow{OD}=\frac{2}{3}\overrightarrow{OB}=\frac{2}{3}b，\therefore \overrightarrow{DA}=\overrightarrow{OA}-\overrightarrow{OD}=2a-b-\frac{2}{3}b=2a-\frac{5}{3}b.$

Final answer: $\overrightarrow{OA}=2a-b$, $\overrightarrow{DA}=2a-\frac{5}{3}b$
Subfield: Plane Geometry


Question: $如图所示，在\triangle OAB中，C为AB的中点，点D在边OB上，且\overrightarrow{OD}=2\overrightarrow{DB}，OC，AD相交于点E，设\overrightarrow{OC}=a，\overrightarrow{OB}=b.$<img_2925>
$若\overrightarrow{OE}=\lambda \overrightarrow{OC}，求实数\lambda 的值.$

Solution:
$\because \overrightarrow{OE}=\lambda \overrightarrow{OC}，\therefore \overrightarrow{EA}=\overrightarrow{OA}-\overrightarrow{OE}=2a-b-\lambda a=(2-\lambda )a-b.\because A，E，D三点共线，\therefore 存在实数t，使\overrightarrow{EA}=t\overrightarrow{DA}，即(2-\lambda )a-b=2ta-\frac{5}{3}tb,即(2-\lambda -2t)a=\left(1-\frac{5}{3}t\right)b,\therefore \left\{\begin{matrix}2-\lambda -2t=0,\\ 1-\frac{5}{3}t=0,\end{matrix}\right.解得\left\{\begin{matrix}t=\frac{3}{5},\\ \lambda =\frac{4}{5}.\end{matrix}\right.$

Final answer: $\frac{4}{5}$
Subfield: Plane Geometry


Question: $如图所示，在\triangle OAB中，C为AB的中点，点D在边OB上，且\overrightarrow{OD}=2\overrightarrow{DB}，OC，AD相交于点E，设\overrightarrow{OC}=a，\overrightarrow{OB}=b.$![Image](https://cdntk.53zaixian.com/tq/1703664185721778178/img/compression/M_84f868e8d0bfdee4f3e191900e0c9f3b.jpg)
$若\overrightarrow{OE}=\lambda \overrightarrow{OC}，求实数\lambda 的值.$

Solution:
$\because \overrightarrow{OE}=\lambda \overrightarrow{OC}，\therefore \overrightarrow{EA}=\overrightarrow{OA}-\overrightarrow{OE}=2a-b-\lambda a=(2-\lambda )a-b.\because A，E，D三点共线，\therefore 存在实数t，使\overrightarrow{EA}=t\overrightarrow{DA}，即(2-\lambda )a-b=2ta-\frac{5}{3}tb,即(2-\lambda -2t)a=\left(1-\frac{5}{3}t\right)b,\therefore \left\{\begin{matrix}2-\lambda -2t=0,\\ 1-\frac{5}{3}t=0,\end{matrix}\right.解得\left\{\begin{matrix}t=\frac{3}{5},\\ \lambda =\frac{4}{5}.\end{matrix}\right.$

Final answer: $\frac{4}{5}$
Subfield: Plane Geometry


Question: $如图，D，E，F分别是等边三角形ABC各边的中点.$![Image](https://cdntk.53zaixian.com/tq/1703664159965573123/img/compression/M_9939585213033cb5c8f43a12e64926fb.jpg)
$写出图中与向量\overrightarrow{DE}长度相等的向量；$

Solution:
$与\overrightarrow{DE}长度相等的向量为\overrightarrow{EF},\overrightarrow{FD}，\overrightarrow{AF}，\overrightarrow{FC},\overrightarrow{BD},\overrightarrow{DA},\overrightarrow{CE},\overrightarrow{EB}.$

Final answer: $\overrightarrow{EF},\overrightarrow{FD}，\overrightarrow{AF}，\overrightarrow{FC},\overrightarrow{BD},\overrightarrow{DA},\overrightarrow{CE},\overrightarrow{EB}$
Subfield: Plane Geometry


Question: $如图，D，E，F分别是等边三角形ABC各边的中点.$<img_3124>
$写出图中与向量\overrightarrow{DE}长度相等的向量；$

Solution:
$与\overrightarrow{DE}长度相等的向量为\overrightarrow{EF},\overrightarrow{FD}，\overrightarrow{AF}，\overrightarrow{FC},\overrightarrow{BD},\overrightarrow{DA},\overrightarrow{CE},\overrightarrow{EB}.$





Final answer: $\overrightarrow{EF},\overrightarrow{FD}，\overrightarrow{AF}，\overrightarrow{FC},\overrightarrow{BD},\overrightarrow{DA},\overrightarrow{CE},\overrightarrow{EB}$
Subfield: Plane Geometry


Question: $如图，D，E，F分别是等边三角形ABC各边的中点.$![Image](https://cdntk.53zaixian.com/tq/1703664159965573123/img/compression/M_9939585213033cb5c8f43a12e64926fb.jpg)
$写出图中与向量\overrightarrow{FD}相等的向量；$

Solution:
$与\overrightarrow{FD}相等的向量为\overrightarrow{CE},\overrightarrow{EB}.$

Final answer: $\overrightarrow{CE},\overrightarrow{EB}$
Subfield: Plane Geometry


Question: $如图，D，E，F分别是等边三角形ABC各边的中点.$<img_3124>
$写出图中与向量\overrightarrow{FD}相等的向量；$

Solution:
$与\overrightarrow{FD}相等的向量为\overrightarrow{CE},\overrightarrow{EB}.$



Final answer: $\overrightarrow{CE},\overrightarrow{EB}$
Subfield: Plane Geometry


Question: $如图，D，E，F分别是等边三角形ABC各边的中点.$![Image](https://cdntk.53zaixian.com/tq/1703664159965573123/img/compression/M_9939585213033cb5c8f43a12e64926fb.jpg)
$分别写出图中与向量\overrightarrow{DE}，\overrightarrow{FD}共线的向量.$

Solution:
$与\overrightarrow{DE}共线的向量为\overrightarrow{AC},\overrightarrow{AF},\overrightarrow{FC}；与\overrightarrow{FD}共线的向量为\overrightarrow{CE},\overrightarrow{EB},\overrightarrow{CB}.$

Final answer: $与\overrightarrow{DE}共线的向量为\overrightarrow{AC},\overrightarrow{AF},\overrightarrow{FC}；与\overrightarrow{FD}共线的向量为\overrightarrow{CE},\overrightarrow{EB},\overrightarrow{CB}.$
Subfield: Plane Geometry


Question: $如图，D，E，F分别是等边三角形ABC各边的中点.$<img_3124>
$分别写出图中与向量\overrightarrow{DE}共线的向量.$

Solution:
$与\overrightarrow{DE}共线的向量为\overrightarrow{AC},\overrightarrow{AF},\overrightarrow{FC}；与\overrightarrow{FD}共线的向量为\overrightarrow{CE},\overrightarrow{EB},\overrightarrow{CB}.$



Final answer: \overrightarrow{AC},\overrightarrow{AF},\overrightarrow{FC}
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_2992>
$若\overrightarrow{AO}=\lambda \overrightarrow{AB}+\mu \overrightarrow{AC}，求\lambda ，\mu 的值；$

Solution:
$因为\overrightarrow{AO}=2\overrightarrow{OD}，所以\overrightarrow{AO}=\frac{2}{3}\overrightarrow{AD}，因为D为BC的中点，所以\overrightarrow{AD}=\frac{1}{2}(\overrightarrow{AB}+\overrightarrow{AC})，所以\overrightarrow{AO}=\frac{2}{3}\overrightarrow{AD}=\frac{1}{3}\overrightarrow{AB}+\frac{1}{3}\overrightarrow{AC}，又\overrightarrow{AO}=\lambda \overrightarrow{AB}+\mu \overrightarrow{AC}，\overrightarrow{AB}，\overrightarrow{AC}不共线，所以\lambda =\frac{1}{3}，\mu =\frac{1}{3}.$

Final answer: $\frac{1}{3}$，$\frac{1}{3}$
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/1703664200271818754/img/compression/M_0d5e93b7e69803ce9d98c1e40aa80f24.jpg)
$若\overrightarrow{AO}=\lambda \overrightarrow{AB}+\mu \overrightarrow{AC}，求\lambda ，\mu 的值；$

Solution:
$因为\overrightarrow{AO}=2\overrightarrow{OD}，所以\overrightarrow{AO}=\frac{2}{3}\overrightarrow{AD}，因为D为BC的中点，所以\overrightarrow{AD}=\frac{1}{2}(\overrightarrow{AB}+\overrightarrow{AC})，所以\overrightarrow{AO}=\frac{2}{3}\overrightarrow{AD}=\frac{1}{3}\overrightarrow{AB}+\frac{1}{3}\overrightarrow{AC}，又\overrightarrow{AO}=\lambda \overrightarrow{AB}+\mu \overrightarrow{AC}，\overrightarrow{AB}，\overrightarrow{AC}不共线，所以\lambda =\frac{1}{3}，\mu =\frac{1}{3}.$

Final answer: $\frac{1}{3}$，$\frac{1}{3}$
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_2992>
$设\overrightarrow{AM}=m\overrightarrow{AB}，\overrightarrow{AN}=n\overrightarrow{AC}，m>0，n>0，求\frac{1}{m}+\frac{1}{n}的值.$

Solution:
$因为\overrightarrow{AM}=m\overrightarrow{AB}，\overrightarrow{AN}=n\overrightarrow{AC}，m>0，n>0，所以\overrightarrow{AB}=\frac{1}{m}\overrightarrow{AM}，\overrightarrow{AC}=\frac{1}{n}\overrightarrow{AN}，由(1)知\overrightarrow{AO}=\frac{1}{3}\overrightarrow{AB}+\frac{1}{3}\overrightarrow{AC}，所以\overrightarrow{AO}=\frac{1}{3m}\overrightarrow{AM}+\frac{1}{3n}\overrightarrow{AN}，因为M，O，N三点共线，所以\frac{1}{3m}+\frac{1}{3n}=1，即\frac{1}{m}+\frac{1}{n}=3.$

Final answer: $3$
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/1703664200271818754/img/compression/M_0d5e93b7e69803ce9d98c1e40aa80f24.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:
$因为\overrightarrow{AM}=m\overrightarrow{AB}，\overrightarrow{AN}=n\overrightarrow{AC}，m>0，n>0，所以\overrightarrow{AB}=\frac{1}{m}\overrightarrow{AM}，\overrightarrow{AC}=\frac{1}{n}\overrightarrow{AN}，由(1)知\overrightarrow{AO}=\frac{1}{3}\overrightarrow{AB}+\frac{1}{3}\overrightarrow{AC}，所以\overrightarrow{AO}=\frac{1}{3m}\overrightarrow{AM}+\frac{1}{3n}\overrightarrow{AN}，因为M，O，N三点共线，所以\frac{1}{3m}+\frac{1}{3n}=1，即\frac{1}{m}+\frac{1}{n}=3.$

Final answer: $3$
Subfield: Plane Geometry


Question: $如图，在\triangle ABC中，\angle BAC=\frac{2\pi }{3}，\overrightarrow{AD}=3\overrightarrow{DB}，P为CD上一点，且满足\overrightarrow{AP}=m\overrightarrow{AC}+\frac{1}{2}\overrightarrow{AB}，若\overrightarrow{AB}\cdot \overrightarrow{AC}=-4.$![Image](https://cdntk.53zaixian.com/tq/1703664202490605570/img/compression/M_01672ac9745b5a09b7e394dadbaea3db.jpg)
$求m的值；$

Solution:
$由题意得\overrightarrow{AP}=m\overrightarrow{AC}+\frac{1}{2}\overrightarrow{AB}=m\overrightarrow{AC}+\frac{2}{3}\overrightarrow{AD}，因为C，P，D三点共线，所以m+\frac{2}{3}=1，解得m=\frac{1}{3}.$

Final answer: $\frac{1}{3}$
Subfield: Plane Geometry


Question: $如图，在\triangle ABC中，\angle BAC=\frac{2\pi }{3}，\overrightarrow{AD}=3\overrightarrow{DB}，P为CD上一点，且满足\overrightarrow{AP}=m\overrightarrow{AC}+\frac{1}{2}\overrightarrow{AB}，若\overrightarrow{AB}\cdot \overrightarrow{AC}=-4.$<img_2744>
$求m的值；$

Solution:
$由题意得\overrightarrow{AP}=m\overrightarrow{AC}+\frac{1}{2}\overrightarrow{AB}=m\overrightarrow{AC}+\frac{2}{3}\overrightarrow{AD}，因为C，P，D三点共线，所以m+\frac{2}{3}=1，解得m=\frac{1}{3}.$

Final answer: $\frac{1}{3}$
Subfield: Plane Geometry


Question: $如图，在\triangle ABC中，\angle BAC=\frac{2\pi }{3}，\overrightarrow{AD}=3\overrightarrow{DB}，P为CD上一点，且满足\overrightarrow{AP}=m\overrightarrow{AC}+\frac{1}{2}\overrightarrow{AB}，若\overrightarrow{AB}\cdot \overrightarrow{AC}=-4.$<img_2744>
$求|\overrightarrow{AP}|的最小值.$

Solution:
$因为\overrightarrow{AB}\cdot \overrightarrow{AC}=-4，所以|\overrightarrow{AB}||\overrightarrow{AC}|\\cos  \frac{2\pi }{3}=-4,所以|\overrightarrow{AB}||\overrightarrow{AC}|=8.由(1)可知\overrightarrow{AP}=\frac{1}{3}\overrightarrow{AC}+\frac{1}{2}\overrightarrow{AB}，所以|\overrightarrow{AP}|^{2}=\left(\frac{1}{3}\overrightarrow{AC}+\frac{1}{2}\overrightarrow{AB}\right)^{2}=\frac{|\overrightarrow{AC}|^2}{9}+\frac{|\overrightarrow{AB}|^2}{4}+\frac{1}{3}\overrightarrow{AC}\cdot \overrightarrow{AB}=\frac{|\overrightarrow{AC}|^2}{9}+\frac{|\overrightarrow{AB}|^2}{4}-\frac{4}{3}\geq 2\times \frac{|\overrightarrow{AC}||\overrightarrow{AB}|}{6}-\frac{4}{3}=\frac{4}{3}，当且仅当|\overrightarrow{AC}|=2\sqrt{3},|\overrightarrow{AB}|=\frac{4\sqrt{3}}{3}时取等号.故|\overrightarrow{AP}|\geq \frac{2\sqrt{3}}{3}，则|\overrightarrow{AP}|的最小值为\frac{2\sqrt{3}}{3}.$

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


Question: $如图，在\triangle ABC中，\angle BAC=\frac{2\pi }{3}，\overrightarrow{AD}=3\overrightarrow{DB}，P为CD上一点，且满足\overrightarrow{AP}=m\overrightarrow{AC}+\frac{1}{2}\overrightarrow{AB}，若\overrightarrow{AB}\cdot \overrightarrow{AC}=-4.$![Image](https://cdntk.53zaixian.com/tq/1703664202490605570/img/compression/M_01672ac9745b5a09b7e394dadbaea3db.jpg)
$求|\overrightarrow{AP}|的最小值.$

Solution:
$因为\overrightarrow{AB}\cdot \overrightarrow{AC}=-4，所以|\overrightarrow{AB}||\overrightarrow{AC}|\\cos  \frac{2\pi }{3}=-4,所以|\overrightarrow{AB}||\overrightarrow{AC}|=8.由(1)可知\overrightarrow{AP}=\frac{1}{3}\overrightarrow{AC}+\frac{1}{2}\overrightarrow{AB}，所以|\overrightarrow{AP}|^{2}=\left(\frac{1}{3}\overrightarrow{AC}+\frac{1}{2}\overrightarrow{AB}\right)^{2}=\frac{|\overrightarrow{AC}|^2}{9}+\frac{|\overrightarrow{AB}|^2}{4}+\frac{1}{3}\overrightarrow{AC}\cdot \overrightarrow{AB}=\frac{|\overrightarrow{AC}|^2}{9}+\frac{|\overrightarrow{AB}|^2}{4}-\frac{4}{3}\geq 2\times \frac{|\overrightarrow{AC}||\overrightarrow{AB}|}{6}-\frac{4}{3}=\frac{4}{3}，当且仅当|\overrightarrow{AC}|=2\sqrt{3},|\overrightarrow{AB}|=\frac{4\sqrt{3}}{3}时取等号.故|\overrightarrow{AP}|\geq \frac{2\sqrt{3}}{3}，则|\overrightarrow{AP}|的最小值为\frac{2\sqrt{3}}{3}.$

Final answer: $\frac{2\sqrt{3}}{3}$