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


Question: $如图，已知五边形ABCDE,\overrightarrow{AB}=a,\overrightarrow{BC}=b,\overrightarrow{CD}=c,\overrightarrow{DE}=d,\overrightarrow{EA}=e.$![Image](https://cdntk.53zaixian.com/tq/1704016041487839235/img/compression/M_af11b79bf137434a315eaca1ab6d5599.jpg)
$用a，d，e表示\overrightarrow{DB}；$

Solution:
$\overrightarrow{DB}=\overrightarrow{DE}＋\overrightarrow{EA}＋\overrightarrow{AB}=d＋e＋a.$

Final answer: $d+e+a$
Subfield: Plane Geometry


Question: $如图，已知五边形ABCDE,\overrightarrow{AB}=a,\overrightarrow{BC}=b,\overrightarrow{CD}=c,\overrightarrow{DE}=d,\overrightarrow{EA}=e.$<img_2466>
$用a，d，e表示\overrightarrow{DB}；$

Solution:
$\overrightarrow{DB}=\overrightarrow{DE}＋\overrightarrow{EA}＋\overrightarrow{AB}=d＋e＋a.$

Final answer: $d+e+a$
Subfield: Plane Geometry


Question: $如图，已知五边形ABCDE,\overrightarrow{AB}=a,\overrightarrow{BC}=b,\overrightarrow{CD}=c,\overrightarrow{DE}=d,\overrightarrow{EA}=e.$<img_2466>
$用b，c表示\overrightarrow{DB}；$

Solution:
$\overrightarrow{DB}=\overrightarrow{CB}-\overrightarrow{CD}=-\overrightarrow{BC}-\overrightarrow{CD}=-b-c.$

Final answer: $-\overrightarrow{BC}-\overrightarrow{CD}=-b-c$
Subfield: Plane Geometry


Question: $如图，已知五边形ABCDE,\overrightarrow{AB}=a,\overrightarrow{BC}=b,\overrightarrow{CD}=c,\overrightarrow{DE}=d,\overrightarrow{EA}=e.$![Image](https://cdntk.53zaixian.com/tq/1704016041487839235/img/compression/M_af11b79bf137434a315eaca1ab6d5599.jpg)
$用b，c表示\overrightarrow{DB}；$

Solution:
$\overrightarrow{DB}=\overrightarrow{CB}-\overrightarrow{CD}=-\overrightarrow{BC}-\overrightarrow{CD}=-b-c.$

Final answer: $-\overrightarrow{BC}-\overrightarrow{CD}=-b-c$
Subfield: Plane Geometry


Question: $如图，已知五边形ABCDE,\overrightarrow{AB}=a,\overrightarrow{BC}=b,\overrightarrow{CD}=c,\overrightarrow{DE}=d,\overrightarrow{EA}=e.$<img_2466>
$用a，b，e表示\overrightarrow{EC}；$

Solution:
$\overrightarrow{EC}=\overrightarrow{EA}＋\overrightarrow{AB}＋\overrightarrow{BC}=e+a＋b.$

Final answer: $\overrightarrow{EC}=e+a+b$
Subfield: Plane Geometry


Question: $如图，已知五边形ABCDE,\overrightarrow{AB}=a,\overrightarrow{BC}=b,\overrightarrow{CD}=c,\overrightarrow{DE}=d,\overrightarrow{EA}=e.$![Image](https://cdntk.53zaixian.com/tq/1704016041487839235/img/compression/M_af11b79bf137434a315eaca1ab6d5599.jpg)
$用a，b，e表示\overrightarrow{EC}；$

Solution:
$\overrightarrow{EC}=\overrightarrow{EA}＋\overrightarrow{AB}＋\overrightarrow{BC}=e+a＋b.$

Final answer: $\overrightarrow{EC}=e+a+b$
Subfield: Plane Geometry


Question: $如图，已知五边形ABCDE,\overrightarrow{AB}=a,\overrightarrow{BC}=b,\overrightarrow{CD}=c,\overrightarrow{DE}=d,\overrightarrow{EA}=e.$<img_2466>
$用d，c表示\overrightarrow{EC}.$

Solution:
$\overrightarrow{EC}=-\overrightarrow{CE}=-(\overrightarrow{CD}＋\overrightarrow{DE})=-c-d.$

Final answer: $-\overrightarrow{EC}=c+d$
Subfield: Plane Geometry


Question: $如图，已知五边形ABCDE,\overrightarrow{AB}=a,\overrightarrow{BC}=b,\overrightarrow{CD}=c,\overrightarrow{DE}=d,\overrightarrow{EA}=e.$![Image](https://cdntk.53zaixian.com/tq/1704016041487839235/img/compression/M_af11b79bf137434a315eaca1ab6d5599.jpg)
$用d，c表示\overrightarrow{EC}.$

Solution:
$\overrightarrow{EC}=-\overrightarrow{CE}=-(\overrightarrow{CD}＋\overrightarrow{DE})=-c-d.$

Final answer: $-\overrightarrow{EC}=c+d$
Subfield: Plane Geometry


Question: $如图所示，在平行四边形ABCD中，E，F分别为边AB和BC的中点，G为AC与BD的交点.$![Image](https://cdntk.53zaixian.com/tq/1704016041919852547/img/compression/M_b7490da2195edd4770e8f4c85488accc.jpg)
$若|\overrightarrow{AB}|=|\overrightarrow{AB}+\overrightarrow{BC}+\overrightarrow{CD}|，问四边形ABCD是什么特殊的平行四边形？$

Solution:
$|\overrightarrow{AB}|=|\overrightarrow{AB}+\overrightarrow{BC}+\overrightarrow{CD}|=|\overrightarrow{AD}|，故平行四边形ABCD是菱形.$

Final answer: $菱形$
Subfield: Plane Geometry


Question: $如图，按下列要求作答.$![Image](https://cdntk.53zaixian.com/tq/1704016042955845633/img/compression/M_b7f9e2f6a7207c14cc7a3d1971f1fb3f.jpg)
$以A为始点，作出向量a+b；$

Solution:
$将a,b的起点同时平移到A点，利用向量加法的平行四边形法则作出a+b，如图所示.$

Final answer: $a+b可以通过向量加法的平行四边形法则作出。$
Subfield: Plane Geometry


Question: $如图，按下列要求作答.$![Image](https://cdntk.53zaixian.com/tq/1704016042955845633/img/compression/M_b7f9e2f6a7207c14cc7a3d1971f1fb3f.jpg)
$以B为始点，作出向量c+d+e；$

Solution:
$先将共线向量c,d的起点同时平移到B点,得出c+d,再将向量e的起点与c+d的终点相接，利用向量加法的三角形法则即可作出c+d+e，如图所示.$![Image](https://cdntk.53zaixian.com/tq/1704016043287195649/img/compression/M_40d0b3d405c8b5a943fff8f236c11eef.jpg)

Final answer: $c+d+e$
Subfield: Plane Geometry


Question: $如图，按下列要求作答.$<img_2322>
$以B为始点，作出向量c+d+e；$

Solution:
$先将共线向量c,d的起点同时平移到B点,得出c+d,再将向量e的起点与c+d的终点相接，利用向量加法的三角形法则即可作出c+d+e，如图所示.$<img_2885>

Final answer: $c+d+e$
Subfield: Plane Geometry


Question: $如图，按下列要求作答.$![Image](https://cdntk.53zaixian.com/tq/1704016042955845633/img/compression/M_b7f9e2f6a7207c14cc7a3d1971f1fb3f.jpg)
$若a为单位向量，求|a+b|，|c+d|和|c+d+e|.$

Solution:
$由a是单位向量可知|a|=1，根据作出的向量结合勾股定理可知，|a+b|=\sqrt{1^2+2^2}=\sqrt{5}.由c,d是共线向量可知|c+d|=|-c|=|c|=1.利用图中所示的向量和勾股定理可知|c+d+e|=\sqrt{2^2+3^2}=\sqrt{13}.$

Final answer: $5$, $1$, $\sqrt{13}$
Subfield: Plane Geometry


Question: $如图，按下列要求作答.$<img_2322>
$若a为单位向量，求|a+b|，|c+d|和|c+d+e|.$

Solution:
$由a是单位向量可知|a|=1，根据作出的向量结合勾股定理可知，|a+b|=\sqrt{1^2+2^2}=\sqrt{5}.由c,d是共线向量可知|c+d|=|-c|=|c|=1.利用图中所示的向量和勾股定理可知|c+d+e|=\sqrt{2^2+3^2}=\sqrt{13}.$

Final answer: $5$, $1$, $\sqrt{13}$
Subfield: Plane Geometry


Question: $如图，在\triangle ABC中，CA=1，CB=2，\angle ACB=60^\circ ．$![Image](https://cdntk.53zaixian.com/tq/1706233168848433154/img/compression/M_bbd56a80c580d7e0c57783bf41d741f0.jpg)
$求|\overrightarrow{AB}|；$

Solution:
$易得\overrightarrow{AB}=\overrightarrow{CB}-\overrightarrow{CA}，\overrightarrow{CB}^{2}=4，\overrightarrow{CA}^{2}=1，\overrightarrow{CB}\cdot \overrightarrow{CA}=2\times 1\times \\cos   60^\circ =1，\therefore |\overrightarrow{AB}|=|\overrightarrow{CB}-\overrightarrow{CA}|=\sqrt{(\overrightarrow{CB}-\overrightarrow{CA})^2}=\sqrt{{\overrightarrow{CB}}^2-2\overrightarrow{CB}\cdot \overrightarrow{CA}+{\overrightarrow{CA}}^2}=\sqrt{3}．$

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


Question: $如图，在\triangle ABC中，CA=1，CB=2，\angle ACB=60^\circ ．$<img_1515>
$求|\overrightarrow{AB}|；$

Solution:
$易得\overrightarrow{AB}=\overrightarrow{CB}-\overrightarrow{CA}，\overrightarrow{CB}^{2}=4，\overrightarrow{CA}^{2}=1，\overrightarrow{CB}\cdot \overrightarrow{CA}=2\times 1\times \\cos   60^\circ =1，\therefore |\overrightarrow{AB}|=|\overrightarrow{CB}-\overrightarrow{CA}|=\sqrt{(\overrightarrow{CB}-\overrightarrow{CA})^2}=\sqrt{{\overrightarrow{CB}}^2-2\overrightarrow{CB}\cdot \overrightarrow{CA}+{\overrightarrow{CA}}^2}=\sqrt{3}．$

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


Question: $如图，在\triangle ABC中，CA=1，CB=2，\angle ACB=60^\circ ．$<img_1515>
$已知D是AB上一点，满足\overrightarrow{AD}=\lambda \overrightarrow{AB}，E是CB上一点，满足\overrightarrow{BE}=\lambda \overrightarrow{BC}．①当\lambda =\frac{1}{2}时,求\overrightarrow{AE}\cdot \overrightarrow{CD}；②是否存在非零实数\lambda ，使得\overrightarrow{AE}\perp \overrightarrow{CD}？若存在，求出\lambda 的值；若不存在，请说明理由.$

Solution:
$①当\lambda =\frac{1}{2}时，\overrightarrow{AD}=\frac{1}{2}\overrightarrow{AB}，\overrightarrow{BE}=\frac{1}{2}\overrightarrow{BC}，\therefore D，E分别是AB，BC的中点，\therefore \overrightarrow{AE}=\overrightarrow{AC}+\overrightarrow{CE}=\overrightarrow{AC}+\frac{1}{2}\overrightarrow{CB}，\overrightarrow{CD}=\frac{1}{2}(\overrightarrow{CA}+\overrightarrow{CB})，\therefore \overrightarrow{AE}\cdot \overrightarrow{CD}=\left(\overrightarrow{AC}+\frac{1}{2}\overrightarrow{CB}\right)\cdot \frac{1}{2}(\overrightarrow{CA}+\overrightarrow{CB}) =\frac{1}{2}\overrightarrow{AC}\cdot \overrightarrow{CA}+\frac{1}{2}\overrightarrow{AC}\cdot \overrightarrow{CB}+\frac{1}{4}\overrightarrow{CB}\cdot \overrightarrow{CA}+\frac{1}{4}\overrightarrow{CB}^{2}=-\frac{1}{2}\times 1^{2}+\frac{1}{2}\times 1\times 2\times \\cos   120^\circ +\frac{1}{4}\times 2\times 1\times \\cos   60^\circ +\frac{1}{4}\times 2^{2}=\frac{1}{4}.②存在．理由如下：假设存在非零实数\lambda ，使得\overrightarrow{AE}\perp \overrightarrow{CD}.由\overrightarrow{AD}=\lambda \overrightarrow{AB}，得\overrightarrow{AD}=\lambda (\overrightarrow{CB}-\overrightarrow{CA})，\therefore \overrightarrow{CD}=\overrightarrow{CA}+\overrightarrow{AD}=\overrightarrow{CA}+\lambda (\overrightarrow{CB}-\overrightarrow{CA})=\lambda \overrightarrow{CB}+(1-\lambda )\overrightarrow{CA}．\because \overrightarrow{BE}=\lambda \overrightarrow{BC}，\therefore \overrightarrow{AE}=\overrightarrow{AB}+\overrightarrow{BE}=(\overrightarrow{CB}-\overrightarrow{CA})+\lambda \overrightarrow{BC}=(1-\lambda )\overrightarrow{CB}-\overrightarrow{CA}.\therefore \overrightarrow{AE}\cdot \overrightarrow{CD}=\lambda (1-\lambda )\overrightarrow{CB}^{2}-\lambda \overrightarrow{CB}\cdot \overrightarrow{CA}+(1-\lambda )^{2}\overrightarrow{CB}\cdot \overrightarrow{CA}-(1-\lambda )\overrightarrow{CA}^{2}=4\lambda (1-\lambda )-\lambda +(1-\lambda )^{2}-(1-\lambda )=-3\lambda ^{2}+2\lambda =0，解得\lambda =\frac{2}{3}或\lambda =0(不合题意，舍去)，故存在非零实数\lambda =\frac{2}{3},使得\overrightarrow{AE}\perp \overrightarrow{CD}.$



Final answer: $问题①：当\lambda =1/2时，求⟠AE\cdot ⟠CD的值。
答案：⟠AE\cdot ⟠CD=1/4。
问题②：存在非零实数\lambda ，使得⟠AE\perp ⟠CD吗？若存在，求出\lambda 的值；若不存在，请说明理由。
答案：存在非零实数\lambda =2/3，使得⟠AE\perp ⟠CD。$
Subfield: Plane Geometry


Question: $如图，在\triangle ABC中，CA=1，CB=2，\angle ACB=60^\circ ．$![Image](https://cdntk.53zaixian.com/tq/1706233168848433154/img/compression/M_bbd56a80c580d7e0c57783bf41d741f0.jpg)
$已知D是AB上一点，满足\overrightarrow{AD}=\lambda \overrightarrow{AB}，E是CB上一点，满足\overrightarrow{BE}=\lambda \overrightarrow{BC}．①当\lambda =\frac{1}{2}时,求\overrightarrow{AE}\cdot \overrightarrow{CD}；②是否存在非零实数\lambda ，使得\overrightarrow{AE}\perp \overrightarrow{CD}？若存在，求出\lambda 的值；若不存在，请说明理由.$

Solution:
$①当\lambda =\frac{1}{2}时，\overrightarrow{AD}=\frac{1}{2}\overrightarrow{AB}，\overrightarrow{BE}=\frac{1}{2}\overrightarrow{BC}，\therefore D，E分别是AB，BC的中点，\therefore \overrightarrow{AE}=\overrightarrow{AC}+\overrightarrow{CE}=\overrightarrow{AC}+\frac{1}{2}\overrightarrow{CB}，\overrightarrow{CD}=\frac{1}{2}(\overrightarrow{CA}+\overrightarrow{CB})，\therefore \overrightarrow{AE}\cdot \overrightarrow{CD}=\left(\overrightarrow{AC}+\frac{1}{2}\overrightarrow{CB}\right)\cdot \frac{1}{2}(\overrightarrow{CA}+\overrightarrow{CB}) =\frac{1}{2}\overrightarrow{AC}\cdot \overrightarrow{CA}+\frac{1}{2}\overrightarrow{AC}\cdot \overrightarrow{CB}+\frac{1}{4}\overrightarrow{CB}\cdot \overrightarrow{CA}+\frac{1}{4}\overrightarrow{CB}^{2}=-\frac{1}{2}\times 1^{2}+\frac{1}{2}\times 1\times 2\times \\cos   120^\circ +\frac{1}{4}\times 2\times 1\times \\cos   60^\circ +\frac{1}{4}\times 2^{2}=\frac{1}{4}.②存在．理由如下：假设存在非零实数\lambda ，使得\overrightarrow{AE}\perp \overrightarrow{CD}.由\overrightarrow{AD}=\lambda \overrightarrow{AB}，得\overrightarrow{AD}=\lambda (\overrightarrow{CB}-\overrightarrow{CA})，\therefore \overrightarrow{CD}=\overrightarrow{CA}+\overrightarrow{AD}=\overrightarrow{CA}+\lambda (\overrightarrow{CB}-\overrightarrow{CA})=\lambda \overrightarrow{CB}+(1-\lambda )\overrightarrow{CA}．\because \overrightarrow{BE}=\lambda \overrightarrow{BC}，\therefore \overrightarrow{AE}=\overrightarrow{AB}+\overrightarrow{BE}=(\overrightarrow{CB}-\overrightarrow{CA})+\lambda \overrightarrow{BC}=(1-\lambda )\overrightarrow{CB}-\overrightarrow{CA}.\therefore \overrightarrow{AE}\cdot \overrightarrow{CD}=\lambda (1-\lambda )\overrightarrow{CB}^{2}-\lambda \overrightarrow{CB}\cdot \overrightarrow{CA}+(1-\lambda )^{2}\overrightarrow{CB}\cdot \overrightarrow{CA}-(1-\lambda )\overrightarrow{CA}^{2}=4\lambda (1-\lambda )-\lambda +(1-\lambda )^{2}-(1-\lambda )=-3\lambda ^{2}+2\lambda =0，解得\lambda =\frac{2}{3}或\lambda =0(不合题意，舍去)，故存在非零实数\lambda =\frac{2}{3},使得\overrightarrow{AE}\perp \overrightarrow{CD}.$

Final answer: $问题①：当\lambda =1/2时，求⟠AE\cdot ⟠CD的值。
答案：⟠AE\cdot ⟠CD=1/4。
问题②：存在非零实数\lambda ，使得⟠AE\perp ⟠CD吗？若存在，求出\lambda 的值；若不存在，请说明理由。
答案：存在非零实数\lambda =2/3，使得⟠AE\perp ⟠CD。$
Subfield: Plane Geometry


Question: $在平行四边形ABCD中，AC与BD相交于点O，E是线段OD的中点，连接AE并延长，与CD交于点F，\overrightarrow{AF}=\lambda \overrightarrow{AB}+\mu \overrightarrow{AD}(\lambda ，\mu \in R).$![Image](https://cdntk.53zaixian.com/tq/1706233175873261569/img/compression/M_11ce11db529d01eca855a170553a34cc.jpg)
$求\lambda +\mu 的值；$

Solution:
$由题意得\triangle DEF∽\triangle BEA，DE=\frac{1}{3}EB，\therefore \frac{DF}{AB}=\frac{DE}{EB}=\frac{1}{3}，\therefore \overrightarrow{AF}=\overrightarrow{AD}+\overrightarrow{DF}=\overrightarrow{AD}+\frac{1}{3}\overrightarrow{AB}=\lambda \overrightarrow{AB}+\mu \overrightarrow{AD}，\because \overrightarrow{AB},\overrightarrow{AD}不共线，\therefore \lambda =\frac{1}{3},\mu =1,故\lambda +\mu =\frac{4}{3}.$

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


Question: $在平行四边形ABCD中，AC与BD相交于点O，E是线段OD的中点，连接AE并延长，与CD交于点F，\overrightarrow{AF}=\lambda \overrightarrow{AB}+\mu \overrightarrow{AD}(\lambda ，\mu \in R).$<img_2528>
$求\lambda +\mu 的值；$

Solution:
$由题意得\triangle DEF∽\triangle BEA，DE=\frac{1}{3}EB，\therefore \frac{DF}{AB}=\frac{DE}{EB}=\frac{1}{3}，\therefore \overrightarrow{AF}=\overrightarrow{AD}+\overrightarrow{DF}=\overrightarrow{AD}+\frac{1}{3}\overrightarrow{AB}=\lambda \overrightarrow{AB}+\mu \overrightarrow{AD}，\because \overrightarrow{AB},\overrightarrow{AD}不共线，\therefore \lambda =\frac{1}{3},\mu =1,故\lambda +\mu =\frac{4}{3}.$

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


Question: $在平行四边形ABCD中，AC与BD相交于点O，E是线段OD的中点，连接AE并延长，与CD交于点F，\overrightarrow{AF}=\lambda \overrightarrow{AB}+\mu \overrightarrow{AD}(\lambda ，\mu \in R).$<img_2528>
$若\overrightarrow{AC}=a，\overrightarrow{BD}=b，试用基底{a,b}表示\overrightarrow{AF}.$

Solution:
$设\overrightarrow{AB}=m,\overrightarrow{AD}=n，则\overrightarrow{AC}=\overrightarrow{AB}+\overrightarrow{AD}=m+n=a,\overrightarrow{BD}=\overrightarrow{AD}-\overrightarrow{AB}=n-m=b,\therefore m=\frac{1}{2}a-\frac{1}{2}b，n=\frac{1}{2}a+\frac{1}{2}b.由(1)知\overrightarrow{AF}=\overrightarrow{AD}+\frac{1}{3}\overrightarrow{AB}，\therefore \overrightarrow{AF}=n+\frac{1}{3}m=\frac{1}{2}a+\frac{1}{2}b+\frac{1}{3}\left(\frac{1}{2}\boldsymbol{a}-\frac{1}{2}\boldsymbol{b}\right)=\frac{2}{3}a+\frac{1}{3}b.$

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


Question: $在平行四边形ABCD中，AC与BD相交于点O，E是线段OD的中点，连接AE并延长，与CD交于点F，\overrightarrow{AF}=\lambda \overrightarrow{AB}+\mu \overrightarrow{AD}(\lambda ，\mu \in R).$![Image](https://cdntk.53zaixian.com/tq/1706233175873261569/img/compression/M_11ce11db529d01eca855a170553a34cc.jpg)
$若\overrightarrow{AC}=a，\overrightarrow{BD}=b，试用基底{a,b}表示\overrightarrow{AF}.$

Solution:
$设\overrightarrow{AB}=m,\overrightarrow{AD}=n，则\overrightarrow{AC}=\overrightarrow{AB}+\overrightarrow{AD}=m+n=a,\overrightarrow{BD}=\overrightarrow{AD}-\overrightarrow{AB}=n-m=b,\therefore m=\frac{1}{2}a-\frac{1}{2}b，n=\frac{1}{2}a+\frac{1}{2}b.由(1)知\overrightarrow{AF}=\overrightarrow{AD}+\frac{1}{3}\overrightarrow{AB}，\therefore \overrightarrow{AF}=n+\frac{1}{3}m=\frac{1}{2}a+\frac{1}{2}b+\frac{1}{3}\left(\frac{1}{2}\boldsymbol{a}-\frac{1}{2}\boldsymbol{b}\right)=\frac{2}{3}a+\frac{1}{3}b.$

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


Question: $如图，在平行四边形ABCD中，AP\perp BD，垂足为P.$<img_1969>
$若\overrightarrow{AP}\cdot \overrightarrow{AC}=8，求AP的长；$

Solution:
$由题意得\overrightarrow{AP}\cdot \overrightarrow{AC}=\overrightarrow{AP}\cdot 2\overrightarrow{AO}=2\overrightarrow{AP}\cdot (\overrightarrow{AP}+\overrightarrow{PO})=2\overrightarrow{AP}\cdot \overrightarrow{AP}+0=8，\therefore \overrightarrow{AP}^2=|\overrightarrow{AP}|^{2}=4，解得|\overrightarrow{AP}|=2，故AP的长为2.$

Final answer: $2$
Subfield: Plane Geometry


Question: $如图，在平行四边形ABCD中，AP\perp BD，垂足为P.$![Image](https://cdntk.53zaixian.com/tq/1706233177123164161/img/compression/M_386cf62f2bc59835059311daa430c762.jpg)
$若\overrightarrow{AP}\cdot \overrightarrow{AC}=8，求AP的长；$

Solution:
$由题意得\overrightarrow{AP}\cdot \overrightarrow{AC}=\overrightarrow{AP}\cdot 2\overrightarrow{AO}=2\overrightarrow{AP}\cdot (\overrightarrow{AP}+\overrightarrow{PO})=2\overrightarrow{AP}\cdot \overrightarrow{AP}+0=8，\therefore \overrightarrow{AP}^2=|\overrightarrow{AP}|^{2}=4，解得|\overrightarrow{AP}|=2，故AP的长为2.$

Final answer: $2$
Subfield: Plane Geometry


Question: $如图，在平行四边形ABCD中，AP\perp BD，垂足为P.$![Image](https://cdntk.53zaixian.com/tq/1706233177123164161/img/compression/M_386cf62f2bc59835059311daa430c762.jpg)
$若|\overrightarrow{AB}|=6,|\overrightarrow{AC}|=8,\angle BAC=\frac{\pi }{3},\overrightarrow{AP}=x\overrightarrow{AB}+y\overrightarrow{AC}(x,y\in R)，求y-x的值.$

Solution:
$\because \overrightarrow{AP}=x\overrightarrow{AB}+y\overrightarrow{AC}=x\overrightarrow{AB}+2y\overrightarrow{AO}，且B,P,O三点共线，\therefore x+2y=1①.\because |\overrightarrow{AB}|=6,|\overrightarrow{AC}|=8,\angle BAC=\frac{\pi }{3}，\therefore \overrightarrow{AB}\cdot \overrightarrow{AO}=|\overrightarrow{AB}|\cdot \frac{1}{2}|\overrightarrow{AC}|\\cos  \angle BAC=12，由AP\perp BD可知\overrightarrow{AP}\cdot \overrightarrow{BO}=(x\overrightarrow{AB}+2y\overrightarrow{AO})\cdot (\overrightarrow{AO}-\overrightarrow{AB})=0，即2y\overrightarrow{AO}^{2}-x\overrightarrow{AB}^{2}+(x-2y)\overrightarrow{AB}\cdot \overrightarrow{AO}=0，\therefore y=3x②，联立①②解得x=\frac{1}{7},y=\frac{3}{7}，故y-x=\frac{2}{7}.$

Final answer: $\frac{2}{7}$
Subfield: Plane Geometry


Question: $如图，在平行四边形ABCD中，AP\perp BD，垂足为P.$<img_1969>
$若|\overrightarrow{AB}|=6,|\overrightarrow{AC}|=8,\angle BAC=\frac{\pi }{3},\overrightarrow{AP}=x\overrightarrow{AB}+y\overrightarrow{AC}(x,y\in R)，求y-x的值.$

Solution:
$\because \overrightarrow{AP}=x\overrightarrow{AB}+y\overrightarrow{AC}=x\overrightarrow{AB}+2y\overrightarrow{AO}，且B,P,O三点共线，\therefore x+2y=1①.\because |\overrightarrow{AB}|=6,|\overrightarrow{AC}|=8,\angle BAC=\frac{\pi }{3}，\therefore \overrightarrow{AB}\cdot \overrightarrow{AO}=|\overrightarrow{AB}|\cdot \frac{1}{2}|\overrightarrow{AC}|\\cos  \angle BAC=12，由AP\perp BD可知\overrightarrow{AP}\cdot \overrightarrow{BO}=(x\overrightarrow{AB}+2y\overrightarrow{AO})\cdot (\overrightarrow{AO}-\overrightarrow{AB})=0，即2y\overrightarrow{AO}^{2}-x\overrightarrow{AB}^{2}+(x-2y)\overrightarrow{AB}\cdot \overrightarrow{AO}=0，\therefore y=3x②，联立①②解得x=\frac{1}{7},y=\frac{3}{7}，故y-x=\frac{2}{7}.$

Final answer: $\frac{2}{7}$
Subfield: Plane Geometry


Question: $如图，已知\triangle ABC是边长为2的正三角形，点P_{1}，P_{2}，P_{3}是BC边的四等分点.$![Image](https://cdntk.53zaixian.com/tq/1706233178586976257/img/compression/M_43a2bc1b0a974064e60864ff8b19174a.jpg)
$求\overrightarrow{AB}\cdot \overrightarrow{AP_1}+\overrightarrow{AP_1}\cdot \overrightarrow{AC}的值；$

Solution:
$因为P_{2}为BC边的中点，所以\overrightarrow{AB}+\overrightarrow{AC}=2\overrightarrow{AP_2}，故\overrightarrow{AB}\cdot \overrightarrow{AP_1}+\overrightarrow{AP_1}\cdot \overrightarrow{AC}=\overrightarrow{AP_1}\cdot (\overrightarrow{AB}+\overrightarrow{AC})=2\overrightarrow{AP_1}\cdot \overrightarrow{AP_2}.因为AP_{2}\perp BC，所以2\overrightarrow{AP_1}\cdot \overrightarrow{AP_2}=2(\overrightarrow{AP_2}+\overrightarrow{P_2P_1})\cdot \overrightarrow{AP_2}=2\overrightarrow{AP_2}^{2}=6.因此\overrightarrow{AB}\cdot \overrightarrow{AP_1}+\overrightarrow{AP_1}\cdot \overrightarrow{AC}=6.$

Final answer: $6$
Subfield: Plane Geometry


Question: $如图，已知\triangle ABC是边长为2的正三角形，点P_{1}，P_{2}，P_{3}是BC边的四等分点.$<img_1799>
$求\overrightarrow{AB}\cdot \overrightarrow{AP_1}+\overrightarrow{AP_1}\cdot \overrightarrow{AC}的值；$

Solution:
$因为P_{2}为BC边的中点，所以\overrightarrow{AB}+\overrightarrow{AC}=2\overrightarrow{AP_2}，故\overrightarrow{AB}\cdot \overrightarrow{AP_1}+\overrightarrow{AP_1}\cdot \overrightarrow{AC}=\overrightarrow{AP_1}\cdot (\overrightarrow{AB}+\overrightarrow{AC})=2\overrightarrow{AP_1}\cdot \overrightarrow{AP_2}.因为AP_{2}\perp BC，所以2\overrightarrow{AP_1}\cdot \overrightarrow{AP_2}=2(\overrightarrow{AP_2}+\overrightarrow{P_2P_1})\cdot \overrightarrow{AP_2}=2\overrightarrow{AP_2}^{2}=6.因此\overrightarrow{AB}\cdot \overrightarrow{AP_1}+\overrightarrow{AP_1}\cdot \overrightarrow{AC}=6.$

Final answer: $6$
Subfield: Plane Geometry


Question: $如图，已知\triangle ABC是边长为2的正三角形，点P_{1}，P_{2}，P_{3}是BC边的四等分点.$![Image](https://cdntk.53zaixian.com/tq/1706233178586976257/img/compression/M_43a2bc1b0a974064e60864ff8b19174a.jpg)
$若Q为线段AP_{1}上一点，且\overrightarrow{AQ}=m\overrightarrow{AB}+\frac{1}{12}\overrightarrow{AC}，求实数m的值；$

Solution:
$由\overrightarrow{BP_1}=\frac{1}{4}\overrightarrow{BC}，得\overrightarrow{AP_1}=\overrightarrow{AB}+\overrightarrow{BP_1}=\overrightarrow{AB}+\frac{1}{4}\overrightarrow{BC}=\overrightarrow{AB}+\frac{1}{4}(\overrightarrow{BA}+\overrightarrow{AC})=\frac{3}{4}\overrightarrow{AB}+\frac{1}{4}\overrightarrow{AC}.设\overrightarrow{AQ}=n\overrightarrow{AP_1}(0\leq n\leq 1)，则\overrightarrow{AQ}=\frac{3n}{4}\overrightarrow{AB}+\frac{n}{4}\overrightarrow{AC}=m\overrightarrow{AB}+\frac{1}{12}\overrightarrow{AC}.由平面向量基本定理得\left\{\begin{matrix}\frac{3n}{4}=m,\\ \frac{n}{4}=\frac{1}{12},\end{matrix}\right.解得\left\{\begin{matrix}n=\frac{1}{3},\\ m=\frac{1}{4},\end{matrix}\right.因此m=\frac{1}{4}.$

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


Question: $如图，已知\triangle ABC是边长为2的正三角形，点P_{1}，P_{2}，P_{3}是BC边的四等分点.$<img_1799>
$若Q为线段AP_{1}上一点，且\overrightarrow{AQ}=m\overrightarrow{AB}+\frac{1}{12}\overrightarrow{AC}，求实数m的值；$

Solution:
$由\overrightarrow{BP_1}=\frac{1}{4}\overrightarrow{BC}，得\overrightarrow{AP_1}=\overrightarrow{AB}+\overrightarrow{BP_1}=\overrightarrow{AB}+\frac{1}{4}\overrightarrow{BC}=\overrightarrow{AB}+\frac{1}{4}(\overrightarrow{BA}+\overrightarrow{AC})=\frac{3}{4}\overrightarrow{AB}+\frac{1}{4}\overrightarrow{AC}.设\overrightarrow{AQ}=n\overrightarrow{AP_1}(0\leq n\leq 1)，则\overrightarrow{AQ}=\frac{3n}{4}\overrightarrow{AB}+\frac{n}{4}\overrightarrow{AC}=m\overrightarrow{AB}+\frac{1}{12}\overrightarrow{AC}.由平面向量基本定理得\left\{\begin{matrix}\frac{3n}{4}=m,\\ \frac{n}{4}=\frac{1}{12},\end{matrix}\right.解得\left\{\begin{matrix}n=\frac{1}{3},\\ m=\frac{1}{4},\end{matrix}\right.因此m=\frac{1}{4}.$

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


Question: $如图，已知\triangle ABC是边长为2的正三角形，点P_{1}，P_{2}，P_{3}是BC边的四等分点.$<img_1799>
$若P为线段AP_{3}上的动点，求\overrightarrow{PA}\cdot \overrightarrow{PC}的最小值，并指出当\overrightarrow{PA}\cdot \overrightarrow{PC}取得最小值时点P的位置.$

Solution:
$记\overrightarrow{AB}=a，\overrightarrow{AC}=b，由\overrightarrow{BP_3}=\frac{3}{4}\overrightarrow{BC}得\overrightarrow{AP_3}=\overrightarrow{AB}+\overrightarrow{BP_3}=\overrightarrow{AB}+\frac{3}{4}\overrightarrow{BC}=\overrightarrow{AB}+\frac{3}{4}(\overrightarrow{AC}-\overrightarrow{AB})=\frac{1}{4}\overrightarrow{AB}+\frac{3}{4}\overrightarrow{AC}=\frac{1}{4}a+\frac{3}{4}b.设\overrightarrow{AP}=t\overrightarrow{AP_3}，0\leq t\leq 1，则\overrightarrow{PA}=-\frac{t}{4}a-\frac{3t}{4}b，\overrightarrow{PC}=\overrightarrow{PA}+\overrightarrow{AC}=-\frac{t}{4}a+\left(1-\frac{3t}{4}\right)b,则\overrightarrow{PA}\cdot \overrightarrow{PC}=\left(-\frac{t}{4}\boldsymbol{a}-\frac{3t}{4}\boldsymbol{b}\right)\cdot \left[-\frac{t}{4}\boldsymbol{a}+\left(1-\frac{3t}{4}\right)\boldsymbol{b}\right]=\frac{t}{16}[ta^{2}+(6t-4)a\cdot b+3(3t-4)b^{2}]=\frac{1}{4}(13t^{2}-14t)=\frac{13}{4}\left(t-\frac{7}{13}\right)^2-\frac{49}{52}，t\in [0,1].所以当t=\frac{7}{13}，即AP=\frac{7}{13}AP_{3}时，\overrightarrow{PA}\cdot \overrightarrow{PC}取得最小值，为-\frac{49}{52}.$

Final answer: $\frac{-49}{52}$