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


Question: $如图，在正方形ABCD中，E是BC的中点，点F在CD边上运动(含C，D点).$<img_2490>
$若点F是CD上靠近C的三等分点，设\overrightarrow{EF}=\lambda \overrightarrow{AB}+\mu \overrightarrow{AD}，求\lambda +\mu 的值；$

Solution:
$\because E是BC的中点，点F是CD上靠近C的三等分点，\therefore \overrightarrow{EC}=\frac{1}{2}\overrightarrow{BC}=\frac{1}{2}\overrightarrow{AD}，\overrightarrow{CF}=-\frac{1}{3}\overrightarrow{DC}=-\frac{1}{3}\overrightarrow{AB}，\therefore \overrightarrow{EF}=\overrightarrow{EC}+\overrightarrow{CF}=\frac{1}{2}\overrightarrow{AD}-\frac{1}{3}\overrightarrow{AB}，又\overrightarrow{EF}=\lambda \overrightarrow{AB}+\mu \overrightarrow{AD},\therefore \left(\lambda +\frac{1}{3}\right)\overrightarrow{AB}=\left(\frac{1}{2}-\mu \right)\overrightarrow{AD},\because \overrightarrow{AB},\overrightarrow{AD}不共线，\therefore \lambda +\frac{1}{3}=\frac{1}{2}-\mu =0,\therefore \lambda =-\frac{1}{3}，\mu =\frac{1}{2}，故\lambda +\mu =-\frac{1}{3}+\frac{1}{2}=\frac{1}{6}.$

Final answer: $\frac{1}{6}$
Subfield: Plane Geometry


Question: $如图，在正方形ABCD中，E是BC的中点，点F在CD边上运动(含C，D点).$![Image](https://cdntk.53zaixian.com/tq/1703664192421224451/img/compression/M_a3237b805e8a95fec0308c2f0eb7a36b.jpg)
$若点F是CD上靠近C的三等分点，设\overrightarrow{EF}=\lambda \overrightarrow{AB}+\mu \overrightarrow{AD}，求\lambda +\mu 的值；$

Solution:
$\because E是BC的中点，点F是CD上靠近C的三等分点，\therefore \overrightarrow{EC}=\frac{1}{2}\overrightarrow{BC}=\frac{1}{2}\overrightarrow{AD}，\overrightarrow{CF}=-\frac{1}{3}\overrightarrow{DC}=-\frac{1}{3}\overrightarrow{AB}，\therefore \overrightarrow{EF}=\overrightarrow{EC}+\overrightarrow{CF}=\frac{1}{2}\overrightarrow{AD}-\frac{1}{3}\overrightarrow{AB}，又\overrightarrow{EF}=\lambda \overrightarrow{AB}+\mu \overrightarrow{AD},\therefore \left(\lambda +\frac{1}{3}\right)\overrightarrow{AB}=\left(\frac{1}{2}-\mu \right)\overrightarrow{AD},\because \overrightarrow{AB},\overrightarrow{AD}不共线，\therefore \lambda +\frac{1}{3}=\frac{1}{2}-\mu =0,\therefore \lambda =-\frac{1}{3}，\mu =\frac{1}{2}，故\lambda +\mu =-\frac{1}{3}+\frac{1}{2}=\frac{1}{6}.$

Final answer: $\frac{1}{6}$
Subfield: Plane Geometry


Question: $如图，在正方形ABCD中，E是BC的中点，点F在CD边上运动(含C，D点).$![Image](https://cdntk.53zaixian.com/tq/1703664192421224451/img/compression/M_a3237b805e8a95fec0308c2f0eb7a36b.jpg)
$若AB=2，当\overrightarrow{AE}\cdot \overrightarrow{BF}=1时，求\\cos  \angle EAF的值.$

Solution:
$设\overrightarrow{CF}=m\overrightarrow{CD}(0\leq m\leq 1)，则\overrightarrow{BF}=\overrightarrow{BC}+\overrightarrow{CF}=\overrightarrow{AD}-m\overrightarrow{AB}，又\overrightarrow{AE}=\overrightarrow{AB}+\overrightarrow{BE}=\overrightarrow{AB}+\frac{1}{2}\overrightarrow{AD}，\overrightarrow{AB}\cdot \overrightarrow{AD}=0，\therefore \overrightarrow{AE}\cdot \overrightarrow{BF}=\left(\overrightarrow{AB}+\frac{1}{2}\overrightarrow{AD}\right)\cdot (\overrightarrow{AD}-m\overrightarrow{AB})=-m\overrightarrow{AB}^{2}+\frac{1}{2}\overrightarrow{AD}^{2}=-4m+2=1，故m=\frac{1}{4}.\therefore \overrightarrow{AE}\cdot \overrightarrow{AF}=\left(\overrightarrow{AB}+\frac{1}{2}\overrightarrow{AD}\right)\cdot \left(\overrightarrow{AD}+\frac{3}{4}\overrightarrow{AB}\right)=\frac{3}{4}\overrightarrow{AB}^{2}+\frac{1}{2}\overrightarrow{AD}^{2}=3+2=5,易得|\overrightarrow{AE}|=\sqrt{5}，|\overrightarrow{AF}|=\frac{5}{2}，\therefore \\cos  \angle EAF=\frac{\overrightarrow{AE}\cdot \overrightarrow{AF}}{|\overrightarrow{AE}||\overrightarrow{AF}|}=\frac{5}{\sqrt{5}\times \frac{5}{2}}=\frac{2\sqrt{5}}{5}.$

Final answer: $\frac{2\sqrt{5}}{5}$
Subfield: Plane Geometry


Question: $如图，在正方形ABCD中，E是BC的中点，点F在CD边上运动(含C，D点).$<img_2490>
$若AB=2，当\overrightarrow{AE}\cdot \overrightarrow{BF}=1时，求\\cos  \angle EAF的值.$

Solution:
$设\overrightarrow{CF}=m\overrightarrow{CD}(0\leq m\leq 1)，则\overrightarrow{BF}=\overrightarrow{BC}+\overrightarrow{CF}=\overrightarrow{AD}-m\overrightarrow{AB}，又\overrightarrow{AE}=\overrightarrow{AB}+\overrightarrow{BE}=\overrightarrow{AB}+\frac{1}{2}\overrightarrow{AD}，\overrightarrow{AB}\cdot \overrightarrow{AD}=0，\therefore \overrightarrow{AE}\cdot \overrightarrow{BF}=\left(\overrightarrow{AB}+\frac{1}{2}\overrightarrow{AD}\right)\cdot (\overrightarrow{AD}-m\overrightarrow{AB})=-m\overrightarrow{AB}^{2}+\frac{1}{2}\overrightarrow{AD}^{2}=-4m+2=1，故m=\frac{1}{4}.\therefore \overrightarrow{AE}\cdot \overrightarrow{AF}=\left(\overrightarrow{AB}+\frac{1}{2}\overrightarrow{AD}\right)\cdot \left(\overrightarrow{AD}+\frac{3}{4}\overrightarrow{AB}\right)=\frac{3}{4}\overrightarrow{AB}^{2}+\frac{1}{2}\overrightarrow{AD}^{2}=3+2=5,易得|\overrightarrow{AE}|=\sqrt{5}，|\overrightarrow{AF}|=\frac{5}{2}，\therefore \\cos  \angle EAF=\frac{\overrightarrow{AE}\cdot \overrightarrow{AF}}{|\overrightarrow{AE}||\overrightarrow{AF}|}=\frac{5}{\sqrt{5}\times \frac{5}{2}}=\frac{2\sqrt{5}}{5}.$

Final answer: $\frac{2\sqrt{5}}{5}$
Subfield: Plane Geometry


Question: $如图，在直角梯形ABCD中，已知AB\parallel CD，\angle DAB＝90^\circ ，AB＝4，AD＝CD＝2，对角线AC，BD交于点O，点M在AB上，且满足OM\perp BD.$<img_2854>
$求\overrightarrow{AM}\cdot \overrightarrow{BD}的值；$

Solution:
$解法一：因为AB\parallel CD，AB=2CD，所以AO=2OC，则\overrightarrow{AM}\cdot \overrightarrow{BD}=(\overrightarrow{AO}+\overrightarrow{OM})\cdot \overrightarrow{BD}=\overrightarrow{AO}\cdot \overrightarrow{BD}+\overrightarrow{OM}\cdot \overrightarrow{BD}=\overrightarrow{AO}\cdot \overrightarrow{BD}=\frac{2}{3}\overrightarrow{AC}\cdot \overrightarrow{BD}=\frac{2}{3}(\overrightarrow{AD}+\overrightarrow{DC})\cdot (\overrightarrow{AD}-\overrightarrow{AB})=\frac{2}{3}(\overrightarrow{AD}^{2}-\overrightarrow{DC}\cdot \overrightarrow{AB})=\frac{2}{3}\times (4-2\times 4\times 1)=-\frac{8}{3}.解法二：以A为坐标原点，\overrightarrow{AB}，\overrightarrow{AD}的方向为x轴，y轴的正方向建立平面直角坐标系(图略)，则A(0,0)，B(4,0)，C(2,2)，D(0,2)，所以\overrightarrow{BD}=(-4,2)，因为AB\parallel CD,AB=2CD,所以AO=2OC，所以\overrightarrow{AO}=2\overrightarrow{OC}，易得O\left(\frac{4}{3},\frac{4}{3}\right).设M(m,0),0因为OM\perp BD，所以\overrightarrow{OM}\cdot \overrightarrow{BD}=\left(m-\frac{4}{3}\right)\times (-4)+\left(-\frac{4}{3}\right)\times 2=-4m+\frac{8}{3}=0，解得m=\frac{2}{3}.所以\overrightarrow{AM}=\left(\frac{2}{3},0\right)，所以\overrightarrow{AM}\cdot \overrightarrow{BD}=\frac{2}{3}\times (-4)+0\times 2=-\frac{8}{3}.$

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


Question: $如图，在直角梯形ABCD中，已知AB\parallel CD，\angle DAB＝90^\circ ，AB＝4，AD＝CD＝2，对角线AC，BD交于点O，点M在AB上，且满足OM\perp BD.$![Image](https://cdntk.53zaixian.com/tq/1703664227295719426/img/compression/M_21be4bdc95125c359763245b636a2cd9.jpg)
$求\overrightarrow{AM}\cdot \overrightarrow{BD}的值；$

Solution:
$解法一：因为AB\parallel CD，AB=2CD，所以AO=2OC，则\overrightarrow{AM}\cdot \overrightarrow{BD}=(\overrightarrow{AO}+\overrightarrow{OM})\cdot \overrightarrow{BD}=\overrightarrow{AO}\cdot \overrightarrow{BD}+\overrightarrow{OM}\cdot \overrightarrow{BD}=\overrightarrow{AO}\cdot \overrightarrow{BD}=\frac{2}{3}\overrightarrow{AC}\cdot \overrightarrow{BD}=\frac{2}{3}(\overrightarrow{AD}+\overrightarrow{DC})\cdot (\overrightarrow{AD}-\overrightarrow{AB})=\frac{2}{3}(\overrightarrow{AD}^{2}-\overrightarrow{DC}\cdot \overrightarrow{AB})=\frac{2}{3}\times (4-2\times 4\times 1)=-\frac{8}{3}.解法二：以A为坐标原点，\overrightarrow{AB}，\overrightarrow{AD}的方向为x轴，y轴的正方向建立平面直角坐标系(图略)，则A(0,0)，B(4,0)，C(2,2)，D(0,2)，所以\overrightarrow{BD}=(-4,2)，因为AB\parallel CD,AB=2CD,所以AO=2OC，所以\overrightarrow{AO}=2\overrightarrow{OC}，易得O\left(\frac{4}{3},\frac{4}{3}\right).设M(m,0),0因为OM\perp BD，所以\overrightarrow{OM}\cdot \overrightarrow{BD}=\left(m-\frac{4}{3}\right)\times (-4)+\left(-\frac{4}{3}\right)\times 2=-4m+\frac{8}{3}=0，解得m=\frac{2}{3}.所以\overrightarrow{AM}=\left(\frac{2}{3},0\right)，所以\overrightarrow{AM}\cdot \overrightarrow{BD}=\frac{2}{3}\times (-4)+0\times 2=-\frac{8}{3}.$

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


Question: $如图，在直角梯形ABCD中，已知AB\parallel CD，\angle DAB＝90^\circ ，AB＝4，AD＝CD＝2，对角线AC，BD交于点O，点M在AB上，且满足OM\perp BD.$<img_2854>
$若N为线段AC上任意一点(不含端点)，求\overrightarrow{AN}\cdot \overrightarrow{MN}的最小值.$

Solution:
$解法一：设\overrightarrow{AM}=\lambda \overrightarrow{AB}(0<\lambda <1)，由(1)可得\overrightarrow{AM}\cdot \overrightarrow{BD}=\lambda \overrightarrow{AB}\cdot \overrightarrow{BD}=\lambda \overrightarrow{AB}\cdot (\overrightarrow{AD}-\overrightarrow{AB})=-\lambda \overrightarrow{AB}^{2}=-16\lambda =-\frac{8}{3}，解得\lambda =\frac{1}{6}，即\overrightarrow{AM}=\frac{1}{6}\overrightarrow{AB}.易知\angle CAB=45^\circ ,所以\overrightarrow{AN}\cdot \overrightarrow{MN}=\overrightarrow{AN}\cdot (\overrightarrow{AN}-\overrightarrow{AM})=\overrightarrow{AN}^{2}-\overrightarrow{AN}\cdot \overrightarrow{AM}=\overrightarrow{AN}^{2}-|\overrightarrow{AN}|\times |\overrightarrow{AM}|\times \\cos  \angle CAB=\overrightarrow{AN}^{2}-\frac{1}{6}\times |\overrightarrow{AN}|\times |\overrightarrow{AB}|\times \\cos   45^\circ =|\overrightarrow{AN}|^{2}-\frac{\sqrt{2}}{3}|\overrightarrow{AN}|.令|\overrightarrow{AN}|=t，则0则\overrightarrow{AN}\cdot \overrightarrow{MN}=t^{2}-\frac{\sqrt{2}}{3}t=\left(t-\frac{\sqrt{2}}{6}\right)^{2}-\frac{1}{18}，所以当t=\frac{\sqrt{2}}{6}时，\overrightarrow{AN}\cdot \overrightarrow{MN}有最小值，为-\frac{1}{18}.
解法二：易知\angle CAB=45^\circ ，故可设N(a,a),0<a<2，则\overrightarrow{AN}\cdot \overrightarrow{MN}=(a,a)\cdot \left(a-\frac{2}{3},a\right)=2a^{2}-\frac{2}{3}a=2\left(a-\frac{1}{6}\right)^{2}-\frac{1}{18}，所以当a=\frac{1}{6}时，\overrightarrow{AN}\cdot \overrightarrow{MN}有最小值，为-\frac{1}{18}.$

Final answer: $-\frac{1}{18}$
Subfield: Plane Geometry


Question: $如图，在直角梯形ABCD中，已知AB\parallel CD，\angle DAB＝90^\circ ，AB＝4，AD＝CD＝2，对角线AC，BD交于点O，点M在AB上，且满足OM\perp BD.$![Image](https://cdntk.53zaixian.com/tq/1703664227295719426/img/compression/M_21be4bdc95125c359763245b636a2cd9.jpg)
$若N为线段AC上任意一点(不含端点)，求\overrightarrow{AN}\cdot \overrightarrow{MN}的最小值.$

Solution:
$解法一：设\overrightarrow{AM}=\lambda \overrightarrow{AB}(0<\lambda <1)，由(1)可得\overrightarrow{AM}\cdot \overrightarrow{BD}=\lambda \overrightarrow{AB}\cdot \overrightarrow{BD}=\lambda \overrightarrow{AB}\cdot (\overrightarrow{AD}-\overrightarrow{AB})=-\lambda \overrightarrow{AB}^{2}=-16\lambda =-\frac{8}{3}，解得\lambda =\frac{1}{6}，即\overrightarrow{AM}=\frac{1}{6}\overrightarrow{AB}.易知\angle CAB=45^\circ ,所以\overrightarrow{AN}\cdot \overrightarrow{MN}=\overrightarrow{AN}\cdot (\overrightarrow{AN}-\overrightarrow{AM})=\overrightarrow{AN}^{2}-\overrightarrow{AN}\cdot \overrightarrow{AM}=\overrightarrow{AN}^{2}-|\overrightarrow{AN}|\times |\overrightarrow{AM}|\times \\cos  \angle CAB=\overrightarrow{AN}^{2}-\frac{1}{6}\times |\overrightarrow{AN}|\times |\overrightarrow{AB}|\times \\cos   45^\circ =|\overrightarrow{AN}|^{2}-\frac{\sqrt{2}}{3}|\overrightarrow{AN}|.令|\overrightarrow{AN}|=t，则0则\overrightarrow{AN}\cdot \overrightarrow{MN}=t^{2}-\frac{\sqrt{2}}{3}t=\left(t-\frac{\sqrt{2}}{6}\right)^{2}-\frac{1}{18}，所以当t=\frac{\sqrt{2}}{6}时，\overrightarrow{AN}\cdot \overrightarrow{MN}有最小值，为-\frac{1}{18}.
解法二：易知\angle CAB=45^\circ ，故可设N(a,a),0<a<2，则\overrightarrow{AN}\cdot \overrightarrow{MN}=(a,a)\cdot \left(a-\frac{2}{3},a\right)=2a^{2}-\frac{2}{3}a=2\left(a-\frac{1}{6}\right)^{2}-\frac{1}{18}，所以当a=\frac{1}{6}时，\overrightarrow{AN}\cdot \overrightarrow{MN}有最小值，为-\frac{1}{18}.$

Final answer: $-\frac{1}{18}$
Subfield: Plane Geometry


Question: $如图，在\triangle ABC中，AC=2，AB=4，点D在边BC上，且\overrightarrow{CD}=t\overrightarrow{CB}(t\in R).$![Image](https://cdntk.53zaixian.com/tq/1703664231059025923/img/compression/M_3662aee75d5ff9abcc21687e818466fc.jpg)
$若t=\frac{1}{2}，\angle BAC=\frac{2\pi }{3}，求|\overrightarrow{AD}|；$

Solution:
$若t=\frac{1}{2}，则\overrightarrow{CD}=\frac{1}{2}\overrightarrow{CB}，即点D为BC的中点，所以\overrightarrow{AD}=\frac{1}{2}(\overrightarrow{AB}+\overrightarrow{AC})，因为\angle BAC=\frac{2\pi }{3}，AC=2，AB=4，所以|\overrightarrow{AD}|=\sqrt{\frac{1}{4}({\overrightarrow{AB}}^2+{\overrightarrow{AC}}^2+2\overrightarrow{AB}\cdot \overrightarrow{AC})}==\sqrt{3}.$

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


Question: $如图，在\triangle ABC中，AC=2，AB=4，点D在边BC上，且\overrightarrow{CD}=t\overrightarrow{CB}(t\in R).$<img_1620>
$若t=\frac{1}{2}，\angle BAC=\frac{2\pi }{3}，求|\overrightarrow{AD}|；$

Solution:
$若t=\frac{1}{2}，则\overrightarrow{CD}=\frac{1}{2}\overrightarrow{CB}，即点D为BC的中点，所以\overrightarrow{AD}=\frac{1}{2}(\overrightarrow{AB}+\overrightarrow{AC})，因为\angle BAC=\frac{2\pi }{3}，AC=2，AB=4，所以|\overrightarrow{AD}|=\sqrt{\frac{1}{4}({\overrightarrow{AB}}^2+{\overrightarrow{AC}}^2+2\overrightarrow{AB}\cdot \overrightarrow{AC})}==\sqrt{3}.$

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


Question: $如图，在\triangle ABC中，AC=2，AB=4，点D在边BC上，且\overrightarrow{CD}=t\overrightarrow{CB}(t\in R).$![Image](https://cdntk.53zaixian.com/tq/1703664231059025923/img/compression/M_3662aee75d5ff9abcc21687e818466fc.jpg)
$若t=\frac{1}{5}，AD恰为BC边上的高，求\angle BAC；$

Solution:
$若t=\frac{1}{5}，则\overrightarrow{CD}=\frac{1}{5}\overrightarrow{CB}=\frac{1}{5}(\overrightarrow{AB}-\overrightarrow{AC})，因为AD恰为BC边上的高，所以\overrightarrow{AD}\perp \overrightarrow{BC}，因为\overrightarrow{AD}=\overrightarrow{AC}+\overrightarrow{CD}=\overrightarrow{AC}+\frac{1}{5}(\overrightarrow{AB}-\overrightarrow{AC})=\frac{4}{5}\overrightarrow{AC}+\frac{1}{5}\overrightarrow{AB}，\overrightarrow{BC}=\overrightarrow{AC}-\overrightarrow{AB}，所以\overrightarrow{AD}\cdot \overrightarrow{BC}=\left(\frac{4}{5}\overrightarrow{AC}+\frac{1}{5}\overrightarrow{AB}\right)\cdot (\overrightarrow{AC}-\overrightarrow{AB})=\frac{4}{5}\overrightarrow{AC}^{2}-\frac{3}{5}\overrightarrow{AC}\cdot \overrightarrow{AB}-\frac{1}{5}\overrightarrow{AB}^{2}=\frac{4}{5}\times 2^{2}-\frac{3}{5}\times 2\times 4\times \\cos  \angle BAC-\frac{1}{5}\times 4^{2}=0，所以\\cos  \angle BAC=0，则\angle BAC=\frac{\pi }{2}.$

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


Question: $如图，在\triangle ABC中，AC=2，AB=4，点D在边BC上，且\overrightarrow{CD}=t\overrightarrow{CB}(t\in R).$<img_1620>
$若t=\frac{1}{5}，AD恰为BC边上的高，求\angle BAC；$

Solution:
$若t=\frac{1}{5}，则\overrightarrow{CD}=\frac{1}{5}\overrightarrow{CB}=\frac{1}{5}(\overrightarrow{AB}-\overrightarrow{AC})，因为AD恰为BC边上的高，所以\overrightarrow{AD}\perp \overrightarrow{BC}，因为\overrightarrow{AD}=\overrightarrow{AC}+\overrightarrow{CD}=\overrightarrow{AC}+\frac{1}{5}(\overrightarrow{AB}-\overrightarrow{AC})=\frac{4}{5}\overrightarrow{AC}+\frac{1}{5}\overrightarrow{AB}，\overrightarrow{BC}=\overrightarrow{AC}-\overrightarrow{AB}，所以\overrightarrow{AD}\cdot \overrightarrow{BC}=\left(\frac{4}{5}\overrightarrow{AC}+\frac{1}{5}\overrightarrow{AB}\right)\cdot (\overrightarrow{AC}-\overrightarrow{AB})=\frac{4}{5}\overrightarrow{AC}^{2}-\frac{3}{5}\overrightarrow{AC}\cdot \overrightarrow{AB}-\frac{1}{5}\overrightarrow{AB}^{2}=\frac{4}{5}\times 2^{2}-\frac{3}{5}\times 2\times 4\times \\cos  \angle BAC-\frac{1}{5}\times 4^{2}=0，所以\\cos  \angle BAC=0，则\angle BAC=\frac{\pi }{2}.$

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


Question: $如图，在\triangle ABC中，AC=2，AB=4，点D在边BC上，且\overrightarrow{CD}=t\overrightarrow{CB}(t\in R).$![Image](https://cdntk.53zaixian.com/tq/1703664231059025923/img/compression/M_3662aee75d5ff9abcc21687e818466fc.jpg)
$(3)若AD=3，求t的取值范围.$

Solution:
$由题意得\overrightarrow{AD}=\overrightarrow{AC}+\overrightarrow{CD}=\overrightarrow{AC}+t\overrightarrow{CB}=\overrightarrow{AC}+t(\overrightarrow{AB}-\overrightarrow{AC})=t\overrightarrow{AB}+(1-t)\overrightarrow{AC}，所以\overrightarrow{AD}^2=t^{2}\overrightarrow{AB}^2+(1-t)^{2}\overrightarrow{AC}^2+2t(1-t)\overrightarrow{AB}\cdot \overrightarrow{AC}，即9=16t^{2}+4(1-2t+t^{2})+(16t-16t^{2})\\cos  \angle BAC，易知0(用t表示\\cos  \angle BAC，利用\\cos  \angle BAC的有界性来求t的范围)因为\angle BAC\in (0,\pi ),所以-1又0<t<1，所以\frac{1}{2}<t<\frac{5}{6}.$

Final answer: $(\frac{1}{2}, \frac{5}{6})$
Subfield: Plane Geometry


Question: $如图，在\triangle ABC中，AC=2，AB=4，点D在边BC上，且\overrightarrow{CD}=t\overrightarrow{CB}(t\in R).$<img_1620>
$(3)若AD=3，求t的取值范围.$

Solution:
$由题意得\overrightarrow{AD}=\overrightarrow{AC}+\overrightarrow{CD}=\overrightarrow{AC}+t\overrightarrow{CB}=\overrightarrow{AC}+t(\overrightarrow{AB}-\overrightarrow{AC})=t\overrightarrow{AB}+(1-t)\overrightarrow{AC}，所以\overrightarrow{AD}^2=t^{2}\overrightarrow{AB}^2+(1-t)^{2}\overrightarrow{AC}^2+2t(1-t)\overrightarrow{AB}\cdot \overrightarrow{AC}，即9=16t^{2}+4(1-2t+t^{2})+(16t-16t^{2})\\cos  \angle BAC，易知0(用t表示\\cos  \angle BAC，利用\\cos  \angle BAC的有界性来求t的范围)因为\angle BAC\in (0,\pi ),所以-1又0<t<1，所以\frac{1}{2}<t<\frac{5}{6}.$

Final answer: $(\frac{1}{2}, \frac{5}{6})$
Subfield: Plane Geometry


Question: $如图，已知点C为山顶P在水平面上的射影，一艘巡逻船由南向北沿AD方向行驶，在水平面上的A处测得\angle BAC=15^\circ ，匀速向北航行20分钟到达B处，测得\angle DBC=60^\circ ，此时测得山顶P的仰角为60^\circ ，已知山高为2\sqrt{3}千米.$<img_1573>
$求巡逻船的航行速度；$

Solution:
$由题意知\angle BCP=90^\circ ，\angle PBC=60^\circ ，PC=2\sqrt{3}，在Rt\triangle BCP中，\\tan  \angle PBC=\frac{PC}{BC}，则BC=\frac{2\sqrt{3}}{\\\tan    60^{\circ}}=2，在\triangle ABC中，可得\angle BCA=60^\circ -15^\circ =45^\circ ，由正弦定理得\frac{BC}{\\\sin   \angle BAC}=\frac{AB}{\\\sin   \angle BCA}，即\frac{2}{\\\sin    15^{\circ}}=\frac{AB}{\\\sin    45^{\circ}}，所以AB=\frac{2\\\sin    45^{\circ}}{\\\sin    15^{\circ}}=2(\sqrt{3}+1)，故巡逻船的航行速度是\frac{2(\sqrt{3}+1)}{\frac{20}{60}}=6(\sqrt{3}+1)(千米/时).$

Final answer: $6(\sqrt{3}+1)$ 千米/时
Subfield: Plane Geometry


Question: $如图，已知点C为山顶P在水平面上的射影，一艘巡逻船由南向北沿AD方向行驶，在水平面上的A处测得\angle BAC=15^\circ ，匀速向北航行20分钟到达B处，测得\angle DBC=60^\circ ，此时测得山顶P的仰角为60^\circ ，已知山高为2\sqrt{3}千米.$![Image](https://cdntk.53zaixian.com/tq/1703664254284562435/img/compression/M_6f7f6ac8beaf50161ebcad35ec4ce1fa.jpg)
$求巡逻船的航行速度；$

Solution:
$由题意知\angle BCP=90^\circ ，\angle PBC=60^\circ ，PC=2\sqrt{3}，在Rt\triangle BCP中，\\tan  \angle PBC=\frac{PC}{BC}，则BC=\frac{2\sqrt{3}}{\\\tan    60^{\circ}}=2，在\triangle ABC中，可得\angle BCA=60^\circ -15^\circ =45^\circ ，由正弦定理得\frac{BC}{\\\sin   \angle BAC}=\frac{AB}{\\\sin   \angle BCA}，即\frac{2}{\\\sin    15^{\circ}}=\frac{AB}{\\\sin    45^{\circ}}，所以AB=\frac{2\\\sin    45^{\circ}}{\\\sin    15^{\circ}}=2(\sqrt{3}+1)，故巡逻船的航行速度是\frac{2(\sqrt{3}+1)}{\frac{20}{60}}=6(\sqrt{3}+1)(千米/时).$

Final answer: $6(\sqrt{3}+1)$ 千米/时
Subfield: Plane Geometry


Question: $如图，已知点C为山顶P在水平面上的射影，一艘巡逻船由南向北沿AD方向行驶，在水平面上的A处测得\angle BAC=15^\circ ，匀速向北航行20分钟到达B处，测得\angle DBC=60^\circ ，此时测得山顶P的仰角为60^\circ ，已知山高为2\sqrt{3}千米.$![Image](https://cdntk.53zaixian.com/tq/1703664254284562435/img/compression/M_6f7f6ac8beaf50161ebcad35ec4ce1fa.jpg)
$(2)若该船继续航行10分钟到达D处，此时点C位于D处的南偏东什么方向？$

Solution:
$易得BD=\sqrt{3}+1，在\triangle BCD中，由余弦定理得CD^{2}=BC^{2}+BD^{2}-2BC\times BD\\cos  \angle DBC=2^{2}+(\sqrt{3}+1)^{2}-2\times 2\times (\sqrt{3}+1)\times \frac{1}{2}=6，故CD=\sqrt{6}，在\triangle BCD中，由正弦定理得\frac{CD}{\\\sin   \angle DBC}=\frac{BC}{\\\sin   \angle CDB}，即\frac{\sqrt{6}}{\\\sin    60^{\circ}}=\frac{2}{\\\sin   \angle CDB}，解得\\sin  \angle CDB=\frac{\sqrt{2}}{2}，因为CD>BC，所以\angle CDB为锐角，所以\angle CDB=45^\circ ，故点C位于D处的南偏东45^\circ 方向.$

Final answer: $点C位于D处的南偏东45^\circ 方向。$
Subfield: Plane Geometry


Question: $某地修建了一座避暑山庄O(如图)，为吸引游客，准备在门前两条夹角为\frac{\pi }{6}(即\angle AOB)的小路之间修建一处弓形花园，已知弓形花园的弦长AB=2\sqrt{3}，且点A，B落在小路上，记弓形花园的顶点为M，且\angle MAB=\angle MBA=\frac{\pi }{6}，设\angle OBA=\theta .$![Image](https://cdntk.53zaixian.com/tq/1703664282136244227/img/compression/M_5b51934317275f146482e398b7ea19d4.jpg)
$将OA，OB用含\theta 的式子表示出来；$

Solution:
$在\triangle ABO中，由正弦定理得\frac{OA}{\\\sin    \theta }=\frac{OB}{\\\sin   \angle OAB}=\frac{AB}{\\\sin    \frac{\pi }{6}}，则OA=4\sqrt{3}\\sin   \theta ，OB=4\sqrt{3}\\sin  \angle OAB=4\sqrt{3}\\sin  \left(\theta +\frac{\pi }{6}\right).$

Final answer: $OA=4\sqrt 3\sin \theta$ , $OB=4\sqrt 3\sin (\theta +\pi /6)$
Subfield: Plane Geometry


Question: $某地修建了一座避暑山庄O(如图)，为吸引游客，准备在门前两条夹角为\frac{\pi }{6}(即\angle AOB)的小路之间修建一处弓形花园，已知弓形花园的弦长AB=2\sqrt{3}，且点A，B落在小路上，记弓形花园的顶点为M，且\angle MAB=\angle MBA=\frac{\pi }{6}，设\angle OBA=\theta .$<img_3085>
$将OA，OB用含\theta 的式子表示出来；$

Solution:
$在\triangle ABO中，由正弦定理得\frac{OA}{\\\sin    \theta }=\frac{OB}{\\\sin   \angle OAB}=\frac{AB}{\\\sin    \frac{\pi }{6}}，则OA=4\sqrt{3}\\sin   \theta ，OB=4\sqrt{3}\\sin  \angle OAB=4\sqrt{3}\\sin  \left(\theta +\frac{\pi }{6}\right).$

Final answer: $OA=4\sqrt 3\sin \theta$ , $OB=4\sqrt 3\sin (\theta +\pi /6)$
Subfield: Plane Geometry


Question: $某地修建了一座避暑山庄O(如图)，为吸引游客，准备在门前两条夹角为\frac{\pi }{6}(即\angle AOB)的小路之间修建一处弓形花园，已知弓形花园的弦长AB=2\sqrt{3}，且点A，B落在小路上，记弓形花园的顶点为M，且\angle MAB=\angle MBA=\frac{\pi }{6}，设\angle OBA=\theta .$![Image](https://cdntk.53zaixian.com/tq/1703664282136244227/img/compression/M_5b51934317275f146482e398b7ea19d4.jpg)
$该山庄准备在M点处修建喷泉，如何规划花园(即OA，OB的长度)，才能使喷泉M与山庄O的距离(即OM的长度)最大？$

Solution:
$\because AB=2\sqrt{3}，\angle MAB=\angle MBA=\frac{\pi }{6}，\therefore AM=BM=2，在\triangle OMB中，由余弦定理得OM^{2}=OB^{2}+BM^{2}-2OB\cdot BM\cdot \\cos  \left(\theta +\frac{\pi }{6}\right)=48\\sin  ^{2}\left(\theta +\frac{\pi }{6}\right)+4-16\sqrt{3}\\sin  \left(\theta +\frac{\pi }{6}\right)\\cos  \left(\theta +\frac{\pi }{6}\right)=24\left[1-\\\cos   \left(\frac{\pi }{3}+2\theta \right)\right]+4-8\sqrt{3}\\sin  \left(\frac{\pi }{3}+2\theta \right)=-16\sqrt{3}\left[\frac{1}{2}\\\sin   \left(\frac{\pi }{3}+2\theta \right)+\frac{\sqrt{3}}{2}\\\cos   \left(\frac{\pi }{3}+2\theta \right)\right]+28=-16\sqrt{3}\\sin  \left(2\theta +\frac{2\pi }{3}\right)+28.由题意可得\theta \in \left(0,\frac{5\pi }{6}\right)，\therefore 2\theta +\frac{2\pi }{3}\in \left(\frac{2\pi }{3},\frac{7\pi }{3}\right)，\therefore \\sin  \left(2\theta +\frac{2\pi }{3}\right)\in \left[-1,\frac{\sqrt{3}}{2}\right)，\therefore 当\\sin  \left(2\theta +\frac{2\pi }{3}\right)=-1，即\theta =\frac{5\pi }{12}时，OM的长度取得最大值，为\sqrt{28+16\sqrt{3}}=4+2\sqrt{3}，此时OA=4\sqrt{3}\\sin   \frac{5\pi }{12}=4\sqrt{3}\times \left(\right.\\sin   \frac{\pi }{4}\\cos   \frac{\pi }{6}+\\cos   \frac{\pi }{4}\cdot \\sin   \frac{\pi }{6}\left)\right.=\sqrt{6}+3\sqrt{2}，OB=4\sqrt{3}\\sin  \left(\frac{5\pi }{12}+\frac{\pi }{6}\right)=4\sqrt{3}\\sin  \left(\pi -\frac{5\pi }{12}\right)=4\sqrt{3}\\sin   \frac{5\pi }{12}=\sqrt{6}+3\sqrt{2}，故当OB=OA=\sqrt{6}+3\sqrt{2}时，OM的长度取得最大值，即喷泉M与山庄O的距离最大.$

Final answer: $OB=OA=\sqrt{6}+3\sqrt{2}。$
Subfield: Plane Geometry


Question: $某地修建了一座避暑山庄O(如图)，为吸引游客，准备在门前两条夹角为\frac{\pi }{6}(即\angle AOB)的小路之间修建一处弓形花园，已知弓形花园的弦长AB=2\sqrt{3}，且点A，B落在小路上，记弓形花园的顶点为M，且\angle MAB=\angle MBA=\frac{\pi }{6}，设\angle OBA=\theta .$<img_3085>
$该山庄准备在M点处修建喷泉，如何规划花园(即OA，OB的长度)，才能使喷泉M与山庄O的距离(即OM的长度)最大？$

Solution:
$\because AB=2\sqrt{3}，\angle MAB=\angle MBA=\frac{\pi }{6}，\therefore AM=BM=2，在\triangle OMB中，由余弦定理得OM^{2}=OB^{2}+BM^{2}-2OB\cdot BM\cdot \\cos  \left(\theta +\frac{\pi }{6}\right)=48\\sin  ^{2}\left(\theta +\frac{\pi }{6}\right)+4-16\sqrt{3}\\sin  \left(\theta +\frac{\pi }{6}\right)\\cos  \left(\theta +\frac{\pi }{6}\right)=24\left[1-\\\cos   \left(\frac{\pi }{3}+2\theta \right)\right]+4-8\sqrt{3}\\sin  \left(\frac{\pi }{3}+2\theta \right)=-16\sqrt{3}\left[\frac{1}{2}\\\sin   \left(\frac{\pi }{3}+2\theta \right)+\frac{\sqrt{3}}{2}\\\cos   \left(\frac{\pi }{3}+2\theta \right)\right]+28=-16\sqrt{3}\\sin  \left(2\theta +\frac{2\pi }{3}\right)+28.由题意可得\theta \in \left(0,\frac{5\pi }{6}\right)，\therefore 2\theta +\frac{2\pi }{3}\in \left(\frac{2\pi }{3},\frac{7\pi }{3}\right)，\therefore \\sin  \left(2\theta +\frac{2\pi }{3}\right)\in \left[-1,\frac{\sqrt{3}}{2}\right)，\therefore 当\\sin  \left(2\theta +\frac{2\pi }{3}\right)=-1，即\theta =\frac{5\pi }{12}时，OM的长度取得最大值，为\sqrt{28+16\sqrt{3}}=4+2\sqrt{3}，此时OA=4\sqrt{3}\\sin   \frac{5\pi }{12}=4\sqrt{3}\times \left(\right.\\sin   \frac{\pi }{4}\\cos   \frac{\pi }{6}+\\cos   \frac{\pi }{4}\cdot \\sin   \frac{\pi }{6}\left)\right.=\sqrt{6}+3\sqrt{2}，OB=4\sqrt{3}\\sin  \left(\frac{5\pi }{12}+\frac{\pi }{6}\right)=4\sqrt{3}\\sin  \left(\pi -\frac{5\pi }{12}\right)=4\sqrt{3}\\sin   \frac{5\pi }{12}=\sqrt{6}+3\sqrt{2}，故当OB=OA=\sqrt{6}+3\sqrt{2}时，OM的长度取得最大值，即喷泉M与山庄O的距离最大.$

Final answer: $OB=OA=\sqrt{6}+3\sqrt{2}。$
Subfield: Plane Geometry


Question: $如图，在四棱锥P-ABCD中，底面ABCD是矩形，AB=2，BC=a，PA\perp 底面ABCD.$![Image](https://cdntk.53zaixian.com/tq/1703664477550010370/img/compression/M_2eb72df11b8ef500172a1237d2f902a1.jpg)
$当a为何值时，BD\perp 平面PAC？证明你的结论；$

Solution:
$当a=2时，BD\perp 平面PAC.证明如下：当a=2时，矩形ABCD为正方形，则BD\perp AC.\because PA\perp 平面ABCD，BD\subset 平面ABCD，\therefore BD\perp PA.又AC\cap PA=A，AC，PA\subset 平面PAC，\therefore BD\perp 平面PAC.故当a=2时，BD\perp 平面PAC.$

Final answer: $2$
Subfield: Plane Geometry


Question: $如图，在四棱锥P-ABCD中，底面ABCD是矩形，AB=2，BC=a，PA\perp 底面ABCD.$<img_2759>
$当a为何值时，BD\perp 平面PAC？证明你的结论；$

Solution:
$当a=2时，BD\perp 平面PAC.证明如下：当a=2时，矩形ABCD为正方形，则BD\perp AC.\because PA\perp 平面ABCD，BD\subset 平面ABCD，\therefore BD\perp PA.又AC\cap PA=A，AC，PA\subset 平面PAC，\therefore BD\perp 平面PAC.故当a=2时，BD\perp 平面PAC.$

Final answer: $2$