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


Question: $如图，三棱柱ABC-A_{1}B_{1}C_{1}中，侧面BB_{1}C_{1}C为菱形，B_{1}C的中点为O，且AO\perp 平面BB_{1}C_{1}C.$<img_2463>
$若AC\perp AB_{1}，\angle CBB_{1}＝60^\circ ，BC＝1，求三棱柱ABC-A_{1}B_{1}C_{1}的高．$

Solution:
$作OD\perp BC，垂足为D，连接AD，作OH\perp AD，垂足为H，\because BC\perp AO，BC\perp OD，AO\cap OD＝O，\therefore BC\perp 平面AOD，\because OH\subset 平面AOD，\therefore OH\perp BC，\because OH\perp AD，BC\cap AD＝D，BC，AD\subset 平面ABC，\therefore OH\perp 平面ABC，\because \angle CBB_{1}＝60^\circ ，\therefore \triangle CBB_{1}为等边三角形，\because BC＝1，\therefore OD＝\frac{\sqrt{3}}{4}，\because AC\perp AB_{1}，\therefore OA＝\frac{1}{2}B_{1}C＝\frac{1}{2}，在\triangle AOD中，由等面积法可得OH\cdot AD＝OD\cdot OA，又AD＝\sqrt{OD^2+OA^2}=\frac{\sqrt{7}}{4}，故OH＝\frac{\sqrt{21}}{14}，\because O为B_{1}C的中点，\therefore B_{1}到平面ABC的距离为\frac{\sqrt{21}}{7}，\therefore 三棱柱ABC-A_{1}B_{1}C_{1}的高为\frac{\sqrt{21}}{7}．$<img_2399>

Final answer: $\frac{\sqrt{21}}{7}$
Subfield: Plane Geometry


Question: $如图，三棱柱ABC-A_{1}B_{1}C_{1}中，侧面BB_{1}C_{1}C为菱形，B_{1}C的中点为O，且AO\perp 平面BB_{1}C_{1}C.$![Image](https://cdntk.53zaixian.com/tq/1706223468084920322/img/compression/M_9ce7697b515e435d38d7478eba759a31.jpg)
$若AC\perp AB_{1}，\angle CBB_{1}＝60^\circ ，BC＝1，求三棱柱ABC-A_{1}B_{1}C_{1}的高．$

Solution:
$作OD\perp BC，垂足为D，连接AD，作OH\perp AD，垂足为H，\because BC\perp AO，BC\perp OD，AO\cap OD＝O，\therefore BC\perp 平面AOD，\because OH\subset 平面AOD，\therefore OH\perp BC，\because OH\perp AD，BC\cap AD＝D，BC，AD\subset 平面ABC，\therefore OH\perp 平面ABC，\because \angle CBB_{1}＝60^\circ ，\therefore \triangle CBB_{1}为等边三角形，\because BC＝1，\therefore OD＝\frac{\sqrt{3}}{4}，\because AC\perp AB_{1}，\therefore OA＝\frac{1}{2}B_{1}C＝\frac{1}{2}，在\triangle AOD中，由等面积法可得OH\cdot AD＝OD\cdot OA，又AD＝\sqrt{OD^2+OA^2}=\frac{\sqrt{7}}{4}，故OH＝\frac{\sqrt{21}}{14}，\because O为B_{1}C的中点，\therefore B_{1}到平面ABC的距离为\frac{\sqrt{21}}{7}，\therefore 三棱柱ABC-A_{1}B_{1}C_{1}的高为\frac{\sqrt{21}}{7}．$![Image](https://cdntk.53zaixian.com/tq/1706223468395298818/img/compression/M_c33eff5675e5186df4b9b57032b09f2d.jpg)

Final answer: $\frac{\sqrt{21}}{7}$
Subfield: Plane Geometry


Question: $如图，在\triangle ABC中，已知\angle B=30^\circ ，D是BC边上的一点，AD=5，AC=7，DC=3.$![Image](https://cdntk.53zaixian.com/tq/1706200462365659138/img/compression/M_bffa4d606811429ee3a9f3b219e546a8.jpg)
$求\triangle ADC的面积；$

Solution:
$在\triangle ADC中，由余弦定理的推论得\\cos  \angle ADC=\frac{AD^2+DC^2-AC^2}{2AD\cdot DC}=\frac{5^2+3^2-7^2}{2\times 5\times 3}=-\frac{1}{2}，(2分)\because \angle ADC为三角形的内角，\therefore \angle ADC=120^\circ ，\therefore \\sin  \angle ADC=\frac{\sqrt{3}}{2}，(4分)\therefore S_{\triangle ADC}=\frac{1}{2}AD\cdot DC\cdot \\sin  \angle ADC=\frac{1}{2}\times 5\times 3\times \frac{\sqrt{3}}{2}=\frac{15\sqrt{3}}{4}.(6分)$

Final answer: $\frac{15\sqrt{3}}{4}$
Subfield: Plane Geometry


Question: $如图，在\triangle ABC中，已知\angle B=30^\circ ，D是BC边上的一点，AD=5，AC=7，DC=3.$<img_2126>
$求\triangle ADC的面积；$

Solution:
$在\triangle ADC中，由余弦定理的推论得\\cos  \angle ADC=\frac{AD^2+DC^2-AC^2}{2AD\cdot DC}=\frac{5^2+3^2-7^2}{2\times 5\times 3}=-\frac{1}{2}，(2分)\because \angle ADC为三角形的内角，\therefore \angle ADC=120^\circ ，\therefore \\sin  \angle ADC=\frac{\sqrt{3}}{2}，(4分)\therefore S_{\triangle ADC}=\frac{1}{2}AD\cdot DC\cdot \\sin  \angle ADC=\frac{1}{2}\times 5\times 3\times \frac{\sqrt{3}}{2}=\frac{15\sqrt{3}}{4}.(6分)$

Final answer: $\frac{15\sqrt{3}}{4}$
Subfield: Plane Geometry


Question: $如图，在\triangle ABC中，已知\angle B=30^\circ ，D是BC边上的一点，AD=5，AC=7，DC=3.$![Image](https://cdntk.53zaixian.com/tq/1706200462365659138/img/compression/M_bffa4d606811429ee3a9f3b219e546a8.jpg)
$求边AB的长.$

Solution:
$在\triangle ABD中，\angle ADB=60^\circ ，由正弦定理得\frac{AB}{\\\sin   \angle ADB}=\frac{AD}{\\\sin    B}，\therefore AB=5\sqrt{3}.(12分)$

Final answer: $5\sqrt 3$
Subfield: Plane Geometry


Question: $如图，在\triangle ABC中，已知\angle B=30^\circ ，D是BC边上的一点，AD=5，AC=7，DC=3.$<img_2126>
$求边AB的长.$

Solution:
$在\triangle ABD中，\angle ADB=60^\circ ，由正弦定理得\frac{AB}{\\\sin   \angle ADB}=\frac{AD}{\\\sin    B}，\therefore AB=5\sqrt{3}.(12分)$

Final answer: $5\sqrt 3$
Subfield: Plane Geometry


Question: $如图，位于海岸边A点处的观测站发现在南偏西30^\circ 方向上，距离A点20千米的C处有一艘走私船，于是立刻通知停在A点的正东方向上，且距离A点10(\sqrt{3}-1)千米的B处的缉私艇前往追截，缉私艇立刻奉命以10\sqrt{3}千米/小时的速度追截走私船，此时走私船正以10千米/小时的速度从C处沿南偏东15^\circ 方向逃窜.$![Image](https://cdntk.53zaixian.com/tq/1706200462848004098/img/compression/M_330d21844f47de0fdbb34e0771e01fa7.jpg)
$刚发现走私船时，走私船距离缉私艇多远？在缉私艇的什么方向？$

Solution:
$由题意可知AB=10(\sqrt{3}-1)，AC=20，\angle BAC=120^\circ .连接BC，在\triangle ABC中，由余弦定理可得BC=\sqrt{AB^2+AC^2-2AB\cdot AC\cdot \\\cos    120^{\circ}}=10\sqrt{6}.(2分)由正弦定理得\frac{AC}{\\\sin   \angle ABC}=\frac{BC}{\\\sin   \angle BAC}，解得\\sin  \angle ABC=\frac{\sqrt{2}}{2}，(3分)所以\angle ABC=45^\circ .(4分)故刚发现走私船时，走私船距离缉私艇10\sqrt{6}千米，在缉私艇的西南方向上.　(6分)$![Image](https://cdntk.53zaixian.com/tq/1706200463259045889/img/compression/M_347cbece0b7de6a8cb7b2a4bbf381854.jpg)

Final answer: $刚发现走私船时，走私船距离缉私艇10\sqrt 6千米，在缉私艇的西南方向上.$
Subfield: Plane Geometry


Question: $如图，位于海岸边A点处的观测站发现在南偏西30^\circ 方向上，距离A点20千米的C处有一艘走私船，于是立刻通知停在A点的正东方向上，且距离A点10(\sqrt{3}-1)千米的B处的缉私艇前往追截，缉私艇立刻奉命以10\sqrt{3}千米/小时的速度追截走私船，此时走私船正以10千米/小时的速度从C处沿南偏东15^\circ 方向逃窜.$<img_1529>
$缉私艇至少需要多长时间才能追上走私船？$

Solution:
$设t小时后缉私艇在D处追上走私船，连接BD，则CD=10t，BD=10 \sqrt{3}t.易得\angle BCD=45^\circ +75^\circ =120^\circ .在\triangle BCD中，由正弦定理得\frac{BD}{\\\sin   \angle BCD}=\frac{CD}{\\\sin   \angle CBD}，即\frac{10\sqrt{3}t}{\\\sin    120^{\circ}}=\frac{10t}{\\\sin   \angle CBD}，(8分)解得\\sin  \angle CBD=\frac{1}{2}，则\angle CBD=30^\circ ，所以\angle CDB=180^\circ -\angle BCD-\angle CBD=30^\circ =\angle CBD,(10分)则CD=BC，即10t=10\sqrt{6}，\therefore t=\sqrt{6}.(11分)故缉私艇至少需要\sqrt{6}小时才能追上走私船.(12分)$

Final answer: $\sqrt{6}$ 小时
Subfield: Plane Geometry


Question: $如图，位于海岸边A点处的观测站发现在南偏西30^\circ 方向上，距离A点20千米的C处有一艘走私船，于是立刻通知停在A点的正东方向上，且距离A点10(\sqrt{3}-1)千米的B处的缉私艇前往追截，缉私艇立刻奉命以10\sqrt{3}千米/小时的速度追截走私船，此时走私船正以10千米/小时的速度从C处沿南偏东15^\circ 方向逃窜.$![Image](https://cdntk.53zaixian.com/tq/1706200462848004098/img/compression/M_330d21844f47de0fdbb34e0771e01fa7.jpg)
$缉私艇至少需要多长时间才能追上走私船？$

Solution:
$设t小时后缉私艇在D处追上走私船，连接BD，则CD=10t，BD=10 \sqrt{3}t.易得\angle BCD=45^\circ +75^\circ =120^\circ .在\triangle BCD中，由正弦定理得\frac{BD}{\\\sin   \angle BCD}=\frac{CD}{\\\sin   \angle CBD}，即\frac{10\sqrt{3}t}{\\\sin    120^{\circ}}=\frac{10t}{\\\sin   \angle CBD}，(8分)解得\\sin  \angle CBD=\frac{1}{2}，则\angle CBD=30^\circ ，所以\angle CDB=180^\circ -\angle BCD-\angle CBD=30^\circ =\angle CBD,(10分)则CD=BC，即10t=10\sqrt{6}，\therefore t=\sqrt{6}.(11分)故缉私艇至少需要\sqrt{6}小时才能追上走私船.(12分)$

Final answer: $\sqrt{6}$ 小时
Subfield: Plane Geometry


Question: $如图，在\triangle ABC中，D，F分别是BC，AC的中点，AE＝\frac{2}{3}AD，\overrightarrow{AB}＝a，\overrightarrow{AC}＝b.$<img_1634>
$用a，b分别表示向量\overrightarrow{AE}，\overrightarrow{BF}；$

Solution:
$\because \overrightarrow{AD}＝\frac{1}{2}(\overrightarrow{AB}＋\overrightarrow{AC})＝\frac{1}{2}(a＋b)，\therefore \overrightarrow{AE}＝\frac{2}{3}\overrightarrow{AD}＝\frac{1}{3}(a＋b).\because \overrightarrow{AF}＝\frac{1}{2}\overrightarrow{AC}＝\frac{1}{2}b，\therefore \overrightarrow{BF}＝\overrightarrow{AF}－\overrightarrow{AB}＝－a＋\frac{1}{2}b.$

Final answer: $\overrightarrow{AE}=\frac{1}{3}(a+b)$, $\overrightarrow{BF}=-a+\frac{1}{2}b$
Subfield: Plane Geometry


Question: $如图，在\triangle ABC中，D，F分别是BC，AC的中点，AE＝\frac{2}{3}AD，\overrightarrow{AB}＝a，\overrightarrow{AC}＝b.$![Image](https://cdntk.53zaixian.com/tq/1706236151845572610/img/compression/M_028728be87b0c0cef097b5af5d5a6bf7.jpg)
$用a，b分别表示向量\overrightarrow{AE}，\overrightarrow{BF}；$

Solution:
$\because \overrightarrow{AD}＝\frac{1}{2}(\overrightarrow{AB}＋\overrightarrow{AC})＝\frac{1}{2}(a＋b)，\therefore \overrightarrow{AE}＝\frac{2}{3}\overrightarrow{AD}＝\frac{1}{3}(a＋b).\because \overrightarrow{AF}＝\frac{1}{2}\overrightarrow{AC}＝\frac{1}{2}b，\therefore \overrightarrow{BF}＝\overrightarrow{AF}－\overrightarrow{AB}＝－a＋\frac{1}{2}b.$

Final answer: $\overrightarrow{AE}=\frac{1}{3}(a+b)$, $\overrightarrow{BF}=-a+\frac{1}{2}b$
Subfield: Plane Geometry


Question: $如图，在\triangle ABC中，D，F分别是BC，AC的中点，AE＝\frac{2}{3}AD，\overrightarrow{AB}＝a，\overrightarrow{AC}＝b.$<img_1634>
$求证：B，E，F三点共线.$

Solution:
$证明：由(1)知\overrightarrow{BF}＝－a＋\frac{1}{2}b，\overrightarrow{AE}=\frac{1}{3}(a+b),则\overrightarrow{BE}＝\overrightarrow{AE}－\overrightarrow{AB}＝\frac{1}{3}(a＋b)－a＝－\frac{2}{3}a＋\frac{1}{3}b＝\frac{2}{3}\left(-\boldsymbol{a}+\frac{1}{2}\boldsymbol{b}\right)，\therefore \overrightarrow{BE}＝\frac{2}{3}\overrightarrow{BF}，\therefore \overrightarrow{BE}与\overrightarrow{BF}共线.又BE，BF有公共点B，\therefore B，E，F三点共线.$
Subfield: Plane Geometry


Question: $如图，在\triangle ABC中，D，F分别是BC，AC的中点，AE＝\frac{2}{3}AD，\overrightarrow{AB}＝a，\overrightarrow{AC}＝b.$![Image](https://cdntk.53zaixian.com/tq/1706236151845572610/img/compression/M_028728be87b0c0cef097b5af5d5a6bf7.jpg)
$求证：B，E，F三点共线.$

Solution:
$证明：由(1)知\overrightarrow{BF}＝－a＋\frac{1}{2}b，\overrightarrow{AE}=\frac{1}{3}(a+b),则\overrightarrow{BE}＝\overrightarrow{AE}－\overrightarrow{AB}＝\frac{1}{3}(a＋b)－a＝－\frac{2}{3}a＋\frac{1}{3}b＝\frac{2}{3}\left(-\boldsymbol{a}+\frac{1}{2}\boldsymbol{b}\right)，\therefore \overrightarrow{BE}＝\frac{2}{3}\overrightarrow{BF}，\therefore \overrightarrow{BE}与\overrightarrow{BF}共线.又BE，BF有公共点B，\therefore B，E，F三点共线.$

Final answer: 证明题
Subfield: Plane Geometry


Question: $如图所示,O是正三角形ABC的中心，四边形AOCD和四边形AOBE均为平行四边形,则：$![Image](https://cdntk.53zaixian.com/tq/1706234262248329219/img/compression/M_542bccf7e16f246720da987b59c913e5.jpg)
$与向量\overrightarrow{AD}相等的向量有　　　　;$

Solution:
$\because O是正三角形ABC的中心,\therefore OA=OB=OC.结合相等向量及相反向量的定义及模的概念可知:与向量\overrightarrow{AD}相等的向量有\overrightarrow{OC};$

Final answer: $\overrightarrow{OC}$
Subfield: Plane Geometry


Question: $如图所示,O是正三角形ABC的中心，四边形AOCD和四边形AOBE均为平行四边形,则：$<img_3096>
$与向量\overrightarrow{AD}相等的向量有　　　　;$

Solution:
$\because O是正三角形ABC的中心,\therefore OA=OB=OC.结合相等向量及相反向量的定义及模的概念可知:与向量\overrightarrow{AD}相等的向量有\overrightarrow{OC};$

Final answer: $\overrightarrow{OC}$
Subfield: Plane Geometry


Question: $如图所示,O是正三角形ABC的中心，四边形AOCD和四边形AOBE均为平行四边形,则：$<img_3096>
$与向量\overrightarrow{OA}相反的向量有　　　　;$

Solution:
$与向量\overrightarrow{OA}相反的向量有\overrightarrow{DC},\overrightarrow{EB};$

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


Question: $如图所示,O是正三角形ABC的中心，四边形AOCD和四边形AOBE均为平行四边形,则：$![Image](https://cdntk.53zaixian.com/tq/1706234262248329219/img/compression/M_542bccf7e16f246720da987b59c913e5.jpg)
$与向量\overrightarrow{OA}相反的向量有　　　　;$

Solution:
$与向量\overrightarrow{OA}相反的向量有\overrightarrow{DC},\overrightarrow{EB};$

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


Question: $如图所示,O是正三角形ABC的中心，四边形AOCD和四边形AOBE均为平行四边形,则：$<img_3096>
$与向量\overrightarrow{OA}的模相等的向量有　　　　.(填图中所画出的向量)$

Solution:
$与向量\overrightarrow{OA}的模相等的向量有\overrightarrow{OB},\overrightarrow{OC},\overrightarrow{DC},\overrightarrow{EB},\overrightarrow{AD}.$

Final answer: $\overrightarrow{OB},\overrightarrow{OC},\overrightarrow{DC},\overrightarrow{EB},\overrightarrow{AD}$
Subfield: Plane Geometry


Question: $如图所示,O是正三角形ABC的中心，四边形AOCD和四边形AOBE均为平行四边形,则：$![Image](https://cdntk.53zaixian.com/tq/1706234262248329219/img/compression/M_542bccf7e16f246720da987b59c913e5.jpg)
$与向量\overrightarrow{OA}的模相等的向量有　　　　.(填图中所画出的向量)$

Solution:
$与向量\overrightarrow{OA}的模相等的向量有\overrightarrow{OB},\overrightarrow{OC},\overrightarrow{DC},\overrightarrow{EB},\overrightarrow{AD}.$

Final answer: $\overrightarrow{OB},\overrightarrow{OC},\overrightarrow{DC},\overrightarrow{EB},\overrightarrow{AD}$
Subfield: Plane Geometry


Question: $在如图所示的平面图形中，已知OM=1，ON=2，\overrightarrow{BM}=2\overrightarrow{MA}，\overrightarrow{CN}=2\overrightarrow{NA}.$![Image](https://cdntk.53zaixian.com/tq/1706236201422499841/img/compression/M_ddb90c1ce7eb1b6b17b2189ddc913a7c.jpg)
$设\overrightarrow{BC}=x\overrightarrow{OM}+y\overrightarrow{ON}，求x+y的值；$

Solution:
$因为\overrightarrow{BM}=2\overrightarrow{MA}，\overrightarrow{CN}=2\overrightarrow{NA}，所以\overrightarrow{AB}=3\overrightarrow{AM}，\overrightarrow{AC}=3\overrightarrow{AN}，所以\overrightarrow{BC}=\overrightarrow{AC}－\overrightarrow{AB}=3\overrightarrow{AN}－3\overrightarrow{AM}=3\overrightarrow{MN}=－3\overrightarrow{OM}+3\overrightarrow{ON}，又\overrightarrow{BC}=x\overrightarrow{OM}+y\overrightarrow{ON},所以x=－3,y=3，所以x+y=0.$

Final answer: $0$
Subfield: Plane Geometry


Question: $在如图所示的平面图形中，已知OM=1，ON=2，\overrightarrow{BM}=2\overrightarrow{MA}，\overrightarrow{CN}=2\overrightarrow{NA}.$<img_1638>
$设\overrightarrow{BC}=x\overrightarrow{OM}+y\overrightarrow{ON}，求x+y的值；$

Solution:
$因为\overrightarrow{BM}=2\overrightarrow{MA}，\overrightarrow{CN}=2\overrightarrow{NA}，所以\overrightarrow{AB}=3\overrightarrow{AM}，\overrightarrow{AC}=3\overrightarrow{AN}，所以\overrightarrow{BC}=\overrightarrow{AC}－\overrightarrow{AB}=3\overrightarrow{AN}－3\overrightarrow{AM}=3\overrightarrow{MN}=－3\overrightarrow{OM}+3\overrightarrow{ON}，又\overrightarrow{BC}=x\overrightarrow{OM}+y\overrightarrow{ON},所以x=－3,y=3，所以x+y=0.$

Final answer: $0$
Subfield: Plane Geometry


Question: $在如图所示的平面图形中，已知OM=1，ON=2，\overrightarrow{BM}=2\overrightarrow{MA}，\overrightarrow{CN}=2\overrightarrow{NA}.$<img_1638>
$若\overrightarrow{OM}\parallel \overrightarrow{CN}，且\langle \overrightarrow{OM},\overrightarrow{ON}\rangle \in \left[\frac{\pi }{6},\frac{\pi }{3}\right]，求\overrightarrow{AB}\cdot \overrightarrow{AC}的最小值及此时\overrightarrow{OM}与\overrightarrow{ON}的夹角.$

Solution:
$设\overrightarrow{OM}与\overrightarrow{ON}的夹角为\theta ,则\theta \in \left[\frac{\pi }{6},\frac{\pi }{3}\right]，因为\overrightarrow{OM}\parallel \overrightarrow{CN}，所以\angle CNO=\angle MON=\theta ，设\overrightarrow{CA}=\lambda \overrightarrow{OM}(\lambda \in R)，则\overrightarrow{AB}=\overrightarrow{AC}+\overrightarrow{CB}=－\lambda \overrightarrow{OM}+3(\overrightarrow{OM}－\overrightarrow{ON})=(3－\lambda )\overrightarrow{OM}－3\overrightarrow{ON}，所以\overrightarrow{AB}\cdot \overrightarrow{AC}=[(3－\lambda )\overrightarrow{OM}－3\overrightarrow{ON}]\cdot (－\lambda \overrightarrow{OM})=－\lambda (3－\lambda )\overrightarrow{OM}^{2}+3\lambda \overrightarrow{OM}\cdot \overrightarrow{ON}=(\lambda ^{2}－3\lambda )|\overrightarrow{OM}|^{2}+3\lambda |\overrightarrow{OM}||\overrightarrow{ON}|\\cos   \theta =\lambda ^{2}－3\lambda +6\lambda \\cos   \theta =\lambda ^{2}+(6\\cos   \theta －3)\lambda .当\lambda =－\frac{6\\\cos    \theta -3}{2}时，\lambda ^{2}+(6\\cos   \theta －3)\lambda 取得最小值，且最小值为－\frac{(6\\\cos    \theta -3)^2}{4}，又\theta \in \left[\frac{\pi }{6},\frac{\pi }{3}\right]，所以6\\cos   \theta －3\in [0,3\sqrt{3}－3]，所以－\frac{(6\\\cos    \theta -3)^2}{4}\in \left[-\frac{(3\sqrt{3}-3)^2}{4},0\right]，即－\frac{(6\\\cos    \theta -3)^2}{4}\in \left[\frac{9\sqrt{3}-18}{2},0\right]，所以\overrightarrow{AB}\cdot \overrightarrow{AC}的最小值为\frac{9\sqrt{3}-18}{2}，此时\overrightarrow{OM}与\overrightarrow{ON}的夹角为\frac{\pi }{6}.$

Final answer: $\frac{9\sqrt{3}-18}{2}$, $\frac{\pi }{6}$
Subfield: Plane Geometry


Question: $在如图所示的平面图形中，已知OM=1，ON=2，\overrightarrow{BM}=2\overrightarrow{MA}，\overrightarrow{CN}=2\overrightarrow{NA}.$![Image](https://cdntk.53zaixian.com/tq/1706236201422499841/img/compression/M_ddb90c1ce7eb1b6b17b2189ddc913a7c.jpg)
$若\overrightarrow{OM}\parallel \overrightarrow{CN}，且\langle \overrightarrow{OM},\overrightarrow{ON}\rangle \in \left[\frac{\pi }{6},\frac{\pi }{3}\right]，求\overrightarrow{AB}\cdot \overrightarrow{AC}的最小值及此时\overrightarrow{OM}与\overrightarrow{ON}的夹角.$

Solution:
$设\overrightarrow{OM}与\overrightarrow{ON}的夹角为\theta ,则\theta \in \left[\frac{\pi }{6},\frac{\pi }{3}\right]，因为\overrightarrow{OM}\parallel \overrightarrow{CN}，所以\angle CNO=\angle MON=\theta ，设\overrightarrow{CA}=\lambda \overrightarrow{OM}(\lambda \in R)，则\overrightarrow{AB}=\overrightarrow{AC}+\overrightarrow{CB}=－\lambda \overrightarrow{OM}+3(\overrightarrow{OM}－\overrightarrow{ON})=(3－\lambda )\overrightarrow{OM}－3\overrightarrow{ON}，所以\overrightarrow{AB}\cdot \overrightarrow{AC}=[(3－\lambda )\overrightarrow{OM}－3\overrightarrow{ON}]\cdot (－\lambda \overrightarrow{OM})=－\lambda (3－\lambda )\overrightarrow{OM}^{2}+3\lambda \overrightarrow{OM}\cdot \overrightarrow{ON}=(\lambda ^{2}－3\lambda )|\overrightarrow{OM}|^{2}+3\lambda |\overrightarrow{OM}||\overrightarrow{ON}|\\cos   \theta =\lambda ^{2}－3\lambda +6\lambda \\cos   \theta =\lambda ^{2}+(6\\cos   \theta －3)\lambda .当\lambda =－\frac{6\\\cos    \theta -3}{2}时，\lambda ^{2}+(6\\cos   \theta －3)\lambda 取得最小值，且最小值为－\frac{(6\\\cos    \theta -3)^2}{4}，又\theta \in \left[\frac{\pi }{6},\frac{\pi }{3}\right]，所以6\\cos   \theta －3\in [0,3\sqrt{3}－3]，所以－\frac{(6\\\cos    \theta -3)^2}{4}\in \left[-\frac{(3\sqrt{3}-3)^2}{4},0\right]，即－\frac{(6\\\cos    \theta -3)^2}{4}\in \left[\frac{9\sqrt{3}-18}{2},0\right]，所以\overrightarrow{AB}\cdot \overrightarrow{AC}的最小值为\frac{9\sqrt{3}-18}{2}，此时\overrightarrow{OM}与\overrightarrow{ON}的夹角为\frac{\pi }{6}.$

Final answer: $\frac{9\sqrt{3}-18}{2}$, $\frac{\pi }{6}$
Subfield: Plane Geometry


Question: $已知平行四边形ABCD中，\overrightarrow{EC}=2\overrightarrow{DE}，\overrightarrow{FC}=2\overrightarrow{BF}，\overrightarrow{FG}=2\overrightarrow{GE}.$
$用\overrightarrow{AB}，\overrightarrow{AD}表示\overrightarrow{AG}；$

Solution:
$连接AE,AF,易得\overrightarrow{AE}=\overrightarrow{AD}+\frac{1}{3}\overrightarrow{AB}，\overrightarrow{AF}=\frac{1}{3}\overrightarrow{AD}+\overrightarrow{AB}，又\overrightarrow{FG}=2\overrightarrow{GE}，所以\overrightarrow{AG}－\overrightarrow{AF}=2(\overrightarrow{AE}－\overrightarrow{AG})，所以\overrightarrow{AG}=\frac{2}{3}\overrightarrow{AE}+\frac{1}{3}\overrightarrow{AF}=\frac{5}{9}\overrightarrow{AB}+\frac{7}{9}\overrightarrow{AD}.$

Final answer: $\overrightarrow{AG}=\frac{5}{9}\overrightarrow{AB}+\frac{7}{9}\overrightarrow{AD}$
Subfield: Plane Geometry


Question: $已知平行四边形ABCD中，\overrightarrow{EC}=2\overrightarrow{DE}，\overrightarrow{FC}=2\overrightarrow{BF}，\overrightarrow{FG}=2\overrightarrow{GE}.$
$若|\overrightarrow{AB}|=6，|\overrightarrow{AD}|=3\sqrt{2}，\angle BAD=45^\circ ，建立如图所示的平面直角坐标系，求\overrightarrow{GB}和\overrightarrow{DF}的坐标.
$![Image](https://cdntk.53zaixian.com/tq/1706236171067691010/img/compression/M_e0711c0ba2bf88db3fd93abe6976c509.jpg)![Image](https://cdntk.53zaixian.com/tq/1706236171067691010/img/compression/M_ac3a7b4135544e32d1aeed13e4062964.jpg)

Solution:
$过点D作AB的垂线交AB于点D'，如图所示：$![Image](https://cdntk.53zaixian.com/tq/1706236171067691010/img/compression/M_cf098598fa17b45d1c9e3fe0ef243df7.jpg)$在Rt\triangle ADD'中，\angle D'AD=45^\circ ，|\overrightarrow{AD}|=3\sqrt{2}，所以AD'=DD'=3，易得A(0,0)，B(6,0)，C(9,3)，D(3,3)，E(5,3)，F(7,1)，所以\overrightarrow{DF}=(4，-2)，所以\overrightarrow{AG}=\frac{5}{9}\overrightarrow{AB}+\frac{7}{9}\overrightarrow{AD}=\frac{5}{9}(6，0)+\frac{7}{9}(3,3)=\left(\frac{17}{3},\frac{7}{3}\right)，所以G\left(\frac{17}{3},\frac{7}{3}\right).所以\overrightarrow{GB}=\overrightarrow{AB}－\overrightarrow{AG}=(6，0)-\left(\frac{17}{3},\frac{7}{3}\right)=\left(\frac{1}{3},-\frac{7}{3}\right)，故\overrightarrow{GB}的坐标为\left(\frac{1}{3},-\frac{7}{3}\right)，\overrightarrow{DF}的坐标为(4,－2).$

Final answer: $\left(\frac{1}{3},-\frac{7}{3}\right)$,$(4, -2)$
Subfield: Plane Geometry


Question: $已知平行四边形ABCD中，\overrightarrow{EC}=2\overrightarrow{DE}，\overrightarrow{FC}=2\overrightarrow{BF}，\overrightarrow{FG}=2\overrightarrow{GE}.$
$若|\overrightarrow{AB}|=6，|\overrightarrow{AD}|=3\sqrt{2}，\angle BAD=45^\circ ，建立如图所示的平面直角坐标系，求\overrightarrow{GB}和\overrightarrow{DF}的坐标.
$<img_2595><img_1682>

Solution:
$过点D作AB的垂线交AB于点D'，如图所示：$<img_2265>$在Rt\triangle ADD'中，\angle D'AD=45^\circ ，|\overrightarrow{AD}|=3\sqrt{2}，所以AD'=DD'=3，易得A(0,0)，B(6,0)，C(9,3)，D(3,3)，E(5,3)，F(7,1)，所以\overrightarrow{DF}=(4，-2)，所以\overrightarrow{AG}=\frac{5}{9}\overrightarrow{AB}+\frac{7}{9}\overrightarrow{AD}=\frac{5}{9}(6，0)+\frac{7}{9}(3,3)=\left(\frac{17}{3},\frac{7}{3}\right)，所以G\left(\frac{17}{3},\frac{7}{3}\right).所以\overrightarrow{GB}=\overrightarrow{AB}－\overrightarrow{AG}=(6，0)-\left(\frac{17}{3},\frac{7}{3}\right)=\left(\frac{1}{3},-\frac{7}{3}\right)，故\overrightarrow{GB}的坐标为\left(\frac{1}{3},-\frac{7}{3}\right)，\overrightarrow{DF}的坐标为(4,－2).$



Final answer: $\left(\frac{1}{3},-\frac{7}{3}\right)$,$(4, -2)$
Subfield: Plane Geometry


Question: $如图，在\triangle ABC中，已知AB=2，AC=6，\angle BAC=60^\circ ，点D，E分别在边AB，AC上，且\overrightarrow{AB}=2\overrightarrow{AD}，\overrightarrow{AC}=5\overrightarrow{AE}.$<img_2811>
$若点F为DE的中点，用向量\overrightarrow{AB}和\overrightarrow{AC}表示\overrightarrow{BF}；$

Solution:
$\overrightarrow{BF}=\overrightarrow{BD}+\overrightarrow{DF}=-\frac{1}{2}\overrightarrow{AB}+\frac{1}{2}\overrightarrow{DE}=-\frac{1}{2}\overrightarrow{AB}+\frac{1}{2}(\overrightarrow{AE}-\overrightarrow{AD})=-\frac{1}{2}\overrightarrow{AB}+\frac{1}{2}\left(\frac{1}{5}\overrightarrow{AC}-\frac{1}{2}\overrightarrow{AB}\right)=-\frac{3}{4}\overrightarrow{AB}+\frac{1}{10}\overrightarrow{AC}.$

Final answer: $-\frac{3}{4}\overrightarrow{AB}+\frac{1}{10}\overrightarrow{AC}$
Subfield: Plane Geometry


Question: $如图，在\triangle ABC中，已知AB=2，AC=6，\angle BAC=60^\circ ，点D，E分别在边AB，AC上，且\overrightarrow{AB}=2\overrightarrow{AD}，\overrightarrow{AC}=5\overrightarrow{AE}.$![Image](https://cdntk.53zaixian.com/tq/1706236181318946817/img/compression/M_9b894d125e6ca150c40d6960d7cc7dcf.jpg)
$若点F为DE的中点，用向量\overrightarrow{AB}和\overrightarrow{AC}表示\overrightarrow{BF}；$

Solution:
$\overrightarrow{BF}=\overrightarrow{BD}+\overrightarrow{DF}=-\frac{1}{2}\overrightarrow{AB}+\frac{1}{2}\overrightarrow{DE}=-\frac{1}{2}\overrightarrow{AB}+\frac{1}{2}(\overrightarrow{AE}-\overrightarrow{AD})=-\frac{1}{2}\overrightarrow{AB}+\frac{1}{2}\left(\frac{1}{5}\overrightarrow{AC}-\frac{1}{2}\overrightarrow{AB}\right)=-\frac{3}{4}\overrightarrow{AB}+\frac{1}{10}\overrightarrow{AC}.$

Final answer: $-\frac{3}{4}\overrightarrow{AB}+\frac{1}{10}\overrightarrow{AC}$
Subfield: Plane Geometry


Question: $如图，在\triangle ABC中，已知AB=2，AC=6，\angle BAC=60^\circ ，点D，E分别在边AB，AC上，且\overrightarrow{AB}=2\overrightarrow{AD}，\overrightarrow{AC}=5\overrightarrow{AE}.$![Image](https://cdntk.53zaixian.com/tq/1706236181318946817/img/compression/M_9b894d125e6ca150c40d6960d7cc7dcf.jpg)
$在(1)的条件下，求\overrightarrow{BA}\cdot \overrightarrow{EF}的值.$

Solution:
$\overrightarrow{EF}=\frac{1}{2}\overrightarrow{ED}=\frac{1}{2}(\overrightarrow{AD}-\overrightarrow{AE})，\because \overrightarrow{AB}=2\overrightarrow{AD}，\overrightarrow{AC}=5\overrightarrow{AE}，\therefore \overrightarrow{EF}=\frac{1}{4}\overrightarrow{AB}-\frac{1}{10}\overrightarrow{AC}，\therefore \overrightarrow{BA}\cdot \overrightarrow{EF}=-\overrightarrow{AB}\cdot \left(\frac{1}{4}\overrightarrow{AB}-\frac{1}{10}\overrightarrow{AC}\right)=-\frac{1}{4}\overrightarrow{AB}^{2}+\frac{1}{10}\overrightarrow{AB}\cdot \overrightarrow{AC} =-\frac{1}{4}\times 4+\frac{1}{10}\times 2\times 6\times \\cos   60^\circ =-\frac{2}{5}.$

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


Question: $如图，在\triangle ABC中，已知AB=2，AC=6，\angle BAC=60^\circ ，点D，E分别在边AB，AC上，且\overrightarrow{AB}=2\overrightarrow{AD}，\overrightarrow{AC}=5\overrightarrow{AE}.$<img_2811>
$在(1)的条件下，求\overrightarrow{BA}\cdot \overrightarrow{EF}的值.$

Solution:
$\overrightarrow{EF}=\frac{1}{2}\overrightarrow{ED}=\frac{1}{2}(\overrightarrow{AD}-\overrightarrow{AE})，\because \overrightarrow{AB}=2\overrightarrow{AD}，\overrightarrow{AC}=5\overrightarrow{AE}，\therefore \overrightarrow{EF}=\frac{1}{4}\overrightarrow{AB}-\frac{1}{10}\overrightarrow{AC}，\therefore \overrightarrow{BA}\cdot \overrightarrow{EF}=-\overrightarrow{AB}\cdot \left(\frac{1}{4}\overrightarrow{AB}-\frac{1}{10}\overrightarrow{AC}\right)=-\frac{1}{4}\overrightarrow{AB}^{2}+\frac{1}{10}\overrightarrow{AB}\cdot \overrightarrow{AC} =-\frac{1}{4}\times 4+\frac{1}{10}\times 2\times 6\times \\cos   60^\circ =-\frac{2}{5}.$

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


Question: $如图，在O处有一港口，两艘海轮B，C同时从港口O处出发向正北方向匀速航行，海轮B的航行速度为20海里/时，海轮C的航行速度大于海轮B的航行速度.在港口O北偏东60^\circ 方向上的A处有一观测站，1小时后在A处测得与海轮B的距离为30海里，且A处对两艘海轮B，C的视角为30^\circ .$![Image](https://cdntk.53zaixian.com/tq/1706236228278374403/img/compression/M_71622c7303d33c3495dffb8325a4cddc.jpg)
$求观测站A到港口O的距离；$

Solution:
$因为海轮B的航行速度为20海里/时，所以1小时后，OB=20海里，又AB=30海里，\angle AOB=60^\circ ，所以在\triangle AOB中，由余弦定理，得AB^{2}=OA^{2}+OB^{2}-2\times OA\times OB\times \\cos  \angle AOB,即30^{2}=OA^{2}+20^{2}-2\times OA\times 20\times \\cos   60^\circ ,即OA^{2}-20OA-500=0，所以OA=(10+10\sqrt{6})海里.故观测站A到港口O的距离为(10+10\sqrt{6})海里.$

Final answer: $(10+10\sqrt 6)$ 海里