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


Question: $随着生活水平的不断提高，人们更加关注健康，重视锻炼.通过“小步道”，走出“大健康”，健康步道成为引领健康生活的一道亮丽风景线.如图，A-B-C-A为某区的一条健康步道，AB,AC为线段，\overset{\LARGE{\frown}}{BC}是以BC为直径的半圆，AB=2\sqrt{3} km,AC=4 km,\angle BAC=\frac{\pi }{6}.$![Image](https://cdntk.53zaixian.com/tq/1715283813142585346/img/compression/M_55c03612fc1bc656f33c710e4dc78965.jpg)
$求\overset{\LARGE{\frown}}{BC}的长度;$

Solution:
$连接BC，在\triangle ABC中，由余弦定理可得，BC=\sqrt{AC^2+AB^2-2AC\cdot AB\cdot \\\cos   \angle BAC}=\sqrt{16+12-2\times 4\times 2\sqrt{3}\times \frac{\sqrt{3}}{2}}=2(km)，所以\overset{\LARGE{\frown}}{BC}的长度为\frac{1}{2}\times 2\times \pi \times 1=\pi (km).$

Final answer: $\pi$
Subfield: Plane Geometry


Question: $随着生活水平的不断提高，人们更加关注健康，重视锻炼.通过“小步道”，走出“大健康”，健康步道成为引领健康生活的一道亮丽风景线.如图，A-B-C-A为某区的一条健康步道，AB,AC为线段，\overset{\LARGE{\frown}}{BC}是以BC为直径的半圆，AB=2\sqrt{3} km,AC=4 km,\angle BAC=\frac{\pi }{6}.$<img_1684>
$为满足市民健康生活需要，提升城市品位，改善人居环境，现计划新建健康步道A-D-C(B、D在AC两侧)，其中AD, CD为线段.若\angle ADC=\frac{\pi }{3},求新建的健康步道A-D-C的路程最多可比原有健康步道A-B-C的路程增加多少?$

Solution:
$记AD=a km,CD=b km，则在\triangle ACD中，由余弦定理可得a^{2}+b^{2}-2ab\\cos  \frac{\pi }{3}=16,即a^{2}+b^{2}-ab=16，从而(a+b)^{2}=16+3ab\leq 16+3\left(\frac{a+b}{2}\right)^2,所以\frac{1}{4}(a+b)^{2}\leq 16，则a+b\leq 8，当且仅当a=b=4 km时，等号成立.新建健康步道A-D-C的最长路程为8 km,故新建的健康步道A-D-C的路程最多可比原有健康步道A-B-C的路程增加8-\pi -2\sqrt{3} km.$

Final answer: $8-\pi -2\sqrt{3} $ km
Subfield: Plane Geometry


Question: $随着生活水平的不断提高，人们更加关注健康，重视锻炼.通过“小步道”，走出“大健康”，健康步道成为引领健康生活的一道亮丽风景线.如图，A-B-C-A为某区的一条健康步道，AB,AC为线段，\overset{\LARGE{\frown}}{BC}是以BC为直径的半圆，AB=2\sqrt{3} km,AC=4 km,\angle BAC=\frac{\pi }{6}.$![Image](https://cdntk.53zaixian.com/tq/1715283813142585346/img/compression/M_55c03612fc1bc656f33c710e4dc78965.jpg)
$为满足市民健康生活需要，提升城市品位，改善人居环境，现计划新建健康步道A-D-C(B、D在AC两侧)，其中AD, CD为线段.若\angle ADC=\frac{\pi }{3},求新建的健康步道A-D-C的路程最多可比原有健康步道A-B-C的路程增加多少?$

Solution:
$记AD=a km,CD=b km，则在\triangle ACD中，由余弦定理可得a^{2}+b^{2}-2ab\\cos  \frac{\pi }{3}=16,即a^{2}+b^{2}-ab=16，从而(a+b)^{2}=16+3ab\leq 16+3\left(\frac{a+b}{2}\right)^2,所以\frac{1}{4}(a+b)^{2}\leq 16，则a+b\leq 8，当且仅当a=b=4 km时，等号成立.新建健康步道A-D-C的最长路程为8 km,故新建的健康步道A-D-C的路程最多可比原有健康步道A-B-C的路程增加8-\pi -2\sqrt{3} km.$

Final answer: $8-\pi -2\sqrt{3} $ km
Subfield: Plane Geometry


Question: $如图，D为圆锥的顶点，O是圆锥底面的圆心，AE为底面直径，AE=AD.\triangle ABC是底面的内接正三角形，P为DO上一点，PO=\frac{\sqrt{6}}{6}DO.$<img_2621>
$证明：PA\perp 平面PBC；$

Solution:
$证明：设DO=a,由题设可得PO=\frac{\sqrt{6}}{6}a,AO=\frac{\sqrt{3}}{3}a,AB=a，PA=PB=PC=\frac{\sqrt{2}}{2}a.因此PA^{2}+PB^{2}=AB^{2}，从而PA\perp PB.又PA^{2}+PC^{2}=AC^{2}，故PA\perp PC.所以PA\perp 平面PBC.$
Subfield: Plane Geometry


Question: $如图，D为圆锥的顶点，O是圆锥底面的圆心，AE为底面直径，AE=AD.\triangle ABC是底面的内接正三角形，P为DO上一点，PO=\frac{\sqrt{6}}{6}DO.$![Image](https://cdntk.53zaixian.com/tq/1715283918572306434/img/compression/M_fa2785ceeb7784ada317ae8f0710e744.jpg)
$证明：PA\perp 平面PBC；$

Solution:
$证明：设DO=a,由题设可得PO=\frac{\sqrt{6}}{6}a,AO=\frac{\sqrt{3}}{3}a,AB=a，PA=PB=PC=\frac{\sqrt{2}}{2}a.因此PA^{2}+PB^{2}=AB^{2}，从而PA\perp PB.又PA^{2}+PC^{2}=AC^{2}，故PA\perp PC.所以PA\perp 平面PBC.$

Final answer: 证明题
Subfield: Plane Geometry


Question: $如图，D为圆锥的顶点，O是圆锥底面的圆心，AE为底面直径，AE=AD.\triangle ABC是底面的内接正三角形，P为DO上一点，PO=\frac{\sqrt{6}}{6}DO.$![Image](https://cdntk.53zaixian.com/tq/1715283918572306434/img/compression/M_fa2785ceeb7784ada317ae8f0710e744.jpg)
$求二面角B-PC-E的余弦值.$

Solution:
$以O为坐标原点，\overrightarrow{OE}的方向为y轴正方向，|\overrightarrow{OE}|为单位长，建立如图所示的空间直角坐标系O-xyz.$![Image](https://cdntk.53zaixian.com/tq/1715283918870102018/img/compression/M_875e0b740a62b0c26ff77012801c6cd9.jpg)$由题设可得E(0，1，0)，A(0，-1，0)，C\left(-\frac{\sqrt{3}}{2},\frac{1}{2},0\right)，P\left(0,0,\frac{\sqrt{2}}{2}\right).所以\overrightarrow{EC}=\left(-\frac{\sqrt{3}}{2},-\frac{1}{2},0\right)，\overrightarrow{EP}=\left(0,-1,\frac{\sqrt{2}}{2}\right).设m=(x,y,z)是平面PCE的法向量，则\left\{\begin{matrix}\boldsymbol{m}\cdot \overrightarrow{EP}=0,\\ \boldsymbol{m}\cdot \overrightarrow{EC}=0,\end{matrix}\right.即\left\{\begin{matrix}-y+\frac{\sqrt{2}}{2}z=0,\\ -\frac{\sqrt{3}}{2}x-\frac{1}{2}y=0.\end{matrix}\right.可取m=\left(-\frac{\sqrt{3}}{3},1,\sqrt{2}\right).由(1)知\overrightarrow{AP}=\left(0,1,\frac{\sqrt{2}}{2}\right)是平面PCB的一个法向量，记n=\overrightarrow{AP},则\\cos  \langle n,m\rangle =\frac{\boldsymbol{n}\cdot \boldsymbol{m}}{|\boldsymbol{n}|\cdot |\boldsymbol{m}|}=\frac{2\sqrt{5}}{5}.易知二面角B-PC-E的平面角为锐角，所以二面角B-PC-E的余弦值为\frac{2\sqrt{5}}{5}.$

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


Question: $如图，D为圆锥的顶点，O是圆锥底面的圆心，AE为底面直径，AE=AD.\triangle ABC是底面的内接正三角形，P为DO上一点，PO=\frac{\sqrt{6}}{6}DO.$<img_2621>
$求二面角B-PC-E的余弦值.$

Solution:
$以O为坐标原点，\overrightarrow{OE}的方向为y轴正方向，|\overrightarrow{OE}|为单位长，建立如图所示的空间直角坐标系O-xyz.$<img_2306>$由题设可得E(0，1，0)，A(0，-1，0)，C\left(-\frac{\sqrt{3}}{2},\frac{1}{2},0\right)，P\left(0,0,\frac{\sqrt{2}}{2}\right).所以\overrightarrow{EC}=\left(-\frac{\sqrt{3}}{2},-\frac{1}{2},0\right)，\overrightarrow{EP}=\left(0,-1,\frac{\sqrt{2}}{2}\right).设m=(x,y,z)是平面PCE的法向量，则\left\{\begin{matrix}\boldsymbol{m}\cdot \overrightarrow{EP}=0,\\ \boldsymbol{m}\cdot \overrightarrow{EC}=0,\end{matrix}\right.即\left\{\begin{matrix}-y+\frac{\sqrt{2}}{2}z=0,\\ -\frac{\sqrt{3}}{2}x-\frac{1}{2}y=0.\end{matrix}\right.可取m=\left(-\frac{\sqrt{3}}{3},1,\sqrt{2}\right).由(1)知\overrightarrow{AP}=\left(0,1,\frac{\sqrt{2}}{2}\right)是平面PCB的一个法向量，记n=\overrightarrow{AP},则\\cos  \langle n,m\rangle =\frac{\boldsymbol{n}\cdot \boldsymbol{m}}{|\boldsymbol{n}|\cdot |\boldsymbol{m}|}=\frac{2\sqrt{5}}{5}.易知二面角B-PC-E的平面角为锐角，所以二面角B-PC-E的余弦值为\frac{2\sqrt{5}}{5}.$

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


Question: $如图，在正四棱柱ABCD-A_{1}B_{1}C_{1}D_{1}中，AB=2，AA_{1}=4，点A_{2},B_{2},C_{2}，D_{2}分别在棱AA_{1},BB_{1},CC_{1},DD_{1}上，AA_{2}=1，BB_{2}=DD_{2}=2，CC_{2}=3.$<img_3044>
$证明：B_{2}C_{2}\parallel A_{2}D_{2};$

Solution:
$证明：以C为原点，\overrightarrow{CD}，\overrightarrow{CB},\overrightarrow{CC_1}的方向分别为x轴,y轴,z轴的正方向，建立如图所示的空间直角坐标系，由题意知，A_{2}(2,2,1),B_{2}(0,2,2),C_{2}(0,0,3),D_{2}(2,0,2),则\overrightarrow{B_2C_2}=(0，-2，1)，\overrightarrow{A_2D_2}=(0,-2,1),\therefore \overrightarrow{B_2C_2}=\overrightarrow{A_2D_2}，\therefore \overrightarrow{B_2C_2}\parallel \overrightarrow{A_2D_2},又知B_{2}C_{2}与A_{2}D_{2}无公共点，\therefore B_{2}C_{2}\parallel A_{2}D_{2}.$<img_1670>
Subfield: Plane Geometry


Question: $如图，在正四棱柱ABCD-A_{1}B_{1}C_{1}D_{1}中，AB=2，AA_{1}=4，点A_{2},B_{2},C_{2}，D_{2}分别在棱AA_{1},BB_{1},CC_{1},DD_{1}上，AA_{2}=1，BB_{2}=DD_{2}=2，CC_{2}=3.$![Image](https://cdntk.53zaixian.com/tq/1715283940768563203/img/compression/M_b2fb22a9a4b6a7b1dd37c5e2deebef03.jpg)
$证明：B_{2}C_{2}\parallel A_{2}D_{2};$

Solution:
$证明：以C为原点，\overrightarrow{CD}，\overrightarrow{CB},\overrightarrow{CC_1}的方向分别为x轴,y轴,z轴的正方向，建立如图所示的空间直角坐标系，由题意知，A_{2}(2,2,1),B_{2}(0,2,2),C_{2}(0,0,3),D_{2}(2,0,2),则\overrightarrow{B_2C_2}=(0，-2，1)，\overrightarrow{A_2D_2}=(0,-2,1),\therefore \overrightarrow{B_2C_2}=\overrightarrow{A_2D_2}，\therefore \overrightarrow{B_2C_2}\parallel \overrightarrow{A_2D_2},又知B_{2}C_{2}与A_{2}D_{2}无公共点，\therefore B_{2}C_{2}\parallel A_{2}D_{2}.$![Image](https://cdntk.53zaixian.com/tq/1715283941108301826/img/compression/M_df204ba2bd9062cf121894dc131de51e.jpg)

Final answer: 证明题
Subfield: Plane Geometry


Question: $如图，在正四棱柱ABCD-A_{1}B_{1}C_{1}D_{1}中，AB=2，AA_{1}=4，点A_{2},B_{2},C_{2}，D_{2}分别在棱AA_{1},BB_{1},CC_{1},DD_{1}上，AA_{2}=1，BB_{2}=DD_{2}=2，CC_{2}=3.$![Image](https://cdntk.53zaixian.com/tq/1715283940768563203/img/compression/M_b2fb22a9a4b6a7b1dd37c5e2deebef03.jpg)
$点P在棱BB_{1}上，当二面角P-A_{2}C_{2}-D_{2}为150^\circ 时，求B_{2}P.$

Solution:
$\because 点P在棱BB_{1}上，\therefore 设P(0，2，a)(0\leq a\leq 4)，结合(1)可知\overrightarrow{A_2C_2}=(-2，-2，2)，\overrightarrow{A_2D_2}=(0,-2,1),\overrightarrow{PA_2}=(2,0,1-a),\overrightarrow{PC_2}=(0,-2,3-a).设平面A_{2}C_{2}D_{2}的法向量为n_{1}=(x_{1},y_{1},z_{1}),则\left\{\begin{matrix}\overrightarrow{A_2C_2}\cdot \boldsymbol{n}_1=0,\\ \overrightarrow{A_2D_2}\cdot \boldsymbol{n}_1=0,\end{matrix}\right.即\left\{\begin{matrix}-2x_1-2y_1+2z_1=0,\\ -2y_1+z_1=0,\end{matrix}\right.令z_{1}=2,则n_{1}=(1,1,2).设平面PA_{2}C_{2}的法向量为n_{2}=(x_{2},y_{2},z_{2}),则\left\{\begin{matrix}\overrightarrow{PA_2}\cdot \boldsymbol{n}_2=0,\\ \overrightarrow{PC_2}\cdot \boldsymbol{n}_2=0,\end{matrix}\right.即\left\{\begin{matrix}2x_2+(1-a)z_2=0,\\ -2y_2+(3-a)z_2=0,\end{matrix}\right.令z_{2}=2,则n_{2}=(a-1,3-a,2),又\because 二面角P-A_{2}C_{2}-D_{2}为150^\circ ，\therefore |\\cos   150^\circ |=|\\cos  \langle n_{1},n_{2}\rangle |=\frac{|\boldsymbol{n}_1\cdot \boldsymbol{n}_2|}{|\boldsymbol{n}_1|\cdot |\boldsymbol{n}_2|}=\frac{|a-1+3-a+4|}{\sqrt{6}\cdot \sqrt{(a-1)^2+(3-a)^2+4}}=\frac{\sqrt{6}}{\sqrt{2a^2-8a+14}},即\frac{\sqrt{3}}{2}=\frac{\sqrt{3}}{\sqrt{a^2-4a+7}},化简得a^{2}-4a+3=0,解得a=1或a=3,当a=1时，B_{2}P=1;当a=3时，B_{2}P=1.综上，B_{2}P=1.$

Final answer: $1$
Subfield: Plane Geometry


Question: $如图，在正四棱柱ABCD-A_{1}B_{1}C_{1}D_{1}中，AB=2，AA_{1}=4，点A_{2},B_{2},C_{2}，D_{2}分别在棱AA_{1},BB_{1},CC_{1},DD_{1}上，AA_{2}=1，BB_{2}=DD_{2}=2，CC_{2}=3.$<img_3044>
$点P在棱BB_{1}上，当二面角P-A_{2}C_{2}-D_{2}为150^\circ 时，求B_{2}P.$

Solution:
$\because 点P在棱BB_{1}上，\therefore 设P(0，2，a)(0\leq a\leq 4)，结合(1)可知\overrightarrow{A_2C_2}=(-2，-2，2)，\overrightarrow{A_2D_2}=(0,-2,1),\overrightarrow{PA_2}=(2,0,1-a),\overrightarrow{PC_2}=(0,-2,3-a).设平面A_{2}C_{2}D_{2}的法向量为n_{1}=(x_{1},y_{1},z_{1}),则\left\{\begin{matrix}\overrightarrow{A_2C_2}\cdot \boldsymbol{n}_1=0,\\ \overrightarrow{A_2D_2}\cdot \boldsymbol{n}_1=0,\end{matrix}\right.即\left\{\begin{matrix}-2x_1-2y_1+2z_1=0,\\ -2y_1+z_1=0,\end{matrix}\right.令z_{1}=2,则n_{1}=(1,1,2).设平面PA_{2}C_{2}的法向量为n_{2}=(x_{2},y_{2},z_{2}),则\left\{\begin{matrix}\overrightarrow{PA_2}\cdot \boldsymbol{n}_2=0,\\ \overrightarrow{PC_2}\cdot \boldsymbol{n}_2=0,\end{matrix}\right.即\left\{\begin{matrix}2x_2+(1-a)z_2=0,\\ -2y_2+(3-a)z_2=0,\end{matrix}\right.令z_{2}=2,则n_{2}=(a-1,3-a,2),又\because 二面角P-A_{2}C_{2}-D_{2}为150^\circ ，\therefore |\\cos   150^\circ |=|\\cos  \langle n_{1},n_{2}\rangle |=\frac{|\boldsymbol{n}_1\cdot \boldsymbol{n}_2|}{|\boldsymbol{n}_1|\cdot |\boldsymbol{n}_2|}=\frac{|a-1+3-a+4|}{\sqrt{6}\cdot \sqrt{(a-1)^2+(3-a)^2+4}}=\frac{\sqrt{6}}{\sqrt{2a^2-8a+14}},即\frac{\sqrt{3}}{2}=\frac{\sqrt{3}}{\sqrt{a^2-4a+7}},化简得a^{2}-4a+3=0,解得a=1或a=3,当a=1时，B_{2}P=1;当a=3时，B_{2}P=1.综上，B_{2}P=1.$

Final answer: $1$
Subfield: Plane Geometry


Question: $已知椭圆C的方程为\frac{x^2}{a^2}+\frac{y^2}{3}=1，斜率为k(k\neq 0)的直线与C相交于M，N两点.$![Image](https://cdntk.53zaixian.com/tq/1715284008124891138/img/compression/M_ada47ac1fe8c46c4f77e04992d76fc43.jpg)
$若G为MN的中点，且k_{OG}=-\frac{3}{4k}，求椭圆C的方程；$

Solution:
$设M(x_{1},y_{1}),N(x_{2},y_{2}),G(x_{0},y_{0})，将M，N代入椭圆方程得\left\{\begin{matrix}\frac{x^2_1}{a^2}+\frac{y^2_1}{3}=1,{\text{①}}\\ \frac{x^2_2}{a^2}+\frac{y^2_2}{3}=1,{\text{②}}\end{matrix}\right.①-②得\frac{(x_1+x_2)(x_1-x_2)}{a^2}+\frac{(y_1+y_2)(y_1-y_2)}{3}=0,可得k=\frac{y_1-y_2}{x_1-x_2}=-\frac{3}{a^2}\times \frac{x_1+x_2}{y_1+y_2}=-\frac{3}{a^2}\times \frac{2x_0}{2y_0}=-\frac{3}{a^2}\times \frac{x_0}{y_0},因为k_{OG}=-\frac{3}{4k}=\frac{y_0}{x_0}，所以-\frac{3}{a^2}\times \left(-\frac{4k}{3}\right)=k,所以a^{2}=4,椭圆C的方程为\frac{x^2}{4}+\frac{y^2}{3}=1.$

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


Question: $已知椭圆C的方程为\frac{x^2}{a^2}+\frac{y^2}{3}=1，斜率为k(k\neq 0)的直线与C相交于M，N两点.$<img_2371>
$若G为MN的中点，且k_{OG}=-\frac{3}{4k}，求椭圆C的方程；$

Solution:
$设M(x_{1},y_{1}),N(x_{2},y_{2}),G(x_{0},y_{0})，将M，N代入椭圆方程得\left\{\begin{matrix}\frac{x^2_1}{a^2}+\frac{y^2_1}{3}=1,{\text{①}}\\ \frac{x^2_2}{a^2}+\frac{y^2_2}{3}=1,{\text{②}}\end{matrix}\right.①-②得\frac{(x_1+x_2)(x_1-x_2)}{a^2}+\frac{(y_1+y_2)(y_1-y_2)}{3}=0,可得k=\frac{y_1-y_2}{x_1-x_2}=-\frac{3}{a^2}\times \frac{x_1+x_2}{y_1+y_2}=-\frac{3}{a^2}\times \frac{2x_0}{2y_0}=-\frac{3}{a^2}\times \frac{x_0}{y_0},因为k_{OG}=-\frac{3}{4k}=\frac{y_0}{x_0}，所以-\frac{3}{a^2}\times \left(-\frac{4k}{3}\right)=k,所以a^{2}=4,椭圆C的方程为\frac{x^2}{4}+\frac{y^2}{3}=1.$

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


Question: $已知椭圆C的方程为\frac{x^2}{a^2}+\frac{y^2}{3}=1，斜率为k(k\neq 0)的直线与C相交于M，N两点.$![Image](https://cdntk.53zaixian.com/tq/1715284008124891138/img/compression/M_ada47ac1fe8c46c4f77e04992d76fc43.jpg)
$在(1)的条件下，若P是椭圆C的左顶点，k_{PM}k_{PN}=-\frac{1}{4}，F是椭圆的左焦点，要使F在以MN为直径的圆内，求k的取值范围.$

Solution:
$设MN方程为y=kx+m,则\left\{\begin{matrix}3x^2+4y^2=12,\\ y=kx+m,\end{matrix}\right.(3+4k^{2})x^{2}+8kmx+4m^{2}-12=0,x_{1}+x_{2}=\frac{-8km}{3+4k^2},x_{1}x_{2}=\frac{4m^2-12}{3+4k^2}，y_{1}+y_{2}=k(x_{1}+x_{2})+2m=k\times \left(\frac{-8km}{3+4k^2}\right)+2m=\frac{6m}{3+4k^2},y_{1}y_{2}=(kx_{1}+m)\cdot (kx_{2}+m)=k^{2}x_{1}x_{2}+km(x_{1}+x_{2})+m^{2}=k^{2}\cdot \frac{4m^2-12}{3+4k^2}+km\left(\frac{-8km}{3+4k^2}\right)+m^{2}=\frac{3m^2-12k^2}{3+4k^2}，k_{PM}k_{PN}=\frac{y_1}{x_1+2}\cdot \frac{y_2}{x_2+2}=\frac{y_1y_2}{(x_1+2)(x_2+2)}=\frac{y_1y_2}{x_1x_2+2(x_1+x_2)+4}=\frac{3m^2-12k^2}{4m^2-16km+16k^2}=-\frac{1}{4}，解得m=2k(舍)或m=-k，若F在以MN为直径的圆内，则\overrightarrow{FM}\cdot \overrightarrow{FN}<0,即\overrightarrow{FM}\cdot \overrightarrow{FN}=(x_{1}+1,y_{1})\cdot (x_{2}+1,y_{2})=x_{1}x_{2}+x_{1}+x_{2}+1+y_{1}y_{2}<0,\frac{4m^2-12}{3+4k^2}-\frac{8km}{3+4k^2}+\frac{3m^2-12k^2}{3+4k^2}+1<0，即4k^{2}-12+8k^{2}+3k^{2}-12k^{2}+3+4k^{2}<0，即7k^{2}-9<0且k\neq 0，解得-\frac{3\sqrt{7}}{7}所以k的取值范围为\left(\frac{-3\sqrt{7}}{7},0\right)\cup \left(0,\frac{3\sqrt{7}}{7}\right).$

Final answer: $\left(\frac{-3\sqrt{7}}{7},0\right)\cup \left(0,\frac{3\sqrt{7}}{7}\right)$
Subfield: Plane Geometry


Question: $已知椭圆C的方程为\frac{x^2}{a^2}+\frac{y^2}{3}=1，斜率为k(k\neq 0)的直线与C相交于M，N两点.$<img_2371>
$在(1)的条件下，若P是椭圆C的左顶点，k_{PM}k_{PN}=-\frac{1}{4}，F是椭圆的左焦点，要使F在以MN为直径的圆内，求k的取值范围.$

Solution:
$设MN方程为y=kx+m,则\left\{\begin{matrix}3x^2+4y^2=12,\\ y=kx+m,\end{matrix}\right.(3+4k^{2})x^{2}+8kmx+4m^{2}-12=0,x_{1}+x_{2}=\frac{-8km}{3+4k^2},x_{1}x_{2}=\frac{4m^2-12}{3+4k^2}，y_{1}+y_{2}=k(x_{1}+x_{2})+2m=k\times \left(\frac{-8km}{3+4k^2}\right)+2m=\frac{6m}{3+4k^2},y_{1}y_{2}=(kx_{1}+m)\cdot (kx_{2}+m)=k^{2}x_{1}x_{2}+km(x_{1}+x_{2})+m^{2}=k^{2}\cdot \frac{4m^2-12}{3+4k^2}+km\left(\frac{-8km}{3+4k^2}\right)+m^{2}=\frac{3m^2-12k^2}{3+4k^2}，k_{PM}k_{PN}=\frac{y_1}{x_1+2}\cdot \frac{y_2}{x_2+2}=\frac{y_1y_2}{(x_1+2)(x_2+2)}=\frac{y_1y_2}{x_1x_2+2(x_1+x_2)+4}=\frac{3m^2-12k^2}{4m^2-16km+16k^2}=-\frac{1}{4}，解得m=2k(舍)或m=-k，若F在以MN为直径的圆内，则\overrightarrow{FM}\cdot \overrightarrow{FN}<0,即\overrightarrow{FM}\cdot \overrightarrow{FN}=(x_{1}+1,y_{1})\cdot (x_{2}+1,y_{2})=x_{1}x_{2}+x_{1}+x_{2}+1+y_{1}y_{2}<0,\frac{4m^2-12}{3+4k^2}-\frac{8km}{3+4k^2}+\frac{3m^2-12k^2}{3+4k^2}+1<0，即4k^{2}-12+8k^{2}+3k^{2}-12k^{2}+3+4k^{2}<0，即7k^{2}-9<0且k\neq 0，解得-\frac{3\sqrt{7}}{7}所以k的取值范围为\left(\frac{-3\sqrt{7}}{7},0\right)\cup \left(0,\frac{3\sqrt{7}}{7}\right).$



Final answer: $\left(\frac{-3\sqrt{7}}{7},0\right)\cup \left(0,\frac{3\sqrt{7}}{7}\right)$
Subfield: Plane Geometry


Question: $如图，在\triangle ABC中，A=\frac{2\pi }{3}，AC=\sqrt{2}，CD平分\angle ACB交AB于点D，CD=\sqrt{3}.$![Image](https://cdntk.53zaixian.com/tq/1716267809374924801/img/compression/M_88fd6729a2449795a6d88fdcdd3b105f.jpg)
$求\angle ADC的值；$

Solution:
$在\triangle ADC中，由正弦定理得\frac{CD}{\\\sin    A}=\frac{AC}{\\\sin   \angle ADC}，所以\\sin  \angle ADC=\frac{AC\cdot \\\sin    A}{CD}=\frac{\sqrt{2}\times \\\sin    \frac{2\pi }{3}}{\sqrt{3}}=\frac{\sqrt{2}}{2}，因为0<\angle ADC<\frac{\pi }{3}，所以\angle ADC=\frac{\pi }{4}.$

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


Question: $如图，在\triangle ABC中，A=\frac{2\pi }{3}，AC=\sqrt{2}，CD平分\angle ACB交AB于点D，CD=\sqrt{3}.$<img_1501>
$求\angle ADC的值；$

Solution:
$在\triangle ADC中，由正弦定理得\frac{CD}{\\\sin    A}=\frac{AC}{\\\sin   \angle ADC}，所以\\sin  \angle ADC=\frac{AC\cdot \\\sin    A}{CD}=\frac{\sqrt{2}\times \\\sin    \frac{2\pi }{3}}{\sqrt{3}}=\frac{\sqrt{2}}{2}，因为0<\angle ADC<\frac{\pi }{3}，所以\angle ADC=\frac{\pi }{4}.$

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


Question: $如图，在\triangle ABC中，A=\frac{2\pi }{3}，AC=\sqrt{2}，CD平分\angle ACB交AB于点D，CD=\sqrt{3}.$![Image](https://cdntk.53zaixian.com/tq/1716267809374924801/img/compression/M_88fd6729a2449795a6d88fdcdd3b105f.jpg)
$求\triangle BCD的面积.$

Solution:
$由(1)得\angle BCD=\angle ACD=\pi -\frac{2\pi }{3}-\frac{\pi }{4}=\frac{\pi }{12}，所以\angle ACB=\frac{\pi }{6}.又B=\pi -\frac{2\pi }{3}-\frac{\pi }{6}=\frac{\pi }{6},所以B=\angle ACB，即\triangle ABC为等腰三角形.所以BC=2\times AC\times \\cos   \frac{\pi }{6}=\sqrt{6}，\\sin   \frac{\pi }{12}=\\sin  \left(\frac{\pi }{3}-\frac{\pi }{4}\right)=\frac{\sqrt{3}}{2}\times \frac{\sqrt{2}}{2}-\frac{1}{2}\times \frac{\sqrt{2}}{2}=\frac{\sqrt{6}-\sqrt{2}}{4}，所以S_{\triangle BCD}=\frac{1}{2}BC\cdot CD\cdot \\sin  \angle BCD=\frac{1}{2}\times \sqrt{6}\times \sqrt{3}\times \\sin   \frac{\pi }{12}=\frac{3(\sqrt{3}-1)}{4}.$

Final answer: $\frac{3(\sqrt{3}-1)}{4}$
Subfield: Plane Geometry


Question: $如图，在\triangle ABC中，A=\frac{2\pi }{3}，AC=\sqrt{2}，CD平分\angle ACB交AB于点D，CD=\sqrt{3}.$<img_1501>
$求\triangle BCD的面积.$

Solution:
$由(1)得\angle BCD=\angle ACD=\pi -\frac{2\pi }{3}-\frac{\pi }{4}=\frac{\pi }{12}，所以\angle ACB=\frac{\pi }{6}.又B=\pi -\frac{2\pi }{3}-\frac{\pi }{6}=\frac{\pi }{6},所以B=\angle ACB，即\triangle ABC为等腰三角形.所以BC=2\times AC\times \\cos   \frac{\pi }{6}=\sqrt{6}，\\sin   \frac{\pi }{12}=\\sin  \left(\frac{\pi }{3}-\frac{\pi }{4}\right)=\frac{\sqrt{3}}{2}\times \frac{\sqrt{2}}{2}-\frac{1}{2}\times \frac{\sqrt{2}}{2}=\frac{\sqrt{6}-\sqrt{2}}{4}，所以S_{\triangle BCD}=\frac{1}{2}BC\cdot CD\cdot \\sin  \angle BCD=\frac{1}{2}\times \sqrt{6}\times \sqrt{3}\times \\sin   \frac{\pi }{12}=\frac{3(\sqrt{3}-1)}{4}.$

Final answer: $\frac{3(\sqrt{3}-1)}{4}$
Subfield: Plane Geometry


Question: $如图，在四边形ABCD中，AB=1，CD=DA=2，BC=3，从条件①、条件②中选择一个作为已知，解决下列问题.
条件①：\\cos  \angle DBC=\frac{\sqrt{5}}{3}；条件②：\angle DCB+\angle DAB=\pi .注：如果选择条件①和条件②分别解答，按第一个解答计分.$<img_2300>
$求BD的长；$

Solution:
$选条件①
由余弦定理的推论得\\cos  \angle DBC=\frac{BD^2+BC^2-CD^2}{2BD\cdot BC}=\frac{BD^2+9-4}{6BD}=\frac{\sqrt{5}}{3}，所以BD=\sqrt{5}.
选条件②
在\triangle ABD中，由余弦定理的推论得\\cos  \angle BAD=\frac{AB^2+AD^2-BD^2}{2AB\cdot AD}=\frac{1+4-BD^2}{4}=\frac{5-BD^2}{4}，在\triangle BCD中，由余弦定理的推论得\\cos  \angle BCD=\frac{BC^2+CD^2-BD^2}{2BC\cdot CD}=\frac{9+4-BD^2}{12}=\frac{13-BD^2}{12}，因为\angle DCB+\angle DAB=\pi ，所以\\cos  \angle BAD+\cos  \angle BCD=0，即\frac{5-BD^2}{4}+\frac{13-BD^2}{12}=0，所以BD=\sqrt{7}(舍负).$

Final answer: $\sqrt{5}$, $\sqrt{7}$
Subfield: Plane Geometry


Question: $如图，在四边形ABCD中，AB=1，CD=DA=2，BC=3，从条件①、条件②中选择一个作为已知，解决下列问题.
条件①：\\cos  \angle DBC=\frac{\sqrt{5}}{3}；条件②：\angle DCB+\angle DAB=\pi .注：如果选择条件①和条件②分别解答，按第一个解答计分.$![Image](https://cdntk.53zaixian.com/tq/1716267810087956482/img/compression/M_62155b514d2e0a362c46a3465c8a898e.jpg)
$求BD的长；$

Solution:
$选条件①
由余弦定理的推论得\\cos  \angle DBC=\frac{BD^2+BC^2-CD^2}{2BD\cdot BC}=\frac{BD^2+9-4}{6BD}=\frac{\sqrt{5}}{3}，所以BD=\sqrt{5}.
选条件②
在\triangle ABD中，由余弦定理的推论得\\cos  \angle BAD=\frac{AB^2+AD^2-BD^2}{2AB\cdot AD}=\frac{1+4-BD^2}{4}=\frac{5-BD^2}{4}，在\triangle BCD中，由余弦定理的推论得\\cos  \angle BCD=\frac{BC^2+CD^2-BD^2}{2BC\cdot CD}=\frac{9+4-BD^2}{12}=\frac{13-BD^2}{12}，因为\angle DCB+\angle DAB=\pi ，所以\\cos  \angle BAD+\cos  \angle BCD=0，即\frac{5-BD^2}{4}+\frac{13-BD^2}{12}=0，所以BD=\sqrt{7}(舍负).$

Final answer: $\sqrt{5}$, $\sqrt{7}$
Subfield: Plane Geometry


Question: $如图，在四边形ABCD中，AB=1，CD=DA=2，BC=3，从条件①、条件②中选择一个作为已知，解决下列问题.
条件①：\\cos  \angle DBC=\frac{\sqrt{5}}{3}；条件②：\angle DCB+\angle DAB=\pi .注：如果选择条件①和条件②分别解答，按第一个解答计分.$<img_2300>
$求四边形ABCD的面积.$

Solution:
$选条件①
由\angle DBC\in (0,\pi ),\\cos  \angle DBC=\frac{\sqrt{5}}{3},得\\sin  \angle DBC=\sqrt{1-\\\cos   ^2\angle DBC}=\frac{2}{3}，故S_{\triangle BCD}=\frac{1}{2}BD\cdot BC\\sin  \angle DBC=\frac{1}{2}\times \sqrt{5}\times 3\times \frac{2}{3}=\sqrt{5}.在\triangle ABD中，AB^{2}+AD^{2}=BD^{2}，所以AB\perp AD，故S_{\triangle ABD}=\frac{1}{2}AB\cdot AD=\frac{1}{2}\times 1\times 2=1，所以四边形ABCD的面积S=S_{\triangle BCD}+S_{\triangle ABD}=\sqrt{5}+1.
选条件②
由余弦定理的推论得\\cos  \angle BAD=\frac{2^2+1^2-(\sqrt{7})^2}{2\times 2\times 1}=-\frac{1}{2}，得\angle BAD=\frac{2\pi }{3}，因为\angle DCB+\angle DAB=\pi ，所以\angle BCD=\frac{\pi }{3}，故S_{\triangle ABD}=\frac{1}{2}AB\cdot AD\\sin  \angle BAD=\frac{1}{2}\times 1\times 2\times \frac{\sqrt{3}}{2}=\frac{\sqrt{3}}{2}，S_{\triangle BCD}=\frac{1}{2}BC\cdot CD\\sin  \angle BCD=\frac{1}{2}\times 3\times 2\times \frac{\sqrt{3}}{2}=\frac{3\sqrt{3}}{2}，故四边形ABCD的面积S=S_{\triangle ABD}+S_{\triangle BCD}=\frac{\sqrt{3}}{2}+\frac{3\sqrt{3}}{2}=2\sqrt{3}.$

Final answer: $\sqrt{5}$, $2\sqrt{3}$
Subfield: Plane Geometry


Question: $如图，在四边形ABCD中，AB=1，CD=DA=2，BC=3，从条件①、条件②中选择一个作为已知，解决下列问题.
条件①：\\cos  \angle DBC=\frac{\sqrt{5}}{3}；条件②：\angle DCB+\angle DAB=\pi .注：如果选择条件①和条件②分别解答，按第一个解答计分.$![Image](https://cdntk.53zaixian.com/tq/1716267810087956482/img/compression/M_62155b514d2e0a362c46a3465c8a898e.jpg)
$求四边形ABCD的面积.$

Solution:
$选条件①
由\angle DBC\in (0,\pi ),\\cos  \angle DBC=\frac{\sqrt{5}}{3},得\\sin  \angle DBC=\sqrt{1-\\\cos   ^2\angle DBC}=\frac{2}{3}，故S_{\triangle BCD}=\frac{1}{2}BD\cdot BC\\sin  \angle DBC=\frac{1}{2}\times \sqrt{5}\times 3\times \frac{2}{3}=\sqrt{5}.在\triangle ABD中，AB^{2}+AD^{2}=BD^{2}，所以AB\perp AD，故S_{\triangle ABD}=\frac{1}{2}AB\cdot AD=\frac{1}{2}\times 1\times 2=1，所以四边形ABCD的面积S=S_{\triangle BCD}+S_{\triangle ABD}=\sqrt{5}+1.
选条件②
由余弦定理的推论得\\cos  \angle BAD=\frac{2^2+1^2-(\sqrt{7})^2}{2\times 2\times 1}=-\frac{1}{2}，得\angle BAD=\frac{2\pi }{3}，因为\angle DCB+\angle DAB=\pi ，所以\angle BCD=\frac{\pi }{3}，故S_{\triangle ABD}=\frac{1}{2}AB\cdot AD\\sin  \angle BAD=\frac{1}{2}\times 1\times 2\times \frac{\sqrt{3}}{2}=\frac{\sqrt{3}}{2}，S_{\triangle BCD}=\frac{1}{2}BC\cdot CD\\sin  \angle BCD=\frac{1}{2}\times 3\times 2\times \frac{\sqrt{3}}{2}=\frac{3\sqrt{3}}{2}，故四边形ABCD的面积S=S_{\triangle ABD}+S_{\triangle BCD}=\frac{\sqrt{3}}{2}+\frac{3\sqrt{3}}{2}=2\sqrt{3}.$

Final answer: $\sqrt{5}$, $2\sqrt{3}$
Subfield: Plane Geometry


Question: $如图，在四棱锥P-ABCD中，PC\perp 平面ABCD，AB\parallel DC，DC\perp AC.$<img_1757>
$求证：DC\perp 平面PAC；$

Solution:
$证明：因为PC\perp 平面ABCD，且DC\subset 平面ABCD，所以PC\perp DC.又因为DC\perp AC，AC\cap PC=C，AC，PC\subset 平面PAC，所以DC\perp 平面PAC.$
Subfield: Plane Geometry


Question: $如图，在四棱锥P-ABCD中，PC\perp 平面ABCD，AB\parallel DC，DC\perp AC.$![Image](https://cdntk.53zaixian.com/tq/1716268011443908610/img/compression/M_5ce588d6b6725207da4f9125deb30d96.jpg)
$求证：DC\perp 平面PAC；$

Solution:
$证明：因为PC\perp 平面ABCD，且DC\subset 平面ABCD，所以PC\perp DC.又因为DC\perp AC，AC\cap PC=C，AC，PC\subset 平面PAC，所以DC\perp 平面PAC.$

Final answer: 证明题
Subfield: Plane Geometry


Question: $如图，在四棱锥P-ABCD中，PC\perp 平面ABCD，AB\parallel DC，DC\perp AC.$<img_1757>
$求证：平面PAB\perp 平面PAC；$

Solution:
$证明：因为AB\parallel DC，DC\perp AC，所以AB\perp AC.因为PC\perp 平面ABCD，AB\subset 平面ABCD，所以PC\perp AB.又AC\cap PC=C，AC，PC\subset 平面PAC，所以AB\perp 平面PAC.又AB\subset 平面PAB，所以平面PAB\perp 平面PAC.$
Subfield: Plane Geometry


Question: $如图，在四棱锥P-ABCD中，PC\perp 平面ABCD，AB\parallel DC，DC\perp AC.$![Image](https://cdntk.53zaixian.com/tq/1716268011443908610/img/compression/M_5ce588d6b6725207da4f9125deb30d96.jpg)
$求证：平面PAB\perp 平面PAC；$

Solution:
$证明：因为AB\parallel DC，DC\perp AC，所以AB\perp AC.因为PC\perp 平面ABCD，AB\subset 平面ABCD，所以PC\perp AB.又AC\cap PC=C，AC，PC\subset 平面PAC，所以AB\perp 平面PAC.又AB\subset 平面PAB，所以平面PAB\perp 平面PAC.$

Final answer: 证明题
Subfield: Plane Geometry


Question: $如图，在四棱锥P-ABCD中，PC\perp 平面ABCD，AB\parallel DC，DC\perp AC.$<img_1757>
$设点E为AB的中点.在棱PB上是否存在点F，使得PA\parallel 平面CEF？说明理由.$

Solution:
$在棱PB上存在点F，使得PA\parallel 平面CEF.理由如下：如图，取PB的中点F，连接EF，CE，CF.$<img_1949>$因为E为AB的中点，所以EF\parallel PA.又因为PA\not\subset 平面CEF，EF\subset 平面CEF，所以PA\parallel 平面CEF.$



Final answer: $在棱PB上存在点F，使得PA\parallel 平面CEF。$
Subfield: Plane Geometry


Question: $如图，在四棱锥P-ABCD中，PC\perp 平面ABCD，AB\parallel DC，DC\perp AC.$![Image](https://cdntk.53zaixian.com/tq/1716268011443908610/img/compression/M_5ce588d6b6725207da4f9125deb30d96.jpg)
$设点E为AB的中点.在棱PB上是否存在点F，使得PA\parallel 平面CEF？说明理由.$

Solution:
$在棱PB上存在点F，使得PA\parallel 平面CEF.理由如下：如图，取PB的中点F，连接EF，CE，CF.$![Image](https://cdntk.53zaixian.com/tq/1716268012018528258/img/compression/M_d223a47c448d0497d4d56e4d734a1d89.jpg)$因为E为AB的中点，所以EF\parallel PA.又因为PA\not\subset 平面CEF，EF\subset 平面CEF，所以PA\parallel 平面CEF.$

Final answer: $在棱PB上存在点F，使得PA\parallel 平面CEF。$
Subfield: Plane Geometry


Question: $如图，直角三角形ABC和等边三角形ABD所在平面互相垂直，AB=AC=2，E是线段AD上一点.$![Image](https://cdntk.53zaixian.com/tq/1716268034375544835/img/compression/M_549f66340c6ba88b4f30e820fae38055.jpg)
$设E为AD的中点，求证：BE\perp CD；$

Solution:
$证明：由题意知AB\perp AC.因为平面ABC\perp 平面ABD，平面ABC\cap 平面ABD=AB，所以AC\perp 平面ABD.因为BE\subset 平面ABD，所以AC\perp BE.因为\triangle ABD为等边三角形，E是AD的中点，所以AD\perp BE.因为AC\cap AD=A，所以BE\perp 平面ACD.因为CD\subset 平面ACD,所以BE\perp CD.$

Final answer: 证明题