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


Question: $如图，椭圆C：\frac{x^2}{a^2}+\frac{y^2}{b^2}=1(a>b>0)的离心率是\frac{1}{2}，短轴长为2\sqrt{3}，椭圆的左、右顶点分别为A_{1}，A_{2}，过椭圆C与抛物线E:y^{2}=2px(p>0)的公共焦点F的直线l与椭圆相交于A,B两点，与抛物线E相交于P,Q两点，点M为PQ的中点.$![Image](https://cdntk.53zaixian.com/tq/1658643478329769985/img/compression/M_c31a562730fc76d0fb519e8c44705c00.jpg)
$记\triangle ABA_{1}的面积为S_{1}，\triangle MA_{2}Q的面积为S_{2}，若S_{1}\geq 3S_{2}，求直线l在y轴上的截距的取值范围.$

Solution:
$由题可设l:x=ty+1(t\neq 0),A(x_{1},y_{1}),B(x_{2},y_{2}),P(x_{3},y_{3}),Q(x_{4},y_{4})，联立l与椭圆C的方程，得\left\{\begin{matrix}x=ty+1,\\ \frac{x^2}{4}+\frac{y^2}{3}=1,\end{matrix}\right.消去x并整理，得(3t^{2}+4)y^{2}+6ty－9=0，则\Delta =(6t)^{2}－4(3t^{2}+4)\times (－9)=144(t^{2}+1)>0,y_{1}+y_{2}=\frac{-6t}{3t^2+4},y_{1}y_{2}=\frac{-9}{3t^2+4},由弦长公式得|AB|=\sqrt{1+t^2}|y_{1}－y_{2}|=\sqrt{1+t^2}\cdot \sqrt{(y_1+y_2)^2-4y_1y_2}=\sqrt{1+t^2}\cdot \frac{\sqrt{144(t^2+1)}}{3t^2+4}=\frac{12(t^2+1)}{3t^2+4},点A_{1}(－2,0)到直线l的距离为\frac{3}{\sqrt{1+t^2}},所以S_{1}=\frac{1}{2}|AB|\cdot \frac{3}{\sqrt{1+t^2}}=\frac{18\sqrt{1+t^2}}{3t^2+4}.联立l与抛物线E的方程，得\left\{\begin{matrix}y^2=4x,\\ x=ty+1,\end{matrix}\right.消去x并整理，得y^{2}－4ty－4=0，则\Delta =(－4t)^{2}－4\times (－4)=16(t^{2}+1)>0,y_{3}+y_{4}=4t,y_{3}y_{4}=-4,由弦长公式得|PQ|=\sqrt{1+t^2}|y_{3}－y_{4}|=\sqrt{1+t^2}\cdot \sqrt{(y_3+y_4)^2-4y_3y_4}=\sqrt{1+t^2}\sqrt{16(1+t^2)}=4(1+t^{2})，点A_{2}(2,0)到直线l的距离为\frac{1}{\sqrt{1+t^2}},所以S_{2}=\frac{1}{2}S_{\triangle PQA_{2}}=\frac{1}{2}\cdot \frac{1}{2}\cdot |PQ|\cdot \frac{1}{\sqrt{1+t^2}}=\sqrt{1+t^2}.因为S_{1}\geq 3S_{2}，所以\frac{18\sqrt{1+t^2}}{3t^2+4}\geq 3\sqrt{1+t^2}，解得－\frac{\sqrt{6}}{3}\leq t\leq \frac{\sqrt{6}}{3}.易知直线l在y轴上的截距为-\frac{1}{t},且－\frac{1}{t}\leq －\frac{\sqrt{6}}{2}或－\frac{1}{t}\geq \frac{\sqrt{6}}{2}，所以直线l在y轴上的截距的取值范围为\left(-\infty ,-\frac{\sqrt{6}}{2}\right]\cup \left[\frac{\sqrt{6}}{2},+\infty \right).$

Final answer: $(-\infty ,-\frac{\sqrt{6}}{2}]\cup [\frac{\sqrt{6}}{2},+\infty )$
Subfield: Plane Geometry


Question: $如图，椭圆C：\frac{x^2}{a^2}+\frac{y^2}{b^2}=1(a>b>0)的离心率是\frac{1}{2}，短轴长为2\sqrt{3}，椭圆的左、右顶点分别为A_{1}，A_{2}，过椭圆C与抛物线E:y^{2}=2px(p>0)的公共焦点F的直线l与椭圆相交于A,B两点，与抛物线E相交于P,Q两点，点M为PQ的中点.$<img_2928>
$记\triangle ABA_{1}的面积为S_{1}，\triangle MA_{2}Q的面积为S_{2}，若S_{1}\geq 3S_{2}，求直线l在y轴上的截距的取值范围.$

Solution:
$由题可设l:x=ty+1(t\neq 0),A(x_{1},y_{1}),B(x_{2},y_{2}),P(x_{3},y_{3}),Q(x_{4},y_{4})，联立l与椭圆C的方程，得\left\{\begin{matrix}x=ty+1,\\ \frac{x^2}{4}+\frac{y^2}{3}=1,\end{matrix}\right.消去x并整理，得(3t^{2}+4)y^{2}+6ty－9=0，则\Delta =(6t)^{2}－4(3t^{2}+4)\times (－9)=144(t^{2}+1)>0,y_{1}+y_{2}=\frac{-6t}{3t^2+4},y_{1}y_{2}=\frac{-9}{3t^2+4},由弦长公式得|AB|=\sqrt{1+t^2}|y_{1}－y_{2}|=\sqrt{1+t^2}\cdot \sqrt{(y_1+y_2)^2-4y_1y_2}=\sqrt{1+t^2}\cdot \frac{\sqrt{144(t^2+1)}}{3t^2+4}=\frac{12(t^2+1)}{3t^2+4},点A_{1}(－2,0)到直线l的距离为\frac{3}{\sqrt{1+t^2}},所以S_{1}=\frac{1}{2}|AB|\cdot \frac{3}{\sqrt{1+t^2}}=\frac{18\sqrt{1+t^2}}{3t^2+4}.联立l与抛物线E的方程，得\left\{\begin{matrix}y^2=4x,\\ x=ty+1,\end{matrix}\right.消去x并整理，得y^{2}－4ty－4=0，则\Delta =(－4t)^{2}－4\times (－4)=16(t^{2}+1)>0,y_{3}+y_{4}=4t,y_{3}y_{4}=-4,由弦长公式得|PQ|=\sqrt{1+t^2}|y_{3}－y_{4}|=\sqrt{1+t^2}\cdot \sqrt{(y_3+y_4)^2-4y_3y_4}=\sqrt{1+t^2}\sqrt{16(1+t^2)}=4(1+t^{2})，点A_{2}(2,0)到直线l的距离为\frac{1}{\sqrt{1+t^2}},所以S_{2}=\frac{1}{2}S_{\triangle PQA_{2}}=\frac{1}{2}\cdot \frac{1}{2}\cdot |PQ|\cdot \frac{1}{\sqrt{1+t^2}}=\sqrt{1+t^2}.因为S_{1}\geq 3S_{2}，所以\frac{18\sqrt{1+t^2}}{3t^2+4}\geq 3\sqrt{1+t^2}，解得－\frac{\sqrt{6}}{3}\leq t\leq \frac{\sqrt{6}}{3}.易知直线l在y轴上的截距为-\frac{1}{t},且－\frac{1}{t}\leq －\frac{\sqrt{6}}{2}或－\frac{1}{t}\geq \frac{\sqrt{6}}{2}，所以直线l在y轴上的截距的取值范围为\left(-\infty ,-\frac{\sqrt{6}}{2}\right]\cup \left[\frac{\sqrt{6}}{2},+\infty \right).$

Final answer: $(-\infty ,-\frac{\sqrt{6}}{2}]\cup [\frac{\sqrt{6}}{2},+\infty )$
Subfield: Plane Geometry


Question: $在一张纸上有一个圆C：(x+\sqrt{5})^{2}+y^{2}=4，定点G(\sqrt{5},0)，折叠纸片使圆C上某一点G_{1}恰好与点G重合，这样每次折叠都会留下一条折痕PQ，设直线PQ与直线G_{1}C的交点为T.$![Image](https://cdntk.53zaixian.com/tq/1658391115434115073/img/compression/M_c024a94943d7526abd09fc204d246911.jpg)
$连接TG,求证：||TC|－|TG||为定值，并求出点T的轨迹C_{1}的方程；$

Solution:
$由题意得|TG|=|TG_{1}|，所以||TC|－|TG||=||TC|－|TG_{1}||=2,故||TC|-|TG||为定值.因为2<|CG|=2\sqrt{5},所以点T的轨迹是以C，G为焦点，实轴长为2的双曲线，(2分)则c=\sqrt{5}，a=1，所以b=\sqrt{c^2-a^2}=2，(4分)所以C_{1}的方程为x^{2}－\frac{y^2}{4}=1.(5分)$

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


Question: $在一张纸上有一个圆C：(x+\sqrt{5})^{2}+y^{2}=4，定点G(\sqrt{5},0)，折叠纸片使圆C上某一点G_{1}恰好与点G重合，这样每次折叠都会留下一条折痕PQ，设直线PQ与直线G_{1}C的交点为T.$<img_1626>
$连接TG,求证：||TC|－|TG||为定值，并求出点T的轨迹C_{1}的方程；$

Solution:
$由题意得|TG|=|TG_{1}|，所以||TC|－|TG||=||TC|－|TG_{1}||=2,故||TC|-|TG||为定值.因为2<|CG|=2\sqrt{5},所以点T的轨迹是以C，G为焦点，实轴长为2的双曲线，(2分)则c=\sqrt{5}，a=1，所以b=\sqrt{c^2-a^2}=2，(4分)所以C_{1}的方程为x^{2}－\frac{y^2}{4}=1.(5分)$

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


Question: $在一张纸上有一个圆C：(x+\sqrt{5})^{2}+y^{2}=4，定点G(\sqrt{5},0)，折叠纸片使圆C上某一点G_{1}恰好与点G重合，这样每次折叠都会留下一条折痕PQ，设直线PQ与直线G_{1}C的交点为T.$![Image](https://cdntk.53zaixian.com/tq/1658391115434115073/img/compression/M_c024a94943d7526abd09fc204d246911.jpg)
$设A(－1,0)，M为曲线C_{1}上一点，N为圆x^{2}+y^{2}=1上一点(M，N均不在x轴上),直线AM，AN的斜率分别记为k_{1}，k_{2}，且k_{2}=－\frac{1}{4}k_{1}，求证：直线MN过定点，并求出此定点的坐标.$

Solution:
$设A(x_{A},y_{A}),M(x_{M},y_{M}),N(x_{N},y_{N}).由已知得l_{AM}：y=k_{1}(x+1)，l_{AN}：y=k_{2}(x+1)，易得点A在曲线C_{1}上，联立\left\{\begin{matrix}y=k_1(x+1),\\ x^2-\frac{y^2}{4}=1,\end{matrix}\right.消去y，整理得(4－k^2_1)x^{2}－2k^2_1x－k^2_1－4=0，则x_{A}x_{M}=\frac{-k^2_1-4}{4-k^2_1}，所以x_{M}=\frac{k^2_1+4}{4-k^2_1}，故y_{M}=k_{1}(x_{M}+1)=\frac{8k_1}{4-k^2_1}，所以M\left(\frac{k^2_1+4}{4-k^2_1},\frac{8k_1}{4-k^2_1}\right).(8分)易得点A在圆x^{2}+y^{2}=1上，联立\left\{\begin{matrix}y=k_2(x+1),\\ x^2+y^2=1,\end{matrix}\right.消去y，整理得(1+k^2_2)x^{2}+2k^2_2x+k^2_2－1=0，则x_{A}x_{N}=\frac{k^2_2-1}{1+k^2_2}，所以x_{N}=\frac{-k^2_2+1}{1+k^2_2}，故y_{N}=k_{2}(x_{N}+1)=\frac{2k_2}{1+k^2_2}，因为k_{2}=－\frac{1}{4}k_{1}，所以N\left(\frac{-k^2_1+16}{16+k^2_1},\frac{-8k_1}{16+k^2_1}\right).(10分)若直线MN过定点，则由对称性得定点在x轴上，设定点为S(s,0)，由三点共线得k_{MS}=k_{NS}，即\frac{\frac{8k_1}{4-k^2_1}}{\frac{k^2_1+4}{4-k^2_1}-s}=\frac{\frac{-8k_1}{16+k^2_1}}{\frac{-k^2_1+16}{16+k^2_1}-s},整理得k^2_1+4+(k^2_1－4)s=k^2_1－16+(k^2_1+16)s,解得s=1，即S(1,0).所以直线MN过定点S(1,0).(12分)$

Final answer: 证明题
Subfield: Plane Geometry


Question: $在一张纸上有一个圆C：(x+\sqrt{5})^{2}+y^{2}=4，定点G(\sqrt{5},0)，折叠纸片使圆C上某一点G_{1}恰好与点G重合，这样每次折叠都会留下一条折痕PQ，设直线PQ与直线G_{1}C的交点为T.$<img_1626>
$设A(－1,0)，M为曲线C_{1}上一点，N为圆x^{2}+y^{2}=1上一点(M，N均不在x轴上),直线AM，AN的斜率分别记为k_{1}，k_{2}，且k_{2}=－\frac{1}{4}k_{1}，求证：直线MN过定点，并求出此定点的坐标.$

Solution:
$设A(x_{A},y_{A}),M(x_{M},y_{M}),N(x_{N},y_{N}).由已知得l_{AM}：y=k_{1}(x+1)，l_{AN}：y=k_{2}(x+1)，易得点A在曲线C_{1}上，联立\left\{\begin{matrix}y=k_1(x+1),\\ x^2-\frac{y^2}{4}=1,\end{matrix}\right.消去y，整理得(4－k^2_1)x^{2}－2k^2_1x－k^2_1－4=0，则x_{A}x_{M}=\frac{-k^2_1-4}{4-k^2_1}，所以x_{M}=\frac{k^2_1+4}{4-k^2_1}，故y_{M}=k_{1}(x_{M}+1)=\frac{8k_1}{4-k^2_1}，所以M\left(\frac{k^2_1+4}{4-k^2_1},\frac{8k_1}{4-k^2_1}\right).(8分)易得点A在圆x^{2}+y^{2}=1上，联立\left\{\begin{matrix}y=k_2(x+1),\\ x^2+y^2=1,\end{matrix}\right.消去y，整理得(1+k^2_2)x^{2}+2k^2_2x+k^2_2－1=0，则x_{A}x_{N}=\frac{k^2_2-1}{1+k^2_2}，所以x_{N}=\frac{-k^2_2+1}{1+k^2_2}，故y_{N}=k_{2}(x_{N}+1)=\frac{2k_2}{1+k^2_2}，因为k_{2}=－\frac{1}{4}k_{1}，所以N\left(\frac{-k^2_1+16}{16+k^2_1},\frac{-8k_1}{16+k^2_1}\right).(10分)若直线MN过定点，则由对称性得定点在x轴上，设定点为S(s,0)，由三点共线得k_{MS}=k_{NS}，即\frac{\frac{8k_1}{4-k^2_1}}{\frac{k^2_1+4}{4-k^2_1}-s}=\frac{\frac{-8k_1}{16+k^2_1}}{\frac{-k^2_1+16}{16+k^2_1}-s},整理得k^2_1+4+(k^2_1－4)s=k^2_1－16+(k^2_1+16)s,解得s=1，即S(1,0).所以直线MN过定点S(1,0).(12分)$
Subfield: Plane Geometry


Question: $如图，已知圆C_{1}:(x－1)^{2}+(y+1)^{2}=\frac{1}{4}和抛物线C_{2}:x^{2}=4y,P(x_{0},y_{0})是圆C_{1}上一点，M是抛物线C_{2}上一点，F是抛物线C_{2}的焦点.$<img_1864>
$当直线PM与圆C_{1}相切，且|PM|=|FM|时，求x_{0}的值；$

Solution:
$易得焦点F的坐标为(0,1)，设M\left(x_M,\frac{x^2_M}{4}\right)，则|PM|=\sqrt{|MC_1|^2-|PC_1|^2}=\sqrt{(x_M-1)^2+\left(\frac{x^2_M}{4}+1\right)^2-\frac{1}{4}}，由抛物线的定义，得M到焦点F的距离等于其到抛物线准线y=－1的距离，所以|FM|=\frac{x^2_M}{4}+1，(2分)由|PM|=|FM|，得\sqrt{(x_M-1)^2+\left(\frac{x^2_M}{4}+1\right)^2-\frac{1}{4}}=\frac{x^2_M}{4}+1，所以x_{M}=\frac{1}{2}或x_{M}=\frac{3}{2}，所以M\left(\frac{1}{2},\frac{1}{16}\right)或M\left(\frac{3}{2},\frac{9}{16}\right)，此时直线PM与准线y=－1垂直，所以x_{0}=\frac{1}{2}或x_{0}=\frac{3}{2}.(5分)$

Final answer: $\frac{1}{2}, \frac{3}{2}$
Subfield: Plane Geometry


Question: $如图，已知圆C_{1}:(x－1)^{2}+(y+1)^{2}=\frac{1}{4}和抛物线C_{2}:x^{2}=4y,P(x_{0},y_{0})是圆C_{1}上一点，M是抛物线C_{2}上一点，F是抛物线C_{2}的焦点.$![Image](https://cdntk.53zaixian.com/tq/1658391115761270786/img/compression/M_e9527e6113faf830c1e23d01ff089048.jpg)
$当直线PM与圆C_{1}相切，且|PM|=|FM|时，求x_{0}的值；$

Solution:
$易得焦点F的坐标为(0,1)，设M\left(x_M,\frac{x^2_M}{4}\right)，则|PM|=\sqrt{|MC_1|^2-|PC_1|^2}=\sqrt{(x_M-1)^2+\left(\frac{x^2_M}{4}+1\right)^2-\frac{1}{4}}，由抛物线的定义，得M到焦点F的距离等于其到抛物线准线y=－1的距离，所以|FM|=\frac{x^2_M}{4}+1，(2分)由|PM|=|FM|，得\sqrt{(x_M-1)^2+\left(\frac{x^2_M}{4}+1\right)^2-\frac{1}{4}}=\frac{x^2_M}{4}+1，所以x_{M}=\frac{1}{2}或x_{M}=\frac{3}{2}，所以M\left(\frac{1}{2},\frac{1}{16}\right)或M\left(\frac{3}{2},\frac{9}{16}\right)，此时直线PM与准线y=－1垂直，所以x_{0}=\frac{1}{2}或x_{0}=\frac{3}{2}.(5分)$

Final answer: $\frac{1}{2}, \frac{3}{2}$
Subfield: Plane Geometry


Question: $如图，已知圆C_{1}:(x－1)^{2}+(y+1)^{2}=\frac{1}{4}和抛物线C_{2}:x^{2}=4y,P(x_{0},y_{0})是圆C_{1}上一点，M是抛物线C_{2}上一点，F是抛物线C_{2}的焦点.$<img_1864>
$过P作抛物线C_{2}的两条切线PA，PB，A，B为切点，证明：存在两个x_{0}，使得\triangle PAB的面积等于\frac{3\sqrt{3}}{2}.$

Solution:
$证明：由P(x_{0},y_{0})是圆C_{1}上一点，得(x_{0}－1)^{2}+(y_{0}+1)^{2}=\frac{1}{4}，设直线PA的方程为y－y_{0}=k_{1}(x－x_{0})，与x^{2}=4y联立，得x^{2}－4k_{1}x－4(y_{0}－k_{1}x_{0})=0(*),令\Delta =16k^2_1+16(y_{0}－k_{1}x_{0})=0，整理得k^2_1－k_{1}x_{0}+y_{0}=0①，设直线PB的方程为y－y_{0}=k_{2}(x－x_{0})，同理得k^2_2－k_{2}x_{0}+y_{0}=0②，由①②知，k_{1},k_{2}是方程k^{2}－kx_{0}+y_{0}=0的两根，所以k_{1}+k_{2}=x_{0},k_{1}k_{2}=y_{0}，(7分)由(*)式得，x_{A}+x_{A}=4k_{1}，则x_{A}=2k_{1}，所以A(2k_{1},k^2_1)，同理B(2k_{2},k^2_2),所以k_{AB}=\frac{k^2_1-k^2_2}{2k_1-2k_2}=\frac{k_1+k_2}{2}，(8分)所以直线AB的方程为y－k^2_1=\frac{k_1+k_2}{2}(x－2k_{1})，即y=\frac{k_1+k_1}{2}x－k_{1}k_{2}，即y=\frac{x_0}{2}x－y_{0}.所以|AB|=\sqrt{1+\frac{x^2_0}{4}}|2k_{1}－2k_{2}|=\sqrt{4+x^2_0}\cdot \sqrt{(k_1+k_2)^2-4k_1k_2}=\sqrt{4+x^2_0}\cdot \sqrt{x^2_0-4y_0}，P(x_{0},y_{0})到直线AB的距离d=\frac{|x^2_0-4y_0|}{\sqrt{4+x^2_0}},所以S_{\triangle PAB}=\frac{1}{2}|AB|\cdot d=\frac{1}{2}\sqrt{(x^2_0-4y_0)^3}=\frac{3\sqrt{3}}{2}，(10分)整理得x^2_0－4y_{0}=3，联立\left\{\begin{matrix}(x_0-1)^2+(y_0+1)^2=\frac{1}{4},\\ x^2_0-4y_0=3,\end{matrix}\right.得(x_{0}－1)(x^3_0+x^2_0+19x_{0}－13)=0,所以x_{0}=1或x^3_0+x^2_0+19x_{0}－13=0，设f(x_{0})=x^3_0+x^2_0+19x_{0}－13，\frac{1}{2}\leq x_{0}\leq \frac{3}{2},显然f\left(\frac{1}{2}\right)<0,f(1)>0,f\left(\frac{3}{2}\right)>0，易知f(x_{0})在\left[\frac{1}{2},\frac{3}{2}\right]上单调递增，所以f(x_{0})=x^3_0+x^2_0+19x_{0}－13在\left(\frac{1}{2},1\right)上有唯一零点.所以存在两个x_{0}，使得\triangle PAB的面积等于\frac{3\sqrt{3}}{2}.(12分)$
Subfield: Plane Geometry


Question: $如图，已知圆C_{1}:(x－1)^{2}+(y+1)^{2}=\frac{1}{4}和抛物线C_{2}:x^{2}=4y,P(x_{0},y_{0})是圆C_{1}上一点，M是抛物线C_{2}上一点，F是抛物线C_{2}的焦点.$![Image](https://cdntk.53zaixian.com/tq/1658391115761270786/img/compression/M_e9527e6113faf830c1e23d01ff089048.jpg)
$过P作抛物线C_{2}的两条切线PA，PB，A，B为切点，证明：存在两个x_{0}，使得\triangle PAB的面积等于\frac{3\sqrt{3}}{2}.$

Solution:
$证明：由P(x_{0},y_{0})是圆C_{1}上一点，得(x_{0}－1)^{2}+(y_{0}+1)^{2}=\frac{1}{4}，设直线PA的方程为y－y_{0}=k_{1}(x－x_{0})，与x^{2}=4y联立，得x^{2}－4k_{1}x－4(y_{0}－k_{1}x_{0})=0(*),令\Delta =16k^2_1+16(y_{0}－k_{1}x_{0})=0，整理得k^2_1－k_{1}x_{0}+y_{0}=0①，设直线PB的方程为y－y_{0}=k_{2}(x－x_{0})，同理得k^2_2－k_{2}x_{0}+y_{0}=0②，由①②知，k_{1},k_{2}是方程k^{2}－kx_{0}+y_{0}=0的两根，所以k_{1}+k_{2}=x_{0},k_{1}k_{2}=y_{0}，(7分)由(*)式得，x_{A}+x_{A}=4k_{1}，则x_{A}=2k_{1}，所以A(2k_{1},k^2_1)，同理B(2k_{2},k^2_2),所以k_{AB}=\frac{k^2_1-k^2_2}{2k_1-2k_2}=\frac{k_1+k_2}{2}，(8分)所以直线AB的方程为y－k^2_1=\frac{k_1+k_2}{2}(x－2k_{1})，即y=\frac{k_1+k_1}{2}x－k_{1}k_{2}，即y=\frac{x_0}{2}x－y_{0}.所以|AB|=\sqrt{1+\frac{x^2_0}{4}}|2k_{1}－2k_{2}|=\sqrt{4+x^2_0}\cdot \sqrt{(k_1+k_2)^2-4k_1k_2}=\sqrt{4+x^2_0}\cdot \sqrt{x^2_0-4y_0}，P(x_{0},y_{0})到直线AB的距离d=\frac{|x^2_0-4y_0|}{\sqrt{4+x^2_0}},所以S_{\triangle PAB}=\frac{1}{2}|AB|\cdot d=\frac{1}{2}\sqrt{(x^2_0-4y_0)^3}=\frac{3\sqrt{3}}{2}，(10分)整理得x^2_0－4y_{0}=3，联立\left\{\begin{matrix}(x_0-1)^2+(y_0+1)^2=\frac{1}{4},\\ x^2_0-4y_0=3,\end{matrix}\right.得(x_{0}－1)(x^3_0+x^2_0+19x_{0}－13)=0,所以x_{0}=1或x^3_0+x^2_0+19x_{0}－13=0，设f(x_{0})=x^3_0+x^2_0+19x_{0}－13，\frac{1}{2}\leq x_{0}\leq \frac{3}{2},显然f\left(\frac{1}{2}\right)<0,f(1)>0,f\left(\frac{3}{2}\right)>0，易知f(x_{0})在\left[\frac{1}{2},\frac{3}{2}\right]上单调递增，所以f(x_{0})=x^3_0+x^2_0+19x_{0}－13在\left(\frac{1}{2},1\right)上有唯一零点.所以存在两个x_{0}，使得\triangle PAB的面积等于\frac{3\sqrt{3}}{2}.(12分)$

Final answer: 证明题
Subfield: Plane Geometry


Question: $已知点A,B在直线l：y=x+2上(B在A的上方)，P(2,0)，|AB|=\sqrt{2}，斜率为k_{1}的直线AP交抛物线\Gamma ：y^{2}=4x于点M,N，直线BP交\Gamma 于点R,S，如图所示.$![Image](https://cdntk.53zaixian.com/tq/1658391411285364737/img/compression/M_49ff6444e9720cf006f14b555ab1c22d.jpg)
$求k_{1}的取值范围；$

Solution:
$由题意可知直线AP：y=k_{1}(x-2).因为直线AP与抛物线\Gamma 有两个交点，与直线l有一个交点，所以k_{1}\neq 1，且k_{1}\neq 0.由\left\{\begin{matrix}y=k_1(x-2),\\ y=x+2,\end{matrix}\right.得\left\{\begin{matrix}x=\frac{2(k_1+1)}{k_1-1},\\ y=\frac{4k_1}{k_1-1},\end{matrix}\right.所以A\left(\frac{2(k_1+1)}{k_1-1},\frac{4k_1}{k_1-1}\right).因为|AB|=\sqrt{2}，k_{l}=1,所以B\left(\frac{2(k_1+1)}{k_1-1}+1,\frac{4k_1}{k_1-1}+1\right)，即B\left(\frac{3k_1+1}{k_1-1},\frac{5k_1-1}{k_1-1}\right).(2分)当直线BP的斜率不存在时，\frac{3k_1+1}{k_1-1}=2,解得k_{1}=-3,满足题意.当直线BP的斜率存在时，设为k_{2},则k_{2}=\frac{\frac{5k_1-1}{k_1-1}}{\frac{3k_1+1}{k_1-1}-2}=\frac{5k_1-1}{k_1+3}(k_{1}\neq -3)，所以\left\{\begin{matrix}\frac{5k_1-1}{k_1+3} \ne 1,\\ \frac{5k_1-1}{k_1+3} \ne 0,\end{matrix}\right.所以k_{1}\neq -3且k_{1}\neq \frac{1}{5}且k_{1}\neq 1.(4分)综上，k_{1}\in (-\infty ,0)\cup \left(0,\frac{1}{5}\right)\cup \left(\frac{1}{5},1\right)\cup (1,+\infty ).(5分)$

Final answer: $(-\infty ,0)\cup \left(0,\frac{1}{5}\right)\cup \left(\frac{1}{5},1\right)\cup (1,+\infty )$
Subfield: Plane Geometry


Question: $已知点A,B在直线l：y=x+2上(B在A的上方)，P(2,0)，|AB|=\sqrt{2}，斜率为k_{1}的直线AP交抛物线\Gamma ：y^{2}=4x于点M,N，直线BP交\Gamma 于点R,S，如图所示.$<img_1654>
$求k_{1}的取值范围；$

Solution:
$由题意可知直线AP：y=k_{1}(x-2).因为直线AP与抛物线\Gamma 有两个交点，与直线l有一个交点，所以k_{1}\neq 1，且k_{1}\neq 0.由\left\{\begin{matrix}y=k_1(x-2),\\ y=x+2,\end{matrix}\right.得\left\{\begin{matrix}x=\frac{2(k_1+1)}{k_1-1},\\ y=\frac{4k_1}{k_1-1},\end{matrix}\right.所以A\left(\frac{2(k_1+1)}{k_1-1},\frac{4k_1}{k_1-1}\right).因为|AB|=\sqrt{2}，k_{l}=1,所以B\left(\frac{2(k_1+1)}{k_1-1}+1,\frac{4k_1}{k_1-1}+1\right)，即B\left(\frac{3k_1+1}{k_1-1},\frac{5k_1-1}{k_1-1}\right).(2分)当直线BP的斜率不存在时，\frac{3k_1+1}{k_1-1}=2,解得k_{1}=-3,满足题意.当直线BP的斜率存在时，设为k_{2},则k_{2}=\frac{\frac{5k_1-1}{k_1-1}}{\frac{3k_1+1}{k_1-1}-2}=\frac{5k_1-1}{k_1+3}(k_{1}\neq -3)，所以\left\{\begin{matrix}\frac{5k_1-1}{k_1+3} \ne 1,\\ \frac{5k_1-1}{k_1+3} \ne 0,\end{matrix}\right.所以k_{1}\neq -3且k_{1}\neq \frac{1}{5}且k_{1}\neq 1.(4分)综上，k_{1}\in (-\infty ,0)\cup \left(0,\frac{1}{5}\right)\cup \left(\frac{1}{5},1\right)\cup (1,+\infty ).(5分)$



Final answer: $(-\infty ,0)\cup \left(0,\frac{1}{5}\right)\cup \left(\frac{1}{5},1\right)\cup (1,+\infty )$
Subfield: Plane Geometry


Question: $已知点A,B在直线l：y=x+2上(B在A的上方)，P(2,0)，|AB|=\sqrt{2}，斜率为k_{1}的直线AP交抛物线\Gamma ：y^{2}=4x于点M,N，直线BP交\Gamma 于点R,S，如图所示.$![Image](https://cdntk.53zaixian.com/tq/1658391411285364737/img/compression/M_49ff6444e9720cf006f14b555ab1c22d.jpg)
$若0<k_{1}<\frac{1}{5}，求S_{\triangle MPR}\cdot S_{\triangle SPN}的取值范围.$

Solution:
$设M(x_{1},y_{1})，N(x_{2},y_{2})，R(x_{3},y_{3})，S(x_{4},y_{4})，l_{AP}：x=m_{1}y+2，m_{1}=\frac{1}{k_1}\in (5,+\infty )，l_{BP}：x=m_{2}y+2，m_{2}=\frac{1}{k_2}.连接NR，MS.因为\frac{S_{\vartriangle MPR}}{S_{\vartriangle NMR}}=\frac{|MP|}{|MN|}=\frac{y_1}{y_1-y_2}，\frac{S_{\vartriangle SPN}}{S_{\vartriangle SMN}}=\frac{|PN|}{|MN|}=\frac{-y_2}{y_1-y_2},所以S_{\triangle MPR}=\frac{y_1}{y_1-y_2}\cdot S_{\triangle NMR}，S_{\triangle SPN}=\frac{-y_2}{y_1-y_2}\cdot S_{\triangle SMN}，故S_{\triangle MPR}\cdot S_{\triangle SPN}=-\frac{y_1y_2}{(y_1-y_2)^2}\cdot S_{\triangle SMN}\cdot S_{\triangle NMR}.(6分)记R,S到直线AP的距离分别为d_{3},d_{4}，则S_{\triangle NMR}=\frac{1}{2}|MN|\cdot d_{3}，S_{\triangle SMN}=\frac{1}{2}|MN|\cdot d_{4}.由弦长公式得|MN|=\sqrt{m^2_1+1}\cdot |y_{1}-y_{2}|，则S_{\triangle MPR}\cdot S_{\triangle SPN}=-\frac{y_1y_2}{(y_1-y_2)^2}\cdot \frac{1}{4}|MN|^{2}\cdot d_{3}\cdot d_{4}=-\frac{y_1y_2(1+m^2_1)}{4}\cdot d_{3}\cdot d_{4}.(8分)由\left\{\begin{matrix}x=m_1y+2,\\ y^2=4x,\end{matrix}\right.消去x,得y^{2}-4m_{1}y-8=0，所以y_{1}y_{2}=-8.由\left\{\begin{matrix}x=m_2y+2,\\ y^2=4x,\end{matrix}\right.消去x,得y^{2}-4m_{2}y-8=0,所以y_{3}+y_{4}=4m_{2},y_{3}y_{4}=-8.故S_{\triangle MPR}\cdot S_{\triangle SPN}=2(1+m^2_1)\cdot \frac{\left|\frac{y^2_3}{4}-m_1y_3-2\right| \cdot \left|\frac{y^2_4}{4}-m_1y_4-2\right|}{1+m^2_1}=2\left|\frac{y^2_3y^2_4}{16}-\frac{m_1}{4}y_3y_4(y_3+y_4)-\frac{y^2_3+y^2_4}{2}+m^2_1y_3y_4+2m_1(y_3+y_4)+4\right|=16(m_{1}-m_{2})^{2}.(10分)由(1)可知，k_{2}=\frac{5k_1-1}{k_1+3},又k_{2}=\frac{1}{m_2},k_{1}=\frac{1}{m_1},所以m_{2}=\frac{3m_1+1}{5-m_1}，故S_{\triangle MPR}\cdot S_{\triangle SPN}=16\left(m_1+\frac{3m_1+1}{m_1-5}\right)^2=16\left(m_1-5+\frac{16}{m_1-5}+8\right)^2\geq 16\times \left(2\sqrt{(m_1-5) \cdot \frac{16}{m_1-5}}+8\right)^2=4 096，当且仅当m_{1}-5=\frac{16}{m_1-5},即m_{1}=9，即k_{1}=\frac{1}{9}时取等号.故S_{\triangle MPR}\cdot S_{\triangle SPN}的取值范围是[4 096,+\infty ).(12分)$

Final answer: $[4 096,+\infty )$
Subfield: Plane Geometry


Question: $已知点A,B在直线l：y=x+2上(B在A的上方)，P(2,0)，|AB|=\sqrt{2}，斜率为k_{1}的直线AP交抛物线\Gamma ：y^{2}=4x于点M,N，直线BP交\Gamma 于点R,S，如图所示.$<img_1654>
$若0<k_{1}<\frac{1}{5}，求S_{\triangle MPR}\cdot S_{\triangle SPN}的取值范围.$

Solution:
$设M(x_{1},y_{1})，N(x_{2},y_{2})，R(x_{3},y_{3})，S(x_{4},y_{4})，l_{AP}：x=m_{1}y+2，m_{1}=\frac{1}{k_1}\in (5,+\infty )，l_{BP}：x=m_{2}y+2，m_{2}=\frac{1}{k_2}.连接NR，MS.因为\frac{S_{\vartriangle MPR}}{S_{\vartriangle NMR}}=\frac{|MP|}{|MN|}=\frac{y_1}{y_1-y_2}，\frac{S_{\vartriangle SPN}}{S_{\vartriangle SMN}}=\frac{|PN|}{|MN|}=\frac{-y_2}{y_1-y_2},所以S_{\triangle MPR}=\frac{y_1}{y_1-y_2}\cdot S_{\triangle NMR}，S_{\triangle SPN}=\frac{-y_2}{y_1-y_2}\cdot S_{\triangle SMN}，故S_{\triangle MPR}\cdot S_{\triangle SPN}=-\frac{y_1y_2}{(y_1-y_2)^2}\cdot S_{\triangle SMN}\cdot S_{\triangle NMR}.(6分)记R,S到直线AP的距离分别为d_{3},d_{4}，则S_{\triangle NMR}=\frac{1}{2}|MN|\cdot d_{3}，S_{\triangle SMN}=\frac{1}{2}|MN|\cdot d_{4}.由弦长公式得|MN|=\sqrt{m^2_1+1}\cdot |y_{1}-y_{2}|，则S_{\triangle MPR}\cdot S_{\triangle SPN}=-\frac{y_1y_2}{(y_1-y_2)^2}\cdot \frac{1}{4}|MN|^{2}\cdot d_{3}\cdot d_{4}=-\frac{y_1y_2(1+m^2_1)}{4}\cdot d_{3}\cdot d_{4}.(8分)由\left\{\begin{matrix}x=m_1y+2,\\ y^2=4x,\end{matrix}\right.消去x,得y^{2}-4m_{1}y-8=0，所以y_{1}y_{2}=-8.由\left\{\begin{matrix}x=m_2y+2,\\ y^2=4x,\end{matrix}\right.消去x,得y^{2}-4m_{2}y-8=0,所以y_{3}+y_{4}=4m_{2},y_{3}y_{4}=-8.故S_{\triangle MPR}\cdot S_{\triangle SPN}=2(1+m^2_1)\cdot \frac{\left|\frac{y^2_3}{4}-m_1y_3-2\right| \cdot \left|\frac{y^2_4}{4}-m_1y_4-2\right|}{1+m^2_1}=2\left|\frac{y^2_3y^2_4}{16}-\frac{m_1}{4}y_3y_4(y_3+y_4)-\frac{y^2_3+y^2_4}{2}+m^2_1y_3y_4+2m_1(y_3+y_4)+4\right|=16(m_{1}-m_{2})^{2}.(10分)由(1)可知，k_{2}=\frac{5k_1-1}{k_1+3},又k_{2}=\frac{1}{m_2},k_{1}=\frac{1}{m_1},所以m_{2}=\frac{3m_1+1}{5-m_1}，故S_{\triangle MPR}\cdot S_{\triangle SPN}=16\left(m_1+\frac{3m_1+1}{m_1-5}\right)^2=16\left(m_1-5+\frac{16}{m_1-5}+8\right)^2\geq 16\times \left(2\sqrt{(m_1-5) \cdot \frac{16}{m_1-5}}+8\right)^2=4 096，当且仅当m_{1}-5=\frac{16}{m_1-5},即m_{1}=9，即k_{1}=\frac{1}{9}时取等号.故S_{\triangle MPR}\cdot S_{\triangle SPN}的取值范围是[4 096,+\infty ).(12分)$

Final answer: $[4 096,+\infty )$
Subfield: Plane Geometry


Question: $如图所示，在三棱锥S-ABC中，SA=AB=AC=BC=\sqrt{2}SB=\sqrt{2}SC，O为BC的中点.$<img_2604>
$求证：SO\perp 平面ABC；$

Solution:
$证明：在三棱锥S-ABC中，BC=\sqrt{2}SB=\sqrt{2}SC，O为BC的中点，显然SO\perp BC，连接OA，设SB＝a，则SA＝\sqrt{2}a，SO＝\frac{\sqrt{2}a}{2}，OA＝\frac{\sqrt{6}a}{2}，\therefore SO^{2}+OA^{2}＝SA^{2}，\therefore SO\perp OA，又\because BC\cap OA＝O，BC,OA\subset 平面ABC，\therefore SO\perp 平面ABC.$
Subfield: Plane Geometry


Question: $如图所示，在三棱锥S-ABC中，SA=AB=AC=BC=\sqrt{2}SB=\sqrt{2}SC，O为BC的中点.$![Image](https://cdntk.53zaixian.com/tq/1659094426553167874/img/compression/M_c2d9cb326a1a45d0e7c9cbeb234bbb3e.jpg)
$求证：SO\perp 平面ABC；$

Solution:
$证明：在三棱锥S-ABC中，BC=\sqrt{2}SB=\sqrt{2}SC，O为BC的中点，显然SO\perp BC，连接OA，设SB＝a，则SA＝\sqrt{2}a，SO＝\frac{\sqrt{2}a}{2}，OA＝\frac{\sqrt{6}a}{2}，\therefore SO^{2}+OA^{2}＝SA^{2}，\therefore SO\perp OA，又\because BC\cap OA＝O，BC,OA\subset 平面ABC，\therefore SO\perp 平面ABC.$

Final answer: 证明题
Subfield: Plane Geometry


Question: $如图所示，在三棱锥S-ABC中，SA=AB=AC=BC=\sqrt{2}SB=\sqrt{2}SC，O为BC的中点.$![Image](https://cdntk.53zaixian.com/tq/1659094426553167874/img/compression/M_c2d9cb326a1a45d0e7c9cbeb234bbb3e.jpg)
$求异面直线SC与AB所成角的余弦值；$

Solution:
$以O为原点，\overrightarrow{OA}，\overrightarrow{OB}，\overrightarrow{OS}的方向分别为x轴，y轴，z轴的正方向，建立如图所示的空间直角坐标系，$![Image](https://cdntk.53zaixian.com/tq/1659094426955821057/img/compression/M_8533286f51c90f9d441123c3c868228e.jpg)$则S\left(0,0,\frac{\sqrt{2}a}{2}\right)，C\left(0,- \frac{\sqrt{2}a}{2},0\right)，A\left(\frac{\sqrt{6}a}{2},0,0\right)，B\left(0,\frac{\sqrt{2}a}{2},0\right)，\therefore \overrightarrow{SC}=\left(0,-\frac{\sqrt{2}a}{2},-\frac{\sqrt{2}a}{2}\right)，\overrightarrow{AB}=\left(-\frac{\sqrt{6}a}{2},\frac{\sqrt{2}a}{2},0\right)，\therefore \\cos  \langle \overrightarrow{SC},\overrightarrow{AB}\rangle =\frac{\overrightarrow{SC} \cdot \overrightarrow{AB}}{|\overrightarrow{SC}||\overrightarrow{AB}|}=\frac{-\frac{a^2}{2}}{\sqrt{2}a^2}=-\frac{\sqrt{2}}{4}，\therefore 异面直线SC与AB所成角的余弦值为\frac{\sqrt{2}}{4}.$

Final answer: $\frac{\sqrt{2}}{4}$
Subfield: Plane Geometry


Question: $如图所示，在三棱锥S-ABC中，SA=AB=AC=BC=\sqrt{2}SB=\sqrt{2}SC，O为BC的中点.$<img_2604>
$求异面直线SC与AB所成角的余弦值；$

Solution:
$以O为原点，\overrightarrow{OA}，\overrightarrow{OB}，\overrightarrow{OS}的方向分别为x轴，y轴，z轴的正方向，建立如图所示的空间直角坐标系，$<img_1793>$则S\left(0,0,\frac{\sqrt{2}a}{2}\right)，C\left(0,- \frac{\sqrt{2}a}{2},0\right)，A\left(\frac{\sqrt{6}a}{2},0,0\right)，B\left(0,\frac{\sqrt{2}a}{2},0\right)，\therefore \overrightarrow{SC}=\left(0,-\frac{\sqrt{2}a}{2},-\frac{\sqrt{2}a}{2}\right)，\overrightarrow{AB}=\left(-\frac{\sqrt{6}a}{2},\frac{\sqrt{2}a}{2},0\right)，\therefore \\cos  \langle \overrightarrow{SC},\overrightarrow{AB}\rangle =\frac{\overrightarrow{SC} \cdot \overrightarrow{AB}}{|\overrightarrow{SC}||\overrightarrow{AB}|}=\frac{-\frac{a^2}{2}}{\sqrt{2}a^2}=-\frac{\sqrt{2}}{4}，\therefore 异面直线SC与AB所成角的余弦值为\frac{\sqrt{2}}{4}.$

Final answer: $\frac{\sqrt{2}}{4}$
Subfield: Plane Geometry


Question: $如图所示，在三棱锥S-ABC中，SA=AB=AC=BC=\sqrt{2}SB=\sqrt{2}SC，O为BC的中点.$<img_2604>
$在线段AB上存在一点E，使二面角B-SC-E的余弦值为\frac{\sqrt{15}}{5}，请求出\frac{BE}{BA}的值.$

Solution:
$存在满足条件的点E.由(2)知\overrightarrow{BC}=(0，-\sqrt{2}a,0),\overrightarrow{BA}=\left(\frac{\sqrt{6}a}{2},-\frac{\sqrt{2}a}{2},0\right),\overrightarrow{SC}=\left(0,-\frac{\sqrt{2}a}{2},-\frac{\sqrt{2}a}{2}\right).设\overrightarrow{BE}=\lambda \overrightarrow{BA}(0\leq \lambda \leq 1)，则\overrightarrow{BE}=\left(\frac{\sqrt{6}a\lambda }{2},-\frac{\sqrt{2}a\lambda }{2},0\right)，\therefore \overrightarrow{CE}=\overrightarrow{BE}-\overrightarrow{BC}=\left(\frac{\sqrt{6}a\lambda }{2},-\frac{\sqrt{2}a\lambda }{2},0\right)-(0,-\sqrt{2}a,0)=\left(\frac{\sqrt{6}a\lambda }{2},-\frac{\sqrt{2}(\lambda -2)a}{2},0\right).设平面SCE的一个法向量为n＝(x，y，z)，则\left\{\begin{matrix}\boldsymbol{n} \cdot \overrightarrow{CE}=0,\\ \boldsymbol{n} \cdot \overrightarrow{SC}=0,\end{matrix}\right.即\left\{\begin{matrix}\sqrt{3}\lambda x-(\lambda -2)y=0,\\ y+z=0,\end{matrix}\right.令y=\sqrt{3}\lambda ,得n=(\lambda -2,\sqrt{3}\lambda ,-\sqrt{3}\lambda ).易知OA\perp 平面SBC，\therefore 可取m＝(1，0，0)为平面SBC的一个法向量，\therefore |\\cos  \langle m,n\rangle |=\frac{|\boldsymbol{m} \cdot \boldsymbol{n}|}{|\boldsymbol{m}||\boldsymbol{n}|}=\frac{|\lambda -2|}{\sqrt{{(\lambda -2)}^2+{(\sqrt{3}\lambda )^2}+{(-\sqrt{3}\lambda )^2}}}=\frac{\sqrt{15}}{5}，即2\lambda ^{2}+\lambda -1=0，解得\lambda =\frac{1}{2}或\lambda =-1(舍去)，\therefore 存在满足题意的点E使得二面角B-SC-E的余弦值为\frac{\sqrt{15}}{5}，此时\frac{BE}{BA}=\frac{1}{2}.$



Final answer: $\frac{1}{2}$