用曲线系方法解2023年ELMO预选题G7

\section*{2023年ELMO预选题G7}
已知椭圆$\mathcal E$的焦点为$F_1$和$F_2$，$P$是$\mathcal E$上一点．直线$PF_1$和$PF_2$分别再次交$\mathcal E$于不同的点$A$和点$B$，点$A$和点$B$处的$\mathcal E$的切线交于点$Q$．证明：$PQ$的中点在$\triangle PF_1F_2$的外接圆上．
\begin{proof}设椭圆为：$\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$（$a>b>0$），$c=\sqrt{a^2-b^2}$．
则$F_1(-c,0)$，$F_2(c,0)$，设$P(x_1,y_1)$，$Q(x_2,y_2)$．
$\mathcal E$在点$P$处的切线$PP$方程为：$\dfrac{x_1x}{a^2}+\dfrac{y_1y}{b^2}=1$．
点$Q$关于$\mathcal E$的切点弦方程为：$\dfrac{x_2x}{a^2}+\dfrac{y_2y}{b^2}=1$．
过$P,P,A,B$四点的二次曲线系方程为：\begin{equation}\lambda\left(\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}-1\right)+\mu\left(\frac{x_1x}{a^2}+\frac{y_1y}{b^2}-1\right)\left(\frac{x_2x}{a^2}+\frac{y_2y}{b^2}-1\right)=0.\end{equation}
其中一条为双直线$PA\times PB$：\begin{equation}(y_1x-(x_1+c)y+y_1c)(y_1-(x_1-c)y-y_1c)=0.\end{equation}
对比$(1)$和$(2)$中$x$的系数得：$$\mu\left(\frac{x_{1}}{a^{2}}\cdot(-1)+(-1)\cdot\frac{x_{2}}{a^{2}}\right)=y_{1}\cdot(-y_{1})+y_{1}\cdot y_{1}$$
注意到$\mu\neq 0$（否则为椭圆），于是$x_{1}+x_{2}=0$．
对比$(1)$和$(2)$中$y$的系数得：$$\mu\left(\frac{y_{1}}{b^{2}}\cdot(-1)+(-1)\cdot\frac{y_{2}}{b^{2}}\right)=-(x_{1}+c)\cdot(-y_{1})+y_{1}\cdot(-(x_{1}-c))$$
化简得：$y_{1}+y_{2}=-\dfrac{2b^{2}c^{2}y_{1}}{\mu}$．
对比$(1)$和$(2)$中$xy$的系数得：$$\mu\left(\frac{x_{1}}{a^{2}}\cdot\frac{y_{2}}{b^{2}}+\frac{y_{1}}{b^{2}}\cdot\frac{x_{2}}{a^{2}}\right)=y_{1}\cdot(-(x_{1}-c))+(-(x_{1}+c))\cdot y_{1}$$
代入$x_{1}+x_{2}=0$，并化简得：$y_{1}-y_{2}=\frac{2a^{2}b^{2}y_{1}}{\mu}$．
于是$\dfrac{y_{1}-y_{2}}{y_{1}+y_{2}}=-\dfrac{a^{2}}{c^{2}}$，即$\dfrac{y_{1}+y_{2}}{2}=-\dfrac{c^{2}}{b^{2}}y_{1}$．
因此$PQ$中点$M$的坐标为$\left(0,-\dfrac{c^{2}}{b^{2}}y_{1}\right)$．
由对称性不妨设$y_{1}>0$．
设$\angle F_{1}PF_{2}=\alpha$，$\angle F_{1}MF_{2}=\beta$，则只需证明：$\alpha+\beta=\pi$．
熟知$S_{\triangle PF_{1}F_{2}}=b^{2}\tan\dfrac{\alpha}{2}=\dfrac{1}{2}\cdot 2c\cdot y_{1}$，于是$\tan\dfrac{\alpha}{2}=\dfrac{cy_{1}}{b^{2}}$．
因为$\triangle MF_{1}F_{2}$为等腰三角形，故$\tan\dfrac{\beta}{2}=\dfrac{c}{\frac{c^{2}}{b^{2}}y_{1}}=\dfrac{b^{2}}{cy_{1}}$．
因此$\tan\dfrac{\alpha}{2}\tan\dfrac{\beta}{2}=1$，证毕．\end{proof}