China Mathematical Competition

Let the midpoint of $AB$ be $M(x_0,y_0)$. Then $x_0=\frac{x_1+x_2}{2}=2$ and $y_0=\frac{y_1+y_2}{2}$. We have $$k_{AB}=\frac{y_2-y_1}{x_2-x_1}=\frac{\frac{y_2}{6}-\frac{y_1}{6}}{\frac{y_2}{6}+\frac{y_1}{6}}=\frac{\frac{6}{y_2}+\frac{6}{y_1}}{\frac{6}{y_2}+\frac{6}{y_1}}=\frac{3}{y_0}.$$ The equation of the perpendicular bisector of $AB$ is $$y-y_0=-\frac{y_0}{3}(x-2).\stackrel{◯}{1}$$ It is easy to find that one solution of it is $x=5$, $y=0$. Therefore, the intersection $C$ is a fixed point with coordinate $(5,0)$. From ①, we know the equation of line $AB$ is $y-y_0=\frac{3}{y_0}(x-2)$, or $$x=\frac{y_0}{3}(y-y_0)+2.\stackrel{◯}{2}$$ Substituting ② in $y^2=6x$, we get $y^2=2y_0(y-y_0)+12$, or $$y^2-2y_{0}y+2y_0^2-12=0.\stackrel{◯}{3}$$ As $y_1$ and $y_2$ are two real roots of ③ and $y_1\ne y_2$, we have $$\Delta =4y_0^2-4(2y_0^2-12)=-4y_0^2+48>0.$$ Therefore, $-2\sqrt{3}<y_0<2\sqrt{3}$. Then we have The distance from point $C(5,0)$ to segment $AB$ is $$h=|CM|=\sqrt{(5-2{)}^2+(0-y_0{)}^2}=\sqrt{9+y_0^2}.$$ Therefore, $$\begin{array}{rl}{S}_{△ABC}& =\frac{1}{2}|AB|\cdot h=\frac{1}{3}\sqrt{(9+y_0^2)(12-y_0^2)}\cdot \sqrt{9+y_0^2}\\ & =\frac{1}{3}\sqrt{\frac{1}{2}(9+y_0^2)(24-2y_0^2)(9+y_0^2)}\\ & \le \frac{1}{3}\sqrt{\frac{1}{2}{\left(\frac{9+y_0^2+24-2y_0^2+9+y_0^2}{3}\right)}^3}\\ & =\frac{14}{3}\sqrt{7}.\end{array}$$ The equality holds if and only if $9+y_0^2=24-2y_0^2$, i.e. $$y_0=\pm \sqrt{5}.\text{ Then we get}$$ $$\text{and}A\left(\frac{6+\sqrt{35}}{3},\sqrt{5}+\sqrt{7}\right),B\left(\frac{6-\sqrt{35}}{3},\sqrt{5}-\sqrt{7}\right)$$ $$A\left(\frac{6+\sqrt{35}}{3},-(\sqrt{5}+\sqrt{7})\right),B\left(\frac{6-\sqrt{35}}{3},-\sqrt{5}+\sqrt{7}\right).$$

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Alternative solution.

Similar to Solution 1, we get that $C$, the intersection of the perpendicular bisector of $AB$ and the $x$-axis, is a fixed point with coordinate $(5,0)$. Let $x_1=t_1^2$, $x_2=t_2^2$, $t_1>t_2$, $t_1^2+t_2^2=4$. Then $S_{△ABC}$ is the absolute value of $$\frac{1}{2}\left|\begin{array}{ccc}5& 0& 1\\ t_1^2& \sqrt{6}{t}_1& 1\\ t_2^2& \sqrt{6}{t}_2& 1\end{array}\right|,$$ so $$\begin{array}{rl}{S}_{△ABC}^2& ={\left(\frac{1}{2}(5\sqrt{6}{t}_1+\sqrt{6}{t}_1^2{t}_2-\sqrt{6}{t}_1{t}_2^2-5\sqrt{6}{t}_2)\right)}^2\\ & =\frac{3}{2}(t_1-t_2{)}^2(t_1{t}_2+5{)}^2\\ & =\frac{3}{2}(4-2t_1{t}_2)(t_1{t}_2+5)(t_1{t}_2+5)\\ & \le \frac{3}{2}{\left(\frac{14}{3}\right)}^3.\end{array}$$ Therefore, $S_{△ABC}\le \frac{14}{3}\sqrt{7}$ and the equality holds if and only if $(t_1-t_2{)}^2=t_1{t}_2+5$ and $t_1^2+t_2^2=4$. We then get $t_1=\frac{\sqrt{7}+\sqrt{5}}{\sqrt{6}}$ and $t_2=-\frac{\sqrt{7}-\sqrt{5}}{\sqrt{6}}$, which implies either $$A\left(\frac{6+\sqrt{35}}{3},\sqrt{5}+\sqrt{7}\right),B\left(\frac{6-\sqrt{35}}{3},\sqrt{5}-\sqrt{7}\right)$$ or $$A\left(\frac{6+\sqrt{35}}{3},-(\sqrt{5}+\sqrt{7})\right),B\left(\frac{6-\sqrt{35}}{3},-\sqrt{5}+\sqrt{7}\right).$$