China Mathematical Competition

a. Let $l$ be a straight line such that $y=\frac{1}{3}x+m$, and $A(x_1,y_1),B(x_2,y_2)$. Substituting $y=\frac{1}{3}x+m$ into $\frac{x^2}{36}+\frac{y^2}{4}=1$, and simplifying it, we have $$2x^2+6mx+9m^2-36=0.$$ Then $x_1+x_2=-3m$, $x_1{x}_2=\frac{9m^2-36}{2}$, $k_{PA}=\frac{y_1-\sqrt{2}}{x_1-3\sqrt{2}}$, $k_{PB}=\frac{y_2-\sqrt{2}}{x_2-3\sqrt{2}}$. Therefore, $$\begin{array}{rl}{k}_{PA}+k_{PB}& =\frac{y_1-\sqrt{2}}{x_1-3\sqrt{2}}+\frac{y_2-\sqrt{2}}{x_2-3\sqrt{2}}\\ & =\frac{(y_1-\sqrt{2})(x_2-3\sqrt{2})+(y_2-\sqrt{2})(x_1-3\sqrt{2})}{(x_1-3\sqrt{2})(x_2-3\sqrt{2})}.\end{array}$$ In the expression above, the numerator is equal to $$\begin{array}{rl}& \left(\frac{1}{3}{x}_1+m-\sqrt{2}\right)(x_2-3\sqrt{2})+\left(\frac{1}{3}{x}_2+m-\sqrt{2}\right)(x_1-3\sqrt{2})\\ & =\frac{2}{3}{x}_1{x}_2+(m-2\sqrt{2})(x_1+x_2)-6\sqrt{2}(m-\sqrt{2})\\ & =\frac{2}{3}\cdot \frac{9m^2-36}{2}+(m-2\sqrt{2})(-3m)-6\sqrt{2}(m-\sqrt{2})\\ & =3m^2-12-3m^2+6\sqrt{2}m-6\sqrt{2}m+12=0.\end{array}$$ Therefore, $k_{PA}+k_{PB}=0$. Since $P$ is in the top-left of $l$, we know that the bisector of $\angle APB$ is parallel to the $y$-axis. Therefore, the center of the inscribed circle of $△PAB$ is on line $x=3\sqrt{2}$.

b. When $\angle APB=60^{\circ }$, by the result in (a), we have $k_{PA}=\sqrt{3}$, $k_{PB}=-\sqrt{3}$. Then the equation for line $PA$ is $y-\sqrt{2}=\sqrt{3}(x-3\sqrt{2})$. Substituting it into $\frac{x^2}{36}+\frac{y^2}{4}=1$, and eliminating $y$, we get $$14x^2+9\sqrt{6}(1-3\sqrt{3})x+18(13-3\sqrt{3})=0,$$ which has roots $x_1$ and $3\sqrt{2}$. So $x_1\cdot 3\sqrt{2}=\frac{18(13-3\sqrt{3})}{14}$, i.e. $$x_1=\frac{3\sqrt{2}(13-3\sqrt{3})}{14}.$$ Then we find $$|PA|=\sqrt{1+(\sqrt{3}{)}^2}\cdot |x_1-3\sqrt{2}|=\frac{3\sqrt{2}(3\sqrt{3}+1)}{7}.$$ In the same way, we have $|PB|=\frac{3\sqrt{2}(3\sqrt{3}-1)}{7}$. Therefore, $$\begin{array}{rl}{S}_{△PAB}& =\frac{1}{2}\cdot |PA|\cdot |PB|\cdot \sin 60^{\circ }\\ & =\frac{1}{2}\cdot \frac{3\sqrt{2}(3\sqrt{3}+1)}{7}\cdot \frac{3\sqrt{2}(3\sqrt{3}-1)}{7}\cdot \frac{\sqrt{3}}{2}\\ & =\frac{117\sqrt{3}}{49}.\end{array}$$