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 $x_1 + x_2 = -3m$, $x_1x_2 = \frac{9m^2 - 36}{2}$, $k_{PA} = \frac{y_1 - \sqrt{2}}{x_1 - 3\sqrt{2}}$, $k_{PB} = \frac{y_2 - \sqrt{2}}{x_2 - 3\sqrt{2}}$. Therefore,
$$\begin{aligned} k_{PA} + k_{PB} &= \frac{y_1 - \sqrt{2}}{x_1 - 3\sqrt{2}} + \frac{y_2 - \sqrt{2}}{x_2 - 3\sqrt{2}} \\ &= \frac{(y_1 - \sqrt{2})(x_2 - 3\sqrt{2}) + (y_2 - \sqrt{2})(x_1 - 3\sqrt{2})}{(x_1 - 3\sqrt{2})(x_2 - 3\sqrt{2})}. \end{aligned}$$
In the expression above, the numerator is equal to
$$\begin{aligned} & \left(\frac{1}{3}x_1 + m - \sqrt{2}\right)(x_2 - 3\sqrt{2}) + \left(\frac{1}{3}x_2 + m - \sqrt{2}\right)(x_1 - 3\sqrt{2}) \\ &= \frac{2}{3}x_1x_2 + (m - 2\sqrt{2})(x_1 + x_2) - 6\sqrt{2}(m - \sqrt{2}) \\ &= \frac{2}{3} \cdot \frac{9m^2 - 36}{2} + (m - 2\sqrt{2})(-3m) - 6\sqrt{2}(m - \sqrt{2}) \\ &= 3m^2 - 12 - 3m^2 + 6\sqrt{2}m - 6\sqrt{2}m + 12 = 0. \end{aligned}$$
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 $\triangle PAB$ 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,
$$\begin{align*} S_{\triangle PAB} &= \frac{1}{2} \cdot |PA| \cdot |PB| \cdot \sin 60^\circ \\ &= \frac{1}{2} \cdot \frac{3\sqrt{2}(3\sqrt{3} + 1)}{7} \cdot \frac{3\sqrt{2}(3\sqrt{3} - 1)}{7} \cdot \frac{\sqrt{3}}{2} \\ &= \frac{117\sqrt{3}}{49}. \end{align*}$$