China Mathematical Competition

Euclidean geometry tells us that, $\angle F_1PF_2$ reaches the maximum only if the circle through points $F_1, F_2, P$ is tangent to the line $l$ at $P$. Now suppose $l$ intercepts the $x$-axis at point $A(-8-2\sqrt{3}, 0)$. Then $\angle APF_1 = \angle AF_2P$, and that means $\triangle APF_1 \sim \triangle AF_2P$. So
$$\frac{|PF_1|}{|PF_2|} = \frac{|AP|}{|AF_2|}$$
By using the power of points theorem, we have
$$|AP|^2 = |AF_1| \cdot |AF_2|$$
As $F_1(-2\sqrt{3}, 0)$, $F_2(2\sqrt{3}, 0)$, $A(-8-2\sqrt{3}, 0)$, so
$$|AF_1| = 8, \quad |AF_2| = 8 + 4\sqrt{3}.$$
Then we get
$$$$
$$\begin{aligned} \frac{|PF_1|}{|PF_2|} &= \sqrt{\frac{|AF_1|}{|AF_2|}} = \sqrt{\frac{8}{8+4\sqrt{3}}} \\ &= \sqrt{4-2\sqrt{3}} = \sqrt{3}-1. \end{aligned}$$