China Mathematical Competition

Euclidean geometry tells us that, $\angle F_{1}P{F}_2$ 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 $\angle AP{F}_1=\angle A{F}_{2}P$, and that means $△AP{F}_1\sim △A{F}_{2}P$. So $$\frac{|P{F}_1|}{|P{F}_2|}=\frac{|AP|}{|A{F}_2|}$$ By using the power of points theorem, we have $$|AP{|}^2=|A{F}_1|\cdot |A{F}_2|$$ As $F_1(-2\sqrt{3},0)$, $F_2(2\sqrt{3},0)$, $A(-8-2\sqrt{3},0)$, so $$|A{F}_1|=8,|A{F}_2|=8+4\sqrt{3}.$$ Then we get $$ $$\begin{array}{rl}\frac{|P{F}_1|}{|P{F}_2|}& =\sqrt{\frac{|A{F}_1|}{|A{F}_2|}}=\sqrt{\frac{8}{8+4\sqrt{3}}}\\ & =\sqrt{4-2\sqrt{3}}=\sqrt{3}-1.\end{array}$$