China Mathematical Competition

Suppose $\angle ABF=\theta$ ($0<\theta <\frac{2\pi }{3}$). Then by the Law of Sine, we have $$\frac{|AF|}{\sin \theta }=\frac{|BF|}{\sin (\frac{2\pi }{3}-\theta )}=\frac{|AB|}{\sin \frac{\pi }{3}}$$ And then $$\frac{|AF|+|BF|}{\sin \theta +\sin (\frac{2\pi }{3}-\theta )}=\frac{|AB|}{\sin \frac{\pi }{3}}$$ So $$\frac{|AF|+|BF|}{|AB|}=\frac{\sin \theta +\sin (\frac{2\pi }{3}-\theta )}{\sin \frac{\pi }{3}}=2\cos (\theta -\frac{\pi }{3}).$$ As seen in Fig. 4.1, by using the definition of a parabola and the property of a trapezoid, we have $$|MN|=\frac{|AF|+|BF|}{2}.$$ Then $$\frac{|MN|}{|AB|}=\cos \left(\theta -\frac{\pi }{3}\right).$$ Therefore, $\frac{|MN|}{|AB|}$ reaches the maximum value $1$ when $\theta =\frac{\pi }{3}$.

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Alternative solution.

By using the definition of a parabola and the property of a trapezoid, we have $|MN|=\frac{|AF|+|BF|}{2}$. In $△AFB$, by using the Law of Cosines we have $$\begin{array}{rl}|AB{|}^2& =|AF{|}^2+|BF{|}^2-2|AF|\cdot |BF|\cos \frac{\pi }{3}\\ & =(|AF|+|BF|{)}^2-3|AF|\cdot |BF|\\ & \ge (|AF|+|BF|{)}^2-3{\left(\frac{|AF|+|BF|}{2}\right)}^2\\ & ={\left(\frac{|AF|+|BF|}{2}\right)}^2=|MN{|}^2.\end{array}$$ The equality holds if and only if $|AF|=|BF|$. Therefore, the maximum value of $\frac{|MN|}{|AB|}$ is $1$. $\square$