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 $\theta = \frac{\pi}{3}$.

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