A sphere of radius 4 with center at point $Q$ touches three parallel lines at points $F, G$, and $H$. It is known that the area of triangle $Q G H$ is $4 \sqrt{2}$, and the area of triangle $F G H$ is greater than 16. Find the angle $G F H$.
Let the line $a$ touch the given sphere at point $F$ (Fig.1). Draw through point $Q$ the plane $\alpha$, perpendicular to the line $a$. If the line $a$ intersects this plane at point $F 1$, then $Q F 1 \perp a$, and since a line tangent to a sphere is perpendicular to the radius of the sphere drawn to the point of tangency, $Q F \perp a$. From the uniqueness of the perpendicular drawn to a given line through a given point, it follows that point $F1$ coincides with point $F$. Since the given lines are parallel, the plane $\alpha$ is perpendicular to each of them. Therefore, the plane $\alpha$ also passes through points $G$ and $H$. The triangle $F G H$ is inscribed in the circle of intersection of the sphere with the plane $\alpha$. Let $S_{\triangle F G H}=S$. By the condition,

$$
4 \sqrt{2}=S_{\triangle Q G H}=\frac{1}{2} Q G \cdot Q H \sin \angle G Q H=\frac{1}{2} \cdot 4 \cdot 4 \sin \angle G Q H=8 \sin \angle G Q H
$$

from which $\sin \angle G Q H=\frac{\sqrt{2}}{2}$. Therefore, either $\angle G Q H=45^{\circ}$ or $\angle G Q H=135^{\circ}$. Let $\angle G Q H=45^{\circ}$. If point $F$ lies on the larger arc $G H$, (Fig.2) then the area of triangle $F G H$ is maximized if point $F$ coincides with point $A$, lying on the perpendicular bisector of the chord $G H$, i.e., on the diameter $A B$ of the circle, perpendicular to the chord $G H$. If $C$ is the midpoint of this chord, then

$$
\begin{gathered}
S \leqslant \frac{1}{2} G H \cdot A C \\
>4 \sqrt{2}+4 \cdot 4 \cdot \sin 120^{\circ}=4 \sqrt{2}+8 \sqrt{3}>4+8 \cdot 1.5=4+12=16
\end{gathered}
$$

Thus, $\angle G Q H=135^{\circ}$, and point $F$ lies on the larger arc $G H$. Therefore,

$$
\angle G F H=\frac{1}{2} \angle G Q H=\frac{1}{2} \cdot 135^{\circ}=67.5^{\circ}
$$

## $3 \pi$