7. Three parallel lines touch a sphere with center at point O at points K, L, and M. It is known that the radius of the sphere is $5 \sqrt{2}$, the area of triangle OKL is 7, and the area of triangle KLM is greater than 50. Find the angle KML.

Answer. $\angle \mathrm{KML}=\arccos \frac{1}{5 \sqrt{2}}$.
Solution If a line is tangent to a sphere, then the plane passing through the center of the sphere perpendicular to this line passes through the point of tangency. Consider the plane $\alpha$, passing through point 0 perpendicular to the three given lines. The section of the sphere by the plane $\alpha$ is a circle of radius 5, passing through points $\mathrm{K}, \mathrm{L}$, and $\mathrm{M}$. Denote the height of triangle $0 \mathrm{KL}$, dropped from vertex 0, as $0 \mathrm{H}$, and $\angle \mathrm{KOL}=2 \alpha$. There are two cases.

1) Point $M$ lies on the larger arc KL. The area of the triangle is maximized if $M$ lies on the line $\mathrm{OH}$, and it is equal to

$$
S_{\max }=\frac{1}{2} M H \cdot K L=(5 \sqrt{2}+0 H) K H=(5 \sqrt{2}+5 \sqrt{2} \cos \alpha) 5 \sqrt{2} \sin \alpha=50(1+\cos \alpha) \sin \alpha .
$$

2) Point $M$ lies on the smaller arc $K L$. The area of the triangle is maximized if $M$ lies on the line $\mathrm{OH}$, and it is equal to

$$
S_{\max }=\frac{1}{2} M H \cdot K L=(5-0 H) K H=(5 \sqrt{2}-5 \sqrt{2} \cos \alpha) 5 \sqrt{2} \sin \alpha=50(1-\cos \alpha) \sin \alpha .
$$

The area of triangle KOL is $S_{\mathrm{KOL}}=\frac{1}{2} \mathrm{OK} \cdot \mathrm{OL} \cdot \sin 2 \alpha=25 \sin 2 \alpha$. By the condition $S_{\mathrm{KOL}}=7$, so $\sin 2 \alpha=\frac{7}{25}$. Then $\cos 2 \alpha= \pm \sqrt{1-\frac{49}{625}}= \pm \frac{24}{25}$.

If $\cos 2 \alpha=-\frac{24}{25}$, then $\cos ^{2} \alpha=\frac{1+\cos 2 \alpha}{2}=\frac{1}{50}, \cos \alpha=\frac{1}{5 \sqrt{2}}, \sin \alpha=\frac{7}{5 \sqrt{2}}$. Then in the first case $S_{\max }=50 \cdot\left(1+\frac{1}{5 \sqrt{2}}\right) \cdot \frac{7}{5 \sqrt{2}}=35 \sqrt{2}+7$, and in the second case $-S_{\max }=50 \cdot\left(1-\frac{1}{5 \sqrt{2}}\right) \cdot \frac{7}{5 \sqrt{2}}=35 \sqrt{2}-7$. Only the first case is possible. Then $\angle \mathrm{KML}=\alpha=\arccos \frac{1}{5 \sqrt{2}}$.

If $\cos 2 \alpha=\frac{24}{25}$, then $\cos ^{2} \alpha=\frac{1+\cos 2 \alpha}{2}=\frac{49}{50}, \cos \alpha=\frac{7}{5 \sqrt{2}}, \sin \alpha=\frac{1}{5 \sqrt{2}}$. Then in the first case $S_{\max }=50 \cdot\left(1+\frac{7}{5 \sqrt{2}}\right) \cdot \frac{1}{5 \sqrt{2}}=5 \sqrt{2}+7$, and in the second case $-S_{\max }=50 \cdot\left(1-\frac{7}{5 \sqrt{2}}\right) \cdot \frac{1}{5 \sqrt{2}}=5 \sqrt{2}-7$. Both options are impossible.

Therefore, $\angle \mathrm{KML}=\arccos \frac{1}{5 \sqrt{2}}$.