7. Given a regular prism $K L M N K_{1} L_{1} M_{1} N_{1}$ with base $K L M N$. Planes $\alpha$ and $\beta$ are perpendicular to $L_{1} N$ and pass through vertices $K$ and $N_{1}$, respectively. Let $A$ and $B$ be the points of intersection of planes $\alpha$ and $\beta$ with the diagonal $L_{1} N$, such that $A N < B N$.

a) Find the ratio $L_{1} B: A N$.

b) Suppose additionally that a sphere of radius $\frac{1}{2}$ is tangent to all lateral faces of the prism, as well as to planes $\alpha$ and $\beta$. Find the segment $L_{1} N$ and the volume of the prism $K L M N K_{1} L_{1} M_{1} N_{1}$.
Answer: a) $2: 1$, b) $L_{1} N=\frac{1}{2}(1+\sqrt{13}), V=\frac{1}{2} \sqrt{6+2 \sqrt{13}}$.

Solution. a) By symmetry (with respect to the plane $L N N_{1} L_{1}$), the plane $\alpha$ passes through the point $M$ - and, therefore, through the center $O$ of the face $K L M N$. The segments $L_{1} B$ and $A N$ are projections of the parallel segments $L_{1} N_{1}$ and $N O$ onto the line $L_{1} N$, and $L_{1} N_{1}=2 N O$. Therefore, $L_{1} B: A N=2$.

b) Since the sphere is tangent to all lateral faces of the prism, its projection onto the base is a circle inscribed in this base. Thus, $K N=2 r=1$. Moreover, $\alpha$ and $\beta$ are two parallel planes tangent to the sphere, so the distance between them is equal to the diameter of the sphere, which is 1. Since $L_{1} N \perp \alpha$, this distance is the segment $A B$, so $A B=1$.

Let $L_{1} N=d$. Since $N_{1} B$ is the height of the right triangle $L_{1} N_{1} N$, then $L_{1} B \cdot L_{1} N=L_{1} N_{1}^{2}=2$ and, therefore, $L_{1} B=\frac{2}{d}$. Then $A N=\frac{1}{2} L_{1} B=\frac{1}{d}$ and $A B=L_{1} N-L_{1} B-A N=d-\frac{2}{d}-\frac{1}{d}$. We get the equation $1=d-\frac{3}{d}$, from which $d^{2}-d-3=0, d=\frac{1+\sqrt{13}}{2}$ (since $d>0$).

Finally, the height of the prism is $h=\sqrt{L_{1} N^{2}-L N^{2}}=\sqrt{\frac{1}{4}(14+2 \sqrt{13})-2}=\frac{1}{2} \sqrt{6+2 \sqrt{13}}$; then the volume of the prism is $V=K L^{2} \cdot h=\frac{1}{2} \sqrt{6+2 \sqrt{13}}$.