4.9. Two equal triangles $K L M$ and $K L N$ share a common side $K L$, and $\angle K L M=\angle L K N=60^{\circ}$, $K L=1$ and $L M=K N=6$. The planes $K L M$ and $K L N$ are perpendicular. Find the radius of the sphere that touches the segments $L M$ and $K N$ at their midpoints.
4.9. Let $O_{1}$ and $O_{2}$ be the projections of the center $O$ of a given sphere onto the planes $K L M$ and $K L N$; $P$ and $S$ are the midpoints of the segments $L M$ and $K N$. Since $O P=O S$ and $P K=S L$, then $O K=O L$. Therefore, the projection of points $O_{1}$ and $O_{2}$ onto the line $K L$ is the point $Q$ - the midpoint of the segment $K L$. Since the planes $K L M$ and $K L N$ are perpendicular, then $O O_{1}=O_{2} Q=Q O_{1}$; therefore, the square of the radius of the desired sphere is $P O_{1}^{2}+O O_{1}^{2}=P O_{1}^{2}+Q O_{1}^{2}$. Applying the cosine theorem to the triangle $K L M$, we get $K M^{2}=31$. By the sine theorem, $31=\left(2 R \sin 60^{\circ}\right)^{2}=3 R^{2} . \quad$ Therefore, $P O_{1}^{2}+Q O_{1}^{2}=\left(R^{2}-\right.$ $\left.-P L^{2}\right)+\left(R^{2}-Q L^{2}\right)=62 / 3-9-1 / 4=137 / 12$.