94. The base of the pyramid $ABCD$ is an equilateral triangle $ABC$ with a side length of 12. The edge $BD$ is perpendicular to the plane of the base and equals 10 $\sqrt{3}$. All vertices of this pyramid lie on the lateral surface of a right circular cylinder, whose axis intersects the edge $BD$ and the plane $ABC$. Determine the radius of the cylinder.
94. It can be proven that the axis of the cylinder should pass through the midpoint of edge $B D$ and belong to the plane $B D L$, where $L$ is the midpoint of $A C$. Let the axis of the cylinder form an acute angle $\alpha$ with $B D$. Project the pyramid onto a plane perpendicular to the axis of the cylinder, obtaining a quadrilateral $A_{1} B_{1} C_{1} D_{1}$, in which $\left|A_{1} C_{1}\right|=|A C|=$ $=12$. The diagonals $A_{1} C_{1}$ and $B_{1} D_{1}$ are perpendicular, the point of intersection of the diagonals $F$ bisects $A_{1} C_{1}$, and the diagonal $D_{1} B_{1}$ is divided by point $F$ into segments $6 \sqrt{\overline{3}} \cos \alpha$ and $10 \sqrt{3} \sin \alpha-6 \sqrt{3} \cos \alpha$. From the condition $\left|A_{1} F\right| \cdot\left|F C_{1}\right|=\left|B_{1} F\right| \cdot\left|F D_{1}\right|$, we obtain the equation for $\alpha$:

$$
\sin ^{2} \alpha-5 \sin \alpha \cos \alpha+4 \cos ^{2} \alpha=0
$$

from which we find $\operatorname{tg} \alpha_{1}=1, \operatorname{tg} \alpha_{2}=4$. But $\left|B_{1} D_{1}\right|=10 \sqrt{3} \sin \alpha$ and equals the diameter of the base of the cylinder. We obtain two values for the radius of the base of the cylinder: $\frac{5 \sqrt{6}}{2}$ and $\frac{20 \sqrt{3}}{\sqrt{17}}$.