2. [5 points] Consider all possible tetrahedra $ABCD$ in which $AB=4, AC=CB=5, AD=DB=6$. Each such tetrahedron is inscribed in a cylinder so that all vertices lie on its lateral surface, and the edge $CD$ is parallel to the axis of the cylinder. We select the tetrahedron for which the radius of the cylinder is the smallest among those obtained. What values can the length of $CD$ take in such a tetrahedron?
Answer: $2 \sqrt{7} \pm \sqrt{17}$.

Solution. Let $E$ be the midpoint of $AB$. $CE$ and $DE$ are the medians of the isosceles triangles $ABC$ and $ABD$, and thus are also the angle bisectors and altitudes. Therefore, $AB \perp CE, AB \perp DE$. This means that the segment $AB$ is perpendicular to the plane $CDE$, hence $AB \perp CD$. Thus, $AB$ lies in a plane perpendicular to the axis of the cylinder (denote this plane as $\alpha$). The intersection of the cylinder with this plane is a circle, and $AB$ is a chord of this circle. The radius of the cylinder is minimized if $AB$ is the diameter. Note that this is possible because the segments $DE$ and $CE$ are longer than $\frac{1}{2} AB = 2$. Indeed, from triangles $ACE$ and $ADE$ it follows that $CE = \sqrt{5^2 - 2^2} = \sqrt{21}$, $DE = \sqrt{6^2 - 2^2} = 4 \sqrt{2}$.

Consider the tetrahedron in which $AB$ is the diameter of the cylinder. There are two cases: points $C$ and $D$ lie on the same side or on opposite sides of the plane $\alpha$.

Let $H$ be the projection of points $C$ and $D$ onto the plane $\alpha$. The angle $\angle AHB = 90^\circ$, as it is inscribed in a circle and subtends the diameter. $AH = BH$ due to the congruence of triangles $ACH$ and $BCH$. Therefore, $AH = BH = 2 \sqrt{2}$. By the Pythagorean theorem in the right triangles $AHC$ and $DHC$ respectively: $CH = \sqrt{25 - 8} = \sqrt{17}$, $DH = \sqrt{36 - 8} = 2 \sqrt{7}$.

Thus, if points $C$ and $D$ lie on the same side of the plane $\alpha$, then $CD = DH - CH = 2 \sqrt{7} - \sqrt{17}$. If points $C$ and $D$ lie on opposite sides of the plane $\alpha$, then $CD = DH + CH = 2 \sqrt{7} + \sqrt{17}$.