3.98. The ratio of the volume of a right parallelepiped to the volume of a sphere inscribed in it is $k$. Find the angles at the base of the parallelepiped and the permissible values of $k$.

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3.98. Let's represent the axial section of the figure by a plane passing through the center of the inscribed sphere and parallel to the base. In the section, we obtain a parallelogram \(ABCD\), equal to the base of the right parallelepiped, and the great circle of the inscribed sphere inscribed in it with the center at point \(O\) (Fig. 3.98). From this, it follows that the base of the parallelepiped is a rhombus. We need to find the angles of the base of the parallelepiped if \(V_{\text{par}}: V_{\text{sphere}} = k\).

Let the radius of the sphere be \(r\), then the height of the rhombus is \(2r\). Draw \(BE \perp AD\) and let \(\angle BAD = \alpha\). From \(\triangle BEA\), we find \(AB = \frac{2r}{\sin \alpha}\). Then \(V_{\text{sphere}} = \frac{4}{3} \pi r^3\), \(V_{\text{par}} = S_{ABCD} \cdot 2r = \frac{8r^3}{\sin \alpha}\). By the condition, \(\frac{8r^3}{\sin \alpha} : \frac{4 \pi r^3}{3} = k\), or \(\sin \alpha = \frac{6}{\pi k}\). Therefore, \(\alpha = \arcsin \frac{6}{\pi k}\), \(\angle ADC = \pi - \arcsin \frac{6}{\pi k}\); since \(0 < \sin \alpha \leq 1\), then \(\frac{6}{\pi k} \leq 1\), from which \(k \geq \frac{6}{\pi}\). When \(k = \frac{6}{\pi}\), the base of the parallelepiped is a square, i.e., the parallelepiped is a regular prism.