11.5. Three edges of a tetrahedron, emanating from one vertex, are mutually perpendicular, and have lengths 3, 4, and 4. Find the radii of the inscribed and circumscribed spheres of the tetrahedron.
Answer: $R=\frac{\sqrt{41}}{2}, r=\frac{12(10-\sqrt{34})}{33}$.

Solution. a) The sphere circumscribed around the tetrahedron is determined by 4 points not lying in the same plane - the common point of three perpendicular edges and the ends of these edges. This sphere will also be the circumscribed sphere around the rectangular parallelepiped constructed on these edges. The center of the sphere circumscribed around the rectangular parallelepiped is the point of intersection of its diagonals, which are bisected at the point of intersection. Hence, the radius of the circumscribed sphere $R=\frac{1}{2} \sqrt{3^{2}+4^{2}+4^{2}}=\frac{\sqrt{41}}{2}$

b) Let in the tetrahedron $A B C D A B=3, A C=4$ and $A D=4$. A sphere can be inscribed in any tetrahedron. If the center $O$ of the sphere is connected to the vertices of the tetrahedron, it will divide the tetrahedron into 4 smaller tetrahedra with a common vertex $O$ and bases that are the faces of the original tetrahedron $A B C D$. The volume $V_{A B C D}=\frac{1}{3} \cdot 3 \cdot 4 \cdot 4=16$ and is equal to the sum of the volumes of the four smaller tetrahedra. The heights of all these tetrahedra are equal to the radius $r$ of the inscribed sphere. We will find the areas of the bases of these tetrahedra, i.e., the areas of the faces of the tetrahedron $A B C D$:

$$
S_{A B D}=\frac{1}{2} \cdot 4 \cdot 3=6, S_{A B C}=\frac{1}{2} \cdot 4 \cdot 3=6, S_{A C D}=\frac{1}{2} \cdot 4 \cdot 4=8
$$

By the Pythagorean theorem:

$$
B C=\sqrt{A B^{2}+A C^{2}}=5, B D=\sqrt{A B^{2}+A D^{2}}=5, C D=\sqrt{A D^{2}+A C^{2}}=4 \sqrt{2}
$$

Triangle $B C D$ is isosceles, the height dropped to the base is

$$
\sqrt{5^{2}-(2 \sqrt{2})^{2}}=\sqrt{17}, \text { and } S_{B C D}=\frac{1}{2} \cdot 4 \sqrt{2} \cdot \sqrt{17}=2 \sqrt{34}
$$

Thus, $16=V_{A B C D}=\frac{1}{3} \cdot r \cdot\left(S_{A B D}+S_{A B C}+S_{A C D}+S_{B C D}\right)$, from which

$$
r=\frac{48}{20+2 \sqrt{34}}=\frac{12(10-\sqrt{34})}{33}
$$

Comments. The radius of the circumscribed sphere is worth 2 points, the inscribed - 5 points.