65. In the tetrahedron $S A B C$, the radii of the circumscribed circles of the faces $S A B, S B C$, and $S A C$ are 108. The radius of the inscribed sphere of this tetrahedron is 35, and the distance from the vertex $S$ to its center is 125. What is the radius of the circumscribed sphere of this tetrahedron?
Answer: 112.5. Solution. Let the center of the circumscribed sphere be denoted as $O$, and the center of the inscribed sphere as $I$. Drop perpendiculars from $O$ to the faces of the tetrahedron. Then their feet will be the centers of the circumscribed circles of these faces. Denote the center of the circumscribed circle of triangle $S B C$ as $O_{A}$. Then, since $O_{A} O \perp S B C$, by the Pythagorean theorem we have $S O^{2}=S O_{A}^{2}+O_{A} O^{2}$. Therefore, $O_{A} O=\sqrt{S O^{2}-S O_{A}^{2}}$. Similarly for other centers. Thus, $O$ is equidistant from the faces $S A B, S B C$, and $S A C$. That is, there exists a sphere with center at $O$ and touching the planes $S A B$, $S B C$, and $S C A$. But then this sphere is homothetic to the inscribed sphere of the tetrahedron with center at point $S$. Consequently, triangle $S O O_{A}$ is similar to triangle $S I I_{A}$, where $I_{A}$ is the point of tangency of the inscribed sphere with the face $S B C$. From the similarity, we find $S O=S O_{A} \cdot S I / S I_{A}=\frac{108 \cdot 125}{\sqrt{125^{2}-35^{2}}}=\frac{27 \cdot 4 \cdot 5 \cdot 25}{5 \cdot 8 \cdot 3}=112.5$.