11.6. Edges $S A$, $S B$, and $S C$ of the tetrahedron $S A B C$ are mutually perpendicular, and their lengths are 2, 3, and 6, respectively. a) Find the radius $R$ of the sphere circumscribed around the tetrahedron. b) Does there exist a sphere with a radius smaller than $R$ that contains the given tetrahedron?
Answer. a) $R=7 / 2$; b) exists. Solution. Consider a rectangular coordinate system in space with the origin at point $S$ and coordinate axes along the edges $O A, O B$ and $O C$. The center of the circumscribed sphere can be found in different ways. One way is direct: if we write the equality of the squares of the distances for the desired center $O(x, y, z)$ as $O A^{2}=O B^{2}=O C^{2}=O S^{2} \cdot$ By subtracting the obtained equations, we will have linear equations from which it is easy to find $x, y, z$. Another way is to notice that the center of the sphere should project onto the plane of triangle $A B S$ at the center of the circumscribed circle of this triangle, i.e., at the midpoint of the hypotenuse $A B$ - the point $\left(1 ; \frac{3}{2} ; 0\right)$, and thus the coordinates of the center of the sphere have the form $\left(1 ; \frac{3}{2} ; z\right)$. Similarly, considering triangle $A C S$, we get that the coordinate $z$ should be equal to half the value of $z_{C}$, i.e., equal to 3. As a result, $R=\sqrt{1^{2}+\left(\frac{3}{2}\right)^{2}+3^{2}}=\frac{7}{2}$. The third way is as follows: note that for a rectangular parallelepiped with sides $S A, S B$ and $S C$, its center is equidistant from all vertices of the parallelepiped and, in particular, from $A, B, C$ and $S$. Therefore, the radius $R$ is half the diagonal of the parallelepiped: $R=\frac{1}{2} \sqrt{2^{2}+3^{2}+6^{2}}=\frac{7}{2}$.

b) Consider the center $O_{1}$ of the circle circumscribed around triangle $A B C$. The point $O_{1}$ is the projection of the center $O(x, y, z)$ of the sphere onto the plane $A B C$. If solving the problem in part a) in coordinates, it is not difficult to find the coordinates of point $O_{1}$ as the projection of point $O$ onto the plane: $\frac{x}{2}+\frac{y}{3}+\frac{z}{6}=1$ (the equation of the plane $A B C$) and to verify that the distance from $O_{1}$ to all vertices of the tetrahedron is less than $R$. But we can do without detailed calculations. Indeed, first note that points $S$ and $O$ lie on opposite sides of the plane $A B C$ (this follows, for example, from part a), see the second method for finding the center $O$). Further, the distances $O_{1} A=O_{1} B=O_{1} C$ are strictly less than $R$, since $R$ is the length of the hypotenuse in the corresponding right triangles $O O_{1} A, O O_{1} B, O O_{1} C$. The distance $O_{1} S<O S=R$, since triangle $S O O_{1}$ is obtuse with the obtuse angle $S O_{1} O$; this follows from the equality $\angle O O_{1} M=90^{\circ}$, where $M$ is the point of intersection of the line $S O$ with the plane $A B C$. Thus, all distances from $O_{1}$ to the vertices of the tetrahedron are less than $R$, and therefore the sphere with center $O_{1}$ and radius equal to the minimum of these distances will be the desired one.

## 11th grade