254. Two vertices of a tetrahedron are located on the surface of a sphere with radius $\sqrt{10}$, the other two vertices - on the surface of a sphere with radius 2, concentric with the first. What is the greatest volume of such tetrahedra?
254. Let $ABCD$ be a tetrahedron of maximum volume, and $O$ be the center of the given spheres. Each segment connecting $O$ with a vertex of the tetrahedron must be perpendicular to the face opposite that vertex. For example, if $AO$ is not perpendicular to the plane $BCD$, then on the surface of the sphere on which point $A$ lies, there can be found points that are farther from the plane $BCD$ than point $A$. (This reasoning, obviously, remains valid if $A, B, C$, and $D$ lie on the surfaces of different spheres and are not necessarily concentric.) It follows that the opposite edges of the tetrahedron $ABCD$ are pairwise perpendicular. Suppose further that points $A$ and $B$ lie on a sphere of radius $R=\sqrt{10}$, and $C$ and $D$ lie on a sphere of radius $r=2$. Let $x$ and $y$ be the distances from $O$ to $AB$ and $CD$ respectively.

Draw a section through $AB$ perpendicular to $CD$. Let $K$ be the point of intersection of this plane with $CD$. Considering the properties of our tetrahedron $ABCD$, it is not difficult to prove that $|AK| = |BK|$, and $O$ is the point of intersection of the altitudes of $\triangle ABK$. Draw the altitudes $KL$ and $AM$ (Fig. 51). From the similarity of triangles $ALO$ and $OKM$, we find $|OM| = \frac{xy}{R}$. Further, $|AB| = 2 \sqrt{R^2 - x^2}$, and from the similarity of triangles $ALO$ and $AMB$, we get

![](https://cdn.mathpix.com/cropped/2024_05_21_9b03e22d2b8ca67aa653g-126.jpg?height=367&width=447&top_left_y=228&top_left_x=626)

Fig. 51. 

$$
\frac{R}{\sqrt{R^2 - x^2}} = \frac{2 \sqrt{R^2 - x^2}}{R + \frac{xy}{R}},
$$

from which $2x^2 + xy = R^2$. Similarly, we obtain the equation $2y^2 + xy = r^2$. From the system of equations

$$
\left\{\begin{array}{l}
2x^2 + xy = 10 \\
2y^2 + xy = 4
\end{array}\right.
$$

we find $x = 2, y = 1$. The volume of the tetrahedron $ABCD$ will be equal to $6 \sqrt{2}$.