There are two diverging, perpendicular lines $m$ and $n$ in space. The line that intersects both lines perpendicularly (the so-called normal-transversal or binormal) intersects the given lines at points $M$ and $N$ respectively, with $MN = 2d$. On line $m$, we take a point $A$ at a distance $x$ from $M$, and on line $n$, we take a point $B$ at a distance $y$ from $N$. Determine the center $O$ and the radius of the sphere $G$ circumscribed around the tetrahedron $MABN$.

What is the geometric locus of the points $O$ if the points $A$ and $B$ move on the lines $m$ and $n$ respectively, such that

a) the radius of the sphere $G$ is constantly $2d$;

b) $x = y$ is always true.
Let's complete the tetrahedron $M A B N$ to a parallelepiped, as shown in the figure.

![](https://cdn.mathpix.com/cropped/2024_05_02_1e864e970f37670b528ag-1.jpg?height=559&width=553&top_left_y=222&top_left_x=778)

Since, according to the conditions, $M N \perp M A$, $M N \perp N B$, and $M A \perp N B$, the parallelepiped is a rectangular parallelepiped, and a sphere can be circumscribed around it, which coincides with the sphere circumscribed around the tetrahedron $M A B N$, since four different points that are not in the same plane determine one and only one sphere. The center of the desired sphere is thus the common midpoint of the space diagonals of the rectangular parallelepiped. One of the diagonals is $A B$. Using the Pythagorean theorem twice,

$$
A B = \sqrt{x^2 + (2d)^2 + y^2} = \sqrt{x^2 + y^2 + 4d^2}
$$

The center of the desired sphere is thus the midpoint $O$ of the segment $A B$, and its radius is

$$
O A = O M = O N = O B = \frac{A B}{2} = \frac{1}{2} \sqrt{x^2 + y^2 + 4d^2}
$$

Since $O M = O N$, the point $O$ lies in the plane that perpendicularly bisects the segment $M N$, which intersects the segment $M N$ at its midpoint $F$.

a) If the radius of the sphere is constantly $2d$, that is,

$$
A O = O M = 2d
$$

then in the right triangle $O F M$,

$$
O F = \sqrt{O M^2 + M F^2} = \sqrt{4d^2 - d^2} = d \sqrt{3}
$$

Thus, the points $O$ lie on a circle whose plane perpendicularly bisects the segment $M N$, whose center is the point $F$, and whose radius is $d \sqrt{3}$. Since any point on this circle can be the center of a suitable sphere, the above circle is the geometric locus of the points $O$.

b) If $x = y$, then the distance of the points $O$ from the plane $[A M N]$ is $\left(\frac{y}{2}\right)$, the same as from the plane $[B M N]$ $\left(\frac{x}{2}\right)$, so the points $O$ lie in the bisector planes of these two planes. Based on the above, the points $O$ lie on the intersection line of the plane that perpendicularly bisects the segment $M N$ and these two bisector planes. This intersection line is also the geometric locus of the points $O$, because any point on these lines can be the point $O$.

Note: Both geometric loci can be easily generalized to the case a) when the radius of the sphere $r = \text{constant} (>d)$ and b) when $x = k y$, where $k$ is a constant.

Tibor Harza (Székesfehérvár, József A. g. IV. o. t.)