A regular tetrahedron has a circle inscribed around its base with radius $r$. At what distance from the base, measured along a plane parallel to the base, should the tetrahedron be intersected so that the volume of the tetrahedron is halved?
If the radius of the circle circumscribed around the base of the tetrahedron is $r$, and one edge of the tetrahedron is $a$, then $a=r \sqrt{3}$; the height of the tetrahedron is

$$
m=\sqrt{a^{2}-r^{2}}=r \sqrt{2}
$$

It is known that the volumes of similar pyramids are in the ratio of the cubes of the corresponding heights, or the cubes of the radii of the circles circumscribed around the bases, which is why:

$$
2: 1=r^{3}: r_{1}^{3}
$$

if $r_{1}$ is the radius of the circle circumscribed around the base of the cut-off pyramid. From (1)

$$
r_{1}=r \sqrt[3]{\frac{1}{2}}
$$

But

$$
r_{1}: r=(m-x): m
$$

from which

$$
x=\frac{m\left(r-r^{\prime}\right)}{r}=r(\sqrt{2}-\sqrt[6]{2})
$$

(Gyula Póka, Losoncz.)

Number of solutions: 37