In a tetrahedron, at each vertex, edges of lengths $5$, $\sqrt{41}$, and $\sqrt{34}$ meet. What are the lengths of the segments connecting the midpoints of the opposite edges of the tetrahedron?
Let the tetrahedron be $D E G H$. Let $F_{1}, F_{2}, F_{3}, F_{4}, F_{5}$, and $F_{6}$ be the midpoints of the edges $D H, D E, E G, G H$, $E H$, and $D G$, respectively. The edges meeting at vertex $E$ are of lengths $5, \sqrt{34}$, and $\sqrt{41}$, for example, $E D=5, E H=\sqrt{41}$, and $E G=\sqrt{34}$. If $D G=5$, then two edges of length 5 would emanate from vertex $D$, and similarly, if $D G=\sqrt{34}$, then two edges of length $\sqrt{34}$ would meet at vertex $G$; thus, $D G=\sqrt{41}$. Since $D E=5$ and $D G=\sqrt{41}$, it follows that $D H=\sqrt{34}$. Because $G E=\sqrt{34}$ and $G D=\sqrt{41}$, it follows that $G H=5$.

We use the fact that in any triangle, $s_{a}^{2}=\frac{2 b^{2}+2 c^{2}-a^{2}}{4}$, where the sides of the triangle are $a, b$, and $c$, and $s_{a}$ is the median to side $a$. (This statement is essentially a consequence of problem 289 in the Collection of Geometric Problems II.)

$F_{1} E$ is the median of triangle $D E H$, so $F_{1} E^{2}=\frac{2 D E^{2}+2 H E^{2}-D H^{2}}{4}=24.5$.

$F_{1} G$ is the median of triangle $D G H$, so $F_{1} G^{2}=\frac{2 D G^{2}+2 G H^{2}-D H^{2}}{4}=24.5$.

$F_{1} F_{3}$ is the median of triangle $E F_{1} G$, so $F_{1} F_{3}^{2}=\frac{2 F_{1} E^{2}+2 F_{1} G^{2}-E G^{2}}{4}=16$, hence $F_{1} F_{3}=4$.

Similarly, in triangle $H E G$: $F_{4} E_{2}=31.25$, in triangle $D G H$: $F_{4} D_{2}=31.25$, in triangle $D F_{4} E$: $F_{2} F_{4}=5$, and in triangle $D E H$: $F_{5} D_{2}=19.25$, in triangle $H G E$: $F_{5} G_{2}=19.25$, and in triangle $D F_{5} G$: $F_{5} F_{6}=3$.

The lengths of the segments connecting the midpoints of opposite edges of the tetrahedron are 3, 4, and 5, respectively, specifically, the midpoints of the edges of length $\sqrt{41}$ are connected by a segment of length 3, the midpoints of the edges of length $\sqrt{34}$ are connected by a segment of length 4, and the midpoints of the edges of length 5 are connected by a segment of length 5.

Balázs Szekeres (Szolnok, Verseghy Ferenc Gymnasium, 10th grade)

Note. It can be shown that the parallelepiped containing the tetrahedron in the problem is a rectangular cuboid. If the lengths of the edges of the cuboid are $a, b, c$, then by the Pythagorean theorem: $a^{2}+b^{2}=25, b^{2}+c^{2}=41, c^{2}+a^{2}=34$. From this system of equations: $a=3, b=4, c=5$. These edge lengths now coincide with the lengths of the segments connecting the midpoints of opposite edges of the tetrahedron.