How far are the midpoints of two skew edges of a regular octahedron with edge length $a$ from each other?
I. solution: The labeling is shown in the figure. $M, N, P$ are the midpoints of the respective edges, so $M N$ is the desired distance.

![](https://cdn.mathpix.com/cropped/2024_05_02_ed3e24c4ef364bc81cd9g-1.jpg?height=539&width=501&top_left_y=191&top_left_x=801)

In $\triangle M N P$, $M P$ (as the midline of $\triangle D E A$) $=\frac{a}{2}, P N=a$, and $\angle M P N = \angle E D C = 60^{\circ}$ (since the legs are parallel and in the same direction). It follows that $\triangle M N P$ is a right triangle, with hypotenuse $P N = a$, one leg $M P = \frac{a}{2}$, and thus the desired leg $M N = \frac{a}{2} \sqrt{3}$.

Endre Csáki (Győr, Révai g. IV. o. t.)

II. solution: Connecting the midpoint $Q$ of edge $B E$ with points $M$ and $N$, the resulting trapezoid $P N Q M$ has side $P N = a$, and the other three sides are $\frac{a}{2}$. Thus, this trapezoid is half of a regular hexagon with side length $\frac{a}{2}$ (a cross-section of the octahedron). The radius of the circle circumscribed around this hexagon is $\frac{a}{2}$, and thus the side of the inscribed regular triangle $M N = \frac{a}{2} \sqrt{3}$.

Pál Bártfai (Bp., L, Petófi g. II. o. t.)

III. solution: Let $R$ be the midpoint of edge $D E$, then $M R \| A D \| N C$, and $M R = \frac{a}{2} = N C$, so $N C R M$ is a parallelogram because two sides are parallel and equal. It follows that the other two sides are also equal (and parallel), so $M N = C R$. (In other words, $C R M N$ is a parallel translation). But $C R$ is the height of the equilateral triangle $C D E$, and thus $M N = \frac{a}{2} \sqrt{3}$.