Let's determine a point on each of two skew edges of a regular octahedron with edge length $a$, such that the distance between these two points is minimized. What is this minimal distance?
I. Solution: Let the edges $A B C D E F$ of the octahedron be such that the edges $A D$ and $B C$ are perpendicular to the plane of our drawing, and project the octahedron orthogonally onto this plane (Figure 1).

![](https://cdn.mathpix.com/cropped/2024_05_02_59a0f77142379d80d28eg-1.jpg?height=647&width=493&top_left_y=262&top_left_x=751)

Figure 1

In the projection, the edges $A D$ and $B C$ degenerate into points. The points $M N$ on the edges $A D$ and $E B$ are intersected by the line that (besides intersecting both edges) is perpendicular to both edges (normal transversal). From this determination, it follows that $M N$ is parallel to the projection plane and thus appears in its true size in the projection. Since the projection plane of the edge $E B$ is parallel to the edge $A D$, in the projection $M N \perp E B$. The ratio $E N$ to $N B$ (since we are talking about a parallel projection) is the same in space and in the projection.

In the projection,

$$
N B C_{\triangle} \sim B O E_{\triangle}
$$

as right-angled triangles with a common acute angle, so

$$
M N: A B=E O: E B
$$

But $A B=a, \quad E O=\frac{a}{2} \sqrt{2} \quad$ and $\quad E B=\frac{a}{2} \sqrt{3}, \quad$ and thus

$$
\begin{gathered}
M N=\frac{a \frac{a}{2} \sqrt{2}}{\frac{a}{2} \sqrt{3}}=\frac{a \sqrt{6}}{3} \\
N B=\sqrt{A B^{2}-M N^{2}}=\sqrt{a^{2}-\frac{2 a^{2}}{3}}=\frac{a}{3} \sqrt{3} \\
E N=E B-N B=\frac{a}{2} \sqrt{3}-\frac{a}{3} \sqrt{3}=\frac{a}{6} \sqrt{3}
\end{gathered}
$$

Thus, $E N: N B=1: 2$.

Therefore, the points sought divide the edges in the ratio $1: 2$, and the sought minimum distance is $M N=\frac{a \sqrt{6}}{3}$.

II. Solution: Let the face $A B C$ of the octahedron be in the plane of our drawing. In this case, the face $D E F$, which is parallel to this face, is parallel to the plane of our drawing, and the orthogonal projection of the octahedron in the plane of our drawing is shown in Figure 2.

![](https://cdn.mathpix.com/cropped/2024_05_02_59a0f77142379d80d28eg-1.jpg?height=513&width=463&top_left_y=2134&top_left_x=793)

Figure 2

The normal transversal of the edges $A B$ and $D E$ is $M N$, which is clearly perpendicular to the plane of our drawing and thus its projection degenerates into a point. If we denote the intersection point of $A B$ and $E F$ in the projection by $P$, it is trivial from the angles that $A M=M E=$ $M P=P E=P B$, so the point $M$ trisects the edge $A B$.

The distance $M N$ is nothing other than the distance of the plane $[A B C]$ from the parallel plane $[D E F]$, or the distance of the point $D$ from the projection plane. The latter is a leg of a right-angled triangle, whose hypotenuse is the altitude of the triangle $A C D$ in space, and the other leg is the projection of this hypotenuse, $D Q$. $D Q$ is half the radius of the circle circumscribed around the projection of the octahedron and thus a third of the altitude of the triangle $Q B$.

Thus,

$$
M N=\sqrt{\left(\frac{a}{2} \sqrt{3}\right)^{2}}-\sqrt{\left(\frac{a}{6} \sqrt{3}\right)^{2}}=\sqrt{\frac{3 a^{2}}{4}-\frac{a^{2}}{12}}=\sqrt{\frac{8 a^{2}}{12}}=a \sqrt{\frac{2}{3}}=\frac{a \sqrt{6}}{3} .
$$

László Eördögh (Budapest, VIII., Apáczai Csere g. IV. o. t.)