We construct a regular octahedron paper model with $10 \mathrm{~cm}$ edges from sheets of paper, then remove two adjacent faces. As a result, the remaining faces become movable along the edges. Determine the distance between the endpoints of the removed edge in all possible positions. What are the limits within which the resulting numbers move?
Suspend the octahedron vertically along one of its axes, and label the lower vertex among those on it as $A$, the upper one as $F$, and the other vertices as $B, C, D, E$ in sequence. Leave the faces resting on the $B E$ edge, and move the remaining faces along the edges as the model allows. During the movement, keep the labeling of the vertices and the vertical position of the $A F$ segment. Thus, the $B, C, D, E$ vertices will remain equidistant from the $A, F$ points, so they will always be in the perpendicular bisector plane $S$ of the $A F$ segment. The movement does not impose any constraint if we assume that $S$ does not change its position, and we can even assume that the $C D$ edge also remains stationary. Then the $A, F$ points will be on the $\sqrt{3} / 2$ radius circle $k$ in the perpendicular bisector plane of the $C D$ segment, and the $B, E$ points will be on the unit radius circles $k_{1}, k_{2}$ in $S$ centered at $C$ and $D$ respectively (we choose the edge length, or the decimeter, as the unit. All three circles pass through the third vertices $G$ and $H$ of the two equilateral triangles in $S$ resting on the $C D$ edge.

![](https://cdn.mathpix.com/cropped/2024_05_02_633bfc65200d95a37599g-1.jpg?height=987&width=1423&top_left_y=550&top_left_x=340)

If we move the $A, F$ points on $k$ towards $G$, the $B, E$ points move away from $G$, since the lengths of the $A B, B E, F E, E A$ edges remain unchanged. Since during this $B E \leq B F+F E=2$, the maximum distance between the $B, E$ points is 2. Then $F$ and $A$ coincide with the point $G$, and the faces of the octahedron overlap into $S$. Moving the $A, F$ points in the opposite direction, the $B, E$ points get closer to each other and to $G$, until they finally coincide with it. If $d$ is any number for which

$$
0 \leq d \leq 2
$$

is satisfied, draw lines parallel to the $G H$ line in $S$ at a distance of $d / 2$ from it, and let these intersect the circles $k_{1}, k_{2}$ on the $G$ side of $C D$ at points $B, E$ respectively. Then $B G \leq 1 \leq B H$, so the unit sphere centered at $B$ intersects $k$ at two points, let these be $A$ and $F$. Thus, we have obtained the points of our network, so the distance between the $B, E$ points can be any number $d$ satisfying (1).

Remark. Let the midpoint of $C D$ be $K$, the angle $B C D$ be $\gamma$, and the angle $G K F$ be $\varphi$. Then

$$
B F^{2}=\frac{3}{4} \sin ^{2} \varphi+\left(\cos \gamma-\frac{1}{2}\right)^{2}+\left(\sin \gamma-\frac{\sqrt{3}}{2} \cos \varphi\right)^{2}=1
$$

so

$$
\cos \varphi=\frac{1-\cos \gamma}{\sqrt{3} \sin \gamma}
$$

If, starting from $G$, $B$ rotates around $C$ at a constant angular speed, $F$ does not move at a constant angular speed on $k$, but its position is a continuous function of $\gamma$. This continuity ensures mathematically the physical existence of the movement in question. We do not consider the demonstration of this to be part of the solution, because in the problem we only used the intuitive geometric image of the movement.