An insect lives on the surface of a regular octahedron with edges of length $2$. It travels along the surface from the midpoint of one edge to the midpoint of the opposite edge (the edge symmetric to it about the center of the octahedron). What is the length of the shortest such trip?

1. Unfold three faces that touch in a strip: they form an isosceles trapezoid with bases $4$ and $2$ and legs $2$. The two given edges are the two legs of this trapezoid.
2. A shortest path on the surface is a straight segment in an unfolding. The midpoints of the legs are joined by the midline of the trapezoid, which has length $\dfrac{4+2}2=3$.
3. No face contains points of both edges, and no face through one of the edges shares an edge with a face through the other, so every path crosses at least three faces; by symmetry the three faces around a common vertex give the shortest crossing. So the shortest trip has length $\boxed{3}$.