Five, glue the bases of two congruent regular tetrahedra together. In the resulting hexahedron, all dihedral angles are equal, and the vertices can be divided into two categories: in the first category, each vertex emits three edges; and in the second category, each vertex emits four edges. Try to find the ratio of the length of the line segment connecting two vertices of the first category to the length of the line segment connecting two vertices of the second category.
Let $D-A B C$ and $E-A B C$ be congruent regular tetrahedra, with their base being the equilateral $\triangle A B C$.

If their bases are glued together, then the vertices $D$ and $E$ of the resulting hexahedron should pass through the centroid of $\triangle A B C$, and lie on a line perpendicular to the plane $A B C$. This way, the resulting hexahedron is not only symmetric about the plane $A B C$, but also symmetric about the plane $A D E$.

By symmetry, the perpendicular segments from vertices $D$ and $E$ to the edge $A B$ are equal. Let $F$ be the foot of the perpendicular, then $\angle D F E$ is the plane angle of the dihedral angle $D-A B-E$.

Similarly, the perpendicular segments from vertices $B$ and $C$ to the edge $A D$ are also equal. Let $G$ be the foot of the perpendicular, then $\angle B G C$ is the plane angle of the dihedral angle $B-A D-C$.

From the given conditions, it can be proven that $\triangle A D E$ is also an equilateral triangle, with the ratio of its side length to the side length of $\triangle B C$ equal to the ratio of their heights. That is,
$$
\frac{D E}{B C}={ }_{A H}^{A S} \text {. And } A S=\frac{2}{3} \text {, so } \frac{D E}{B C}=\frac{2}{3} \text {. }
$$