5. In the spatial tetrahedron $ABCD$, each face is a similar right triangle. Let $a$ be the length of the longest edge, and $b$ be the length of the shortest edge. Find $a: b$.

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5. The four faces of a tetrahedron cannot all be congruent.

Otherwise, the trihedral angle formed by three edges meeting at the same vertex can be denoted as $\alpha, \beta, \gamma$. Since all four faces are congruent, $\alpha, \beta, \gamma$ are the three interior angles of a congruent triangle, so $\alpha+\beta+\gamma=\pi$. However, by the property of the trihedral angle, $\alpha+\beta>\gamma$, which contradicts the fact that $\alpha, \beta, \gamma$ are all acute angles.

Therefore, let $DC$ be the longest edge, $DC=a$. By the similarity of the tetrahedron, the shortest edge and $DC$ cannot be coplanar, so it must be $AB$, and let $AB=b$.

As shown in Figure 15-13, Rt $\triangle DBC \sim \operatorname{Rt} \triangle DAC$, and they share the hypotenuse $DC$, so $\triangle DBC \cong \triangle CAD$, and $DB=AC, DA=BC, (DA \neq DB)$.
Let the length of $DC$ be $a$, and the length of $AD$ be $ka$, then $BC=ka$.
From $BC^2=DC \cdot AC$, we get $AC=DB=k^2a$, and from $DB^2=AB \cdot BC$, we get $AB=k^3a$.
In Rt $\triangle DAC$, we have $k^4a^2 + k^2a^2 = a^2$, which simplifies to $k^4 + k^2 - 1 = 0$. Solving this, we get
$$
\begin{aligned}
k & =\sqrt{\frac{\sqrt{5}-1}{2}}, \text{ hence } \\
\frac{a}{b} & =\frac{(\sqrt{5}-1) \sqrt{\sqrt{5}-1}}{4} .
\end{aligned}
$$