6.14. All faces of the tetrahedron are similar right-angled triangles. Find the ratio of the greatest edge to the smallest.
6.14. Let $a$ be the length of the longest edge of the tetrahedron. In both faces adjacent to this edge, it is the hypotenuse. These faces are equal, as similar right triangles with a common hypotenuse are equal; let $m$ and $n$ be the lengths of the legs of these right triangles, and $b$ be the length of the sixth edge of the tetrahedron. There are two possible cases:

1) Edges of length $m$ extend from one end of the edge $a$, and edges of length $n$ extend from the other. In the triangle with sides $m, m$, and $b$, the right angle can only be opposite $b$; moreover, in the triangle with sides $a, m$, and $n$, the legs must not be equal, i.e., $m \neq n$. In the end, we get that all faces of the tetrahedron are equal.
2) From each end of the edge $a$, one edge of length $m$ and one edge of length $n$ extend. In this case, if $a = b$, the tetrahedron will also be equifacial.

Note now that in an equifacial tetrahedron, there cannot be any right dihedral angles (Problem 6.49). Therefore, in reality, only the second case is possible, and $b < a$. Considering that $a^{2} = m^{2} + n^{2}$, we get $\lambda^{4} = \lambda^{2} + 1$, i.e., $\lambda = \sqrt{(1 + \sqrt{5}) / 2}$.