Let $AB,AC$ and $AD$ be the edges of a cube, $AB=\alpha$. Point $E$ was marked on the ray $AC$ so that $AE=\lambda \alpha$, and point $F$ was marked on the ray $AD$ so that $AF=\mu \alpha$ ($\mu> 0, \lambda >0$). Find (characterize) pairs of numbers $\lambda$ and $\mu$ such that the cross-sectional area of a cube by any plane parallel to the plane $BCD$ is equal to the cross-sectional area of the tetrahedron $ABEF$ by the same plane.
Problem
Source: IX All-Ukrainian Tournament of Young Mathematicians, Qualifying p12
Tags: geometry, 3D geometry, tetrahedron, Ukrainian TYM