Routh's theorem in three dimensions The most well known case of Routh's triangle theorem is:  

If the sides BC, CA,and AB are trisected at the points D, E, and F, respectively, then the area of the inside triangle formed by AD, BE, CF is $\dfrac{1}{7}$th of the area of that of the triangle ABC.

Here is my question: can Routh's theorem be generalized to a tetrahedron which is cut by 4 planes through its 4 vertices and cutting the opposite faces appropriately?
As far as I know, this question has never been contemplated in the literature.
 A: Yes, it generalizes. 
For any two nondegenerate tetrahedra A and B, you can find an affine transformation such that T(A)=B. Since affine transformations preserve ratios of line segments and areas and volumes, with a Routh's-theorem-type construction the ratio of volume of original tetrahedron to volume of inner polyhedron will be preserved. 
As a commenter pointed out, with a tetrahedron there may be more than one way of defining a sensible cut. Once you've defined this, though, you can find a convenient tetrahedron (maybe a regular one, maybe one with a lot of right angles) and use it to calculate the proportion you want.
A: The following paper establishes a generalization of Routh's theorem to 3 dimensions:
Semyon Litvinov,
František Marko, Routh’s theorem for tetrahedra.Geom. Dedicata 174 (2015), 155–167
First, a new tetrahedron is determined by its vertices, which are points on 4 edges of the original one. The authors start with a tetrahedron $ABCD$, choose points $M, N, K, L$ on the edges $AB$, $BC$, $CD$, $DA$ which cut these edges in ratios $x, y, z, t$ respectively, and give an explicit formula for the ratio of the volumes of $KLMN$ and $ABCD$ in terms of $x, y, z, t$. Another tetrahedron is enclosed by the planes $ABK, BCL, CDM, DAN$ and the authors find explicitly the corresponding volume ratio.
A follow-up paper is available on the ArXiv and contains $n$-dimensional generalizations and interesting bibliographical remarks.
