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.

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    $\begingroup$ You should tell us how you want the 4 planes to cut the opposite faces. If the tetrahedron has vertices ABCD, and you want to take a plane through A which is parallel to one of the edges of the face BCD, then you need to choose that edge. Note that this will leave at least two edge unchosen however you do it. $\endgroup$ – Andrew Lobb Mar 14 '10 at 17:35

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.

  • $\begingroup$ The main problem is HOW do you define a "sensible cut"?...Do you "trisect" the area of the opposite face, do any three of the cutting planes define a corresponding 1/7 triangle in a face of the tetrahedon? Yes, T(A)=B, but I am looking for an explicit construction. $\endgroup$ – Mark B Villarino Mar 15 '10 at 11:18

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.


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