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Comments on Dividing a cuboid in four

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Dividing a cuboid in four

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Suppose we have an irregular cuboid (that is, a hexahedron with six irregular quadrilateral faces) ABCDEFGH that we wish to divide into four smaller cuboids, such that three of them each fully share a face with one of the outer cuboid's three faces adjacent to point H (so, JKCDNOGH, IBKDMFOH, and LMNOEFGH in the diagram), and the fourth cuboid shares a vertex with opposite point A (thus AIJKLMNO in the diagram), and such that each of the four cuboids has equal volume.

Diagram

How might one determine coordinates for I, J, K, L, M, N and O, that satisfy these constraints?

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Firstly, I assume that your usage of "quadrilateral" implies planarity, and I will largely assume that the hexahedron is convex. Then the hexahedron can be represented as six vectors, being the (unnormalised) normals for the planes under the convention that the plane is given by $v \cdot N = 1$. This gives 18 degrees of freedom for the hexahedron, which is less than the 24 df of 8 vertices in general position because of the planarity constraints.

The smaller hexahedron $AIJKLMNO$ shares three of its planes with $ABCDEFGH$ so it has 9 df. The volume constraints on the subhexahedra give three independent constraints; I think the planarity constraints on $OKDH$, $OMFH$, $ONGH$ are also independent; so that leaves 3 df of flexibility.

To set up all of the volumes, the easiest approach is probably to assume that all the hexahexadra are convex; take the centroid of each one as the point of six pyramids with the hexahedron faces as their bases; if the point of a pyramid is $X$, the base is $ABCD$ in clockwise order, and $P$ is any point in the plane of the base, then the (signed) volume is $\frac16 (X-P) \cdot ((A-C) \times (B-D))$. For symmetry it's probably useful to take $P = \frac{A+B+C+D}{4}$. With the correct assignment of signs you get a quadratic programming setup to which you can add any objective function you want and tackle with standard techniques. It's probably worth trying to take $O = A + \frac{H-A}{\sqrt[3]4}$ as an initial guess to assign the signs.

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Thank you so much for this. Unfortunately, my understanding of mathematics is limited, and, well... R... (3 comments)
Thank you so much for this. Unfortunately, my understanding of mathematics is limited, and, well... R...
Yair Rand‭ wrote over 1 year ago

Thank you so much for this. Unfortunately, my understanding of mathematics is limited, and, well... Regarding the sentence "With the correct assignment of signs you get a quadratic programming setup to which you can add any objective function you want and tackle with standard techniques." - I didn't really understand any of that. Could you please elaborate a bit on that?

Peter Taylor‭ wrote over 1 year ago

"Correct assignment of signs": when I said "signed volume" it's because if the given value is negative you need to negate it to get the correct volume. Assigning the signs is essentially guessing which ones need to be negated. (The last sentence gives an idea for making an intelligent guess; in the worst case, you need to do trial and error over $2^{24}$ possible assignments).

Peter Taylor‭ wrote over 1 year ago

Quadratic programming has a long Wikipedia page which I won't attempt to summarise beyond to say that it's about optimising an "objective function" subject to quadratic constraints. Since all you care about is the existence of a solution, the choice of objective function is arbitrary. The Wikipedia page lists the standard techniques and mentions some software; I think that the details of using the individual software packages is probably off-topic here.