Triangles are easy and tetrahedra are hard | Hacker News Reader

Triangles are easy and tetrahedra are hard

quantamagazine.org

64 pointsrbanffy16 comments

All that reading to learn that one property of triangles doesnt extend to tetrahedron? Blah.

Did you know if you make a 4x4 matrix where the columns are the (x,y,z,1) vertices of a tetrahedron, the rows of its inverse are the coefficients of the plane equations of the faces of that tetrahedron? You can also use the adjoins. Geometry is fun.

They have really failed to maintain the quality of their articles as they have grown.

I found this stimulating. I don’t need to be shown some ancient mysteries of prime numbers to be entertained, it is quite sufficient to put my mind on the track of “why would that not work in higher dimensions?” Why isn’t our universe built to scale with dimensions and where exactly does it say that?

Agree, the article takes us on an intellectual journey. The same could not be said of grandparent comment

That seems like a pretty unfair summary. The author is a high school math teacher and the audience is clearly relatively non-technical.

But I didn’t know that rational tetrahedra (dihedral angles all rational multiples of π) were classified in 2020 https://arxiv.org/abs/2011.14232

If you have a tetrahedron ABCD, you can also get its volume by dividing the scalar triple product of the vectors AB, AC, and AD by 6. This is because a parallelepiped is made of 6 identical tetrahedra

Accidentally discovered that when I was bored during calculus class :]

While your volume formula is correct, I don't see a way to cut a general parallepiped into 6 identical tetrahdtra. In fact, I don't see how it can be accomplished with a cube.

Identical up to reflection.

I can see how to do it with a cube now but I don't think it works for general paralleleliped as it has 3 pair of different faces and a naive way of dividing it into 6 tetrahedra would cut each face into two triangles so we get 3 sets of 4 congruent triangles. You can't allocate 3 sets of 4 congruent triangles into 6 equal groups.

You are right. The tetrahedra are identical up to volume (not reflection).

Start with the cube, and notice that when you deform it into the general parallelepiped, the volumes of all the tetrahedra stay equal to eachother.

Rotate the parallelepiped so that you are staring down its longest diagonal, or any diagonal really. Now the 3 faces of the parallelepiped directly facing you can be cut into 6 congruent triangles. Slice down these 6 triangles to get 6 tetrahedra. it is then easy to show by Cavalieri's principle that that the volume of any of these tets is equal to the original tet. so the original tet has volume 1/6 of the parallepiped.

However these 6 tets are not necessarily congruent, when I commented I was visualizing the case when the parallelepiped has equal side lengths, but in the general case the 6 tetrahedra are not congruent. However each of the 6 is congruent to at least 1 other out of the 6, since the parallepiped has reflection symmetry.

I really want to say the volume of a tetrahedron is something like a constant (maybe 1/6) times the determinant of the 4x4 matrix of points. This would be related to what you said. Take the determinant by expansion along the row of 1's, which results in the sum of 4 sub-determinants which... I don't remember ;-)

>Did you know if you make a 4x4 matrix where the columns are the (x,y,z,1) vertices of a tetrahedron, the rows of its inverse are the coefficients of the plane equations of the faces of that tetrahedron? You can also use the adjoins. Geometry is fun.

Nice! And it's pretty easy to see why:

- in homogenous coordinates the plane equation is just dot(point, plane) = 0

- for every column (i.e. point) of a 4D matrix, exactly 3 rows (i.e. planes) of its inverse will have a dot product of 0

Reading the title I thought this was going to be about the impossibility of decomposing all 3D polyhedra in tetrahedra [1]. As you may know every polygon in 2D can be however decomposed into triangles, but this result doesn't, surprisingly, extend to 3D:

[1] https://en.wikipedia.org/wiki/Sch%C3%B6nhardt_polyhedron

BTW, whether or not a polyhedron is tetrahedralizable is NP-Complete.

You can extend these triangles and tetrahedral analyses into higher dimensional simplexes, which give insights into structure of categorical probability distributions as number of categories increases.

One interesting property is that the volume of the simplex goes something like 1 / n!, which breaks a lot of intuition of what happens as n gets large.

It was pointed out recently (I think in the 'high-dimensional space is pointy' thread) that in high dimensions, the (n-d hyper)cube may not be the most meaningful measure of volume (and that euclidean distance may not be the most meaningful measure of distance).

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