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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
by phkahler 5y ago
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.
- prvc 5y agoThey have really failed to maintain the quality of their articles as they have grown.
- andreskytt 5y agoI 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?
- TheRealNGenius 5y agoAgree, the article takes us on an intellectual journey. The same could not be said of grandparent comment
- jacobolus 5y agoThat 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 https://arxiv.org/abs/2011.14232
- voldacar 5y agoIf 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 :]
- howling 5y agoWhile 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.
- amelius 5y agoIdentical up to reflection.
- howling 5y agoI 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.
- amelius 5y agoYou 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.
- voldacar 5y agoRotate 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.
- 5y ago
- fbanon 5y ago>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