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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 accomplish
by howling 5y ago
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.
- 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.