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Re: Formula For Volume of a Tetrahedron
Posted:
Oct 6, 1998 7:34 PM


I've not really been following much of this thread, but the thing you said about 1/n! being the volume of the inscribed simplex reminded me of something I learned recently that you might enjoy, which is that if you take a cube, in ndimensional space, with vertices at (a_1,...a_n) where a_i are all either 0 or 1, and then make slices, where the cuts pass through all the vertices with exactly k 1s for each k from 0 to n (they are all parallel, given by sum a_i =k; this slices the thing into n parts, and, the ratios of the volumes of the parts is the nth row of the triangle of Eulerian numbers; so for the square, you cut it and get two triangles, areas ratio 1:1; for the cube, you cut off the two opposite of your inscribed tetrahedra, and also there is a bit left in the middle, and you get volume ratios 1:4:1; the sum of the nth row of Eulerican numbers is n!, and the two end numbers are always 1; so this gives your result as a special case. Since going to parallelapipeds is just a linear thing, I think this should work more generally.
By the way  Eulerian numbers are great  I got these numbers from solving a completely different problem, then found what they were from Sloane's integer sequence web page, and then looked up and found out they cropped up in billions of other contexts (well, quite a lot anyway). They are really lovely, I definitely recommend looking them up if you've not met them before. You can find the geometry application above if you search on mathsci net. You can also find the definition on line at Eric W's CRC concise encylopedia of mathematics.
Helena



