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