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Statement: Say we have a Tetrahedron in $n$ dimensional space. Which means just a bunch of $n+1$ points. We want its measure. So, we pick any of the $n+1$ points. We take the remaining $n$ points, which lie on an $n-1$ dimensional sub-space and calculate the $n-1$ dimensional measure of the Tetrahedron formed by them in that sub-space. Then, we take the point we put aside and calculate its perpendicular distance from the $n-1$ dimensional sub-space (height). To get the measure of the original Tetrahedron, we multiply the $n-1$ dimensional measure with the height and divide that product by $n$.

Is the statement above correct? Can we prove it in general? Will particularly appreciate linear algebra approaches.

And are there any elegant proofs that an average teenager could understand for the special case of $n=3$?

Rohit Pandey
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    Can easily prove it with calculus and induction. If the base of an $n$ volume is $v$, an $n-1$ dimensional tetr, then as it tapers to $0$ and a point $s$ along the height, the cross section is $s^{n-1}\times v$ so the volume of the $n$ dimensional tetra is.... – fleablood Dec 08 '24 at 01:01
  • Here are some other elegant arguments: https://math.stackexchange.com/q/1718021/837198 – Mathieu Rundström Dec 10 '24 at 04:24

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