Tartaglia's Formula
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Theorem
Let $T$ be a tetrahedron with vertices $\mathbf d_1, \mathbf d_2, \mathbf d_3$ and $\mathbf d_4$.
For all $i$ and $j$, let the distance between $\mathbf d_i$ and $\mathbf d_j$ be denoted $d_{ij}$.
Then the volume $V_T$ of $T$ satisfies:
- $V_T^2 = \dfrac {1} {288} \det \ \begin{vmatrix} 0 & 1 & 1 & 1 & 1\\ 1 & 0 & d_{12}^2 & d_{13}^2 & d_{14}^2 \\ 1 & d_{12}^2 & 0 & d_{23}^2 & d_{24}^2 \\ 1 & d_{13}^2 & d_{23}^2 & 0 & d_{34}^2 \\ 1 & d_{14}^2 & d_{24}^2 & d_{34}^2 & 0 \end{vmatrix}$
Proof
A proof of Tartaglia's Formula will be found in a proof of the Value of Cayley-Menger Determinant as a tetrahedron is a $3$-simplex.
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Also known as
Tartaglia's Formula is also known as Piero della Francesca's Tetrahedron Formula, for Piero della Francesca.
Also see
Source of Name
This entry was named for Niccolò Fontana Tartaglia.
Sources
- 2009: Karl Wirth and André S. Dreiding: Edge lengths determining tetrahedrons (Elem. Math. Vol. 64: pp. 160 – 170): Theorem $3.1$