I have two tetrahedrons joint by base (let's call it ABC triangle):
tetrahedron ABCD
tetrahedron ABCE
I know lengths (read as distances) of all edges. Is there any reasonably simple way to find length (distance) of DE?
Also, if you know how to find height of the tetrahedron by the lengths of edges, please comment or add answer - this I feel would be 90% of the solution.
PS: those figures are general and are not of special case.
Image just to clarify how it looks like
Any input would be helpful! Thanks!
The question can be solved using the formulas for the $n$-dimensional volume of an $n$-dimensional simplex. It can be found here.
For five points $P_0,\ldots,P_4$ in 4-dimensional space and their mutual distances $d_{ij}=\overline{P_iP_j}$, $i\neq j$ the 4-dimensional volume or content $C$ is given by $$ \begin{align} C^2 & = -\frac{1}{(4!)^2 2^4} \begin{vmatrix} 0 & d_{01}^2 & d_{02}^2 & d_{03}^2 & d_{04}^2 & 1 \\ d_{01}^2 & 0 & d_{12}^2 & d_{13}^2 & d_{14}^2 & 1 \\ d_{02}^2 & d_{12}^2 & 0 & d_{23}^2 & d_{24}^2 & 1 \\ d_{03}^2 & d_{13}^2 & d_{23}^2 & 0 &d_{34}^2 & 1 \\ d_{04}^2 & d_{14}^2 & d_{24}^2 & d_{34}^2& 0 & 1 \\ 1 & 1 & 1 & 1 & 1 & 0 \end{vmatrix}. \end{align}$$ In the question, we have five points. We rename them $P_1=A$, $P_2=B$, $P_3=C$, $P_4=D$ and finally $P_0=E$. The key obervation is that they are contained in a 3-dimensional subspace if and only if $C=0$ (and $P_1$,..,$P_4$ and $P_0$,$P_2$,$P_3$,$P_4$, respectively, form non-degenerate tetrahedra).
This gives an equation for $x=d_{01}^2$: $$ \begin{align} \begin{vmatrix} 0 & x & d_{02}^2 & d_{03}^2 & d_{04}^2 & 1 \\ x & 0 & d_{12}^2 & d_{13}^2 & d_{14}^2 & 1 \\ d_{02}^2 & d_{12}^2 & 0 & d_{23}^2 & d_{24}^2 & 1 \\ d_{03}^2 & d_{13}^2 & d_{23}^2 & 0 &d_{34}^2 & 1 \\ d_{04}^2 & d_{14}^2 & d_{24}^2 & d_{34}^2& 0 & 1 \\ 1 & 1 & 1 & 1 & 1 & 0 \end{vmatrix}=0. \end{align}$$ Using Laplace's expansion, it can be seen that this is a quadratic equation for $x$. The larger solution is $x_1=\overline{DE}^2$, the smaller solution is $x_2=\overline{DE'}^2$, where $E'$ is the symmetric to $E$ with respect to the plane containing $A,B,C$. Observe that $\overline{AE'}=\overline{AE}$, $\overline{BE'}=\overline{BE}$ and $\overline{CE'}=\overline{CE}$.
The coefficient $M$ of $x^2$ in this equation actually is the negative of a subdeterminant: $$M=- \begin{align} \begin{vmatrix} 0 & d_{23}^2 & d_{24}^2 & 1 \\ d_{23}^2 & 0 &d_{34}^2 & 1 \\ d_{24}^2 & d_{34}^2& 0 & 1 \\ 1 & 1 & 1 & 0 \end{vmatrix}. \end{align}$$ Again by the determinant formula, this is $M=16A^2$, where $A$ is the area of the triangle $P_2P_3P_4$.
The heights mentioned in the question can also be expressed using determinants. The volume $V$ of the tetrahedron $P_1P_2P_3P_4$ is given by $$V^2=\frac1{3!^2 2^3} \begin{align} \begin{vmatrix} 0 & d_{12}^2 & d_{13}^2 & d_{14}^2 & 1 \\ d_{12}^2 & 0 & d_{23}^2 & d_{24}^2 & 1 \\ d_{13}^2 & d_{23}^2 & 0 &d_{34}^2 & 1 \\ d_{14}^2 & d_{24}^2 & d_{34}^2& 0 & 1 \\ 1 & 1 & 1 & 1 & 0 \end{vmatrix}. \end{align}$$ One height of this tetrahedron, i.e. the distance $h$ of $P_1$ from $P_2P_3P_4$ is then obtained from the formula $$V=\frac13 Ah$$ where $A$ is the area of the triangle $P_2P_3P_4$ expressed above.