<s>[INST] How can I determine the position of the apex of an irregular tetrahedron I have an irregular tetrahedron the base of which is an equilateral triangle.  Knowing the lengths of all sides I need to then determine the position of the apex.
Is there anyone that can offer a bit of guidance on this.
 [/INST] i will take the equilateral triangle base $ABC$ to have sides of length $2\sqrt3$ and the edges $AD= a, Bd = b, CD = c.$ set up the coordinate system so that $$A=(2,0,0),\, B=(-1, \sqrt 3, 0),\, C = (-1,-\sqrt 3, 0), D=(x,y,z)  $$ we will derive a formula for $x,y,z$ in terms of $a,b,c.$
on equating the squares of length we have $$(x-2)^2 + y^2 + z^2=a^2,\,(x+1)^2+(y-\sqrt 3)^2 + z^2 = b^2,\, (x+1)^2+(y+\sqrt 3)^2 + z^2 = c^2 $$
from these we can derive $$6x - 2\sqrt 3 y = b^2 - a^2,\, 6x + 2\sqrt 3y = c^2 - a^2.$$
the solutions are $$x = \frac{c^2 + b^2 - 2a^2}{12}, \, y = \frac{c^2 - b^2}{4\sqrt3}, z = \pm\sqrt{a^2 - y^2 - (x-2)^2} . $$
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