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.
 A: Doing this in $\mathbb{R}^3$, let the three points of the equilateral triangular base of the tetrahedron be $(0,0,0)$, $(0,2\ell,0)$, and $(\sqrt{3}\ell,a,0)$. Suppose the lengths of the edges of the tetrahedron between each of these point and our apex are $a$, $b$, and $c$ respectively. The apex (if it exists) will be the intersection of the three spheres given by the following equations:
$$\begin{cases}
  x^2 + y^2 + z^2 = a^2 \\
  x^2 + (y-2\ell)^2 + z^2 = b^2 \\
  (x-\sqrt{3}\ell)^2 + (y-\ell)^2 + z^2 = c^2 \\
\end{cases}$$
After some rough scratch work, I ended up getting our apex as the point
$$\left(
  \frac{a^2+b^2-2c^2+4\ell^2}{4\sqrt{3}\ell},\quad
  \frac{a^2-b^2+4\ell^2}{4\ell}, \quad
  \sqrt{a^2-x^2-y^2}
\right)$$
The $z$-coordinate is kinda gross, so I left it here in terms of $x$ and $y$.
A: Let the vertices of the base triangle be A, B and C and let the top vertex be D.
Consider the side triangle ABD. Let P be the point on AB that is directly below D. DP is therefore the perpendicular height of the triangle ABD. You should be able to use cosine rule etc to find P.
Consider the side triangle ACD. Let Q be the point on AC that is directly below D. DQ is therefore the perpendicular height of the triangle ACD. You should be able to use cosine rule etc to find Q.
You will note that DP = DQ. Consider the triangle PDQ. This is isosceles. Let the midpoint of PQ be M. The perpendicular height of this triangle is MD.
Note: this method does not rely on ABC being equilateral.
The apex of the triangle will be directly above M, height MD.
A: 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 $$x = \frac{c^2 + b^2 - 2a^2}{12}, \, y = \frac{c^2 - b^2}{4\sqrt3}, z = \pm\sqrt{a^2 - y^2 - (x-2)^2} . $$
