A pyramid with square base $ABCD$ and apex $E$ has a known coordinates for $A,B,C,D$, they are $A= (0, 0, 0), B = (a, 0,0), C = (a,a,0), D = (0, a,0) $ where $a$ is known. In addition, the distances from the apex to the four vertices of the bases are also given. Now I want to determine the coordinates of the apex $(x, y, z)$.
My Attempt:
Let $s_1 = EA, s_2 = EB, s_3 = EC, s_4 =ED $, then from the distance formula,
$s_1^2 = x^2 + y^2 + z^2$
$s_2^2 = (x - a)^2 + y^2 + z^2$
$s_3^2 = (x - a)^2 + (y - a)^2 + z^2$
$s_4^2 = x^2 + (y - a)^2 + z^2$
Taking the differences,
$ s_1^2 - s_2^2 = 2 a x - a^2 $
$ s_1^2 - s_3^2 = 2 a x + 2 a y - 2 a^2 $
$ s_1^2 - s_4^2 = 2 a y - a^2 $
For this system of equations to be consistent, we must have
$ s_1^2 - s_2^2 + s_1^2 - s_4^2 = s_1^2 - s_3^2 $
That is,
$ s_1^2 + s_3^2 = s_2^2 + s_4^2 $
Otherwise there is no solution.
Assuming this condition is satisfied, we can do away with the second equation. The first and third equations give $x$ and $ y$ because $a$ is known. Plugging $x$ and $y$ into the first original equation gives the value of $z$.
Any comments/alternative solutions are welcome and highly appreciated.
Consider the three spheres $S(A,s_1), S(B,s_2), S(C,s_3)$.
The apex (with coordinates $(x_a,y_a,z_a)$) is one of their two common points of intersection. We will choose the one with positive $z$ coordinate.
The intersection of the spheres with horizontal plane $ABCD$ are circles.
First, determine the radical center (or "power center") $P$ of these circles (intersection of radical axes of these circles 2 by 2). This center has coordinates $(x_a,y_a,0)$.
Now, in order to find $z_a$, erect a vertical line from this point $C$ : the apex is the point on this line which is at distance $s_1$ from $A$ (and therefore at distance $s_2$ from $B$, etc.)
Edit: Explanations about radical center.
When we have 3 circles with resp. equations :
$$\begin{cases} \underbrace{x^2+y^2-2a_1x-2b_1y+c_1}_{p_1(x,y)}&=&0\\ \underbrace{x^2+y^2-2a_2x-2b_2y+c_2}_{p_1(x,y)}&=&0\\ \underbrace{x^2+y^2-2a_3x-2b_3y+c_3}_{p_1(x,y)}&=&0 \end{cases}$$
The radical axis is the point such that
$$p_1(x,y)=p_2(x,y)=p_3(x,y)$$
which implies by taking differences, that :
$$\begin{cases} 2(a_1-a_2)x-2(b_1-b_2)y+(c_2-c_1)&=&0\\ 2(a_1-a_3)x-2(b_1-b_3)y+(c_3-c_1)&=&0 \end{cases}$$
which is a linear system of two equations with two unknowns (representing two "radical axes'), easy to solve.