Does this math mobile work? How to build: two identical ABCD quadrilaterals (paperboard) each divided into two parts (triangles) by the diagonals AC and BD. Two rods of lengths AC and BD that cross perpendicularly at point E. The triangles are hanging at the ends: In A the triangle BCD, in C the triangle ABD, in B the triangle ACD and in D the triangle ABC. A line is tied at point E (intersection of diagonals) to hang the mobile.
intersection point of diagonals:
$\begin{array}{} x(E)=\frac{x(A)·Δ(BCD)+x(C)·Δ(ABD)}{Δ(BCD)+Δ(ABD)} & y(E)=\frac{y(A)·Δ(BCD)+y(C)·Δ(ABD)}{Δ(BCD)+Δ(ABD)} \end{array}$
where:
$Δ(BCD)=\frac{1}{2}·\left| \begin{array}{} x(B) & y(B) & 1 \\ x(C) & y(C) & 1 \\ x(D) & y(D) & 1 \\ \end{array} \right|$
$Δ(ABD)=\frac{1}{2}·\left| \begin{array}{} x(A) & y(A) & 1 \\ x(B) & y(B) & 1 \\ x(D) & y(D) & 1 \\ \end{array} \right|$
formula for mobile (by analogy)
$x(E)=\frac{x(A)·Δ(BCD)+x(B)·Δ(ACD)+x(C)·Δ(ABD)+x(D)·Δ(ABC)}{Δ(BCD)+Δ(ACD)+Δ(ABD)+Δ(ABC)}$
$y(E)=\frac{y(A)·Δ(BCD)+y(B)·Δ(ACD)+y(C)·Δ(ABD)+y(D)·Δ(ABC)}{Δ(BCD)+Δ(ACD)+Δ(ABD)+Δ(ABC)}$
formula for direct calculation (Laplace, 3th column)
$x(E)=\frac{1}{2}·\frac{\left| \begin{array}{} x(A) & y(A) & x(A) & 1 \\ x(B) & y(B) & -x(B) & 1 \\ x(C) & y(C) & x(C) & 1 \\ x(D) & y(D) & -x(D) & 1 \\ \end{array} \right| }{\left| \begin{array}{} x(A) & y(A) & 0 & 1 \\ x(B) & y(B) & -1 & 1 \\ x(C) & y(C) & 0 & 1 \\ x(D) & y(D) & -1 & 1 \\ \end{array} \right| }$
$y(E)=\frac{1}{2}·\frac{\left| \begin{array}{} y(A) & x(A) & y(A) & 1 \\ y(B) & x(B) & -y(B) & 1 \\ y(C) & x(C) & y(C) & 1 \\ y(D) & x(D) & -y(D) & 1 \\ \end{array} \right| }{\left| \begin{array}{} y(A) & x(A) & 0 & 1 \\ y(B) & x(B) & -1 & 1 \\ y(C) & x(C) & 0 & 1 \\ y(D) & x(D) & -1 & 1 \\ \end{array} \right| }$
