A system of three nonlinear equations2 Let $(x_1,y_1), (x_2,y_2), (x_3,y_3)$ be variables and $(a,b),det_1, dot_1,det_2,dot_2,det_3,dot_3$ be given constants. How can I solve the following system of nonlinear equations based on constant values?
\begin{cases} 
x_1x_2-a(x_1+x_2)+a^2+y_1y_2-b(y_1+y_2)+b^2 = dot_1\\
x_1y_2-x_2y_1+b(x_2-x_1)+a(y_1-y_2) = det_1\\
x_2x_3-a(x_2+x_3)+a^2+y_2y_3-b(y_2+y_3)+b^2 = dot_2\\
x_2y_3-x_3y_2+b(x_3-x_2)+a(y_2-y_3) = det_2\\
x_3x_1-a(x_3+x_1)+a^2+y_3y_1-b(y_3+y_1)+b^2 = dot_3\\
x_3y_1-x_1y_3+b(x_1-x_3)+a(y_3-y_1) = det_3\\
\end{cases}
As I know it has an unlimited number of answers but I don't know how to solve it. I'm looking for answers in integer and real numbers.
 A: I will just give a sketch for now how to solve it, I can fill in more details if necessary.
First define vectors $v_i = (x_i-a,y_i-b),\, i = 1,2,3$. Then we can rewrite the equations as 
\begin{align}
\langle v_1, v_2\rangle = a_{12},\ \langle v_2, v_3\rangle = a_{23},\ \langle v_3, v_1\rangle = a_{31},\\
\det[v_1^t\ v_2^t] = b_{12},\ \det[v_2^t\ v_3^t] = b_{23},\ \det[v_3^t\ v_1^t] = b_{31},\\
\end{align} where I changed dot's and det's to a's and b's.
Denote by $\vartheta_{ij}$ the (oriented) angle between vectors $v_i$ and $v_j$. Remember that dot product in $\mathbb R^2$ is given by formula $\|v\|\|w\|\cos\vartheta$ and that determinants above  measure (signed) area of parallelogram formed by the vectors, which can also be expressed as $\|v\|\|w\|\sin\vartheta$, and thus the system becomes
\begin{align}
\|v_1\|\|v_2\|\cos\vartheta_{12} = a_{12},\ \|v_2\|\|v_3\|\cos\vartheta_{23} = a_{23},\ \|v_3\|\|v_1\|\cos\vartheta_{31} = a_{31},\\
\|v_1\|\|v_2\|\sin\vartheta_{12} = b_{12},\ \|v_2\|\|v_3\|\sin\vartheta_{23} = b_{23},\ \|v_3\|\|v_1\|\sin\vartheta_{31} = b_{31}.\\
\end{align}
Then, you can find the lengths $\|v_i\|$ by looking at the system $$\|v_1\|\|v_2\| = c_{12},\ \|v_2\|\|v_3\| = c_{23},\ \|v_3\|\|v_1\| = c_{31},$$ where $c_{ij} = \sqrt{a_{ij}^2+b_{ij}^2}$ obtained by squaring the above equations and using $\sin^2t+\cos^2t = 1$. Solving it gives you $$\|v_1\|=\sqrt{\frac{c_{12}c_{31}}{c_{23}}},\ \|v_2\|=\sqrt{\frac{c_{12}c_{23}}{c_{31}}},\ \|v_3\|=\sqrt{\frac{c_{23}c_{31}}{c_{12}}}.$$
The whole thing is, from geometric perspective, obviously rotationally invariant (rotating all of the vectors won't change the angles between them or the areas), so fix some angle $\vartheta$. Then, the solution can be represented as complex numbers as
$$v_1 = \|v_1\|e^{i\vartheta},\ v_2 = \|v_2\|e^{i(\vartheta +\vartheta_{12})},\ v_3 = \|v_3\|e^{i(\vartheta-\vartheta_{31})}.$$
More explicitly, $e^{i\vartheta_{12}} = \cos\vartheta_{12} + i\sin\vartheta_{12} = \frac{a_{12}}{c_{12}}+\frac{b_{12}}{c_{12}}i$ and $e^{i\vartheta_{31}} = \cos\vartheta_{31} + i\sin\vartheta_{31} = \frac{a_{31}}{c_{31}}+\frac{b_{31}}{c_{31}}i$, so we have
\begin{align}v_1 &= \sqrt{\frac{c_{12}c_{31}}{c_{23}}}(\cos\vartheta + i\sin\vartheta),\\ 
v_2 &= \sqrt{\frac{c_{12}c_{23}}{c_{31}}}(\cos\vartheta + i\sin\vartheta)(\frac{a_{12}}{c_{12}}+i\frac{b_{12}}{c_{12}}),\\ 
v_3 &= \sqrt{\frac{c_{23}c_{31}}{c_{12}}}(\cos\vartheta + i\sin\vartheta)(\frac{a_{31}}{c_{31}}-i\frac{b_{31}}{c_{31}}),\ \vartheta\in\mathbb R.\end{align}
All you have to do now is expand and $x$'s will be the real parts, while $y$'s the imaginary parts of the above complex number representation.
Note, however, that the system is overdetermined since knowing the angles between $v_1$ and $v_2$ and $v_2$ and $v_3$ will also give you the angle between $v_1$ and $v_3$.
A: Hint:
The equations can be seen as the dot and cross products of three vectors formed by the common origin $(a,b)$ and three points.
The ratios $\dfrac{\text{det}}{\text{dot}}$ give you the (tangent of the) angles they form, and the square roots of the pairwise sum-of-squares give you the products of the lengths. From the latter you can retrieve the individual lengths.
The exact positions are undetermined, only the relative directions are known. You can choose one arbitrarily and the others will follow.
