uniqueness of a geometric configuration This question has come up in radar signal processing. A virtual element is defined as the midpoint of the segment between a transmitting (T) point (antenna) and receiving (R) point (antenna). Given a set of $T>1$ and $R> 1$ points in a plane there can be formed $TR$ midpoints (virtual elements) of which we assume that $TR\ge 4$ and are all distinct and do not coincide with any of the T and R points. My question if/when will the set of $TR$ points determine the planar disposition uniquely of the $T$ and $R$ points? Similar question can be asked for the 3D case.
 A: Let $t_i$ be a vector (in two or three dimensions, it doesn't make any difference) giving the position of the $i$-th transmitter ($1\le i\le T$), $r_j$ the position of the $j$-th receiver ($1\le j\le R$) and $a_{ij}$ the position of the midpoint between $t_i$ and $r_j$. You are asking if, knowing all $a_{ij}$, it is possible to find all $t_i$ and $r_j$. 
For every $i$ and $j$ the following relations hold: 
$$t_i+r_j=2a_{ij},
\quad 1\le i\le T,\ 1\le j\le R,$$
which form a system of $T\cdot R$ equations in $T+R$ unknowns. We can have uniqueness if the associated matrix has rank $T+R$, but that is not possible: the rank of that matrix is $T+R-1$. 
To see why, I'll consider for example the case $T=R=3$. The matrix associated with the system of equations has $9$ rows and $6$ columns: 
$$
\pmatrix{
1 & 0 & 0 & 1 & 0 & 0 \\
1 & 0 & 0 & 0 & 1 & 0 \\
1 & 0 & 0 & 0 & 0 & 1 \\
0 & 1 & 0 & 1 & 0 & 0 \\
0 & 1 & 0 & 0 & 1 & 0 \\
0 & 1 & 0 & 0 & 0 & 1 \\
0 & 0 & 1 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 & 1 & 0 \\
0 & 0 & 1 & 0 & 0 & 1 \\
}
$$
Each row has only two non-vanishing entries, both equal to $1$, one of them in the first $T$ columns and the other one in the last $R$ columns.
The first $R$ rows (having $1$ as first entry) are linearly independent. If we add to them the $(R+1)$-th row we get $R+1$ linearly independent rows, but all other rows having $1$ as second entry are linearly dependent from those $R+1$ rows: for instance, the fifth row in the matrix can be obtained by subtracting the first row from the fourth, and adding the second row. For the same reason, we can add to the set of linearly independent rows only one row having $1$ as third entry, and so on. We can thus obtain a maximum of $R+T-1$ rows, which is then the rank of the matrix.
It follows that we don't have a unique solution: we can choose at will one among $t_i$ or $r_j$. Moreover, quantities $a_{ij}$ are constrained by $RT-T-R+1$ equations which must be satisfied, if we want a solution to exist. Such constraints are all the equations of the form: 
$$
a_{xm}+a_{yn}=a_{xn}+a_{ym},
$$
which have a simple geometrical explanation: two transmitters and two receivers form a (possibly self-intersecting) quadrilateral, but the midpoints of the sides of any quadrilateral form a parallelogram, whose diagonals meet at their midpoint.
In the above example ($T=R=3$), there are $9$ such equations, but only four among them are linearly independent. These four constraints can be chosen for instance as follows:
$$
a_{22} + a_{33} = a_{23} + a_{32} \\ 
a_{21} + a_{33} = a_{23} + a_{31} \\ 
a_{12} + a_{33} = a_{13} + a_{32} \\ 
a_{11} + a_{33} = a_{13} + a_{31} \\ 
$$
Of course I assumed that for every given midpoint $a_{ij}$ we know indices $i$ and $j$, that is we know of which couple transmitter-receiver it is the midpoint. If that information is missing, things are even worse, because multiple associations could be possible, leading to even more solutions.
