**Background:** It is not known if any point exists on the XY plane that is at rational distance from all 4 vertices of a unit square lying on the XY plane (https://mathworld.wolfram.com/RationalDistanceProblem.html#:~:text=The%20rational%20distance%20problem%20asks,can%20be%20cleared%20by%20multiplication). 

However, it is trivial to find other quadrilaterals with such a point - indeed, the points (3,0), (0,4), (-3,0), (0,-4) form a quad (a rhombus) with all sides of length 5 and the origin is at rational distances from all its vertices. The points (15,0), (0,25), (-48,0) and (0, -36) form a quadrilateral with all side lengths different integers and all 4 vertices at integer distances from origin; and the line segments from the origin to the vertices cut the quad into 4 mutually unequal right triangles. 

**Question:** Is there a quadrilateral Q with (1) all 4 vertices having rational coordinates, (2) edges having rational lengths and (3) *more than one point P* on the plane has all distances from the 4 vertices rational? Does insisting that the coordinates of the P's are also rational have an impact on the question? If for some Q, more than one such P can be found, is there an upper bound on the number of such special P's?