A rational distance problem with (possibly) multiple solutions

Question

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?

Answer 1

It is known that there are infinite sets $S$ of points on the unit circle with rational coordinates and rational distances. So we can take any four points of $S$, make them the vertices of $Q$, and get infinitely many other points $P \in S$ at rational distances from all four vertices of $Q$.

Explicitly, let $$ S = \{ z^2 : z \in {\bf Q}(i), \, z \bar z = 1 \}, $$ which is infinite because there are infinitely many primitive Pythagorean triples $(a,b,c)$ and each yields $z = (a+bi)/c$ with $z \bar z = 1$. The distance between any two points $w^2,z^2 \in S$ is $$ |w^2 - z^2| = \left| wz \left(w/z - z/w\right) \right| = \left|w\right| \left|z\right| \left| (w/z) - (w/z)^{-1} \right|, $$ and $|w| = |z| = 1$, while $(w/z)^{-1} = \overline{w/z}$ so $$ \left| (w/z) - (w/z)^{-1} \right| = | w/z - \overline{w/z} \, | = \left| 2i \, {\rm Im}(w/z) \right| = 2 \, {\rm Im} (w/z) \in {\bf Q}, $$ QED.

For example, starting from the 3,4,5 triangle and scaling by 25 we find that the rectangle with vertices $(\pm 7, \pm 24)$ has all sides and both diagonals rational, and there are infinitely points on its circumcircle $x^2 + y^2 = 25^2$ at rational distances from all four vertices, starting with $(x,y) = (\pm 25, 0)$.

It is harder to find cases with infinitely many $P$ in the interior of the quadrilateral. (The above construction does give one such $P$, namely the center of the circle.)

Answer 2

The rectangle with vertices $(\pm60,\pm15)$ has rational distances from the corners to both $(\pm52,0)$, because both $(15,8,17)$ and $(15,112,113)$ are Pythagorean triples.

Answer 3

You should consult my joint paper with Andrew Bremner: https://www.sciencedirect.com/science/article/pii/S0022314X15002243 concernig points at rational distances from the vertices of certain geometric objects. In particular, as one of several results, we prove that the set of positive rational numbers $a$ such that there exist infinitely many rational points in the plane which lie at rational distance from the four vertices of the rectangle with vertices $(0, 0), (0, 1), (a, 0), (a, 1)$ is dense in the set of positive real numbers.