Rational distance problem My question is related to kind of problems, called "rational distances problem"(at least by wolfram mathworld).
I couldn't find a specific solution, so it would be a real help if you have an idea or link about the solution. Here is the problem :
Given $n>2$ points on the plane, not all on a line, is there another point, call $X$, which its distance from all the given points is rational number?
 A: The Wolfram Mathworld article (see here) clearly states that the problem is unsolved in the very specific case where $n=4$ and the four points are the vertices of a unit square. This makes it immediately clear that asking the question in such generality as you have is completely pointless, and that it is extremely likely that there exist configurations for which no such $X$ exists.
Indeed it is very easy to find configurations for which no such point exists. For the following configuration of $5$ points the algebra is particularly simple; let
$$P_1:=(1,0),\qquad P_2:=(-1,0),\qquad P_3=(0,1),\qquad P_4=(0,-1),\qquad P_5=(\pi,0).$$
Suppose $X=(x_0,y_0)\in\Bbb{R}^2$ is such that $r_i:=d(X,P_i)$ is rational for all $i$. Then
$$d(X,P_1)^2=x_0^2-2x_0+1+y_0^2=r_1^2,$$
$$d(X,P_2)^2=x_0^2+2x_0+1+y_0^2=r_2^2,$$
from which it follows that $x_0$ is rational because
$$4x_0=r_2^2-r_1^2=(r_2-r_1)(r_2+r_1).$$
By symmetry also $y_0$ is rational. But then
$$d(X,P_5)^2=(x_0-\pi)^2+y_0^2,$$
is not rational, a contradiction. Hence no such point $X\in\Bbb{R}^2$ exists.
A: The OP asks, "Given $n\gt2$ points on the plane, not all on a line, is there another point, call $X$, which its distance from all the given points is rational number?"
The answer is, sometimes yes and sometimes no.
To see that the answer is sometimes yes, consider any $n$ points on a circle of radius $1$. They are obviously all at the rational distance $1$ from the center of the circle.
To see that the answer is sometimes no, start with any two points, and draw all circles of rational radius around each point, then take the set of intersections of pairs of these circles. This set is countable, and it constitutes the set of all points where the future point $X$ must lie.  Now around each of these countably many points, draw all circles of rational radius. The union of all these circles is a set of measure $0$ (because each circle is a set of measure $0$, and there are only countably many of them).  So if your third point (when $n\gt2$) is not in this set of measure $0$, it is at an irrational distance from all the points which are at a rational distance from both of the first two points. Roughly speaking, if you pick your $n$ points "at random," then, with probability $1$, any three of them will fail to be simultaneously at a rational distance from any other point.
