Rational distances to 3 points in ${\bf R}^2$

Question

Consider three points $(0,0)$, $(1,0)$, and $(0, {\alpha})$ in ${\bf R}^2$. If there exists a point $P =(x,y)$ such that $x^2 + y^2 =r^2$, $(x-1)^2 + y^2 =s^2$, and $x^2 + (y-{\alpha})^2 =t^2$ where $r$, $s$, and $t$ are all rational numbers, then some algebra shows that $${{\alpha}^2}(4r^2 - (r^2 - s^2 -1)^2)=((r^2 - t^2)-{\alpha}^2)^2.$$ This suggests that if it is an irrational number, then $\alpha$ must be an algebraic number of degree $2$, or $4$. Thus, for example, no point $P=(x,y)\in{\bf R}^2$ exists for which all three of the distances from $P$ to $(0,0)$, $(1,0)$, and $(0, {\sqrt[3]{2}})$ are rational. My question is whether this holds for all quadratic irrationals. So, for example, does there exist a point $P=(x,y)\in{\bf R}^2$ such that all three of the distances from $P$ to each of $(0,0)$, $(1,0)$, and $(0, {\sqrt{2}})$ are rational numbers? I thank Thomas Preu who gave the answer below for correcting my original post.

Answer 1

This answer contains a mistake. Since the OP thinks the answer is valuable nevertheless I keep it with indications about the mistake.

In general questions as this one can lead to rather involved problems in diophantine geometry.

For general $\alpha$ eliminating $x,y$ yields the degree $4$ equation in $r,s,t\in\mathbb{Q}$ $$\alpha^2\left(4r^2-(r^2-s^2-1)^2\right)=\left(t^2-r^2-\alpha^2\right)^2.$$ If $\alpha^2\in\mathbb{Z}$ this is about rational points on an affine degree $4$ surface, which already is quite hard in general.
We can always rewrite $,r=\frac{R}{U},s=\frac{S}{U},t=\frac{T}{U}$ with $R,S,T,U\in\mathbb{Z}$ where the fractions are not necessarily reduced. We get after substitution and multiplication by $U^4$: $$\alpha^2\left(4R^2U^2-(R^2-S^2-U^2)^2\right)=\left(T^2-R^2-\alpha^2U^2\right)^2.$$ For $\alpha^2\in\mathbb{Z}$ this is about integral points on an affine degree $4$ solid, which is usually even harder.
Look at two special cases: $\alpha^2\in\{2,3\}$:

• $\alpha^2=2$: The equation specializes to $$2\left(4R^2U^2-(R^2-S^2-U^2)^2\right)=\left(T^2-R^2-2U^2\right)^2.$$ Original but wrong argument: Assume we have an integral solution $(R,S,T,U)\in\mathbb{Z}$. LHS is a product with factors $2$ and $F:=4R^2U^2-(R^2-S^2-U^2)^2$, while RHS is a square. Using unique prime factorization we get that $F$ must have an odd multiplicity in the prime factor $2$. The summand $4R^2U^2$ is a square of an even number and thus only has an even, positive multiplicity in the prime factor $2$, which enforces that $(R^2-S^2-U^2)^2$ must have an odd multiplicity in $2$. This is however impossible. Since we have no integral solution to this equation, such a point $P=(x,y)$ cannot exist for $\alpha$ with $\alpha^2=2$.
  Analysis of the mistake: My argument would have worked only for a product but not for a difference of two squares. As student pointed out the term $\left(4R^2U^2-(R^2-S^2-U^2)^2\right)$ could still have odd multiplicity in the prime $2$ as the example $6^2-2^2=2^5$ indicates. Furthermore student found a family of points $P(x,y)=\left(r-\frac{1}{2r},0\right)$ for arbitrary $r\in\mathbb{Q}\setminus\{0\}$ solving the original problem. Another solution is given by $P\left(\frac{1}{2},\sqrt{2}\right)$.
  Actually: any odd positive integer $n=2m-1$ is a difference of two squares: $n=m^2-(m-1)^2$; any positive multiple of $4$, i.e. $n=4m$, is a difference of two squares: $n=(m+1)^2-(m-1)^2$; any positive even integer $n$ not a multiple of $4$ is not the difference of two squares: if it was we had $n=a^2-b^2=(a-b)(a+b)$ for some integers $a,b$, but the factors are either both even (then $n$ is a multiple of $4$) or both odd (then $n$ is not even), which leads to a contradiction.

• $\alpha^2=3$: The three points $(0,0), (1,0), (0,\sqrt{3})$ form a triangle which is "half of an equilateral triangle". From this special geometric situation it is not hard to come up with a solution, e.g. $P=\left(\frac{1}{2},\frac{1}{2}\sqrt{3}\right)$ with $r=s=t=1$.

For more general $\alpha$, degree considerations as indicated by the OP might be helpful, but it is not true, that the degree of $\alpha$ has to be necessarily $2$ in order to have a solution $P$, as the following example shows:

• For $P=\left(\frac{1}{2},\frac{1}{2}\sqrt{3}\right)$ (forcing $r=s=1$) we can set $\alpha=\frac{1}{2}\sqrt{3}+\frac{1}{2}\sqrt{15}$ (implying $t=2$), but this $\alpha$ has degree $4$.

It might be true, that for $\alpha$ of degree different from $\{1,2,4\}$ it is impossible to find a $P$ with the described properties. I leave that as an open problem, as discussion of the three examples already should give considerable insight into this problem to take it from here.