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Consider a right-angled triangle with vertex co-ordinates $(0,0), (a,0)$ and $(0,b)$. The length of the hypotenuse is $c$. Moreover, $a, b$ and $c$ are positive integers i.e., $(a,b,c)$ is a Pythagorean triple.

Following are my conjectures in the increasing order of difficulty.

Conjecture 1: There exists a rational point $(r,0)$ at a rational distance from $(0,b)$, such that $r > 0$, $r \neq a$.

Conjecture 2: There exists a rational point $(r,0)$ at a rational distance from $(0,b)$, such that $0< r < a$.

Conjecture 3: There exists an infinite number of rational points $(r, 0)$ at a rational distance from $(0,b)$, such that $0< r < a$.

Conjecture 4a: For any two rational numbers $0< r_1 < r_2 < a$, there exists a rational point $(r,0)$ at a rational distance from $(0,b)$, such that $0< r_1 < r < r_2 < a$.

Conjecture 4b: For any two rational numbers $0< r_1 < r_2 < a$, there exists an infinite number of rational points $(r,0)$ at a rational distance from $(0,b)$, such that $0< r_1 < r < r_2 < a$.

Note that conjecture 4a implies 4b.

My question: Are the above conjectures known to be true / false ?

Note: In all the above conjectures, I want the existence of the point $(r,0)$ on the $x$-axis, for both cases $a > b$ and $a < b$.

Motivation: An affirmative answer to the above questions will help me generate certain fractals in a clean way.

A known fact: The set of points with rational distances to the vertices of a given triangle with sides of rational length is everywhere dense 1.

1 J.H.J. Almering, Rational quadrilaterals, Indag. Math. 25 (1963) 192–199.

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    $\begingroup$ For conjecture 1 and 2: just choose $r =\frac{ b^2}{a}$, and the hypotenuse will be $\frac{ bc}{a}$. $\endgroup$ Aug 5 at 23:19
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    $\begingroup$ For conjecture 3: Choose two integers $m$ and $n$, so that $\frac{b}{a}<\frac{2mn}{m^2-n^2}$. Use $r =\frac{b(m^2-n^2)}{2mn}$ and the hypotenuse will be $\frac{b(m^2+n^2)}{2mn}$. $\endgroup$ Aug 6 at 0:01
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    $\begingroup$ @RicardoCruz For your choice of $r$, if $b>a$, then $r=\frac{b^2}a > b > a$. The choice works for conjecture 1 but needs more work for conjecture 2. $\endgroup$
    – peterwhy
    Aug 6 at 1:44
  • $\begingroup$ Yes, @peterwhy, you are right. I assumed from the figure that $b<a$. $\endgroup$ Aug 6 at 12:54
  • $\begingroup$ I want the answers for both cases: a > b and a < b. I have added this clarification above. $\endgroup$ Aug 6 at 17:26

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