# Rational point on a right angled triangle 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.

• For conjecture 1 and 2: just choose $r =\frac{ b^2}{a}$, and the hypotenuse will be $\frac{ bc}{a}$. Aug 5 at 23:19
• 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 $\frac{b(m^2+n^2)}{2mn}$. Aug 6 at 0:01
• @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. Aug 6 at 1:44
• Yes, @peterwhy, you are right. I assumed from the figure that $b<a$. Aug 6 at 12:54
• I want the answers for both cases: a > b and a < b. I have added this clarification above. Aug 6 at 17:26