Is the set $\ \left\{ \frac{b-a }{c-a}:\ (a,b,c)\ \text{is a primitive Pythagorean triple with}\ a<b<c\ \right\}\ $ dense in $\ [0,1]\ $ and how do you show this?
It seems likely true based on glancing at the tree in the picture, however I am not very knowledgeable on the properties of Primitive Pythagorean triples...
Let me try an argument that is, perhaps, excessively geometric.
I start with the rational parametrization of the unit circle, \begin{align} x&=\frac{2t}{t^2+1}\\ y&=\frac{t^2-1}{t^2+1}\,. \end{align} You probably have seen this, maybe in a different form. But it’s clear that every rational value of $t$ gives a point on the unit circle whose both coordinates are rational, and equally, every rational point on the circle comes from a rational $t$, via $(x,y)\mapsto\frac{y+1}{x}$. For instance, $t=2$ corresponds to the point $(\frac45,\frac35)$.
And I hope you see at a glance that the (primitive) Pythagorean triangles are in one-to-one correspondence with the first-quadrant rational points on the circle.
Now, the points below the line $y=x$ for $1\le t\le1+\sqrt2$ can give us $a=(t^2-1)/(t^2+1)$, $b=2t/(t^2-1)$, and $a<b<c=1$, and if we calculate your ratio $$ R(t)=\frac{b-a}{c-a}=\frac{1+2t-t^2}2\,, $$ in which $t=1$ gives $R(1)=1$, while $R(1+\sqrt2)=0$. Now in the range $t\in[1,1+\sqrt2]$, rational values of $t$ give rational values of $R(t)$, and though these are not the only rational values of $R$, at least they are dense among the values of $R$. And thus we’re done.
Consider the function $$f(x) = 1-\frac{2}{x^2-2x+1}$$
This function is continuous, maps the interval $(1+\sqrt{2},\infty)$ onto $(0,1)$ and it's increasing.
Choose any interval $(u,v) \subseteq [0,1]$, the preimage of $(u,v)$ under $f$ is the interval $(f^{-1}(u),f^{-1}(v))$. Choose a rational $r/s \in (f^{-1}(u),f^{-1}(v))$ with $r$ and $s$ odd and coprime. (It's known that $\Bbb Q$ is dense in $\Bbb R$, but it's not hard to see that if you restrict to rationals with odd numerators and denominators it's still dense).
Then $f(r/s) \in (u,v)$.
But $$f(r/s) = 1- \frac{2}{(r/s)^2-2 r/s + 1}= \frac{\frac{r^2-s^2}{2}-rs}{\frac{r^2+s^2}{2}-r s} = \frac{b - a}{c-a}$$
where we defined $a=rs$, $b=\frac{r^2-s^2}{2}$ and $c=\frac{r^2+s^2}{2}$.
The triple $(a,b,c)$ is a primitive Pythagorean triple with $a<b<c$ and $\frac{b-a}{c-a} \in (u,v)$. Since $(u,v)$ was any interval contained in $[0,1]$, the set is dense.
A Euclid's formula variation lets us easily see what the differences are that go into $\dfrac{B-A}{C-A}\space$ and how $\space B-A>0 \space$ requires that $\space 2k^2>m.\space$
\begin{align*} A&=\bigg(\big(2n-1+k\big)^2-k^2\bigg)\\ &\qquad\qquad=(2n-1)^2+2(2n-1)k\\ B&=\quad 2\big(2n-1+k\big)k\\ &\qquad\qquad\qquad\qquad= 2(2n-1)k+2k^2\\ C&=\bigg(\big(2n-1+k\big)^2+k^2\bigg)\\ &\qquad\qquad=(2n-1)^2+2(2n-1)k+2k^2 \end{align*}
Examples, using this formula, are \begin{align*} &(n,2k^2)=(1,2)\quad F(n,k)=F(1,1)= (3,4,5)\\ &(n,2k^2)=(2,18)\quad F(n,k)=F(2,3)= (27,36,45)\\ &(n,2k^2)=(3,32)\quad F(n,k)=F(3,4)= (65,72,97)\\ &(n,2k^2)=(4,50)\quad F(n,k)=F(4,5)= (119,120,169)\\ \end{align*} The formula produces non-trivial triples for all pairs of natural numbers.
\begin{equation} n,k\in\mathbb{N} \implies\\ A=\big\{3,5,7,\cdots\big\}\\ B=\big\{4,8,12,\cdots\big\}\\ \qquad C\subset\big\{5,9,13,\cdots,4x+1\big\} \end{equation}
$ (B-A)=2k^2-(2n-1)^2\implies \\ (B-A)\in\big\{1,7,17,23,31,\cdots\}\implies \\ (B-A)\equiv\pm 1 \pmod 8 $
$ (C-A)=2k^2\implies\\ (C-A)\in\big\{2,8,18,32,50,\cdots\big\} $
$\dfrac{B-A}{C-A} =\dfrac{2k^2-(2n -1)^2}{2k^2} =1-\dfrac{(2n-1)^2}{2k^2}$
$(B>A)\implies 2k^2 \ge (2n-1)^2\\ \implies k \ge n\\ \implies 0 < \dfrac{B-A}{C-A} < 1\\$
Note that $\space\ 2k^2 \ge (2n-1)^2\space$ is not sufficient for $\space B>A\space $ as in: $\qquad\qquad (n,2k^2)=(2,8)\quad F(n,k)=F(2,2)= (21,20,29)\\ $
We can see, with arbitrarily skinny or fat Pythagorean triples like $$F(1,5000)=(10001,50010000,50010001)\quad (C-B)=1\\ \text{ or } \\ F(23661,33461)=(5406093003,5406093004,7645370045) \\ (B-A)=1)$$ from this formula, there are infinite triples where $\space \dfrac{B-A}{C-A}\space $ is equal to or comes arbitrarily close to any number in $\space[0,1].$
I developed a formula which generates only triples where $GCD(A,B,C)$ is an odd square. It skips the trivials, the doubles, the even square multiples, and the doubles of these that Euclid's formula generates. Perhaps it might help with this problem. For example, it shows how $\dfrac{B-A}{C-A} =1-\dfrac{(2n-1)^2}{2k^2}$ where $\space n\space$ is a set of triples and $\space k\space $ is the number of the triple in that set. If $2k^2>(2n-1)^2,\space$ then $B>A.$
– poetasis Jan 28 '22 at 23:27