How to find the side lengths of a triangle in the incircle of a Pythagorean triple

Question

Given a Pythagorean triple with an incircle, how do I find the sides of a triangle connecting the triple's tangents to that incircle. In the diagram below, I know how to find side $\,i\,$ but not the other two.

How do I find point $\,H\,$ coordinates or is there some other way to find the lengths of segments $\,j\,$ and $\,k$?

$\textbf{Update: }$ From a comment by Calvin Lin, I have inferred that, if we let vertex circle radii moving counterclockwise from the upper right be $\quad r_1=5\quad r_2=12\quad r_3=3\quad$ we can get the $H$-values as follows.

\begin{align*} X_H =\dfrac{r_2}{r_1+r_2}\cdot(r_2+r_3) =\dfrac{A(A-B+C)}{2C} =\dfrac{180}{17} \\\\Y_H  =\dfrac{r_2}{r_1+r_2}\cdot(r_1+r_3) =\dfrac{B(A-B+C)}{2 C}=\dfrac{96}{17} \end{align*} so

\begin{align*} i&=\sqrt{(15-12)^2+(3-0)^2}=3\sqrt{2}\\ \\ j&=\sqrt{\bigg(12-\dfrac{180}{17}\bigg)^2+\bigg(\dfrac{96}{17}-0\bigg)^2} =\dfrac{24}{\sqrt{17}}\\ \\ k&=\sqrt{\bigg(15-\dfrac{180}{17}\bigg)^2+\bigg(\dfrac{96}{17}-3\bigg)^2} =15\sqrt{\dfrac{2}{17}} \end{align*}

Thank you Calvin Lin.

Answer 1

With $(m,n)$ integers, and $m>n$ then $a = m^2-n^2, b = 2mn, c = m^2+n^2$ form a $Pyhagorean triple.$

For the $(15,8,17)$ triangle $m=4, n=1$

The perimeter of this triangle is
$p = a+b+c = m^2 - n^2 + 2mn + m^2+n^2 = 2m(m + n)\\ s = \frac 12 p = m(m+n)\\ r = \frac {ab}{p} = \frac {(m^2-n^2)(2mn)}{2m(m+n)} = n(m-n)\\ s-a = m(m+n)-m^2+n^2 = n(m+n)\\ s-b = m(m+n) - 2mn = m(m-n)$

$r$ is the radius of the in-circle if that was not clear.

$\tan \frac 12 A = \frac {r}{s-a} = \frac {n(m-n)}{n(m+n)} = \frac {m-n}{n+m}\\ \tan \frac 12 B = \frac {r}{s-b} = \frac {n(m-n)}{m(m-n)} = \frac {n}{m}$

$j = 2(s-a) \sin \frac 12 A = \frac {2n(m+n)(m-n)}{\sqrt {(m-n)^2 + (n+m)^2}} = \frac {2n(m^2-n^2)}{\sqrt {2m^2 + 2n^2}} = \frac {an\sqrt 2}{\sqrt{c}}$

$k = 2(s-b) \sin \frac 12 B = \frac {2mn(m-n)}{\sqrt {m^2 + n^2}} = \frac {b(m-n)}{\sqrt{c}}$

$i = (s-c)\sqrt 2 = n(m-n)\sqrt{2}$

With the $(15,8,17)$ triangle, $(i,j,k) = (3\sqrt 2, \frac {15\sqrt 2}{\sqrt {17}},\frac {24}{\sqrt {17}})$

Answer 2

For a general $\triangle ABC$, define the side lengths $BC = a$, $CA = b$, $AB = c$, the semiperimeter $s = (a+b+c)/2$, and the inradius $IA' = IB' = IC' = r$.

Then since $AB' = AC'$ and $BC' = BA'$ and $CA' = CB'$, we have $s = AB' + BC' + CA' = AC' = BA' + CB'$. Moreover, $$AB' = s - (BC' + CA') = s - (BA' + CA') = s - a,$$ so similarly, $$BC' = s-b, \quad CA' = s-c.$$ Then since $\triangle AB'I \sim \triangle AA''B'$, it follows that $$\frac{A''B'}{B'I} = \frac{AA''}{AB'}.$$ And since $\triangle AA''B'$ is right, $AA'' = \sqrt{(AB')^2 - (A''B')^2}$. Hence $$\frac{A''B'}{r} = \frac{\sqrt{(s-a)^2 - (A''B')^2}}{s-a},$$ and solving for $A''B'$ yields $$A''B' = \frac{r(s-a)}{\sqrt{r^2 + (s-a)^2}}.$$

Since Heron's formula gives $$|\triangle ABC| = \sqrt{s(s-a)(s-b)(s-c)}$$ and we also have $$|\triangle ABC| = |\triangle IAB| + |\triangle IBC| + |\triangle ICA| = \frac{rc + ra + rb}{2} = rs,$$ then $$r = \sqrt{\frac{(s-a)(s-b)(s-c)}{s}}.$$ Consequently, after elimination of $r$ and simplification, $$B'C' = 2A''B' = 2(s-a)\sqrt{\frac{(s-b)(s-c)}{bc}}.$$ Similarly, $$A'B' = 2(s-c)\sqrt{\frac{(s-a)(s-b)}{ab}}, \quad C'A' = 2(s-b)\sqrt{\frac{(s-c)(s-a)}{ca}}.$$ These formulae do not require $\triangle ABC$ to be right, nor that $a,b,c$ are integers. However, if $\angle C$ is right, then some simplification occurs because $s-c = r$. In particular, $$A'B' = \frac{a+b-c}{\sqrt{2}}, \\ B'C' = (b+c-a)\sqrt{\frac{c-b}{2c}}, \\ C'A' = (a+c-b)\sqrt{\frac{c-a}{2c}}. $$