Is there a way to see this geometrically? In my answer to this question -
Finding the no. of possible right angled triangle.
- I derived this result:
If a right triangle
has integer sides
$a, b, c$
and integer inradius $r$,
then all possible values
of $a$ and $b$
can be gotten in terms of $r$
as follows:
For every possible divisor $d$
of $2r^2$,
$a = 2r+d$
and
$b = 2r+\dfrac{2r^2}{d}$.
These are exactly the solutions.
From this, of course,
the number of solutions
depends only on
the prime factorization of $r$.
My answer involved
some annoyingly complicated algebra.
My question is
"is there a geometrical way
to show that the expressions
for $a$ and $b$ are true?"
(Added later)
Another way to phrase this,
without mentioning divisibility:
Take a rectangle of area $2r^2$.
Extend the sides by $2r$.
Then the inradius of the
resulting right triangle
is $r$.
 A: As shown in this answer by using Heron's formula, if a generic triangle has integer inradius $r$ and integer sides $a$, $b$, $c$, then its sides can be written as $a=x+y$, $b=x+z$, $c=y+z$, where positive integers $x$, $y$, $z$ satisfy
$$
r^2(x+y+z)=xyz.
$$
If the triangle is rectangle, Pythagoras' constraint $a^2+b^2=c^2$ translates into the additional relation $x+y+z=yz/x$, which substituted into the above equation gives $x=r$.
One can then set $x=r$ in the same equation and solve for $z$, to find:
$$
z=r{y+r\over y-r},
\quad\hbox{that is:}\quad
z=r+{2r^2\over d},
\ \hbox{where}\ d=y-r.
$$
For $z$ to be integer $d$ must be a divisor of $2r^2$ and substituting $x=r$, $y=r+d$, $z=r+2r^2/d$ into the expressions for $a$, $b$ one finds the expressions reported in the question:
$$
a=2r+d,\quad b=2r+{2r^2\over d}.
$$
EDIT.
There is a simpler geometrical way of getting the same result. As one can see from the diagram below, in a right triangle with legs $a$, $b$, hypotenuse $c$ and inradius $r$ one has:
$$
c=a+b-2r.
$$

Squaring both sides of that equation gives
$$
2r(a+b)=2r^2+ab,
$$
and one can solve for $b$ to obtain:
$$
b=2r{a-r\over a-2r}.
$$
Defining $d=a-2r$ this can be rewritten as
$$
b=2r{d+r\over d}=2r+{2r^2\over d}.
$$
From that it is apparent that $d$ must be a divisor of $2r^2$ for $a$, $b$ and $r$ to be integers, and one recovers the same results obtained above.
