Radius of in-circle The questions from geometry are most fascinating to me. As a parent I love to learn geometry and shapes from my son's textbook. I came across this statement about the in-circle:
Let  $m, n$ that are integers and $m > n$. Let $a = 2mn$, $b = m^2 - n^2$ and $c = m^2 + n^2$ be the sides of the Pythagorean triangle. Then the radius of the in-circle $r$ will be  $nm - n^2$. 
How can this happen?
Thanks in advance.
 A: A useful fact about triangles is $A = rs,$
where $r$ is the inradius and $s$ is the "semiperimeter" 
$$s := \frac{a+b+c}{2}.$$
Why? Draw in the segments between the incenter and the vertices (these are the angle bisectors). This cuts the triangle into three smaller triangles. Each has height $r$ and the bases are $a,b,$ and $c$, respectively. So the areas of these triangles are $\frac{ra}{2}, \frac{rb}{2},$ and $\frac{rc}{2}$; the sum of these areas is $rs$. 
Now that you have this; what's another way of writing the area of a right triangle? We'd have $A = \frac{1}{2}(2mn)(m^2-n^2) = mn(m^2-n^2) = mn(m-n)(m+n)$. For this triangle, $s = \frac{1}{2}(2mn + (m^2-n^2) + (m^2+n^2)) = mn + m^2 = m(n+m)$.
Equating our two expressinos for the area,
$mn(m-n)(m+n) = rm(n+m)$, and cancelling gives $r = n(m-n) = mn - n^2$.
A: Hint :
$$r=\sqrt{\frac{(s-a)(s-b)(s-c)}{s}} ~\text{where}~ s=\frac{a+b+c}{2}$$
relation between inradius and area of triangle 
A: Perhaps firstly convince yourself that the right triangle formed with those sides ($a, b$ and $c$ above) is geometrically allowed. Use Pythagoras to convinced yourself this triangle can be constructed.
Then (I haven't put this on paper this is merely conjecture) construct angle bisectors for each of the three vertices, these bisectors will slice through the inscribed circle. Does this help?
