incircle radius determined by Pythagorean triple

If the sides of a right triangle are integers, then so is the radius of the incircle of this triangle.
For example, the incircle radius of the Egyptian triangle is 1.

Proof.  The sides of such a right triangle may be expressed by the integer parametres $m,n$ with $m>n>0$  as

$a=\mathrm{\hspace{0.33em}2}mn,b=m^2-n^2,c=m^2+n^2;$(1)

the radius of the incircle (http://planetmath.org/Incircle) is

$r={\displaystyle \frac{2A}{a+b+c}},$(2)

where $A$ is the area of the triangle.  Using (1) and (2) we obtain

$$r=\frac{2\cdot 2mn\cdot (m^2-n^2)/2}{2mn+(m^2-n^2)+(m^2+n^2)}=\frac{2mn(m+n)(m-n)}{2m(m+n)}=(m-n)n,$$

which is a positive integer.

Remark.  The corresponding radius of the circumcircle need not to be integer, since by Thales’ theorem, the radius is always half of the hypotenuse which may be odd (e.g. 5).