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).

Title	incircle radius determined by Pythagorean triple
Canonical name	IncircleRadiusDeterminedByPythagoreanTriple
Date of creation	2013-03-22 17:45:46
Last modified on	2013-03-22 17:45:46
Owner	pahio (2872)
Last modified by	pahio (2872)
Numerical id	14
Author	pahio (2872)
Entry type	Feature
Classification	msc 11A05
Synonym	incircle radius of right triangle
Related topic	Triangle
Related topic	PythagoreanTriple
Related topic	DifferenceOfSquares
Related topic	FirstPrimitivePythagoreanTriplets
Related topic	X4Y4z2HasNoSolutionsInPositiveIntegers