# incircle radius determined by Pythagorean triple

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

 $\displaystyle a\;=\;2mn,\quad b\;=\;m^{2}\!-\!n^{2},\quad c\;=\;m^{2}\!+\!n^{2};$ (1)

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

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

 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