# Radius of incircle and Pythagorean triangle

How is the following statement true?

If $m$ and $n$ are positive integers with $m > n$. Let $a = 2mn, b = m^2 - n^2$ and $c = m^2 + n^2$ be the sides of a Pythagorean triangle. Then the radius of the in-circle $r$ is given by the integer $r = n (m-n)$.

Thanks!

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## 1 Answer

This is correct. This can be derived from the following equality for a right angled triangle with sides $a, b, c$ (where $a^2+b^2=c^2$):

$$r(a+b+c)=ab$$

This can be interpreted as an equality of areas when you split up the triangle in three smaller ones by drawing lines from the center of the incircle to each of the three vertices.

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Of course, you have in mind that $c$ is the hypotenuse. –  Gerry Myerson Feb 28 '12 at 10:08
@GerryMyerson: sure. I expressed that as $a^2+b^2=c^2$. But surely the names $a,b,c$ are folklore? ;-) –  WimC Feb 28 '12 at 10:57