# Inradius of Pythagorean Triangle is Integer/Proof 2

## Theorem

Let $\triangle ABC$ be a Pythagorean triangle.

Let $r$ be the inradius of $\triangle ABC$.

Then $r$ is an integer.

## Proof

Let $\triangle ABC$ be such that $\angle C$ is a right angle.

Let:

$A$ be opposite the side $a$
$B$ be opposite the side $b$
$C$ be opposite the side $c$ Let $I$ denote the incenter of $\triangle ABC$.

From Solutions of Pythagorean Equation, we have that:

$c = m^2 + n^2$

for some $m, n \in \Z_{>0}$, and that the other sides are $m^2 - n^2$ and $2 m n$

Without loss of generality, let the sides $a$ and $b$ be such that:

 $\displaystyle a$ $=$ $\displaystyle m^2 - n^2$ $\displaystyle b$ $=$ $\displaystyle 2 m n$

We have that:

 $\displaystyle \map \Area {\triangle ABC}$ $=$ $\displaystyle \map \Area {\triangle BIC} + \map \Area {\triangle AIC} + \map \Area {\triangle AIB}$ $\displaystyle \leadsto \ \$ $\displaystyle \dfrac {a b} 2$ $=$ $\displaystyle \dfrac {a r} 2 + \dfrac {b r} 2 + \dfrac {c r} 2$ Area of Triangle in Terms of Side and Altitude $\displaystyle$ $=$ $\displaystyle \dfrac {r \paren {a + b + c} } 2$ $\displaystyle \leadsto \ \$ $\displaystyle r$ $=$ $\displaystyle \dfrac {a b} {a + b + c}$ $\displaystyle$ $=$ $\displaystyle \dfrac {\paren {m^2 - n^2} \paren {2 m n} } {\paren {m^2 - n^2} + \paren {2 m n} + \paren {m^2 + n^2} }$ $\displaystyle$ $=$ $\displaystyle \dfrac {2 m n \paren {m - n} \paren {m + n} } {2 m \paren {m + n} }$ $\displaystyle$ $=$ $\displaystyle \dfrac {2 m n \paren {m - n} \paren {m + n} } {2 m \paren {m + n} }$ $\displaystyle$ $=$ $\displaystyle n \paren {m - n}$

As $m$ and $n$ are both integers, it follows that $r$ is also an integer.

$\blacksquare$