Inradius of Pythagorean Triangle is Integer/Proof 1

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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 $c$
Inradius of Pythagorean Triangle is Integer.png


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:

\(\displaystyle a\) \(=\) \(\displaystyle m^2 - n^2\)
\(\displaystyle b\) \(=\) \(\displaystyle 2 m n\)


Hence:

\(\displaystyle c\) \(=\) \(\displaystyle \paren {a - r} + \paren {b - r}\)
\(\displaystyle \) \(=\) \(\displaystyle a + b - 2 r\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle r\) \(=\) \(\displaystyle \dfrac {a + b - c} 2\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle r\) \(=\) \(\displaystyle \dfrac {m^2 - n^2 + 2 m n - \paren {m^2 + n^2} } 2\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle r\) \(=\) \(\displaystyle \dfrac {2 m n - 2 n^2} 2\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle r\) \(=\) \(\displaystyle n \paren {m - n}\)


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

$\blacksquare$