Inradius of Pythagorean Triangle is Integer/Proof 1

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 $c$

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:

\(\ds a\)

\(=\)

\(\ds m^2 - n^2\)

\(\ds b\)

\(=\)

\(\ds 2 m n\)

Hence:

\(\ds c\)

\(=\)

\(\ds \paren {a - r} + \paren {b - r}\)

\(\ds \)

\(=\)

\(\ds a + b - 2 r\)

\(\ds \leadsto \ \ \)

\(\ds r\)

\(=\)

\(\ds \dfrac {a + b - c} 2\)

\(\ds \leadsto \ \ \)

\(\ds r\)

\(=\)

\(\ds \dfrac {m^2 - n^2 + 2 m n - \paren {m^2 + n^2} } 2\)

\(\ds \leadsto \ \ \)

\(\ds r\)

\(=\)

\(\ds \dfrac {2 m n - 2 n^2} 2\)

\(\ds \leadsto \ \ \)

\(\ds r\)

\(=\)

\(\ds n \paren {m - n}\)

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

$\blacksquare$