Inradius of Pythagorean Triangle is Integer

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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 1

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.

$\blacksquare$


Proof 2

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


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.

$\blacksquare$