Inradius of Pythagorean Triangle is Integer
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:
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.
$\blacksquare$
Proof 2
Let $\triangle ABC$ be such that $\angle C$ is a right angle.
Let:
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:
\(\ds a\) | \(=\) | \(\ds m^2 - n^2\) | ||||||||||||
\(\ds b\) | \(=\) | \(\ds 2 m n\) |
We have that:
\(\ds \map \Area {\triangle ABC}\) | \(=\) | \(\ds \map \Area {\triangle BIC} + \map \Area {\triangle AIC} + \map \Area {\triangle AIB}\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \dfrac {a b} 2\) | \(=\) | \(\ds \dfrac {a r} 2 + \dfrac {b r} 2 + \dfrac {c r} 2\) | Area of Triangle in Terms of Side and Altitude | ||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {r \paren {a + b + c} } 2\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds r\) | \(=\) | \(\ds \dfrac {a b} {a + b + c}\) | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {\paren {m^2 - n^2} \paren {2 m n} } {\paren {m^2 - n^2} + \paren {2 m n} + \paren {m^2 + n^2} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {2 m n \paren {m - n} \paren {m + n} } {2 m \paren {m + n} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {2 m n \paren {m - n} \paren {m + n} } {2 m \paren {m + n} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds n \paren {m - n}\) |
As $m$ and $n$ are both integers, it follows that $r$ is also an integer.
$\blacksquare$