# 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:

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

\(\displaystyle b\) | \(=\) | \(\displaystyle 2 m n\) |

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:

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.

$\blacksquare$