# 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:

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.

$\blacksquare$