# incircle radius determined by Pythagorean triple

If the sides of a right triangle^{} are integers, then so is the radius of the incircle^{} of this triangle.

For example, the incircle radius of the Egyptian triangle is 1.

Proof. The sides of such a right triangle may be expressed by the integer parametres $m,n$ with $m>n>0$ as

$a=\mathrm{\hspace{0.33em}2}mn,b={m}^{2}-{n}^{2},c={m}^{2}+{n}^{2};$ | (1) |

the radius of the incircle (http://planetmath.org/Incircle) is

$r={\displaystyle \frac{2A}{a+b+c}},$ | (2) |

where $A$ is the area of the triangle. Using (1) and (2) we obtain

$$r=\frac{2\cdot 2mn\cdot ({m}^{2}-{n}^{2})/2}{2mn+({m}^{2}-{n}^{2})+({m}^{2}+{n}^{2})}=\frac{2mn(m+n)(m-n)}{2m(m+n)}=(m-n)n,$$ |

which is a positive integer.

Remark. The corresponding radius of the circumcircle^{} need not to be integer, since by Thales’ theorem, the radius is always half of the hypotenuse^{} which may be odd (e.g. 5).

Title | incircle radius determined by Pythagorean triple |

Canonical name | IncircleRadiusDeterminedByPythagoreanTriple |

Date of creation | 2013-03-22 17:45:46 |

Last modified on | 2013-03-22 17:45:46 |

Owner | pahio (2872) |

Last modified by | pahio (2872) |

Numerical id | 14 |

Author | pahio (2872) |

Entry type | Feature |

Classification | msc 11A05 |

Synonym | incircle radius of right triangle |

Related topic | Triangle |

Related topic | PythagoreanTriple |

Related topic | DifferenceOfSquares |

Related topic | FirstPrimitivePythagoreanTriplets |

Related topic | X4Y4z2HasNoSolutionsInPositiveIntegers |