# Nearly Equilateral Heronian Triangles

Heronian triangles are those whose side lengths and areas are integers. There are no Heronian triangles that are equilateral (all sides equal), but there are some that are *nearly* equilateral. There are three distinct ways this can occur:

**(1)** side lengths n-1, n, and n+1

**(2)** side lengths n, n, and n+1

**(3)** side lengths n, n+1, and n+1

For each case, there are actually infinitely many Heronian triangles which can be generated by recursive formulas.

### Case I: Sides n-1, n, and n+1

Here is a table showing the first five instances of such Heronian triangles:

Sides | Perimeter | Area | ||

n-1 | n | n+1 | ||

3 | 4 | 5 | 12 | 6 |

13 | 14 | 15 | 42 | 84 |

51 | 52 | 53 | 156 | 1170 |

193 | 194 | 195 | 582 | 16296 |

723 | 724 | 725 | 2172 | 226974 |

Define a sequence T(n) where T(1) = 4, T(2) = 14, T(3) = 52,... T(n) = the middle side length of the n^{th} nearly equilateral Heronian triangle in Case I. It turns out that T(n) satisfies a second order linear recursive equation:

T(n+2) = 4T(n+1) - T(n).

Using this formula, we can find the sixth row of entries for this table.

T(6) = 4T(5) - T(4)

= 4*724 - 194

= 2702.

This means that the next Heronian triangle in this list has side lengths 2701, 2702, and 2703.

### Case II: Sides n, n, and n+1

Here is a table showing the first five isosceles Heronian triangles in which the longest side is 1 more than the two shorter sides:

Sides | Perimeter | Area | ||

n | n | n+1 | ||

5 | 5 | 6 | 16 | 12 |

65 | 65 | 66 | 196 | 1848 |

901 | 901 | 902 | 2704 | 351780 |

12545 | 12545 | 12546 | 37636 | 68149872 |

174725 | 174725 | 174726 | 524176 | 13219419708 |

As you can see, the side lengths increase much more rapidly than in the previous case. Define a sequence V(n) where V(1) = 5, V(2) = 65, V(3) = 901,... V(n) = the shorter side length of the n^{th} nearly equilateral Heronian triangle in Case II. V(n) satisfies a third order linear recursive equation:

V(n+3) = 15V(n+2) - 15V(n+1) + V(n).

Using this formula, we can find the sixth row of entries for this table.

V(6) = 15V(5) - 15V(4) + V(3)

= 15*174725 - 15*12545 + 901

= 2433601.

This means that the next Heronian triangle in this list has side lengths 2433601, 2433601, and 2433602.

### Case III: Sides n, n+1, and n+1

Here is a table showing the first five isosceles Heronian triangles in which the longer sides are 1 more than the shortest side:

Sides | Perimeter | Area | ||

n | n+1 | n+1 | ||

16 | 17 | 17 | 50 | 120 |

240 | 241 | 241 | 722 | 25080 |

3360 | 3361 | 3361 | 10082 | 4890480 |

46816 | 46817 | 46817 | 140450 | 949077360 |

652080 | 652081 | 652081 | 1956242 | 184120982760 |

Now define a sequence W(n) where W(1) = 16, W(2) = 240, W(3) = 3360,... W(n) = the shortest side length of the n^{th} nearly equilateral Heronian triangle in Case III. W(n) also satisfies a third order linear recursive equation:

W(n+3) = 15W(n+2) - 15W(n+1) + W(n).

This is the exact same relation as in Case II, except the initial conditions are different. Using this formula, we can find the sixth row of entries for the Case III table.

W(6) = 15W(5) - 15W(4) + W(3)

= 15*652080 - 15*46816 + 3360

= 9082320.

This means that the next Heronian triangle in this list has side lengths 9082320, 9082321, and 9082321.

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