A256584 - OEIS

%I #8 Apr 04 2015 18:05:41

%S 6,12,54,96,108,126,144,180,192,216,234,240,264,270,336,360,378,408,

%T 480,486,504,522,540,594,600,744,750,756,864,900,972,990,1008,1026,

%U 1116,1134,1224,1296,1350,1386,1404,1494,1500,1536,1584,1620,1656,1728,1800,1872

%N Integer areas of integer-sided triangles where at least one of the three altitudes is of perfect square length.

%C a(n) contains A210643.

%C There are triangles with rational square, for instance, with the area 144, we find for (a,b,c)=(6,50,52) the altitudes {Ha,Hb,Hc} = {72/13, 144/25, 48} but with the same area we find also for (a,b,c)=(18,20,34) the altitudes {Ha,Hb,Hc} = {144/17, 72/5, 16}.

%C The corresponding squares are 4, 4, 9, 16, 9, 9, 16, 9, 16, 36, 9, 16, 16, 36, 16, 9, 36, 16, 16, 36, 16, 36, 36, 36, 16, 16, 25, 36, 36, 36, 36, 36, 16, 36, 36, 36, 144, 144, 36, 36, 36, 36, 100, 64, 144, 36, 36, 36, 144, 36, ...

%C The subsequence of the primitive triangles are 6, 12, 126, 144, 180, 216, 234, ...

%C The area A of a triangle whose sides have lengths a, b, and c is given by Heron's formula: A = sqrt(s*(s-a)*(s-b)*(s-c)), where s = (a+b+c)/2.

%C The altitudes of a triangle with sides length a, b, c and area A have length given by Ha= 2A/a, Hb= 2A/b, Hc= 2A/c.

%C Properties of this sequence:

%C - The sequence is infinite because from de initial primitive triangle (3,4,5), the area A’ of the triangle (3*3^m, 4*3^m, 5*3^m) is also in the sequence where A’ = 6*3^2m and {Ha, Hb, Hc} = {4*3^m, 3^(m+1), (4*3^(m+1))/5}. The altitude Ha or Hb is square.

%C - There are three subsets of numbers included into a(n):

%C Case (i): A subset with right triangles (a,b,c) where a^2+b^2 = c^2 with area a2(n) = {6, 54, 96, 180, 240, 270, ...}

%C Case (ii): A subset with isosceles triangles of area a1(n)= {12, 108, 192, 360, 480, 540, ...} = 2*a1(n).

%C Case (iii): A subset with non-isosceles and non-right triangles of area a3(n)= {126, 144, 216, 234, 264, 336, ...}

%C -----------------------------------------------------------

%C |    A |   a |   b  |   c |    Ha    |    Hb   |   Hc     |

%C -----------------------------------------------------------

%C |    6 |   3 |   4  |   5 |   12/5   |    3    |    4     |

%C |   12 |   5 |   5  |   6 |     4    |   24/5  |   24/5   |

%C |   54 |   9 |  12  |  15 |   36/5   |    9    |   12     |

%C |   96 |  12 |  16  |  20 |    48/5  |   12    |   16     |

%C |  108 |  15 |  15  |  24 |     9    |   72/5  |   72/5   |

%C |  126 |  15 |  28  |  41 |   252/41 |    9    |   84/5   |

%C |  144 |  18 |  20  |  34 |  144/17  |   72/5  |   16     |

%C |  180 |   9 |  40  |  41 |  360/41  |    9    |   40     |

%C |  192 |  20 |  20  |  24 |    16    |   96/5  |   96/5   |

%C |  216 |  12 |  39  |  45 |   48/5   |  144/13 |   36     |

%C |  234 |  15 |  41  |  52 |    9     |  468/41 |  156/5   |

%C |  240 |  16 |  30  |  34 |  240/17  |   16    |   30     |

%C |  264 |  33 |  34  |  65 |  528/65  |  264/17 |   16     |

%H Eric Weisstein, <a href="http://mathworld.wolfram.com/Altitude.html">Altitude</a>

%H Eric Weisstein, <a href="http://mathworld.wolfram.com/IsoscelesTriangle.html">Isosceles Triangle</a>

%H Eric Weisstein, <a href="http://mathworld.wolfram.com/RightTriangle.html">Right Triangle</a>

%e 1350 is in the sequence because the altitudes of the triangle (45, 60, 75) are (60, 45, 36).

%t nn=200;lst={};Do[s=(a+b+c)/2;If[IntegerQ[s],area2=s (s-a) (s-b) (s-c);If[0<area2&&IntegerQ[Sqrt[area2]]&&(IntegerQ[Sqrt[2*Sqrt[area2]/a]]||IntegerQ[Sqrt[2*Sqrt[area2]/b]]||IntegerQ[Sqrt[2*Sqrt[area2]/c]]),AppendTo[lst,Sqrt[area2]]]],{a,nn},{b,a},{c,b}];Union[lst]

%Y Cf. A210643.

%K nonn

%O 1,1

%A _Michel Lagneau_, Apr 02 2015