A256695 Integer areas A of integer-sided triangles (a, b, c) such that the area of the triangle (a+b, a+c, b+c) is also an integer.

12, 48, 108, 192, 300, 432, 588, 768, 972, 1008, 1200, 1452, 1728, 2028, 2352, 2448, 2520, 2700, 2772, 3060, 3072, 3468, 3888, 4032, 4332, 4800, 5292, 5808, 6348, 6912, 7500, 8112, 8748, 9072, 9408, 9792, 10080, 10092, 10800, 11088, 11532, 11628, 12240, 12288

Offset: 1

Author

Michel Lagneau, Apr 08 2015

Keywords

• nonn

Comments

The areas of the primitive triangles are 12, 2520, 2772, 3060, 4032, 5808, 9072, 11088, 11628, 17136, 24948, 41580, ...

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.

The sequence A135453 (numbers of the form 12*n^2) is included in the sequence because a(1) = 12 is a primitive triangle of the subsequence k^2*a(1), k=1,2,3,...

The following table gives the first values (A, A', a, b, c) where A is the integer area of the triangle (a, b, c), A' is the integer area of the triangle (a+b, a+c, b+c).

+-------+--------+------+------+------+

| A | A' | a | b | c |

+-------+--------+------+------+------+

| 12 | 60 | 5 | 5 | 8 |

| 48 | 240 | 10 | 10 | 16 |

| 108 | 540 | 15 | 15 | 24 |

| 192 | 960 | 20 | 20 | 32 |

| 300 | 1500 | 25 | 25 | 40 |

| 432 | 2160 | 30 | 30 | 48 |

| 588 | 2940 | 35 | 35 | 56 |

| 768 | 3840 | 40 | 40 | 64 |

| 972 | 4860 | 45 | 45 | 72 |

+-------+--------+------+------+------+

We find a majority of isosceles triangles, but there is a subsequence of non-isosceles triangles with areas 2520, 3060, 10080, 11088, ...

+--------+----------+-------+-------+-------+

| A | A' | a | b | c |

+--------+----------+-------+-------+-------+

| 2520 | 18270 | 29 | 174 | 175 |

| 3060 | 33150 | 39 | 221 | 250 |

| 10080 | 73080 | 58 | 348 | 350 |

| 11088 | 64350 | 150 | 169 | 275 |

| 12240 | 12240 | 78 | 442 | 500 |

| 17136 | 92820 | 168 | 221 | 325 |

| 41580 | 183150 | 250 | 333 | 407 |

+--------+----------+-------+-------+-------+

Examples

			a(1) = 12 because, for (a,b,c) = (5, 5, 8) => s = (5+5+8)/2 = 9, and
A = sqrt(9(9-5)(9-5)(9-8)) = sqrt(144) = 12 and the triangle (5+5, 5+8, 5+8) = (10, 13, 13)=> s1 = (10+13+13)/2 = 18, and A1 = sqrt(18(18-10)(18-13)(18-13)) = sqrt(3600) = 60 is an integer.
		

Crossrefs

Cf. A135453, A188158.

Programs

Mathematica

nn=1000;lst={};Do[s=(a+b+c)/2;If[IntegerQ[s],area2=s(s-a)(s-b)(s-c);u=a+b;v=a+c;w=b+c;s1=(u+v+w)/2;area3=s1(s1-u)(s1-v)(s1-w);If[area2>0&&area3>0&&IntegerQ[Sqrt[area2]]&&IntegerQ[Sqrt[area3]], AppendTo[lst,Sqrt[area2]]]],{a,nn},{b,a},{c,b}];Union[lst]