A228383 Area A of the triangle such that A, the sides, and the inradius are integers.

6, 24, 30, 36, 42, 48, 54, 60, 66, 84, 96, 108, 114, 120, 126, 132, 144, 150, 156, 168, 180, 192, 198, 210, 216, 240, 252, 264, 270, 294, 300, 324, 330, 336, 360, 378, 384, 390, 396, 408, 420, 432, 456, 462, 468, 480, 486, 504, 510, 522, 528, 540, 546, 570

Offset: 1

Author

Michel Lagneau, Aug 21 2013

Keywords

• nonn

Comments

The sequences A208984 and A185210 are subsequences of this sequence. The corresponding inradius r are 1, 2, 2, 2, 2, 3, 3, 3, 3, 4, 4, 4, 3, 4, 3, ...

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 inradius r is given by r = A/s.

a(n) is divisible by 6 and the squares of the form 36k^2 are in the sequence.

Examples

			24 is in the sequence because for (a, b, c) = (6, 8, 10) => s =(6 + 8 + 10)/2 = 12; A = sqrt(12*(12-6)*(12-8)*(12-10)) = sqrt(576)= 24; r = A/s = 2.
		

Links

• Mohammad K. Azarian, Solution of problem 125: Circumradius and Inradius, Math Horizons, Vol. 16, No. 2 (Nov. 2008), pp. 32-34.
• Eric W. Weisstein, MathWorld: Inradius

Crossrefs

Cf. A188158, A120572, A210250, A208984, A185210.

Programs

Mathematica

nn = 1000; lst = {}; Do[s = (a + b + c)/2; If[IntegerQ[s], area2 = s (s - a) (s - b) (s - c); If[0 < area2 <= nn^2 && IntegerQ[Sqrt[area2]] && IntegerQ[Sqrt[area2]/s], AppendTo[lst, Sqrt[area2]]]], {a, nn}, {b, a}, {c, b}]; Union[lst]