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

1,1

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.

Links

Table of n, a(n) for n=1..54.

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

Example

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.

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]

Crossrefs

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

Sequence in context: A344700 A064510 A335736 * A249667 A114274 A335215

Adjacent sequences: A228380 A228381 A228382 * A228384 A228385 A228386

Keyword

nonn

Author

Michel Lagneau, Aug 21 2013

Status

approved