A230195 Integer areas A of the triangles such that A and the sides are integers, and the length of the inradius is a prime number.

24, 30, 36, 42, 48, 54, 60, 66, 84, 96, 114, 120, 126, 150, 156, 198, 210, 270, 294, 330, 336, 390, 420, 462, 504, 510, 546, 570, 630, 714, 726, 756, 810, 840, 924, 930, 1008, 1014, 1056, 1134, 1386, 1428, 1554, 1638, 1680, 1716, 1734, 1848, 1890, 1950, 2016

Offset: 1

Author

Michel Lagneau, Oct 11 2013

Keywords

• nonn

Comments

Subsequence of A228383.

The corresponding inradii r are 2, 2, 2, 2, 3, 3, 2, 3, 3, 3, 3, 3, 3, 5, 3, 3, 3, 5, 7, 5, 7, 5, 7, 7, 7, 5, 7, 5, ...

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.

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 is prime.
		

Links

• Mohammad K. Azarian, Solution of problem 125: Circumradius and Inradius, Math Horizons, Vol. 16, No. 2 (Nov. 2008), p. 32.

Crossrefs

Cf. A228383.

Programs

Mathematica

nn = 1500; 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]] && PrimeQ[Sqrt[area2]/s], AppendTo[lst, Sqrt[area2]]]], {a, nn}, {b, a}, {c, b}]; Union[lst]