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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.
1
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
1,1
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.
LINKS
Mohammad K. Azarian, Solution of problem 125: Circumradius and Inradius, Math Horizons, Vol. 16, No. 2 (Nov. 2008), p. 32.
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 is prime.
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]
CROSSREFS
Cf. A228383.
Sequence in context: A167758 A248693 A075422 * A098030 A290451 A068544
KEYWORD
nonn
AUTHOR
Michel Lagneau, Oct 11 2013
STATUS
approved