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A208984
Areas A of the triangles such that A, the sides, the circumradius and the inradius are integers.
12
24, 96, 120, 168, 216, 240, 336, 384, 432, 480, 600, 624, 672, 720, 768, 840, 864, 960, 1080, 1176, 1320, 1344, 1512, 1536, 1560, 1680, 1728, 1848, 1920, 1944, 2016, 2040, 2160, 2184, 2304, 2376, 2400, 2496, 2520, 2688, 2856, 2880, 2904, 3000, 3024, 3072, 3240
OFFSET
1,1
COMMENTS
a(n) is divisible by 24, and the positive squares A000290(n) are included in the sequence a(n)/24 = {1, 4, 5, 7, 9, 10, 14, 16, 18, 20, 25, 26, 28, 30, 32, 35, 36, 40, 45, 49, 55, 56, 63, 64, 65, ...}.
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 and the circumradius is given by R = abc/4A.
LINKS
Mohammad K. Azarian, Solution of problem 125: Circumradius and Inradius, Math Horizons, Vol. 16, No. 2 (Nov. 2008), p. 32.
Eric W. Weisstein, MathWorld: Circumradius
Eric W. Weisstein, MathWorld: Inradius
EXAMPLE
a(1) = 24 because, for (a,b,c) = (6, 8, 10) => s = (6 + 8 + 10)/2 = 12, and
A = sqrt(12(12-6)(12-8)(12-10)) = sqrt(576) = 24;
R = abc/4A = 480/4*24 = 5;
r = A/p = 24/12 = 2.
MAPLE
with(numtheory):T:=array(1..1000):k:=0:nn:=250: for a from 1
to nn do: for b from a to nn do: for c from b to nn do: p:=(a+b+c)/2 : x:=p*(p-a)*(p-b)*(p-c): if x>0 then s:=sqrt(x) :if s=floor(s) and irem(a*b*c, 4*s) = 0 and irem(s, p)=0 then k:=k+1:T[k]:= s: else fi:fi:od:od:od: L := [seq(T[i], i=1..k)]:L1:=convert(T, set):A:=sort(L1, `<`): print(A):
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[a*b*c/(4* Sqrt[area2])] && IntegerQ[Sqrt[area2]/s], AppendTo[lst, Sqrt[area2]]]], {a, nn}, {b, a}, {c, b}]; Union[lst]
CROSSREFS
Sequence in context: A055671 A090214 A283446 * A103251 A256418 A198387
KEYWORD
nonn
AUTHOR
Michel Lagneau, Mar 04 2012
STATUS
approved