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A210207
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Area A of the non-right triangles such that A, the sides, and the circumradius are integers.
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5
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168, 432, 480, 624, 672, 768, 1320, 1512, 1536, 1560, 1680, 1728, 1848, 1920, 2040, 2304, 2376, 2496, 2520, 2688, 2856, 3024, 3072, 3240, 3696, 3720, 3840, 3864, 3888, 4104, 4200, 4320, 4536, 5280, 5376, 5616, 5712, 6000, 6048, 6144, 6240, 6552, 6720, 6912
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OFFSET
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1,1
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COMMENTS
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A103251 gives the areas of right triangles with the same property (the area, the sides, and the circumradius are integers). Thus the intersection of this sequence with A103251 will give the areas of 2 families of triangles with the same property: one family of right triangles and one family of non-right triangles.
For example a(3) = A103251(8) = 480 generates two triangles whose sides are
(a,b,c) = (32, 50, 78) = > A = 480, R = 65, and 32^2 + 50^2 is no square;
(a,b,c) = (20, 48, 52) = > A = 480, R = 26, and 20^2 + 48^2 = 52^2 is square.
{a(n) intersection A103251} = {480, 1320, 1536, 1920, 2520, 3024, 3696, 3840, ...}
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LINKS
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FORMULA
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Area A = sqrt(s*(s-a)*(s-b)*(s-c)) with s = (a+b+c)/2 (Heron's formula);
Circumradius R = a*b*c/4A.
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EXAMPLE
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168 is in the sequence because, for (a,b,c) = (14,30,40), A = sqrt(42*(42-14)*(42-30)*(42-40)) = 168, and 14^2 + 30^2 is no square.
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MAPLE
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T:=array(1..4000):nn:=400:k:=0: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): u:=a^2+b^2:if x>0 then x1:=sqrt(x) : y:=a*b*c/(4*x1):
else fi:if x1=floor(x1) and y = floor(y) and u <> c^2 then k:=k+1:T[k]:=x1:else fi:od:od:od: L := [seq(T[i], i=1..k)]:L1:=convert(T, set):A:=sort(L1, `<`): print(A):
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MATHEMATICA
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nn=400; lst={}; Do[s=(a+b+c)/2; If[IntegerQ[s], area2=s (s-a) (s-b) (s-c); If[0 < area2 && a^2 != b^2+c^2 && IntegerQ[Sqrt[area2]] && IntegerQ[a*b*c/(4*Sqrt[area2])], AppendTo[lst, Sqrt[area2]]]], {a, nn}, {b, a}, {c, b}]; Union[lst]
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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