A210207 Area A of the non-right triangles such that A, the sides, and the circumradius are integers.
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
Offset: 1
Keywords
Examples
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.
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
Programs
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Maple
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
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]
Formula
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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