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A219192
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Area A of the bicentric quadrilaterals such that A, the sides, the radius of the circumcircle and the radius of the incircle are integers.
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1
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2352, 9408, 21168, 37632, 58800, 69360, 84672, 115248, 150528, 190512, 235200, 253920, 277440, 284592, 338688, 397488, 460992, 529200, 602112, 624240, 645792, 679728, 762048, 849072, 940800, 1015680, 1037232, 1109760, 1138368, 1244208, 1354752, 1470000, 1589952
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OFFSET
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1,1
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COMMENTS
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Also numbers n such that there exists a decomposition n^2 = a*b*c*d where a,b,c,d are the sides of a bicentric quadrilateral with the area, the inradius and the circumradius integers.
In Euclidean geometry, a bicentric quadrilateral is a convex quadrilateral that has both an incircle and a circumcircle. If the sides are a, b, c, d, then the area is given by A = sqrt(a*b*c*d). The inradius r of a bicentric quadrilateral is determined by the sides a, b, c, d according to r = sqrt(a*b*c*d)/(a+c) = sqrt(a*b*c*d)/(b+d). The circumradius R (the radius of the circumcircle) is given by R = sqrt((ab+cd)(ac+bd)(ad+bc))/4A.
If n is in this sequence, so is n*k^2 for any k > 0. Thus this sequence is infinite.
In view of the preceding comment, one might call "primitive" the elements of the sequence for which there is no k>1 such that n/k^2 is again a term of the sequence. These elements are 2352, 69360, 253920, 645792,... are listed in A219193.
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LINKS
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EXAMPLE
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2352 is in the sequence because, with sides (a,b,c,d) = (56,56,42,42) we obtain :
s = (56+56+42+42)/2 = 98;
A = sqrt(56*56*42*42) = 2352 = sqrt((98-56)(98-56)(98-42)(98-42)) (Brahmagupta’s Formula);
r = 2352/(56+42) =24.
R = sqrt((56*56+42*42)(56*42+56*42)(56*42+56*42))/(4*2352) = 35.
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MAPLE
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with(numtheory):nn:=15000:for a from 1 to nn do: b:=a: for c from b to nn do: for d from c to c while(sqrt(a*b*c*d)=floor(sqrt(a*b*c*d))) do:s:=(a+b+c+d)/2:a1:=(s-a)*(s-b)*(s-c)*(s-d):a2:=sqrt(a*b*c*d):r1:=a2/(a+c):r2:=a2/(b+d):rr:= sqrt((a*b+c*d) * (a*c+b*d) * (a*d+b*c))/(4*a2):if a1>0 and floor(sqrt(a1))=sqrt(a1) and a2 =floor(a2) and a2=sqrt(a1) and r1=floor(r1) and r2=floor(r2) and r1=r2 and rr =floor(rr) then printf ( "%d %d %d %d %d %d %d\n", a2, a, b, c, d, r1, rr):else fi:od:od:od:
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MATHEMATICA
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nn=15000; lst={}; Do[s=(2*a+2*d)/2; If[IntegerQ[s], area2=(s-a)*(s-a)*(s-d)*(s-d); area22=a*a*d*d; If[0<area2&&IntegerQ[Sqrt[area2]]&&IntegerQ[Sqrt[area22]&&IntegerQ[Sqrt[area22]/(a+d)]&&IntegerQ[Sqrt[(a*a+d*d)*(a*d+a*d)*(a*d+a*d)/((s-a)*(s-a)*(s-d)*(s-d))]/4]], AppendTo[lst, Sqrt[area22]]]], {a, nn}, {d, a}]; 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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