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A219192 Area A of the bicentric quadrilaterals such that A, the sides, the radius of the circumcircle and the radius of the incircle are integers. 1
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 (list; graph; refs; listen; history; text; internal format)



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.


Table of n, a(n) for n=1..33.

Mohammad K. Azarian, Solution to Problem S125: Circumradius and Inradius, Math Horizons, Vol. 16, Issue 2, November 2008, p. 32.

E. Gürel, Solution to Problem 1472, Maximal Area of Quadrilaterals, Math. Mag. 69, 149, 1996.

Martin Josefsson, The area of a Bicentric Quadrilateral, Forum Geometricum (2011) 11:155-164.

Eric Weisstein's World of Mathematics, Cyclic Quadrilateral

Eric Weisstein's World of Mathematics, Bicentric Quadrilateral


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.


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:


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]


Cf. A210250, A219193.

Sequence in context: A003552 A173628 A130023 * A234078 A219193 A063517

Adjacent sequences:  A219189 A219190 A219191 * A219193 A219194 A219195




Michel Lagneau, Nov 14 2012



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