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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.

%I A219192 #18 Feb 16 2025 08:33:18
%S A219192 2352,9408,21168,37632,58800,69360,84672,115248,150528,190512,235200,
%T A219192 253920,277440,284592,338688,397488,460992,529200,602112,624240,
%U A219192 645792,679728,762048,849072,940800,1015680,1037232,1109760,1138368,1244208,1354752,1470000,1589952
%N 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.
%C A219192 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.
%C A219192 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.
%C A219192 If n is in this sequence, so is n*k^2 for any k > 0. Thus this sequence is infinite.
%C A219192 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.
%H A219192 Mohammad K. Azarian, <a href="http://www.jstor.org/stable/25678790">Solution to Problem S125: Circumradius and Inradius</a>, Math Horizons, Vol. 16, Issue 2, November 2008, p. 32.
%H A219192 E. Gürel, <a href="http://www.jstor.org/stable/2690677">Solution to Problem 1472, Maximal Area of Quadrilaterals</a>, Math. Mag. 69, 149, 1996.
%H A219192 Martin Josefsson, <a href="http://forumgeom.fau.edu/FG2011volume11/FG201116.pdf">The area of a Bicentric Quadrilateral</a>, Forum Geometricum (2011) 11:155-164.
%H A219192 Eric Weisstein's World of Mathematics, <a href="https://mathworld.wolfram.com/CyclicQuadrilateral.html">Cyclic Quadrilateral</a>
%H A219192 Eric Weisstein's World of Mathematics, <a href="https://mathworld.wolfram.com/BicentricQuadrilateral.html">Bicentric Quadrilateral</a>
%e A219192 2352 is in the sequence because, with sides (a,b,c,d) = (56,56,42,42) we obtain :
%e A219192 s = (56+56+42+42)/2 = 98;
%e A219192 A = sqrt(56*56*42*42) = 2352 = sqrt((98-56)(98-56)(98-42)(98-42)) (Brahmagupta’s Formula);
%e A219192 r = 2352/(56+42) =24.
%e A219192 R = sqrt((56*56+42*42)(56*42+56*42)(56*42+56*42))/(4*2352) = 35.
%p A219192 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:
%t A219192 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]
%Y A219192 Cf. A210250, A219193.
%K A219192 nonn
%O A219192 1,1
%A A219192 _Michel Lagneau_, Nov 14 2012