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%I #26 May 28 2016 07:24:36
%S 768,936,1200,2856,3072,3744,4536,4800,5016,5376,6696,6912,7056,7560,
%T 7752,8184,8424,9240,10800,11424,11544,12288,12480,12936,14976,16848,
%U 18144,18696,19200,19200,20064,21504,23040,23400,24024,25080,25704,25944,26784,27048,27648,27648,27648,27864,28224,28560,30000,30240,31008,32736,33696,34560,36960,36960,37632,40392,40560,40824,41064,41184,42240,42840,43200
%N Area A of the cyclic quadrilaterals PQRS with PQ>=QR>=RS>=SP, such that A, the sides, the radius of the circumcircle and the two diagonals are integers.
%C Subsequence of A210250.
%C In Euclidean geometry, a cyclic quadrilateral is a quadrilateral whose vertices all lie on a single circle. This circle is called the circumcircle or circumscribed, and the vertices are said to be concyclic.
%C The area A of a cyclic quadrilateral with sides a, b, c, d is given by Brahmagupta’s formula : A = sqrt((s - a)(s -b)(s - c)(s - d)) where s, the semiperimeter is s= (a+b+c+d)/2.
%C The circumradius R (the radius of the circumcircle) is given by:
%C R = sqrt(ab+cd)(ac+bd)(ad+bc)/4A
%C The diagonals of a cyclic quadrilateral have length:
%C p = sqrt((ab+cd)(ac+bd)/(ad+bc))
%C q = sqrt((ac+bd)(ad+bc)/(ab+cd)).
%D Mohammad K. Azarian, Circumradius and Inradius, Problem S125, Math Horizons, Vol. 15, Issue 4, April 2008, p. 32.
%H 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 E. Gürel, <a href="http://www.jstor.org/stable/2690677?seq=7">Solution to Problem 1472, Maximal Area of Quadrilaterals</a>, Math. Mag. 69 (1996), 149.
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/CyclicQuadrilateral.html">Cyclic Quadrilateral</a>
%e 936 is in the sequence because, with sides (a,b,c,d) = (14,30,40,48) we obtain:
%e s = (14+30+40+48)/2 = 66;
%e A = sqrt((66-14)(66-30)(66-40)(66-48))=936;
%e R = sqrt((14*30+40*48)(14*40+30*48)(14*48+30*40))/(4*936) = 93600/3744 =25;
%e p = sqrt((14*30+40*48)( 14*40+30*48)/( 14*48+30*40)) = 50;
%e q= sqrt((14*40+30*48)( 14*48+30*40)/( 14*30+40*48)) = 40.
%t SMax=10000;
%t Do[
%t Do[
%t x=S^2/(u v w);
%t If[u+v+w+x//OddQ, Continue[]];
%t If[v+w+x<=u, Continue[]];
%t r=Sqrt[v w+u x]Sqrt[u w+v x]Sqrt[u v+w x]/(4S);
%t If[r//IntegerQ//Not, Continue[]];
%t {a, b, c, d}=(u+v+w+x)/2-{u, v, w, x};
%t If[4S r/(a b+c d)//IntegerQ//Not,Continue[]];
%t If[4S r/(a d+b c)//IntegerQ//Not,Continue[]];
%t (*{a, b, c, d, r, S}//Sow*);
%t S//Sow; Break[]; (*to generate a table, comment out this line and uncomment previous line*)
%t , {u, S^2//Divisors//Select[#, S<=#^2&]&}
%t , {v, S^2/u//Divisors//Select[#, S^2<=u#^3&&#<=u&]&}
%t , {w, S^2/(u v)//Divisors//Select[#, S^2<=u v#^2&&#<=v&]&}
%t ]
%t , {S, 24, SMax, 24}
%t ]//Reap//Last//Last
%t {x, r, a, b, c, d}=.; (* _Albert Lau_, May 25 2016 *)
%Y Cf. A210250.
%K nonn
%O 1,1
%A _Michel Lagneau_, Nov 15 2012
%E Incorrect Mathematica program removed by _Albert Lau_, May 25 2016
%E Missing terms 18144, 20064, 21504 and more term from _Albert Lau_, May 25 2016
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