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%I #29 Mar 03 2018 02:29:20
%S 48,192,240,432,480,672,768,936,960,1200,1440,1680,1728,1920,2160,
%T 2352,2640,2688,2856,3072,3744,3840,3864,3888,4032,4320,4368,4536,
%U 4800,5016,5040,5376,5712,5760,5808,5880,6000,6048,6072,6696,6720,6912,7056,7392,7560,7680,7728,7752,7920
%N Area A of the cyclic quadrilaterals such that A, the sides and the radius of the circumcircle are integers.
%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 circle, 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 corresponding R of a(n) are not unique; for example, for a(12) = 1680 => (a,b,c,d) = (24, 24, 70, 70) with R = 37 and (a,b,c,d) = (40, 40, 42,42) with R = 29.
%C The smallest corresponding R of a(n) is {5, 10, 13, 15, 17, 25, 20, 25, 26, 25, 41, 29, ...}.
%C Properties of this sequence:
%C A majority of quadrilaterals [a, b, c, d] have the property that a = b and c = d, and in this case s = a+c, A = a*c and R = sqrt(a^2+c^2)/2. Because a and c are even => a = 2p and c = 2q, then A = 4pq and R = sqrt(p^2+q^2). Consequently, 2*A103251(n) is included in this sequence.
%C Nevertheless, there also exist quadrilaterals whose four sides are distinct, for example [a, b, c, d] = [14, 30, 40, 48] => A = 936 = a(8) and R = 25. The subset of a(n) with this property is {936, 2856, 3744, 3864, 4536, 5016, 5376, 5712, 5880, 6696, 7056, 7560, ...}.
%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 48 is in the sequence because, for (a,b,c,d) = (6,6,8,8),
%e s = (6+6+8+8)/2 = 14;
%e A = sqrt((14-6)(14-6)(14-8)(14-8)) = 48;
%e R = sqrt((6*6+8*8)(6*8+6*8)(6*8+6*8))/(4*48) = 960/192 = 5.
%t SMax=8000;
%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};{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}=.;
%t (* _Albert Lau_, May 25 2016 *)
%Y Cf. A103251, A208984.
%K nonn
%O 1,1
%A _Michel Lagneau_, Mar 19 2012
%E Incorrect Mathematica program removed by _Albert Lau_, May 25 2016
%E Missing term 5880 and more terms from _Albert Lau_, May 25 2016
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