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A273890 Integer area A of the cyclic quadrilaterals such that A, the sides and the two diagonals are integers.

%I A273890 #7 Feb 16 2025 08:33:36
%S A273890 192,234,300,432,714,768,936,1134,1200,1254,1344,1674,1728,1764,1890,
%T A273890 1938,2046,2106,2226,2310,2352,2700,2856,2886,3072,3120,3234,3744,
%U A273890 3888,3990,4092,4212,4368,4536,4674,4800,4914,5016,5292,5376,5760,5850,6006,6270,6426
%N A273890 Integer area A of the cyclic quadrilaterals such that A, the sides and the two diagonals are integers.
%C A273890 The areas of the primitive cyclic quadrilaterals of this sequence are in A273691.
%C A273890 This sequence contains A233315 (768, 936, 1200,...).
%C A273890 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 A273890 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 A273890 In a cyclic quadrilateral with successive vertices A, B, C, D and sides a = AB, b = BC, c = CD, and d = DA, the lengths of the diagonals p = AC and q = BD can be expressed in terms of the sides as
%C A273890 p = sqrt((ac+bd)(ad+bc)/(ab+cd)) and q = sqrt((ac+bd)(ab+cd)/(ad+bc)).
%C A273890 The circumradius R (the radius of the circumcircle) is given by :
%C A273890 R = sqrt((ab+cd)(ac+bd)(ad+bc))/4A.
%C A273890 The corresponding sides of a(n) are not unique, for example for a(6) = 768 => (a,b,c,d) = (25, 25, 25, 39) or (a,b,c,d) = (14, 30, 30, 50).
%C A273890 The following table gives the first values (A, a, b, c, d, p, q, R) where A is the integer area, a, b, c, d are the integer sides of the cyclic quadrilateral, p, q are the integer diagonals, and R .
%C A273890 +--------+-------+-------+-------+--------+-------+------+-------+
%C A273890 |    A   |   a   |   b   |   c   |   d    |   p   |  q   |  R    |
%C A273890 +--------+-------+-------+-------+--------+-------+------+-------+
%C A273890 |   192  |   7   |   15  |   15  |   25   |   20  |  24  | 25/2  |
%C A273890 |   234  |   7   |   15  |   20  |   24   |   20  |  25  | 25/2  |
%C A273890 |   300  |  15   |   15  |   20  |   20   |   24  |  25  | 25/2  |
%C A273890 |   432  |  11   |   25  |   25  |   25   |   30  |  30  | 125/8 |
%C A273890 |   714  |  16   |   25  |   33  |   60   |   39  |  52  | 65/2  |
%C A273890 |   768  |  25   |   25  |   25  |   39   |   40  |  40  | 125/6 |
%C A273890 |   768  |  14   |   30  |   30  |   50   |   40  |  48  | 25    |
%C A273890 |   936  |  14   |   30  |   40  |   48   |   40  |  50  | 25    |
%C A273890 |  1134  |  16   |   25  |   52  |   65   |   39  |  63  | 65/2  |
%C A273890 |  1200  |  30   |   30  |   40  |   40   |   48  |  50  | 25    |
%C A273890 |  1254  |  16   |   25  |   60  |   63   |   39  |  65  | 65/2  |
%C A273890 |  1344  |  25   |   33  |   39  |   65   |   52  |  60  | 65/2  |
%C A273890 ..................................................................
%H A273890 Eric Weisstein's World of Mathematics, <a href="https://mathworld.wolfram.com/CyclicQuadrilateral.html">Cyclic Quadrilateral</a>
%e A273890 192 is in the sequence because, for (a,b,c,d) = (7,15,15,25) we find:
%e A273890 s = (7+15+15+25)/2 = 31;
%e A273890 A = sqrt((31-7)(31-15)(31-15)(31-25)) = 192;
%e A273890 p = sqrt((7*15+15*25)*(7*25+15*15)/(7*15+15*25)) = 20;
%e A273890 q = sqrt((7*15+15*25)*(7*15+15*25)/(7*25+15*15)) = 24.
%t A273890 nn=200; lst={}; Do[s=(a+b+c+d)/2; If[IntegerQ[s], area2=(s-a)*(s-b)*(s-c)*(s-d); d1=Sqrt[(a*c+b*d)*(a*d+b*c)/(a*b+c*d)];d2=Sqrt[(a*c+b*d)*(a*b+c*d)/(a*d+b*c)];If[0<area2 && IntegerQ[Sqrt[area2]] && IntegerQ[d1]&& IntegerQ[d2], AppendTo[lst, Sqrt[area2]]]], {a, nn}, {b, a}, {c, b}, {d, c}]; Union[lst]
%Y A273890 Cf. A210250, A218431, A219225, A230136, A233315, A242778, A273691.
%K A273890 nonn
%O A273890 1,1
%A A273890 _Michel Lagneau_, Jun 02 2016