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

192, 234, 300, 432, 714, 768, 936, 1134, 1200, 1254, 1344, 1674, 1728, 1764, 1890, 1938, 2046, 2106, 2226, 2310, 2352, 2700, 2856, 2886, 3072, 3120, 3234, 3744, 3888, 3990, 4092, 4212, 4368, 4536, 4674, 4800, 4914, 5016, 5292, 5376, 5760, 5850, 6006, 6270, 6426

Offset: 1

Author

Michel Lagneau, Jun 02 2016

Keywords

• nonn

Comments

The areas of the primitive cyclic quadrilaterals of this sequence are in A273691.

This sequence contains A233315 (768, 936, 1200,...).

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.

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.

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

p = sqrt((ac+bd)(ad+bc)/(ab+cd)) and q = sqrt((ac+bd)(ab+cd)/(ad+bc)).

The circumradius R (the radius of the circumcircle) is given by :

R = sqrt((ab+cd)(ac+bd)(ad+bc))/4A.

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).

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 .

+--------+-------+-------+-------+--------+-------+------+-------+

| A | a | b | c | d | p | q | R |

+--------+-------+-------+-------+--------+-------+------+-------+

| 192 | 7 | 15 | 15 | 25 | 20 | 24 | 25/2 |

| 234 | 7 | 15 | 20 | 24 | 20 | 25 | 25/2 |

| 300 | 15 | 15 | 20 | 20 | 24 | 25 | 25/2 |

| 432 | 11 | 25 | 25 | 25 | 30 | 30 | 125/8 |

| 714 | 16 | 25 | 33 | 60 | 39 | 52 | 65/2 |

| 768 | 25 | 25 | 25 | 39 | 40 | 40 | 125/6 |

| 768 | 14 | 30 | 30 | 50 | 40 | 48 | 25 |

| 936 | 14 | 30 | 40 | 48 | 40 | 50 | 25 |

| 1134 | 16 | 25 | 52 | 65 | 39 | 63 | 65/2 |

| 1200 | 30 | 30 | 40 | 40 | 48 | 50 | 25 |

| 1254 | 16 | 25 | 60 | 63 | 39 | 65 | 65/2 |

| 1344 | 25 | 33 | 39 | 65 | 52 | 60 | 65/2 |

..................................................................

Examples

			192 is in the sequence because, for (a,b,c,d) = (7,15,15,25) we find:
s = (7+15+15+25)/2 = 31;
A = sqrt((31-7)(31-15)(31-15)(31-25)) = 192;
p = sqrt((7*15+15*25)*(7*25+15*15)/(7*15+15*25)) = 20;
q = sqrt((7*15+15*25)*(7*15+15*25)/(7*25+15*15)) = 24.
		

Links

• Eric Weisstein's World of Mathematics, Cyclic Quadrilateral

Crossrefs

Cf. A210250, A218431, A219225, A230136, A233315, A242778, A273691.

Programs

Mathematica

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