OFFSET
1,1
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 |
..................................................................
LINKS
Eric Weisstein's World of Mathematics, Cyclic Quadrilateral
EXAMPLE
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.
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<area2 && IntegerQ[Sqrt[area2]] && IntegerQ[d1]&& IntegerQ[d2], AppendTo[lst, Sqrt[area2]]]], {a, nn}, {b, a}, {c, b}, {d, c}]; Union[lst]
CROSSREFS
KEYWORD
nonn
AUTHOR
Michel Lagneau, Jun 02 2016
STATUS
approved