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A230136
Integer areas A of integer-sided cyclic quadrilaterals such that the circumradius is of prime length.
5
48, 240, 480, 1440, 1680, 2640, 5040, 6720, 7920, 10560, 12480, 13680, 18720, 21840, 28560, 31200, 32640, 34320, 36960, 44880, 48720, 53040, 63840, 71760, 77520, 87360, 92400, 100320, 115920, 147840, 187680, 201600, 215280, 236640, 244800, 257040, 277200
OFFSET
1,1
COMMENTS
Subset of A210250. The corresponding prime circumradius are 5, 13, 17, 41, 29, 61, 53, 101, 73, 89, 97, 109, 149, 313, 257, 173,...
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.
The circumradius R (the radius of the circumcircle) is given by R = sqrt(ab+cd)(ac+bd)(ad+bc)/4A.
The corresponding R of a(n) are not unique, for example for a(5) = 1680 => (a,b,c,d) = (24, 24, 70, 70) with R = 37 and (a,b,c,d) = (40, 40, 42, 42) with R = 29.
It seems that the quadrilaterals are of the form (a, a, b, b).
The following table gives the first values (A, R, a, b, c, d) where A is the integer area, R the radius of the circumcircle, and a, b, c, d are the integer sides of the cyclic quadrilateral.
************************************************
* A * R * a * b * c * d *
************************************************
* 48 * 5 * 6 * 6 * 8 * 8 *
* 240 * 13 * 10 * 10 * 24 * 24 *
* 480 * 17 * 16 * 16 * 30 * 30 *
* 1440 * 41 * 18 * 18 * 80 * 80 *
* 1680 * 29 * 24 * 24 * 42 * 42 *
* 2640 * 61 * 22 * 22 * 120 * 120 *
* 5040 * 53 * 56 * 56 * 90 * 90 *
* 7920 * 101 * 40 * 40 * 198 * 198 *
* 10560 * 73 * 96 * 96 * 110 * 110 *
* 12480 * 89 * 78 * 78 * 160 * 160 *
* 18720 * 97 * 130 * 130 * 144 * 144 *
...........................................
LINKS
Mohammad K. Azarian, Solution to Problem S125: Circumradius and Inradius, Math Horizons, Vol. 16, Issue 2, November 2008, p. 32.
E. Gürel, Solution to Problem 1472, Maximal Area of Quadrilaterals, Math. Mag. 69 (1996), 149.
Eric Weisstein's World of Mathematics, Cyclic Quadrilateral
EXAMPLE
48 is in the sequence because, for (a,b,c,d) = (6,6,8,8) and :
s = (6+6+8+8)/2 = 14;
A = sqrt((14-6)(14-6)(14-8)(14-8))=48;
R = sqrt((6*6+8*8)(6*8+6*8)(6*8+6*8))/(4*48) = 960/192 = 5 is prime.
MATHEMATICA
nn=1000; lst={}; Do[s=(a+b+c+d)/2; If[IntegerQ[s], area2=(s-a)*(s-b)*(s-c)*(s-d); If[0 < area2 && IntegerQ[Sqrt[area2]] && PrimeQ[Sqrt[(a*b+c*d)*(a*c+b*d)*(a*d+b*c)/((s-a)*(s-b)*(s-c)*(s-d))]/4], AppendTo[lst, Sqrt[area2]]]], {a, nn}, {b, a}, {c, b}, {d, c}]; Union[lst]
CROSSREFS
Cf. A210250.
Sequence in context: A211750 A211760 A333670 * A157913 A181773 A052683
KEYWORD
nonn
AUTHOR
Michel Lagneau, Oct 10 2013
STATUS
approved