A230136 Integer areas A of integer-sided cyclic quadrilaterals such that the circumradius is of prime length.

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, 85680, 87360, 92400, 100320, 110400, 115920, 118560, 140400, 147840, 182160, 187680, 201600, 215280, 235200, 236640, 244800, 255360, 257040, 265200, 277200

Offset: 1

Author

Michel Lagneau, Oct 10 2013

Keywords

• nonn

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 *

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

Examples

			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.
		

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.
• Kival Ngaokrajang, Illustration of initial terms
• Eric Weisstein's World of Mathematics, Cyclic Quadrilateral

Crossrefs

Cf. A210250.

Programs

Mathematica

SMax = 277300
Do[
  Do[
    x=S^2/(u v w);
    If[u+v+w+x//OddQ, Continue[]];
    If[v+w+x<=u, Continue[]];
    r=Sqrt[v w+u x]Sqrt[u w+v x]Sqrt[u v+w x]/(4S);
    If[r//PrimeQ//Not, Continue[]];
    (*{a, b, c, d}=(u+v+w+x)/2-{u, v, w, x}; {a, b, c, d, r, S}//Sow*);
    S//Sow; Break[]; (*to generate a table, comment out this line and uncomment previous line*)
    , {u, S^2//Divisors//Select[#, S<=#^2&]&}
    , {v, S^2/u//Divisors//Select[#, S^2<=u#^3&&#<=u&]&}
    , {w, S^2/(u v)//Divisors//Select[#, S^2<=u v#^2&&#<=v&]&}
  ]
  , {S, 24, SMax, 24}
]//Reap//Last//Last
{x, r, a, b, c, d}=.;
(* Zachary Sizer, Jan 02 2025, adapted from the program for A210250 by Albert Lau *)

Extensions

Incorrect program removed and missing term 85680 and others added by Zachary Sizer, Jan 02 2025