A210250 Area A of the cyclic quadrilaterals such that A, the sides and the radius of the circumcircle are integers.

48, 192, 240, 432, 480, 672, 768, 936, 960, 1200, 1440, 1680, 1728, 1920, 2160, 2352, 2640, 2688, 2856, 3072, 3744, 3840, 3864, 3888, 4032, 4320, 4368, 4536, 4800, 5016, 5040, 5376, 5712, 5760, 5808, 5880, 6000, 6048, 6072, 6696, 6720, 6912, 7056, 7392, 7560, 7680, 7728, 7752, 7920

Offset: 1

Author

Michel Lagneau, Mar 19 2012

Keywords

• nonn

Comments

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 circle, 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(12) = 1680 => (a,b,c,d) = (24, 24, 70, 70) with R = 37 and (a,b,c,d) = (40, 40, 42,42) with R = 29.

The smallest corresponding R of a(n) is {5, 10, 13, 15, 17, 25, 20, 25, 26, 25, 41, 29, ...}.

Properties of this sequence:

A majority of quadrilaterals [a, b, c, d] have the property that a = b and c = d, and in this case s = a+c, A = a*c and R = sqrt(a^2+c^2)/2. Because a and c are even => a = 2p and c = 2q, then A = 4pq and R = sqrt(p^2+q^2). Consequently, 2*A103251(n) is included in this sequence.

Nevertheless, there also exist quadrilaterals whose four sides are distinct, for example [a, b, c, d] = [14, 30, 40, 48] => A = 936 = a(8) and R = 25. The subset of a(n) with this property is {936, 2856, 3744, 3864, 4536, 5016, 5376, 5712, 5880, 6696, 7056, 7560, ...}.

Examples

			48 is in the sequence because, for (a,b,c,d) = (6,6,8,8),
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.
		

References

• Mohammad K. Azarian, Circumradius and Inradius, Problem S125, Math Horizons, Vol. 15, Issue 4, April 2008, p. 32.

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

Crossrefs

Cf. A103251, A208984.

Programs

Mathematica

SMax=8000;
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//IntegerQ//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}=.;
(* Albert Lau, May 25 2016 *)

Extensions

Incorrect Mathematica program removed by Albert Lau, May 25 2016

Missing term 5880 and more terms from Albert Lau, May 25 2016