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A210250 Area A of the cyclic quadrilaterals such that A, the sides and the radius of the circumcircle are integers. 12
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 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,1

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, ...}.

REFERENCES

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

LINKS

Table of n, a(n) for n=1..49.

Mohammad K. Azarian, Solution to Problem S125: Circumradius and Inradius, Math Horizons, Vol. 16, Issue 2, November 2008, p. 32.

E. G├╝rel, 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),

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.

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 *)

CROSSREFS

Cf. A103251, A208984.

Sequence in context: A233967 A233785 A233960 * A259038 A231174 A259245

Adjacent sequences:  A210247 A210248 A210249 * A210251 A210252 A210253

KEYWORD

nonn

AUTHOR

Michel Lagneau, Mar 19 2012

EXTENSIONS

Incorrect Mathematica program removed by Albert Lau, May 25 2016

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

STATUS

approved

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Last modified September 27 03:21 EDT 2020. Contains 337380 sequences. (Running on oeis4.)