

A218431


Cyclic quadrilateral numbers: numbers n = a*b*c*d such that the integers a,b,c,d are the sides of a cyclic quadrilateral with the area and the circumradius integers.


4



2304, 36864, 57600, 186624, 230400, 451584, 589824, 630000, 806400, 921600, 1440000, 2073600, 2822400, 2985984, 3686400, 4665600, 5531904, 6969600, 7225344, 8960000, 9437184, 10080000, 12672000, 12902400, 14745600, 15116544, 16257024, 18662400, 19079424
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OFFSET

1,1


COMMENTS

A cyclic quadrilateral is a quadrilateral for which a circle can be circumscribed so that it touches each polygon vertex.
A cyclic quadrilateral number n is an integer with at least one decomposition n = a*b*c*d such that the area of the quadrilateral of sides (a,b,c,d) and the circumradius are an integer. Because this property is not always unique, we introduce the notion of "cyclic quadrilateral order" for each cyclic quadrilateral number n, denoted by CQO(n). For example, CQO(2304) = 1 because the decomposition 2304 = 8*8*6*6 is unique with the quadrilateral (8,8,6,6) whose area A is given by Brahmagupta’s formula: A = sqrt((s  a)(s  b)(s  c)(s  d)) where the semiperimeter is s = (a+b+c+d)/2, and where the circumradius R (the radius of the circumcircle) is given by R = sqrt((ab+cd)(ac+bd)(ad+bc))/(4A) => A = sqrt((148)(148)(146)(146)) = 48, and R = 5, but CQO(2822400) = 2 because 2822400 = 24*24*70*70 = 40*40*42*42 and the area of the quadrilateral (24,24,70,70) equals 1680 with R = 37 and the area of the quadrilateral (40,40,42,42) equals also 1680 with R = 29.
The number of ways to write n = a*b*c*d with 1 <= a <= b <= c <= d <= n is given by A218320, thus: CQO(n) <= A218320(n).
If n is in this sequence, so is n*k^4 for any k > 0. Thus this sequence is infinite.
In view of the preceding comment, one might call "primitive" the elements of the sequence for which there is no k > 1 such that n/k^4 is again a term of the sequence. These elements are 2304, 57600, 230400, 451584, 630000, ...


REFERENCES

Mohammad K. Azarian, Circumradius and Inradius, Problem S125, Math Horizons, Vol. 15, Issue 4, April 2008, p. 32. Solution published in Vol. 16, Issue 2, November 2008, p. 32.


LINKS

Table of n, a(n) for n=1..29.
E. Gürel, Solution to Problem 1472, Maximal Area of Quadrilaterals, Math. Mag. 69 (1996), 149.
Eric W. Weisstein, MathWorld: Cyclic Quadrilateral


EXAMPLE

2304 is in the sequence because 2304 = 8*8*6*6 and we obtain:
s = (8+8+6+6)/2 = 14;
A = sqrt((148) (148)(146)(146))=48;
R = sqrt((8*8+6*6)(8*6+8*6)(8*6+8*6))/(4*48) = 5.


MATHEMATICA

nn=200; lst={}; Do[s=(a+b+c+d)/2; If[IntegerQ[s], area2=(sa)*(sb)*(sc)*(sd); If[0<area2&&IntegerQ[Sqrt[area2]]&&IntegerQ[Sqrt[(a*b+c*d)*(a*c+b*d)*(a*d+b*c)/((sa)*(sb)*(sc)*(sd))]/4], AppendTo[lst, a*b*c*d]]], {a, nn}, {b, a}, {c, b}, {d, c}]; Union[lst]


CROSSREFS

Cf. A210250.
Sequence in context: A256729 A204102 A177759 * A187292 A235423 A235419
Adjacent sequences: A218428 A218429 A218430 * A218432 A218433 A218434


KEYWORD

nonn,nice


AUTHOR

Michel Lagneau, Oct 28 2012


EXTENSIONS

Typos in comment fixed by Zak Seidov and M. F. Hasler, Sep 21 2013, Sep 21 2013


STATUS

approved



