A218431 Cyclic quadrilateral numbers: numbers m = a*b*c*d such that the integers a,b,c,d are the sides of a cyclic quadrilateral whose area and circumradius are integers.

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

Offset: 1

Author

Michel Lagneau, Oct 28 2012

Keywords

• nonn
• nice

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 m is an integer with at least one decomposition m = a*b*c*d such that the area of the quadrilateral of sides (a,b,c,d) and the circumradius are integers. Because this property is not always unique, we introduce the notion of "cyclic quadrilateral order" for each cyclic quadrilateral number m, denoted by CQO(m). 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 the circumradius R (the radius of the circumcircle) is given by R = sqrt((ab+cd)*(ac+bd)*(ad+bc))/(4A) => A = sqrt((14-8)*(14-8)*(14-6)*(14-6)) = 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) also equals 1680 with R = 29.

The number of ways to write m = a*b*c*d with 1 <= a <= b <= c <= d <= m is given by A218320, thus: CQO(m) <= A218320(m).

If m is in this sequence, so is m*k^4 for any k > 0. Thus this sequence is infinite.

In view of the preceding comment, one might call "primitive" the terms m of the sequence for which there is no k > 1 such that m/k^4 is again a term of the sequence. These terms are 2304, 57600, 230400, 451584, 630000, ...

Examples

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

Links

• E. Gürel, Solution to Problem 1472, Maximal Area of Quadrilaterals, Math. Mag. 69 (1996), 149.
• Eric W. Weisstein, MathWorld: Cyclic Quadrilateral.

Crossrefs

Cf. A210250.

Programs

Mathematica

nn=200; lst={}; Do[s=(a+b+c+d)/2; If[IntegerQ[s], area2=(s-a)*(s-b)*(s-c)*(s-d); If[0
				

Extensions

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