cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

A219192 Area A of the bicentric quadrilaterals such that A, the sides, the radius of the circumcircle and the radius of the incircle are integers.

Original entry on oeis.org

2352, 9408, 21168, 37632, 58800, 69360, 84672, 115248, 150528, 190512, 235200, 253920, 277440, 284592, 338688, 397488, 460992, 529200, 602112, 624240, 645792, 679728, 762048, 849072, 940800, 1015680, 1037232, 1109760, 1138368, 1244208, 1354752, 1470000, 1589952
Offset: 1

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Author

Michel Lagneau, Nov 14 2012

Keywords

Comments

Also numbers n such that there exists a decomposition n^2 = a*b*c*d where a,b,c,d are the sides of a bicentric quadrilateral with the area, the inradius and the circumradius integers.
In Euclidean geometry, a bicentric quadrilateral is a convex quadrilateral that has both an incircle and a circumcircle. If the sides are a, b, c, d, then the area is given by A = sqrt(a*b*c*d). The inradius r of a bicentric quadrilateral is determined by the sides a, b, c, d according to r = sqrt(a*b*c*d)/(a+c) = sqrt(a*b*c*d)/(b+d). The circumradius R (the radius of the circumcircle) is given by R = sqrt((ab+cd)(ac+bd)(ad+bc))/4A.
If n is in this sequence, so is n*k^2 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^2 is again a term of the sequence. These elements are 2352, 69360, 253920, 645792,... are listed in A219193.

Examples

			2352 is in the sequence because, with sides (a,b,c,d) = (56,56,42,42) we obtain :
s = (56+56+42+42)/2 = 98;
A = sqrt(56*56*42*42) = 2352 = sqrt((98-56)(98-56)(98-42)(98-42)) (Brahmagupta’s Formula);
r = 2352/(56+42) =24.
R = sqrt((56*56+42*42)(56*42+56*42)(56*42+56*42))/(4*2352) = 35.
		

Crossrefs

Programs

  • Maple
    with(numtheory):nn:=15000:for a from 1 to nn do: b:=a: for c from b to nn do: for d from c to c while(sqrt(a*b*c*d)=floor(sqrt(a*b*c*d))) do:s:=(a+b+c+d)/2:a1:=(s-a)*(s-b)*(s-c)*(s-d):a2:=sqrt(a*b*c*d):r1:=a2/(a+c):r2:=a2/(b+d):rr:= sqrt((a*b+c*d) * (a*c+b*d) * (a*d+b*c))/(4*a2):if a1>0 and floor(sqrt(a1))=sqrt(a1) and a2 =floor(a2) and a2=sqrt(a1) and r1=floor(r1) and r2=floor(r2) and r1=r2 and rr =floor(rr) then printf ( "%d %d %d %d %d %d %d\n",a2,a,b,c,d,r1,rr):else fi:od:od:od:
  • Mathematica
    nn=15000;lst={};Do[s=(2*a+2*d)/2;If[IntegerQ[s],area2=(s-a)*(s-a)*(s-d)*(s-d);area22=a*a*d*d;If[0