%I #36 Feb 16 2025 08:33:20
%S 42,168,378,672,1050,1512,2058,2088,2688,3000,3402,4200,5082,6048,
%T 6960,7098,8232,8352,9450,10752,12000,12138,13608,15162,16800,18522,
%U 18792,20328,22218,24192,26250,27000,27840,28392,30618,31416,32928,33408,35322,36000,37800,40362
%N Integer areas A of the integer-sided triangles such that the product of the inradius and the circumradius is a square.
%C The areas of the primitive triangles of sides (a, b, c) and inradius, circumradius equals respectively to r and R are 42, 3000,... The sides of the nonprimitive triangles are of the form (a*k, b*k, c*k) with r’ = r*k and R’=R*k where r’, R’ are respectively the inradius and the circumradius of the nonprimitive triangles. The areas A’ of the nonprimitive triangles are A’ = A*k^2. The set {A016850} (numbers (5n)^2) is included in the set of the products r*R (see the table below).
%C The area A of a triangle whose sides have lengths a, b, and c is given by Heron's formula: A = sqrt(s*(s-a)*(s-b)*(s-c)), where s = (a+b+c)/2. The inradius r is given by r = A/s and the circumradius is given by R = abc/4A.
%C The product r*R is given by r*R = abc/2(a+b+c).
%C The following table gives the first values (A, a, b, c, r, R, r*R).
%C -----------------------------------------------------
%C | A | a | b | c | r | R | r*R |
%C -----------------------------------------------------
%C | 42 | 7 | 15 | 20 | 2 | 25/2 | 5^2 |
%C | 168 | 14 | 30 | 40 | 4 | 25 | 10^2 |
%C | 378 | 21 | 45 | 60 | 6 | 75/2 | 15^2 |
%C | 672 | 28 | 60 | 80 | 8 | 50 | 20^2 |
%C | 1050 | 35 | 75 | 100 | 10 | 125/2 | 25^2 |
%C | 1512 | 42 | 90 | 120 | 12 | 75 | 30^2 |
%C | 2058 | 49 | 105 | 140 | 14 | 175/2 | 35^2 |
%C | 2688 | 56 | 120 | 160 | 16 | 100 | 40^2 |
%C | 3000 | 80 | 85 | 85 | 24 | 289/6 | 34^2 |
%C | 3402 | 63 | 135 | 180 | 18 | 225/2 | 45^2 |
%C | 4200 | 70 | 150 | 200 | 20 | 125 | 50^2 |
%C | 5082 | 77 | 165 | 220 | 22 | 275/2 | 55^2 |
%C | 6048 | 84 | 180 | 240 | 24 | 150 | 60^2 |
%C | 6960 | 58 | 300 | 338 | 20 | 845/4 | 65^2 |
%C | 7098 | 91 | 195 | 260 | 26 | 325/2 | 65^2 |
%C ....................................................
%D Mohammad K. Azarian, Circumradius and Inradius, Problem S125, Math Horizons, Vol. 15, Issue 4, April 2008, p. 32.
%H Zak Seidov, <a href="/A232329/b232329.txt">Table of n, a(n) for n = 1..100</a>
%H Zak Seidov, <a href="/A232329/a232329.txt">Table of 814 values of area A, sides a>=b>=c, semiprime s, and sqrt(Rr), in the order of increasing "a" from 20 up to 10000.</a>
%H Mohammad K. Azarian, <a href="http://www.jstor.org/stable/25678790">Solution to Problem S125: Circumradius and Inradius</a>, Math Horizons, Vol. 16, Issue 2, November 2008, p. 32.
%H Eric W. Weisstein, <a href="https://mathworld.wolfram.com/Circumradius.html">MathWorld: Circumradius</a>
%H Eric W. Weisstein, <a href="https://mathworld.wolfram.com/Inradius.html">MathWorld: Inradius</a>
%e a(1) = 42 because, for (a,b,c) = (7, 15, 20):
%e the semiperimeter s = (7+15+20)/2 =21, and
%e A = sqrt(21*(21-7)*(21-15)*(21-20)) = 42
%e R = abc/4A = 7*15*20/(4*42) = 25/2
%e r = A/s = 42/21 = 2, hence r*R = 25 is a square.
%t nn=800;lst={};Do[s=(a+b+c)/2;rr=a*b*c/(2*(a+b+c))
%t ;If[IntegerQ[s],area2=s(s-a)(s-b)(s-c);If[0<area2&&IntegerQ[Sqrt[area2]] &&IntegerQ[Sqrt[rr]],AppendTo[lst,Sqrt[area2]]]],{a,nn},{b,a},{c,b}];Union[lst]
%o (PARI) lista(nn)=lst=[]; for (a = 1, nn, for (b=1, a, for (c=1, b, s=(a+b+c)/2; rr=a*b*c/(2*(a+b+c)); if ((type(s) == "t_INT") && (type(rr) == "t_INT"), area2=s*(s-a)*(s-b)*(s-c); if ((0<area2) && issquare(area2) && issquare(rr), lst = concat(lst, sqrtint(area2));););););); Set(lst); \\ after Mathematica; _Michel Marcus_, Jun 09 2015
%o (PARI) {for(a=20,10000,forstep(b=a,2,-1,forstep(c=min(b,a+b-1),a-b+1,-1,if((a+b+c)%2<1,s=(a+b+c)/2;if(issquare(s*(s-a)*(s-b)*(s-c),&A),
%o if((a*b*c)%(2*(a+b+c))<1&&if(issquare(a*b*c/(2*(a+b+c)),&d),
%o print([A,a,b,c,s,d]))))))))} \\ Faster version uesd for afile. Zak Seidov, Jun 06 2015
%Y Cf. A016850, A188158, A208984, A232274.
%K nonn,changed
%O 1,1
%A _Michel Lagneau_, Nov 22 2013
%E Missing term 33408 added by _Zak Seidov_, Jun 08 2015