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A230491 Integer areas of the integer-sided triangles such that the length of the inradius is a square.

%I A230491 #12 Feb 16 2025 08:33:20
%S A230491 6,84,96,108,120,132,144,156,168,180,240,264,300,324,396,420,432,468,
%T A230491 486,504,540,594,630,684,720,756,864,990,1026,1080,1116,1134,1152,
%U A230491 1224,1332,1344,1404,1440,1494,1536,1584,1638,1680,1710,1728,1782,1824,1872,1890
%N A230491 Integer areas of the integer-sided triangles such that the length of the inradius is a square.
%C A230491 The primitive areas are 6, 84, 108, 120, 132, 144, 156, 168, ...
%C A230491 The non-primitive areas 16*a(n) are in the sequence because if r is the inradius corresponding to a(n), then 4*r is the inradius corresponding to 16*a(n).
%C A230491 The following table gives the first values (A, r, a, b, c) where A is the integer area, r the inradius and a, b, c are the integer sides of the triangle.
%C A230491 ******************************
%C A230491 *   A *  r  *  a *  b *   c  *
%C A230491 *******************************
%C A230491 *   6 *  1  *  3 *  4 *   5  *
%C A230491 *  84 *  4  * 13 * 14 *  15  *
%C A230491 *  96 *  4  * 12 * 16 *  20  *
%C A230491 * 108 *  4  * 15 * 15 *  24  *
%C A230491 * 120 *  4  * 10 * 24 *  26  *
%C A230491 * 132 *  4  * 11 * 25 *  30  *
%C A230491 * 144 *  4  * 18 * 20 *  34  *
%C A230491 * 156 *  4  * 15 * 26 *  37  *
%C A230491 * 168 *  4  * 10 * 35 *  39  *
%C A230491 * 180 *  4  *  9 * 40 *  41  *
%C A230491 * 240 *  4  * 12 * 50 *  58  *
%C A230491 * 264 *  4  * 33 * 34 *  65  *
%C A230491 * 300 *  4  * 25 * 51 *  74  *
%C A230491 * 324 *  4  *  9 * 75 *  78  *
%C A230491 * 396 *  4  * 11 * 90 *  97  *
%C A230491 * 420 *  4  * 21 * 85 * 104  *
%C A230491 * 432 *  9  * 30 * 30 *  36  *
%C A230491 * 468 *  9  * 25 * 39 *  40  *
%C A230491 .........................
%D A230491 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.
%H A230491 Eric W. Weisstein, <a href="https://mathworld.wolfram.com/Inradius.html">MathWorld: Inradius</a>
%F A230491 Area A = sqrt(s*(s-a)*(s-b)*(s-c)) with s = (a+b+c)/2 (Heron's formula) and inradius r = A/s.
%e A230491 84 is in the sequence because the area of triangle (13, 14, 15) is given by Heron's formula A = sqrt(21*(21-13)*(21-14)*(21-15))= 84 where the number 21 is the semiperimeter and the inradius is given by r = A/s = 84/21 = 4 is a square.
%t A230491 nn = 600; lst = {}; Do[s = (a + b + c)/2; If[IntegerQ[s], area2 = s (s - a) (s - b) (s - c); If[0 < area2 && IntegerQ[Sqrt[area2]] && IntegerQ[Sqrt[Sqrt[area2]/s]], AppendTo[lst, Sqrt[area2]]]], {a, nn}, {b, a}, {c, b}]; Union[lst]
%Y A230491 Cf. A000290, A188158, A228383.
%K A230491 nonn
%O A230491 1,1
%A A230491 _Michel Lagneau_, Oct 20 2013