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A230479 Integer areas of the integer-sided triangles such that the length of the circumradius is a square.

%I A230479 #12 Feb 16 2025 08:33:20
%S A230479 168,336,432,600,768,2688,5376,6000,6912,9600,12288,13608,14280,20280,
%T A230479 27216,28560,30720,32928,34560,34992,38640,43008,46200,48600,62208,
%U A230479 69360,77280,86016,96000,105000,108000,110592,118272,153600,196608
%N A230479 Integer areas of the integer-sided triangles such that the length of the circumradius is a square.
%C A230479 The primitive areas are 168, 338, 432, 600, 768, 13608, 14280, 20280, 27216, ...
%C A230479 The non-primitive areas 16*a(n) are in the sequence because if R is the circumradius corresponding to a(n), then 4*R is the circumradius corresponding to 16*a(n).
%C A230479 Each circumradius belongs to the sequence {25, 100, 169, 225, 289, 400, 625, 676, ...}, and it seems that this last sequence is A198385 (second of a triple of squares in arithmetic progression).
%C A230479 The following table gives the first values (A, R, a, b, c) where A is the integer area, R the radius of the circumcircle, and a, b, c are the integer sides of the triangle.
%C A230479 **************************************
%C A230479 *     A  *    R  *   a  *   b  *  c  *
%C A230479 **************************************
%C A230479 *   168  *   25  *  14  *  30 *  40  *
%C A230479 *   336  *   25  *  14  *  48 *  50  *
%C A230479 *   432  *   25  *  30  *  30 *  48  *
%C A230479 *   600  *   25  *  30  *  40 *  50  *
%C A230479 *   768  *   25  *  40  *  40 *  48  *
%C A230479 *  2688  *  100  *  56  * 120 * 160  *
%C A230479 *  5376  *  100  *  56  * 192 * 200  *
%C A230479 *  6912  *  100  * 120  * 120 * 192  *
%C A230479 *  9600  *  100  * 120  * 160 * 200  *
%C A230479 * 12288  *  100  * 160  * 160 * 192  *
%C A230479 * 13608  *  225  * 126  * 270 * 360  *
%C A230479 * 14280  *  169  * 130  * 238 * 312  *
%C A230479 * 20280  *  169  * 130  * 312 * 338  *
%C A230479 * 27216  *  225  * 126  * 432 * 450  *
%C A230479 .............................
%D A230479 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 A230479 Eric W. Weisstein, <a href="https://mathworld.wolfram.com/Circumradius.html">MathWorld: Circumradius</a>
%F A230479 Area A = sqrt(s*(s-a)*(s-b)*(s-c)) with s = (a+b+c)/2 (Heron's formula);
%F A230479 Circumradius R = a*b*c/4A.
%e A230479 168 is in the sequence because the area of the triangle (14, 30, 40) is given by Heron's formula A = sqrt(42*(42-14)*(42-30)*(42-40))= 168 where the number 42 is the semiperimeter, and the circumcircle is given by R = a*b*c/(4*A) = 14*30*40/(4*168) = 25, which is a square.
%t A230479 nn = 1000; 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[a*b*c/(4*Sqrt[area2])]], AppendTo[lst, Sqrt[area2]]]], {a, nn}, {b, a}, {c, b}]; Union[lst]
%Y A230479 Cf. A188158, A208984, A210207.
%K A230479 nonn
%O A230479 1,1
%A A230479 _Michel Lagneau_, Oct 20 2013