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 A230479 Integer areas of the integer-sided triangles such that the length of the circumradius is a square. 0
 168, 336, 432, 600, 768, 2688, 5376, 6000, 6912, 9600, 12288, 13608, 14280, 20280, 27216, 28560, 30720, 32928, 34560, 34992, 38640, 43008, 46200, 48600, 62208, 69360, 77280, 86016, 96000, 105000, 108000, 110592, 118272, 153600, 196608 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,1 COMMENTS The primitive areas are 168, 338, 432, 600, 768, 13608, 14280, 20280, 27216, ... 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). 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). 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. ************************************** *     A  *    R  *   a  *   b  *  c  * ************************************** *   168  *   25  *  14  *  30 *  40  * *   336  *   25  *  14  *  48 *  50  * *   432  *   25  *  30  *  30 *  48  * *   600  *   25  *  30  *  40 *  50  * *   768  *   25  *  40  *  40 *  48  * *  2688  *  100  *  56  * 120 * 160  * *  5376  *  100  *  56  * 192 * 200  * *  6912  *  100  * 120  * 120 * 192  * *  9600  *  100  * 120  * 160 * 200  * * 12288  *  100  * 160  * 160 * 192  * * 13608  *  225  * 126  * 270 * 360  * * 14280  *  169  * 130  * 238 * 312  * * 20280  *  169  * 130  * 312 * 338  * * 27216  *  225  * 126  * 432 * 450  * ............................. REFERENCES 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. LINKS Eric W. Weisstein, MathWorld: Circumradius FORMULA Area A = sqrt(s*(s-a)*(s-b)*(s-c)) with s = (a+b+c)/2 (Heron's formula); Circumradius R = a*b*c/4A. EXAMPLE 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. MATHEMATICA 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] CROSSREFS Cf. A188158, A208984, A210207. Sequence in context: A302365 A038812 A008890 * A105915 A158219 A273771 Adjacent sequences:  A230476 A230477 A230478 * A230480 A230481 A230482 KEYWORD nonn AUTHOR Michel Lagneau, Oct 20 2013 STATUS approved

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Last modified October 5 18:15 EDT 2022. Contains 357261 sequences. (Running on oeis4.)