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Integer areas of the integer-sided triangles such that the length of the circumradius is a square.
0

%I #11 Nov 08 2016 07:35:24

%S 168,336,432,600,768,2688,5376,6000,6912,9600,12288,13608,14280,20280,

%T 27216,28560,30720,32928,34560,34992,38640,43008,46200,48600,62208,

%U 69360,77280,86016,96000,105000,108000,110592,118272,153600,196608

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

%C The primitive areas are 168, 338, 432, 600, 768, 13608, 14280, 20280, 27216, ...

%C 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 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 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 **************************************

%C * A * R * a * b * c *

%C **************************************

%C * 168 * 25 * 14 * 30 * 40 *

%C * 336 * 25 * 14 * 48 * 50 *

%C * 432 * 25 * 30 * 30 * 48 *

%C * 600 * 25 * 30 * 40 * 50 *

%C * 768 * 25 * 40 * 40 * 48 *

%C * 2688 * 100 * 56 * 120 * 160 *

%C * 5376 * 100 * 56 * 192 * 200 *

%C * 6912 * 100 * 120 * 120 * 192 *

%C * 9600 * 100 * 120 * 160 * 200 *

%C * 12288 * 100 * 160 * 160 * 192 *

%C * 13608 * 225 * 126 * 270 * 360 *

%C * 14280 * 169 * 130 * 238 * 312 *

%C * 20280 * 169 * 130 * 312 * 338 *

%C * 27216 * 225 * 126 * 432 * 450 *

%C .............................

%D 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 Eric W. Weisstein, <a href="http://mathworld.wolfram.com/Circumradius.html">MathWorld: Circumradius</a>

%F Area A = sqrt(s*(s-a)*(s-b)*(s-c)) with s = (a+b+c)/2 (Heron's formula);

%F Circumradius R = a*b*c/4A.

%e 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 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 Cf. A188158, A208984, A210207.

%K nonn

%O 1,1

%A _Michel Lagneau_, Oct 20 2013