A208984 -id:A208984 - cp's OEIS frontend

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

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

Offset: 1

Michel Lagneau, Oct 20 2013

nonn

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  *
.............................

			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.

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.

Eric W. Weisstein, MathWorld: Circumradius

Cf. A188158, A208984, A210207.

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]

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.

A256629 Integer areas A of integer-sided triangles such that the length of the circumradius is a prime number.

Original entry on oeis.org

24, 120, 240, 720, 840, 1320, 2520, 3360, 3960, 5280, 6240, 6840, 9360, 10920, 14280, 15600, 16320, 17160, 18480, 22440, 24360, 26520, 31920, 35880, 38760, 42840, 43680, 46200, 50160, 55200, 57960, 59280, 70200, 73920, 91080, 93840, 100800, 107640, 117600, 118320, 122400

Offset: 1

Michel Lagneau, Apr 05 2015

nonn

Subsequence of A208984.
For the same area, the number of triangles such that the length of the circumradius is a prime number is not unique; for example, from a(5)= 840 we find two triangles of sides (a,b,c)=(40,42,58) and (24,70,74) where R = 29 and 37, respectively.
The circumradius R values corresponding to the terms of the sequence are 5, 13, 17, 41, (29 or 37), 61, 53, 113, 101, 73, 89, 181, 97, (109 or 197), 149, ...
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 circumradius R is given by R = abc/4A.
Observation:
- all sides of the triangles are even;
- the inradius values are also even;
- the first triangle, of sides (6,8,10), is the unique triangle in which the lengths of the inradius and the circumradius are both prime numbers (r = A/p = 24/12 = 2 and R = abc/4A = 480/4*24 = 5).
For the same area, it is possible to find a prime inradius (see A230195), but the corresponding circumradius is generally rational. For example, for a(2) = 120, we find two triangles:
(10,24,26) => r = 4 and R = 13;
(16,25,39) => r = 3 prime and R = 65/2.
The following table gives the first values (A, a, b, c, r, R) where A is the integer area, a,b,c are the sides and r = A/p, R = a*b*c/4*A are respectively the values of the inradius and the circumradius.
+--------+------+-------+-------+------+-------+
|    A   |   a  |   b   |   c   |   r  |   R   |
+--------+------+-------+-------+------+-------+
|    24  |   6  |    8  |   10  |   2  |    5  |
|   120  |  10  |   24  |   26  |   4  |   13  |
|   240  |  16  |   30  |   34  |   6  |   17  |
|   720  |  18  |   80  |   82  |   8  |   41  |
|   840  |  40  |   42  |   58  |  12  |   29  |
|   840  |  24  |   70  |   74  |  10  |   37  |
|  1320  |  22  |  120  |  122  |  10  |   61  |
|  2520  |  56  |   90  |  106  |  20  |   53  |
|  3360  |  30  |  224  |  226  |  14  |  113  |
|  3960  |  40  |  198  |  202  |  18  |  101  |
|  5280  |  96  |  110  |  146  |  30  |   73  |
|  6240  |  78  |  160  |  178  |  30  |   89  |
+--------+------+-------+-------+------+-------+

			a(1) = 24 because, for (a,b,c) = (6, 8, 10) => s= (6+8+10)/2 =12, and
A = sqrt(12(12-6)(12-8)(12-10)) = sqrt(576) = 24;
R = abc/4A = 480/4*24 = 5 is prime.

Cf. A188158, A208984, A230195.

Missing terms 91080 and 117600 added by Zachary Sizer, Jan 02 2025

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

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