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 A227895 Integer areas of integer-sided triangles where at least one median is of prime length. 0
 12, 24, 60, 120, 168, 240, 420, 660, 720, 840, 1092, 1320, 1680, 2448, 2520, 2640, 3360, 3420, 3960, 5280, 5460, 6072, 6240, 6840, 9360, 10920, 12240, 14280, 15600, 15960, 16320, 17160, 18480, 21840, 22440, 24480, 26520, 27720, 31920, 35880, 38760, 43680 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,1 COMMENTS Subset of A181924. Using Heron's formula for the area A of a triangle with sides (a, b, c), the existence of a triangle with three rational medians and integer (or rational) area implies a solution of the Diophantine system: 4x^2 = 2a^2 + 2b^2 - c^2 4y^2 = 2a^2 + 2c^2 - b^2 4z^2 = 2b^2 + 2c^2 - a^2 A^2 = s(s-a)(s-b)(s-c) where s = (a+b+c)/2 is the semiperimeter and  x, y, z the medians. There is no solution known to this system at this time. The problem is similar to the more famous unsolved problem of finding a box with edges, face diagonals and body diagonals all rational. Such a box also involves seven quantities which must satisfy a system of four Diophantine equations: d^2 = a^2 + b^2; e^2 = a^2 + c^2; f^2 = b^2 + c^2; g^2 = a^2 + b^2 + c^2. where a, b and c are the lengths of the edges (see Guy in the reference). Properties of this sequence: There exist three class of triangles (a, b, c): (i) A class of isosceles triangles where a = b < c => the median m = 2*A/c. Example: for a(1) = 12, m = 2*12/8 = 3; (ii) A class of Pythagorean where a^2 + b^2 = c^2, and it is easy to check that the median m = c/2. Example: for a(2) = 24, 6^2 + 8^2 = 10^2 and m = 10/2 = 5; (iii) A class of non-isosceles and non-Pythagorean triangles (a,b,c) having one or two integer medians. Example: for a(4) = 120, (a,b,c) = (10, 24, 26)and m = sqrt((2a^2 + 2b^2 - a^2)/4) = sqrt((2*10^2+2*24^2-26^2)/4) = 13. The following table gives the first values (A, m1, m2, m3, a,b,c) where A is the area, m1, m2, m3 are the medians and a, b, c the integer sides of the triangles. +-----+---------------+---------------+----+----+----+-----+ |   A |    m1         |   m2          | m3 |  a |  b |  c  | +-----+---------------+---------------+----+----+----+-----+ |  12 | 3*sqrt(17)/2  | 3*sqrt(17)/2  |  3 |  5 |  5 |   8 | |  24 |   sqrt(73)    |  2*sqrt(13)   |  5 |  6 |  8 |  10 | |  60 | sqrt(1321)/2  | sqrt(1321)/2  |  5 | 13 | 13 |  24 | | 120 |   sqrt(601)   |  2*sqrt(61)   | 13 | 10 | 24 |  26 | | 168 | sqrt(5233)/2  | sqrt(5233)/2  |  7 | 25 | 25 |  48 | | 240 |  2*sqrt(241)  |   sqrt(481)   | 17 | 16 | 30 |  34 | | 420 | sqrt(8689)/2  | sqrt(6001)/2  | 17 | 25 | 39 |  56 | | 660 | sqrt(32521)/2 | sqrt(32521)/2 | 11 | 61 | 61 | 120 | | 720 |  sqrt(6481)   |  2*sqrt(481)  | 41 | 18 | 80 |  82 | +-----+---------------+---------------+----+----+----+-----+ LINKS Andrew Bremner and Richard K. Guy, A Dozen Difficult Diophantine Dilemmas, American Mathematical Monthly 95(1988) 31-36. Ralph H. Buchholz, On Triangles with rational altitudes, angle bisectors or medians, Newcastle University (1989), 21-22. Ralph H. Buchholz and Randall L. Rathbun, An infinite set of Heron triangles with two rational medians, The American Mathematical Monthly, Vol. 104, No. 2 (Feb., 1997), pp. 107-115. Eric W. Weisstein, MathWorld: HeronianTriangle EXAMPLE 1680 is in the sequence because the triangle (a,b,c) = (52, 102, 146) => A = 1680 and m1 = 4*sqrt(949), m2 = 35 and m3 = 97 is a prime number. MATHEMATICA nn=800; lst={}; Do[s=(a+b+c)/2; If[IntegerQ[s], area2=s (s-a) (s-b) (s-c); m1=(2*b^2+2*c^2-a^2)/4; m2=(2*c^2+2*a^2-b^2)/4; m3=(2*a^2+2*b^2-c^2)/4; If[0

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Last modified November 17 06:06 EST 2019. Contains 329217 sequences. (Running on oeis4.)