A227895 Integer areas of integer-sided triangles where at least one median is of prime length.

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

Offset: 1

Author

Michel Lagneau, Oct 14 2013

Keywords

• nonn

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 - c^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 |

+-----+---------------+---------------+----+----+----+-----+

Examples

			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.
		

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

Crossrefs

Cf. A181924.

Programs

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