A181924 -id:A181924 - cp's OEIS frontend

A181928 Area A of the triangles such that A, the sides and two medians are integers.

1680, 6720, 15120, 26880, 42000, 60480, 82320, 107520, 136080, 168000, 203280, 221760, 241920, 283920, 329280, 378000, 430080, 485520, 544320, 606480, 672000, 740880, 813120, 887040, 888720, 967680, 1050000, 1135680, 1224720

Offset: 1

Michel Lagneau, Apr 02 2012

nonn

The first six primitives triangles (with areas {1680, 221760, 8168160, 95726400, 302793120, 569336866560}) have been discovered by Ralph H. Buchholz and are listed in a table of the chapter 4 of his thesis (see Links).
Later on, Buchholz & Rathbun identified an infinite family of Heronian triangles with 2 integer medians (comprising 4 of the 6 triangles above). The next two primitive triangles in such family have areas 8548588738240320 and 17293367819066194215360. - Giovanni Resta, Apr 05 2017
The areas of non-primitive triangles are of the form {1680*k^2}, {221760*k^2}, {8168160*k^2}, {95726400*k^2}, {302793120*k^2}, ...
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, faces 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).
Theorems (from Ralph H. Buchholz)
(i) Any triangle with two integer medians has an even semiperimeter.
(ii) If a Heron triangle has two integer medians then its area is divisible by 120.
It seems that, for any n, a(n) == 0 (mod 1680). The reverse is not always true: e.g., as mentioned by Giovanni Resta, the triangle with sides (56*k, 61*k, 75*k) has area of the form 1680 * k^2, but it cannot be a term of a(n). - Sergey Pavlov, Mar 31 2017

			1680 is in the sequence because the corresponding triangle (52, 102, 146) contains two integer medians 35 and 97;
221760 is in the sequence because the corresponding triangle (582, 1252, 1750) contains two integer medians 433 and 1144.

Ralph H. Buchholz, On triangles with rational altitudes, angles bisectors or medians, PHD Thesis, University of Newcastle, Nov 1989.

Ralph H. Buchholz, Triangles with two integer medians
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.
Richard K. Guy, A Dozen Difficult Diophantine Dilemmas, American Mathematical Monthly 95(1988) 31-36.
Eric Weisstein's World of Mathematics, Heronian Triangle.

Cf. A188158, A181924.

with(numtheory):nn:=300:for a from 1 to nn do: for b from a to nn do: for c from b to nn do:p:=(a+b+c)/2:s:=p*(p-a)*(p-b)*(p-c):if s>0 then s1:=sqrt(s): m11:=sqrt((2*b^2+2*c^2-a^2)/4): m22:=sqrt((2*c^2+2*a^2-b^2)/4): m33:=sqrt((2*a^2+2*b^2-c^2)/4):if (s1=floor(s1) and m11=floor(m11) and m22=floor(m22)) or (s1=floor(s1) and m11=floor(m11) and m33=floor(m33))  or (s1=floor(s1) and m22=floor(m22) and m33=floor(m33))  then print(s1):print(a):print(b):print(c):print(m11):print(m22):print(m33):else fi:fi:od:od:od:

nn=600; 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 < area2 && (IntegerQ[Sqrt[area2]] && IntegerQ[(Sqrt[m1])] && IntegerQ[Sqrt[m2]]) || (IntegerQ[Sqrt[area2]] && IntegerQ[Sqrt[m1]] && IntegerQ[Sqrt[m3]]) || (IntegerQ[Sqrt[area2]] && IntegerQ[Sqrt[m2]] && IntegerQ[Sqrt[m3]]), AppendTo[lst,Sqrt[area2]]]], {a,nn}, {b,a}, {c,b}]; Union[lst]

A182171 Area A of the triangles such that A, the sides and three perpendicular bisectors are integers.

Original entry on oeis.org

108, 384, 432, 768, 972, 1536, 1728, 2700, 3072, 3456, 3888, 5292, 6144, 6912, 8748, 9600, 10800, 12288, 13068, 13824, 15552, 17280, 18252, 18816, 18900, 19200, 21168, 24300, 24576, 27000, 27648, 31104, 31212, 34560, 34992, 37632

Offset: 1

Michel Lagneau, Apr 16 2012

nonn

Let a triangle with the angles (A, B, C) and the sides opposite the angles labeled (a, b, c). The length of the perpendicular bisectors is given by (x, y, z) where:
x is the perpendicular bisector passing through the midpoint of the segment BC = a;
y is the perpendicular bisector passing through the midpoint of the segment AC = b;
z is the perpendicular bisector passing through the midpoint of the segment AB = c.
We obtain the relations:
x = (a/2)*tg B if x intersects AB or (a/2)* tg C if x intersects AC;
y = (b/2)* tg A if y intersects AB or (b/2)* tg C if y intersects BC;
z = (c/2)*tg A if z intersects AC or (c/2) *tg B if z intersects BC.
The area A of the 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.
Finally, we obtain:
x = (a/2) * min {tg B, tg C }; y = (b/2) * min {tg A, tg C }; z = (c/2) * min {tg A, tg B } with tg A = 4*A/(b^2+c^2-a^2) ; tg B = 4*A/(c^2+a^2-b^2) ; tg C = 4*A/(a^2+b^2-c^2).
Properties of this sequence:
The numbers of the form 108*n^2, 384*n^2, 768*n^2, 17280*n^2, 18900*n^2 are in the sequence because the area of the primitive triangles (15, 15, 18), (24, 32, 40), (40, 40, 64), (120, 288, 312), (150, 255, 315) are 108, 384, 768 , 17280 and 18900 respectively.
There exists three class of numbers included into a(n) :
Case (i) : a subset of isosceles triangles;
Case (ii) : a subset of right triangles;
Case (iii) : other (neither isosceles nor right triangle).

			Primitive solutions follow:
Area,  ( a,   b,   c),  ( x,   y,   z), Case
  108,  (15,  15,  18),  (10,  10,  12), Isosceles,
  384,  (24,  32,  40),  (16,  12,  15), Right,
  768,  (40,  40,  64),  (15,  15,  24), Isosceles,
17280, (120, 288, 312), (144,  60,  65), Right,
18900, (150, 255, 315), (100,  68,  84), Other,
27000, (255, 255, 450),  (68,  68, 120), Isosceles,
34560, (312, 312, 576),  (65,  65, 120), Isosceles,
53760, (272, 400, 448), (255, 150, 168), Other,
54000, (240, 450, 510), (225, 120, 136), Right,
91476, (429, 462, 495), (364, 308, 330), Other,
95256, (252, 819, 945), (168, 104, 120), Other,
96768, (336, 720, 960), (126, 105, 140), Other.

Ray Chandler, Table of n, a(n) for n = 1..65
Eric W. Weisstein, MathWorld: Triangle

Cf. A188158, A210643, A181924, A181925, A181928.

zz:=evalf(1/10^6):k:=0:nn:=350:
for a from 15 to nn do:
   for b from a to nn do:
       for c from b to nn do:
             p:=(a+b+c)/2:s:=p*(p-a)*(p-b)*(p-c):
             u:= a^2+b^2-c^2:v:= b^2+c^2-a^2 :w:=c^2+a^2-b^2:
             if s>0 then s1:=sqrt(s):else fi:
             if u>0 then u0:=u:else u0:=zz:fi:
             if v>0 then v0:=v:else vo:=zz:fi:
             if w>0 then w0:=w:else w0:=zz:fi:
a0:= evalf(2*a*s1/w0):a1:=evalf(2*a*s1/u0): b0:= evalf(2*b*s1/u0):b1:=evalf(2*b*s1/v0): c0:= evalf(2*c*s1/w0):c1:=evalf(2*c*s1/v0):
             if a0

x = Min{2*a*A/(c^2+a^2-b^2) , 2*a*A/(a^2+b^2-c^2)};
y = Min{2*b*A/(a^2+b^2-c^2)  , 2*b*A/(b^2+c^2-a^2)};
z = Min{2*c*A/(c^2+a^2-b^2)  , 2*c*A/(b^2+c^2-a^2)}.

More terms from Ray Chandler, Apr 24 2013

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

Original entry on oeis.org

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

Michel Lagneau, Oct 14 2013

nonn

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 |
+-----+---------------+---------------+----+----+----+-----+

			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.

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

Cf. A181924.

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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A181928 Area A of the triangles such that A, the sides and two medians are integers.

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A182171 Area A of the triangles such that A, the sides and three perpendicular bisectors are integers.

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

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