A231275 Integer areas of the outer vecten triangles of integer-sided triangles.

49, 91, 105, 196, 289, 301, 364, 379, 420, 441, 459, 505, 529, 609, 631, 697, 751, 784, 799, 819, 889, 897, 945, 961, 991, 1009, 1156, 1171, 1204, 1219, 1225, 1351, 1395, 1401, 1456, 1516, 1521, 1611, 1680, 1681, 1689, 1759, 1764, 1836, 1849, 1939, 2020, 2095

Offset: 1

Author

Michel Lagneau, Nov 06 2013

Keywords

• nonn

Comments

Consider the external erection of three squares on the sides of a triangle ABC. These centers form a triangle IJK. The vertices of this triangle satisfy a number of remarkable properties.

The area of the outer Vecten triangle is A' = A + (a^2 + b^2 + c^2)/8, where A is the area of the reference triangle.

Its side lengths are

a' = sqrt((b^2 + c^2 + 4*A)/2),

b' = sqrt((a^2 + c^2 + 4*A)/2),

c' = sqrt((a^2 + b^2 + 4*A)/2).

The circumcircle of the outer Vecten circle is the outer vecten circle.

Properties of this sequence:

The primitive triangles are 49, 91, 105, 289, 301, ...

The nonprimitive triangles of areas 4*a(n), 9*a(n), ..., p^2*a(n), ... are in the sequence.

It appears that if the triangles are isosceles, one of the sides of the outer vecten triangles is an integer (see the table below).

The following table gives the first values (A, A', a, b, c, a', b', c') where A is the area of the initial triangles, A' is the area of the outer vecten triangles, a, b, c are the integer sides of the initial triangles, and a', b', c' are the sides of the outer vecten triangles.

-----------------------------------------------------------------------

| A' | A | a | b | c | a' | b' | c' |

-----------------------------------------------------------------------

| 49 | 24 | 6 | 8 | 10 | sqrt(130) | 2*sqrt(29) | 7*sqrt(2) |

| 91 | 48 | 10 | 10 | 12 | sqrt(218) | sqrt(218) | 14 |

| 105 | 48 | 10 | 10 | 16 | sqrt(274) | sqrt(274) | 14 |

| 196 | 96 | 12 | 16 | 20 | 2*sqrt(130)| 4*sqrt(29)| 14*sqrt(2) |

| 289 | 120 | 10 | 24 | 26 | sqrt(866) | 2*sqrt(157)| 17*sqrt(2) |

| 301 | 96 | 8 | 26 | 30 |14*sqrt(5) | sqrt(674) | sqrt(562) |

| 364 | 192 | 20 | 20 | 24 |2*sqrt(218) | 2*sqrt(218)| 28 |

| 379 | 144 | 18 | 20 | 34 |sqrt(1066) | 2*sqrt(257)| 5*sqrt(26) |

| 420 | 192 | 20 | 20 | 32 |2*sqrt(274) | 2*sqrt(274)| 28 |

| 441 | 216 | 18 | 24 | 30 |3*sqrt(130) | 6*sqrt(29) | 21*sqrt(2) |

| 459 | 240 | 20 | 26 | 26 | 34 | sqrt(1018) | sqrt(1018) |

| 505 | 168 | 14 | 30 | 40 |sqrt(1586) | sqrt(1586) | 2*sqrt(221)|

....................................................................

Examples

			49 is in the sequence. We use two ways:
First way: with the triangle (6, 8, 10) the formula A' = A + (a^2 + b^2 + c^2)/8 gives directly the result: A' = 24 + (6^2 + 8^2 + 10^2)/8 = 24 + 25 = 49 where the area A = 24 is obtained by Heron's formula A = sqrt(s*(s-a)*(s-b)*(s-c)) = sqrt(12*(12-6)*(12-8)*(12-10)) = sqrt(576) = 24, where s is the semiperimeter.
Second way: by calculation of the sides a', b', c' and by use of Heron's formula.
  a' = sqrt((b^2 + c^2 + 4*A)/2) = sqrt((8^2 + 10^2 + 4*24)/2) = sqrt(130);
  b' = sqrt((a^2 + c^2 + 4*A)/2) = sqrt((6^2 + 10^2 + 4*24)/2) = 2*sqrt(29);
  c' = sqrt((a^2 + b^2 + 4*A)/2) = sqrt((6^2 + 8^2 + 4*24)/2) = 7*sqrt(2).
Now we use Heron's formula with (a',b',c').
We find
  A'=sqrt(s1*(s1-a')*(s1-b')*(s1-c')) with:
  s1 = (a'+b'+c')/2 = (sqrt(130) + 2*sqrt(29) + 7*sqrt(2))/2;
We find A'= 49.
		

References

• H. S. M. Coxeter and S. L. Greitzer, Points and Lines Connected with a Triangle, Ch. 1 in Geometry Revisited, Washington, DC, Math. Assoc. Amer., pp. 1-26 and 96-97, 1967.

Links

• Eric Weisstein's World of Mathematics, Outer Vecten Triangle

Crossrefs

Cf. A188158, A231276.

Programs

Mathematica

nn=500; lst={}; Do[s =(a + b + c)/2; If[IntegerQ[s], area2 = s (s-a)(s-b) (s-c); If[area2 > 0 && IntegerQ[Sqrt[area2] + (a^2 + b^2 + c^2)/8], AppendTo[lst,  Sqrt[area2] + (a^2 + b^2 + c^2)/8]]], {a,nn}, {b,a}, {c,b}]; Union[lst]