%I #15 Dec 06 2015 21:16:25
%S 49,91,105,196,289,301,364,379,420,441,459,505,529,609,631,697,751,
%T 784,799,819,889,897,945,961,991,1009,1156,1171,1204,1219,1225,1351,
%U 1395,1401,1456,1516,1521,1611,1680,1681,1689,1759,1764,1836,1849,1939,2020,2095
%N Integer areas of the outer vecten triangles of integer-sided triangles.
%C 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
%C A' = A + (a^2 + b^2 + c^2)/8,
%C where A is the area of the reference triangle. Its side lengths are
%C a' = sqrt((b^2 + c^2 + 4*A)/2),
%C b' = sqrt((a^2 + c^2 + 4*A)/2),
%C c' = sqrt((a^2 + b^2 + 4*A)/2).
%C The circumcircle of the outer Vecten circle is the outer vecten circle.
%C Properties of this sequence:
%C The primitive triangles are 49, 91, 105, 289, 301, ...
%C The nonprimitive triangles of areas 4*a(n), 9*a(n), ..., p^2*a(n), ... are in the sequence.
%C It appears that if the triangles are isosceles, one of the sides of the outer vecten triangles is integer (see the table below).
%C 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.
%C -----------------------------------------------------------------------
%C | A' | A | a | b | c | a' | b' | c' |
%C -----------------------------------------------------------------------
%C | 49 | 24 | 6 | 8 | 10 | sqrt(130) | 2*sqrt(29) | 7*sqrt(2) |
%C | 91 | 48 | 10 | 10 | 12 | sqrt(218) | sqrt(218) | 14 |
%C | 105 | 48 | 10 | 10 | 16 | sqrt(274) | sqrt(274) | 14 |
%C | 196 | 96 | 12 | 16 | 20 | 2*sqrt(130)| 4*sqrt(29)| 14*sqrt(2) |
%C | 289 | 120 | 10 | 24 | 26 | sqrt(866) | 2*sqrt(157)| 17*sqrt(2) |
%C | 301 | 96 | 8 | 26 | 30 |14*sqrt(5) | sqrt(674) | sqrt(562) |
%C | 364 | 192 | 20 | 20 | 24 |2*sqrt(218) | 2*sqrt(218)| 28 |
%C | 379 | 144 | 18 | 20 | 34 |sqrt(1066) | 2*sqrt(257)| 5*sqrt(26) |
%C | 420 | 192 | 20 | 20 | 32 |2*sqrt(274) | 2*sqrt(274)| 28 |
%C | 441 | 216 | 18 | 24 | 30 |3*sqrt(130) | 6*sqrt(29) | 21*sqrt(2) |
%C | 459 | 240 | 20 | 26 | 26 | 34 | sqrt(1018) | sqrt(1018) |
%C | 505 | 168 | 14 | 30 | 40 |sqrt(1586) | sqrt(1586) | 2*sqrt(221)|
%C ....................................................................
%D 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.
%H Eric W. Weisstein, <a href="http://mathworld.wolfram.com/OuterVectenTriangle.html">MathWorld: Outer Vecten Triangle</a>
%e 49 is in the sequence. We use two ways:
%e 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.
%e Second way: by calculation of the sides a', b', c' and by use of Heron's formula.
%e a' = sqrt((b^2 + c^2 + 4*A)/2) = sqrt((8^2 + 10^2 + 4*24)/2) = sqrt(130);
%e b' = sqrt((a^2 + c^2 + 4*A)/2) = sqrt((6^2 + 10^2 + 4*24)/2) = 2*sqrt(29);
%e c' = sqrt((a^2 + b^2 + 4*A)/2) = sqrt((6^2 + 8^2 + 4*24)/2) = 7*sqrt(2).
%e Now we use Heron's formula with (a',b',c'). We find
%e A'=sqrt(s1*(s1-a')*(s1-b')*(s1-c')) with:
%e s1 = (a'+b'+c')/2 = (sqrt(130) + 2*sqrt(29) + 7*sqrt(2))/2;
%e We find A'= 49.
%t 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]
%Y Cf. A188158, A231276.
%K nonn
%O 1,1
%A _Michel Lagneau_, Nov 06 2013
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