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A231275 Integer areas of the outer vecten triangles of integer-sided triangles. 1
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 (list; graph; refs; listen; history; text; internal format)
OFFSET
1,1
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 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)|
....................................................................
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
EXAMPLE
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.
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]
CROSSREFS
Sequence in context: A259766 A273937 A326257 * A158725 A090825 A350704
KEYWORD
nonn
AUTHOR
Michel Lagneau, Nov 06 2013
STATUS
approved

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Last modified July 14 13:51 EDT 2024. Contains 374318 sequences. (Running on oeis4.)