A333391 -id:A333391 - cp's OEIS frontend

A336328 Primitive triples for integer-sided triangles with A < B < C < 2*Pi/3 and such that FA + FB + FC is an integer where F is the Fermat point of the triangle.

57, 65, 73, 73, 88, 95, 43, 147, 152, 127, 168, 205, 97, 185, 208, 111, 221, 280, 49, 285, 296, 95, 312, 343, 296, 315, 361, 152, 343, 387, 323, 392, 407, 147, 377, 437, 285, 464, 469, 255, 343, 473, 247, 408, 485, 469, 589, 624, 403, 725, 728, 871, 901, 931

Offset: 1

Bernard Schott, Jul 17 2020

Inspired by Project Euler, Problem 143 (see link).
The triples are displayed in increasing order of largest side c, and if largest sides coincide then by increasing order of the middle side b; so, each triple (a, b, c) is in increasing order.
If one angle of the triangle, for example C, is >= 2*Pi/3 then the Fermat point F is this vertex C, so, FA + FB + FC becomes CA + CB, while when all angles are < 2*Pi/3, then the Fermat point is inside the triangle (see link Fermat points), this last condition means that c^2 < a^2 + a*b + b^2.
As a < b < c, then FA > FB > FC.
If FA + FB + FC = d, then we have this "beautifully symmetric equation" between a, b, c and d: 3*(a^4 + b^4 + c^4 + d^4) = (a^2 + b^2 + c^2 + d^2)^2 (see Martin Gardner).
Equivalently: if a point M is inside an equilateral triangle A'B'C' and integer distances to vertices are MA' = a = A072054(n), MB' = b = A072053(n), MC' = c = A072052(n), then the side of this equilateral triangle A'B'C' is equal to d = FA + FB + FC = A061281(n) where F is the Fermat point of the triangle ABC with sides (a,b,c) (see Martin Gardner).
+-----+-----+-----+-----------+-----------+-----------+-----+-------+
|  a  |  b  |  c  |    FA     |    FB     |    FC     |  d  | a+b+c |
+-----------+-----+-----------+-----------+-----------+-----+-------+
|  57 |  65 |  73 |   325/7   |   264/7   |   195/7   | 112 |  195  |
|  73 |  88 |  95 |   440/7   |   325/7   |   264/7   | 147 |  256  |
|  43 | 147 | 152 |  5016/37  |  1064/37  |   765/37  | 185 |  342  |
| 127 | 168 | 205 | 39360/283 | 27265/283 | 13464/283 | 283 |  500  |
|  97 | 185 | 208 | 14800/91  |  6528/91  |  3515/91  | 273 |  490  |
| 111 | 221 | 280 | 70720/331 | 34200/331 |  4641/331 | 331 |  612  |
|  49 | 285 | 296 | 91200/331 | 12376/331 |  5985/331 | 331 |  630  |
|  95 | 312 | 343 |  3864/13  |  1015/13  |   360/13  | 403 |  750  |
| 296 | 315 | 361 |  9405/43  |  8512/43  |  6120/43  | 559 |  972  |
| 152 | 343 | 387 | 30429/97  | 11520/97  |  5096/97  | 485 |  882  |
.....................................................................
From the previous table, we observe that every FA, FB, FC is a fraction while FA + FB + FC = d is an integer (A336329). Jinyuan Wang has found that the 37th triple is the first for which the common denominator of these fractions is 1 (A351477).

			The table begins:
   57,  65,  73;
   73,  88,  95;
   43, 147, 152;
  127, 168, 205;
   97, 185, 208;
  111, 221, 280;
   49, 285, 296;
  .............
For first triple (57, 65, 73) and corresponding d = FA + FB + FC = 325/7 + 264/7 + 195/7 = 112, relation gives: 3*(57^4 + 65^4 + 73^4 + 112^4) = (57^2 + 65^2 + 73^2 + 112^2)^2 = 642470409.

Martin Gardner, Mathematical Circus, Elegant triangles, First Vintage Books Edition, 1979, p. 65.

Project Euler, Problem 143 - Investigating the Torricelli point of a triangle.
Eric Weisstein's World of Mathematics, Fermat points.

Cf. A336329 (FA + FB + FC), A336330 (smallest side), A336331 (middle side), A336332 (largest side), A336333 (perimeter), A351801 (FA numerator), A351802 (FB numerator), A351803 (FC numerator), A351477 (common denominator of FA, FB, FC), A351476 (other FA + FB + FC).
Cf. A333391 (with isogonic center).
Cf. A061281, A072052, A072053, A072054.

If FA + FB + FC = d, then
d = sqrt(((a^2 + b^2 + c^2)/2) + (1/2) * sqrt(6*(a^2*b^2 + b^2*c^2 + c^2*a^2) - 3*(a^4 + b^4 + c^4))), or,
d^2 = (1/2) * (a^2 + b^2 + c^2) + 2 * S * sqrt(3) where S = area of triangle ABC.

A336332 Largest side, in increasing order, of primitive integer-sided triangles with A < B < C < 2*Pi/3 and such that FA + FB + FC is an integer where F is the Fermat point of the triangle.

Original entry on oeis.org

73, 95, 152, 205, 208, 280, 296, 343, 361, 387, 407, 437, 469, 473, 485, 624, 728, 931, 1016, 1273, 1311, 1313, 1368, 1387, 1443, 1457, 1463, 1469, 1477, 1519, 1560, 1591, 1687, 1895, 2015, 2045, 2045, 2085, 2197, 2231, 2289, 2347, 2363, 2416, 2465, 2553, 2728, 2821, 2923

Offset: 1

Bernard Schott, Jul 20 2020

nonn

Inspired by Project Euler, Problem 143 (see link).
This sequence is increasing because triples are in increasing order of largest side.
For the corresponding primitive triples and miscellaneous properties and references, see A336328.
If FA + FB + FC = d, then we have this "beautifully symmetric equation" between a, b, c and d (see Martin Gardner): 3*(a^4 + b^4 + c^4 + d^4) = (a^2 + b^2 + c^2 + d^2)^2.

			a(36) = a(37) = 2045 is the smallest largest side that appears twice because:
   (1023, 1387, 2045) is a triple with FA+FB+FC = 2408, and
   (1051, 1744, 2045) is a triple with FA+FB+FC = 2709.

Martin Gardner, Mathematical Circus, Elegant triangles, First Vintage Books Edition, 1979, p. 65

Project Euler, Problem 143 - Investigating the Torricelli point of a triangle.

Cf. A336328 (triples), A336329 (FA + FB + FC), A336330 (smallest side), A336331 (middle side), this sequence (largest side), A336333 (perimeter).

Cf. A072052 (largest sides: primitives and multiples), A333391.

a(n) = A336328(n, 3).

cp's OEIS Frontend

A336328 Primitive triples for integer-sided triangles with A < B < C < 2*Pi/3 and such that FA + FB + FC is an integer where F is the Fermat point of the triangle.

Views

Author

Keywords

Comments

Examples

References

Links

Crossrefs

Formula