The OEIS is supported by the many generous donors to the OEIS Foundation.
%I #34 Mar 13 2022 18:53:47
%S 57,65,73,73,88,95,43,147,152,127,168,205,97,185,208,111,221,280,49,
%T 285,296,95,312,343,296,315,361,152,343,387,323,392,407,147,377,437,
%U 285,464,469,255,343,473,247,408,485,469,589,624,403,725,728,871,901,931
%N 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.
%C Inspired by Project Euler, Problem 143 (see link).
%C 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.
%C 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.
%C As a < b < c, then FA > FB > FC.
%C 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).
%C 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).
%C +-----+-----+-----+-----------+-----------+-----------+-----+-------+
%C | a | b | c | FA | FB | FC | d | a+b+c |
%C +-----------+-----+-----------+-----------+-----------+-----+-------+
%C | 57 | 65 | 73 | 325/7 | 264/7 | 195/7 | 112 | 195 |
%C | 73 | 88 | 95 | 440/7 | 325/7 | 264/7 | 147 | 256 |
%C | 43 | 147 | 152 | 5016/37 | 1064/37 | 765/37 | 185 | 342 |
%C | 127 | 168 | 205 | 39360/283 | 27265/283 | 13464/283 | 283 | 500 |
%C | 97 | 185 | 208 | 14800/91 | 6528/91 | 3515/91 | 273 | 490 |
%C | 111 | 221 | 280 | 70720/331 | 34200/331 | 4641/331 | 331 | 612 |
%C | 49 | 285 | 296 | 91200/331 | 12376/331 | 5985/331 | 331 | 630 |
%C | 95 | 312 | 343 | 3864/13 | 1015/13 | 360/13 | 403 | 750 |
%C | 296 | 315 | 361 | 9405/43 | 8512/43 | 6120/43 | 559 | 972 |
%C | 152 | 343 | 387 | 30429/97 | 11520/97 | 5096/97 | 485 | 882 |
%C .....................................................................
%C 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).
%D Martin Gardner, Mathematical Circus, Elegant triangles, First Vintage Books Edition, 1979, p. 65.
%H Project Euler, <a href="https://projecteuler.net/problem=143">Problem 143 - Investigating the Torricelli point of a triangle</a>.
%H Eric Weisstein's World of Mathematics, <a href="https://mathworld.wolfram.com/FermatPoints.html">Fermat points</a>.
%F If FA + FB + FC = d, then
%F 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,
%F d^2 = (1/2) * (a^2 + b^2 + c^2) + 2 * S * sqrt(3) where S = area of triangle ABC.
%e The table begins:
%e 57, 65, 73;
%e 73, 88, 95;
%e 43, 147, 152;
%e 127, 168, 205;
%e 97, 185, 208;
%e 111, 221, 280;
%e 49, 285, 296;
%e .............
%e 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.
%Y 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).
%Y Cf. A333391 (with isogonic center).
%Y Cf. A061281, A072052, A072053, A072054.
%K nonn,tabf
%O 1,1
%A _Bernard Schott_, Jul 17 2020