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Perimeter 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.
11

%I #11 Feb 19 2022 10:21:58

%S 195,256,342,500,490,612,630,750,972,882,1122,961,1218,1071,1140,1682,

%T 1856,2703,2508,3015,2990,3636,3348,3572,3136,3364,3640,3328,3249,

%U 3362,3312,3330,4530,4250,4921,4455,4840,4565,5054,4945,5307,5655,5440,6440,5746,6561,5588

%N Perimeter 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.

%C Inspired by Project Euler, Problem 143 (see link).

%C For the corresponding primitive triples and miscellaneous properties and references, see A336328.

%C If FA + FB + FC = d, then we have this "beautifully symmetric equation" between a, b, c and d (see Martin Gardner):

%C 3*(a^4 + b^4 + c^4 + d^4) = (a^2 + b^2 + c^2 + d^2)^2.

%C This sequence is not increasing. For example, a(4) = 500 for triangle with largest side = 205 while a(5) = 490 for triangle with largest side = 208.

%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>.

%F a(n) = A336328(n, 1) + A336328(n, 2) + A336328(n, 3).

%F a(n) = A336330(n) + A336331(n) + A336332(n).

%e a(1) = 195 because the first triple is (57, 65, 73) with corresponding d = FA + FB + FC = 264/7 + 195/7 + 325/7 = 112 and 57 + 65 + 73 = 195.

%Y Cf. A336328 (triples), A336329 (FA + FB + FC), A336330 (smallest side), A336331 (middle side), A336332 (largest side), A351476, A351477.

%K nonn

%O 1,1

%A _Bernard Schott_, Jul 21 2020