cp's OEIS Frontend

A336329 When F is the Fermat point of a triangle ABC, this sequence lists the integer total distances FA + FB + FC corresponding to primitive triangles in A336328.

%I A336329 #23 Feb 12 2025 11:55:50
%S A336329 112,147,185,283,273,331,331,403,559,485,645,520,691,592,637,965,1047,
%T A336329 1560,1415,1688,1649,2093,1895,2045,1687,1843,2073,1839,1768,1805,
%U A336329 1729,1729,2593,2337,2792,2408,2709,2696,2813,2704,2960,3192,3007,3681,3217,3752,2855
%N A336329 When F is the Fermat point of a triangle ABC, this sequence lists the integer total distances FA + FB + FC corresponding to primitive triangles in A336328.
%C A336329 Inspired by Project Euler, Problem 143 (see link).
%C A336329 The triples of sides (a,b,c) with a < b < c are in increasing order of largest side.
%C A336329 For the corresponding primitive triples and miscellaneous properties and references, see A336328.
%C A336329 If FA + FB + FC = d, then we have this "beautifully symmetric equation" between a, b, c and d (see Martin Gardner):
%C A336329   3*(a^4 + b^4 + c^4 + d^4) = (a^2 + b^2 + c^2 + d^2)^2.
%C A336329 For the terms of the data, every FA, FB, FC is a fraction but FA + FB + FC is an integer (see example).
%C A336329 This sequence is not increasing. For example, a(5) = 283 for triangle with largest side = 205 while a(6) = 273 for triangle with largest side = 208. Also, a(6) = a(7) = 331 show that two distinct triangles can have the same minimum possible integer distance FA + FB + FC.
%D A336329 Martin Gardner, Mathematical Circus, Elegant triangles, First Vintage Books Edition, 1979, p. 65.
%H A336329 Project Euler, <a href="https://projecteuler.net/problem=143">Problem 143 - Investigating the Torricelli point of a triangle</a>.
%F A336329 For triangle (a, b, c) whose area is S, and d = FA+FB+FC, then
%F A336329 d = sqrt((1/2)*(a^2+b^2+c^2) + 2*S*sqrt(3)), also,
%F A336329 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 A336329 3*(a^4 + b^4 + c^4 + d^4) = (a^2 + b^2 + c^2 + d^2)^2.
%e A336329 For first triple (57, 65, 73), d = 112 is solution of 3*(57^4 + 65^4 + 73^4 + d^4) = (57^2 + 65^2 + 73^2 + d^2)^2, hence, 112 is a term because d = FA + FB + FC = 264/7 + 195/7 + 325/7 = 112.
%o A336329 (PARI) lista(nn) = my(d); for(c=4, nn, for(b=ceil(c/sqrt(3)), c-1, for(a=1+(sqrt(4*c^2-3*b^2)-b)\2, b-1, if(gcd([a, b, c])==1 && issquare(6*(a^2*b^2+b^2*c^2+c^2*a^2)-3*(a^4+b^4+c^4), &d) && issquare((a^2+b^2+c^2+d)/2, &d), print1(d, ", "))))); \\ _Jinyuan Wang_, Jul 20 2020
%Y A336329 Cf. A336328 (triples), A336330 (smallest side), A336331 (middle side), A336332 (largest side), A336333 (perimeter), A351477.
%Y A336329 Cf. A061281 (supersequence with non-primitive terms).
%K A336329 nonn
%O A336329 1,1
%A A336329 _Bernard Schott_, Jul 18 2020
%E A336329 More terms from _Jinyuan Wang_, Jul 20 2020