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A335034 Primitive triples, in nondecreasing order of perimeter, for integer-sided triangles with two perpendicular medians; each triple is in increasing order, and if perimeters coincide then increasing order of the smallest side.

%I A335034 #76 Jan 08 2024 09:03:29
%S A335034 13,19,22,17,22,31,25,38,41,37,58,59,41,58,71,53,62,101,61,82,109,65,
%T A335034 79,122,85,118,149,89,121,158,101,139,178,109,122,211,113,142,209,145,
%U A335034 178,271,145,191,262,149,229,242,157,179,302,173,269,278,181,218,341
%N A335034 Primitive triples, in nondecreasing order of perimeter, for integer-sided triangles with two perpendicular medians; each triple is in increasing order, and if perimeters coincide then increasing order of the smallest side.
%C A335034 The study of these integer triangles that have two perpendicular medians was proposed in the problem of Concours Général in 2007 in France (see links).
%C A335034 If medians drawn from A and B are perpendicular in centroid G, then a^2 + b^2 = 5 * c^2, hence c is always the smallest side.
%C A335034 Some theoretical results and geometrical properties:
%C A335034 -> 1/2 < a/b < 2 and 1 < a/c < 2 (also 1 < b/c < 2).
%C A335034 -> a, b, c are pairwise coprimes.
%C A335034 -> a et b have different parities, so c is always odd.
%C A335034 -> a and b are not divisible nor by 3 nor by 4 neither by 5.
%C A335034 -> The odd prime factors of the even term a' (a' = a or b) are all of the form 10*k +- 1 (see formula for a').
%C A335034 -> The prime factors of the largest odd side b' (b' = a or b) are all of the form 10*k +- 1 (see formula)
%C A335034 -> Consequence: in each increasing triple (c,a,b), c is always the smallest odd side, but a can be either the even side or the largest odd side (see formulas and examples for explanations).
%C A335034 -> cos(C) >= 4/5 (or tan(C) <= 3/4), hence C <= 36.86989...° = A235509 (see Maths Challenge link with picture).
%C A335034 -> CG = c (see Mathematics Stack Exchange link).
%C A335034 -> Area(ABC) = (2/3) * m_a * m_b with m_a (resp. m_b) is the length of median AA' (resp. BB') (see Mathematics Stack Exchange link).
%C A335034 -> cot(A) + cot(B) >= 2/3 (see IMO Compendium link and Doob reference). - _Bernard Schott_, Dec 02 2021
%D A335034 Michael Doob, The Canadian Mathematical Olympiad & L'Olympiade Mathématique du Canada 1969-1993, Canadian Mathematical Society & Société Mathématique du Canada, Problem 3, 1993, page 253, 1993.
%H A335034 Annales Concours Général, <a href="https://www.freemaths.fr/annales-composition-mathematiques-concours-general/concours-general-mathematiques-2007-corrige.pdf">Corrigé Concours Général 2007</a>.
%H A335034 Annales Concours Général, <a href="https://www.freemaths.fr/annales-composition-mathematiques-concours-general/concours-general-mathematiques-2007-sujet.pdf">Sujet Concours Général 2007</a>.
%H A335034 The IMO Compendium, <a href="https://imomath.com/othercomp/Can/CanMO93.pdf">Problem 3</a>, 25-th Canadian Mathematical Olympiad 1993.
%H A335034 Maths Challenge, <a href="https://mathschallenge.net/view/perpendicular_medians">Perpendicular medians</a> (Problem with picture).
%H A335034 Mathematics Stack Exchange, <a href="https://math.stackexchange.com/questions/1890279/perpendicular-medians">Perpendicular medians</a> (Proves that AB = GC).
%H A335034 Mathematics Stack Exchange, <a href="https://math.stackexchange.com/questions/2676657/solving-area-of-a-triangle-where-medians-are-perpendicular">Solving area of a triangle where medians are perpendicular</a>.
%F A335034 There exist two disjoint classes of such triangles, obtained with two distinct families of formulas: let u > v > 0 , u and v with different parities, gcd(u,v) = 1;  a' is the even side and b' the largest odd side.
%F A335034 1st class of triangles: (a',b',c) = (2*(u^2-uv-v^2), u^2+4*u*v-v^2, u^2+v^2) with u/v > 3 and 5 doesn't divide u-3v.
%F A335034 2nd class of triangles: (a',b',c) = (2*(u^2+uv-v^2), -u^2+4*u*v+v^2, u^2+v^2) with 1 < u/v < 2 and 5 doesn't divide u-2v.
%F A335034 For n>=1, a(3n-2) = A335036(n), a(3n-1) = A335347(n), a(3n) = A335348(n).
%e A335034 For 1st class, u/v > 3: (u,v) = (4,1), then (c,a,b) = (c,a',b') = (17,22,31) is the second triple and 22^2 + 31^2 = 5 * 17^2 = 1445.
%e A335034 For 2nd class, 1 < u/v < 2: (u,v) = (3,2), then (c,a,b) = (c,b',a') = (13,19,22) is the first triple and 19^2 + 22^2 = 5 * 13^2 = 845.
%o A335034 (PARI) mycmp(x, y) = {my(xp = vecsum(x), yp = vecsum(y)); if (xp!=yp, return (xp-yp)); return (x[1] - y[1]);}
%o A335034 lista(nn) = {my(vm = List(), vt); for (u=1, nn, for (v=1, nn, if (gcd(u, v) == 1, vt = 0; if ((u/v > 3) && ((u-3*v) % 5), vt = [2*(u^2-u*v-v^2), u^2+4*u*v-v^2, u^2+v^2]); if ((u/v > 1) && (u/v < 2) && ((u-2*v) % 5), vt = [2*(u^2+u*v-v^2), -u^2+4*u*v+v^2, u^2+v^2]); if (gcd(vt) == 1, listput(vm, vt));););); vecsort(apply(vecsort, Vec(vm));, mycmp);} \\ _Michel Marcus_, May 21 2020
%Y A335034 Cf. A103606 (primitive Pythagorean triples), A235509 = arccos(4/5).
%Y A335034 Cf. A335035 (corresponding perimeters), A335036 (smallest side and 1st trisection), A335347 (middle side and 2nd trisection), A335348 (largest side and 3rd trisection), A335273 (even side).
%K A335034 nonn,tabf
%O A335034 1,1
%A A335034 _Bernard Schott_, May 20 2020