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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. 7
13, 19, 22, 17, 22, 31, 25, 38, 41, 37, 58, 59, 41, 58, 71, 53, 62, 101, 61, 82, 109, 65, 79, 122, 85, 118, 149, 89, 121, 158, 101, 139, 178, 109, 122, 211, 113, 142, 209, 145, 178, 271, 145, 191, 262, 149, 229, 242, 157, 179, 302, 173, 269, 278, 181, 218, 341 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,1

COMMENTS

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).

If medians issued of A and B are perpendicular in centroid G, then a^2 + b^2 = 5 * c^2, hence c is always the smallest side.

Some theorical results:

-> 1/2 < a/b < 2 and 1 < a/c < 2 (also 1 < b/c < 2).

-> a, b, c are pairwise coprimes.

-> a et b have different parities, so c is always odd.

-> a and b are not divisible nor by 3 nor by 4 neither by 5.

-> 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').

-> The prime factors of the largest odd side b' (b' = a or b) are all of the form 10*k +- 1 (see formula)

-> 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).

-> cos(C) >= 4/5 (or tan(C) <= 3/4), hence C <= 36.86989...° = A235509 (see Maths Challenge link with picture).

-> CG = c (link Math Stack Exchange).

-> Area(ABC) = (2/3) * m_a * m_b with m_a (resp. m_b) is the length of median AA' (resp. BB') [link Math Stack Exchange].

LINKS

Table of n, a(n) for n=1..57.

Annales Concours Général, Corrigé Concours Général 2007

Annales Concours Général, Sujet Concours Général 2007

Maths Challenge, Perpendicular medians, Problem with picture.

Math Stack Exchange, Perpendicular medians, Prove that AB = GC.

Math Stack Exchange, Solving area of a triangle where medians are perpendicular

FORMULA

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.

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.

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.

For n>=1, a(3n-2) = A335036(n), a(3n-1) = A335037(n), a(3n) = A335038(n).

EXAMPLE

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.

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.

PROG

(PARI) mycmp(x, y) = {my(xp = vecsum(x), yp = vecsum(y)); if (xp!=yp, return (xp-yp)); return (x[1] - y[1]); }

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

CROSSREFS

Cf. A103606 (primitive Pythagorean triples), A235509 = arccos(4/5).

Cf. A335035 (corresponding perimeters), A335036 (smallest side and 1st trisection), A335037 (middle side and 2nd trisection), A335038 (largest side and 3rd trisection), A335273 (even side).

Sequence in context: A171098 A118845 A050717 * A188540 A088184 A274407

Adjacent sequences:  A335031 A335032 A335033 * A335035 A335036 A335037

KEYWORD

nonn,tabf

AUTHOR

Bernard Schott, May 20 2020

STATUS

approved

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Last modified June 16 14:16 EDT 2021. Contains 345057 sequences. (Running on oeis4.)