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.

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

Offset: 1

Author

Bernard Schott, May 20 2020

Keywords

• nonn
• tabf

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 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.

Some theoretical results and geometrical properties:

-> 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 (see Mathematics Stack Exchange link).

-> 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).

-> cot(A) + cot(B) >= 2/3 (see IMO Compendium link and Doob reference). - Bernard Schott, Dec 02 2021

Examples

			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.
		

References

• 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.

Links

• Annales Concours Général, Corrigé Concours Général 2007.
• Annales Concours Général, Sujet Concours Général 2007.
• The IMO Compendium, Problem 3, 25-th Canadian Mathematical Olympiad 1993.
• Maths Challenge, Perpendicular medians (Problem with picture).
• Mathematics Stack Exchange, Perpendicular medians (Proves that AB = GC).
• Mathematics Stack Exchange, Solving area of a triangle where medians are perpendicular.

Crossrefs

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

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).

Programs

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

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) = A335347(n), a(3n) = A335348(n).