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A335034
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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.
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8
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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
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OFFSET
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1,1
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COMMENTS
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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
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REFERENCES
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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.
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LINKS
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The IMO Compendium, Problem 3, 25-th Canadian Mathematical Olympiad 1993.
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FORMULA
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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.
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EXAMPLE
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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.
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PROG
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(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
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CROSSREFS
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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).
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KEYWORD
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nonn,tabf
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AUTHOR
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STATUS
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approved
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