A335348 Largest side of primitive triples for integer-sided triangles that have two perpendicular medians, the triples being ordered by increasing perimeter.

22, 31, 41, 59, 71, 101, 109, 122, 149, 158, 178, 211, 209, 271, 262, 242, 302, 278, 341, 361, 358, 362, 419, 382, 418, 449, 478, 439, 451, 551, 482, 562, 502, 599, 541, 638, 659, 701, 638, 649, 622, 718, 781, 758, 662, 811, 698, 802, 902

Offset: 1

Author

Bernard Schott, Jun 06 2020

Keywords

• nonn

Comments

If medians at A and B are perpendicular at the centroid G, then a^2 + b^2 = 5 * c^2 (see picture in Maths Challenge link), hence c is always the smallest side.

For the corresponding primitive triples and miscellaneous properties, see A335034; such a triangle with sides of integer lengths cannot be isosceles.

The largest side b with c < a < b is not divisible by 3, 4, or 5, and the odd prime factors of this largest side term b are all of the form 10*k +- 1.

In each increasing triple (c,a,b), c is the smallest odd side (A335036), but the largest side b can be either the even side (A335273) or the largest odd side (see formulas and examples for explanations).

This sequence is not increasing: a(12) = 211 for triangle with perimeter = 442 and a(13) = 209 for triangle with perimeter = 464; hence the largest side is not an increasing function of the perimeter of these triangles.

Examples

			-> For 1st class of triangles, u/v > 3:
(u,v) = (4,1), then 3 < u/v < 3+sqrt(10) and (c,a,b) = (c, a',b') = (17,22,31); the relation is 22^2 + 31^2 = 5 * 17^2 = 1445 with a(2) = 31 = b = b'
(u,v) = (10,1), then u/v > 3+sqrt(10) and (c,a,b) = (c, b' ,a') = (101, 139, 178), the relation is 139^2 + 178^2 = 5 * 101^2 = 51005 with a(11) = 178 = b = a'.
-> For 2nd class, 1 < u/v < 2:
(u,v) = (3,2), then (1+sqrt(10))/3 < u/v < 2 and (c,a,b) = (c, b', a') = (13,19,22), the relation is 19^2 + 22^2 = 5 * 13^2 = 845 with a(1) = 22 = b = a'.
(u,v) = (4,3), then 1 < u/v < (1+sqrt(10))/3 and (c,a,b) = (c, a', b') = (25,38,41); the relation is 38^2 + 41^2 = 5 * 25^2 = 3125 with a(3) = 41 = b = b'.
		

Links

• Maths Challenge, Perpendicular medians, Problem with picture.
• Index to sequences related to Olympiads and other Mathematical competitions.

Crossrefs

Cf. A335034 (primitive triples), A335035 (corresponding perimeters), A335036 (smallest side), A335347 (middle side), A335273 (even side).

Cf. A081804 (similar, with hypotenuse for primitive Pythagorean triples).

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, w); 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)); ); ); ); w = vecsort(apply(vecsort, Vec(vm)); , mycmp); vector(#w, k, w[k][3]); }  \\ Michel Marcus, Jun 06 2020

Formula

a(n) = A335034(3n).

a(n) = A335035(n) - A335036(n) - A335347(n).

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.

If 3 < u/v < 3+sqrt(10) then a' (even) < b' and the triple in increasing order is (c, a = a', b = b'),

if u/v > 3+sqrt(10) then a' (even) > b' and the triple in increasing order is (c, a = b', b = a').

--> 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.

If 1 < u/v < (1+sqrt(10))/3 then a' (even) < b' and the triple in increasing order is (c, a = a', b = b'),

If (1+sqrt(10))/3 < u/v < 2 then a' (even) > b' and the triple in increasing order is (c, a = b', b = a').