A335035 Ordered perimeters of primitive integer triangles with two perpendicular medians.

54, 70, 104, 154, 170, 216, 252, 266, 352, 368, 418, 442, 464, 594, 598, 620, 638, 720, 740, 748, 792, 810, 902, 952, 962, 988, 1054, 1102, 1118, 1134, 1148, 1170, 1216, 1274, 1316, 1376, 1426, 1484, 1512, 1564, 1568, 1598, 1600, 1638, 1702, 1710, 1802, 1836, 1862

Offset: 1

Author

Bernard Schott, May 27 2020

Keywords

• nonn

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 link).

If medians drawn from A and B are perpendicular in centroid G, then a^2 + b^2 = 5 * c^2 (see Maths Challenge picture in link).

All terms are even because each triple is composed of one even side and two odd sides.

For the corresponding primitive triples and miscellaneous properties, see A335034.

Examples

			a(1) = 13 + 19 + 22 = 54 with 19^2 + 22^2 = 5 * 13^2 = 845.
		

Links

• Annales Concours Général, Sujet Concours Général 2007
• Maths Challenge, Perpendicular medians, Problem with picture.

Crossrefs

Cf. A024364 (perimeters of primitive Pythagorean triangles).

Cf. A335034 (corresponding primitive triples), A335036 (smallest side), A335347 (middle side), A335348 (largest side), A335273 (even side).

Programs

PARI

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, vecsum(vt)));););); vecsort(vm);} \\ Michel Marcus, May 27 2020

Formula

a(n) = A335036(n) + A335347(n) + A335348(n).