A335034 -id:A335034 - cp's OEIS frontend

A335893 Primitive triples for integer-sided triangles whose angles A < B < C are in arithmetic progression.

3, 7, 8, 5, 7, 8, 7, 13, 15, 8, 13, 15, 5, 19, 21, 16, 19, 21, 11, 31, 35, 24, 31, 35, 7, 37, 40, 33, 37, 40, 13, 43, 48, 35, 43, 48, 16, 49, 55, 39, 49, 55, 9, 61, 65, 56, 61, 65, 32, 67, 77, 45, 67, 77, 17, 73, 80, 63, 73, 80, 40, 79, 91, 51, 79, 91, 11, 91, 96

Offset: 1

Bernard Schott, Jun 29 2020

The triples are displayed in nondecreasing order of middle side, and if middle sides coincide then by increasing order of the largest side, hence, each triple (a, b, c) is in increasing order.
These three properties below are equivalent:
-> integer-sided triangles whose angles A < B < C are in arithmetic progression,
-> integer-sided triangles such that B = (A+C)/2 with A < C,
-> integer-sided triangles such that A < B < C with B = Pi/3.
When A < B < C are in arithmetic progression with B = A + phi and C = B + phi, then 0 < phi < Pi/3.
The corresponding metric relation between sides is b^2 = a^2 - a*c + c^2.
There exists such primitive triangle iff b^2 is an odd square term of A024612. Hence, the first few middle sides b are 7, 13, 19, 31, 37, 43, 49, 61, 67, ... and b is a term of A004611 \ {1}. Indeed, b cannot be even if the triple is primitive.
As B = Pi/3 and C runs from Pi/3 to 2*Pi/3, sin(C) gets a maximum when C = Pi/2 with sin(C) = 1, hence, from law of sines (see link): b/sin(B) = c/sin(C), and c < b/sin(Pi/3) = b * 2/sqrt(3) < 6*b/5. This bound is used in the PARI and Maple programs below.
When triple (a, b, c) is solution, then triple (c-a, b, c) is another solution. Hence, for each b odd solution, there exist 2 triples with same middle side b and same largest side c.
The common tangent to the nine-point circle and the incircle of a triangle ABC is parallel to the Euler line iff angles A < B < C are in arithmetic progression (see Crux Mathematicorum for Indian team selection). - Bernard Schott, Apr 14 2022
These triples are called (primitive) Eisenstein triples (Wikipedia). - Bernard Schott, Sep 21 2022

			(3, 7, 8) is a triple for this sequence because from law of cosines (see link), cos(A) = (7^2 + 8^2 - 3^2)/(2*7*8) = 13/14, cos(B) = (8^2 + 3^2 - 7^2)/(2*8*3) = 1/2 and cos(C) = (3^2 + 7^2 - 8^2)/(2*3*7) = -1/7; then, (A+C)/2 = ( arccos(13/14) + arccos(-1/7) )/2 = Pi/3 = B.
Also, arccos(13/14) ~ 21.787 degrees, arccos(1/2) = 60 degrees, arccos(-1/7) ~ 98.213 degrees, so B-A = C-B ~ 38.213 degrees, hence (A, B, C) are in arithmetic progression.
5^2 - 5*8 + 8^2 = 7^2, hence (5, 7, 8) is another triple for triangle whose angles A < B < C are in arithmetic progression.

V. Lespinard & R. Pernet, Trigonométrie, Classe de Mathématiques élémentaires, programme 1962, problème B-298 p. 124, André Desvigne.

Bill Sands, Indian team selection test 2007, question 5, Crux Mathematicorum, Vol. 36, No. 5 (2010), p. 278.
Eric Weisstein's World of Mathematics, Law of Cosines.
Eric Weisstein's World of Mathematics, Law of Sines.
Wikipedia, Eisenstein triple.
Index to sequences related to Olympiads.

Cf. A004611, A024612.
Cf. A335894 (smallest side), A335895 (middle side), A335896 (largest side), A335897 (perimeter).
Cf. A103606 (primitive Pythagorean triples), A335034 (primitive triples for triangles with two perpendicular medians).

for b from 3 to 250 by 2 do
for c from b+1 to 6*b/5 do
a := (c - sqrt(4*b^2-3*c^2))/2;
if gcd(a,b)=1 and issqr(4*b^2-3*c^2) then print(a,b,c,c-a,b,c); end if;
end do;
end do;

lista(nn) = {forstep(b=1, nn, 2, for(c=b+1, 6*b\5, if (issquare(d=4*b^2 - 3*c^2), my(a = (c - sqrtint(d))/2); if ((denominator(a)==1) && (gcd(a, b) == 1), print(a, ", ", b, ", ", c, ", "); print(c-a, ", ", b, ", ", c, ", ");););););} \\ Michel Marcus, Jul 15 2020

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

Original entry on oeis.org

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

Bernard Schott, May 27 2020

nonn

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.

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

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

Cf. A024364 (perimeters of primitive Pythagorean triangles).

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

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

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

A335036 Smallest side c of the primitive triples (c,a,b) for integer triangles that have two perpendicular medians, ordered by increasing perimeter.

Original entry on oeis.org

13, 17, 25, 37, 41, 53, 61, 65, 85, 89, 101, 109, 113, 145, 145, 149, 157, 173, 181, 185, 193, 197, 221, 229, 233, 241, 257, 265, 269, 281, 277, 289, 293, 313, 317, 337, 349, 365, 365, 377, 377, 389, 397, 401, 409, 421, 433, 445, 461

Offset: 1

Bernard Schott, May 28 2020

nonn

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 odd side (see link Maths Challenge).
c = u^2 + v^2 for some u and v (see formula), so this sequence is subsequence of A004431.
For the corresponding primitive triples and miscellaneous properties, see A335034.
The repetitions for 145, 365, 377,... correspond to smallest sides for triangles with distinct perimeters (see examples).
This sequence is not increasing a(30) = 281 for triangle with perimeter = 1134 and a(31) = 277 for triangle with perimeter = 1148. The smallest side is not an increasing function of the perimeter of these triangles.

			The triples (145, 178, 271) and (145, 191, 262) correspond to triangles with respective perimeters equal to 594 and 598, so a(14) = a(15) = 145.
The triples (365, 418, 701) and (365, 509, 638) correspond to triangles with respective perimeters equal to 1484 and 1512, so a(38) = a(39) = 365.

Maths Challenge, Perpendicular medians, Problem with picture.

Subsequence of A004431.

Cf. A335034 (primitive triples), A335035 (corresponding perimeters), A335347 (middle side), A335348 (largest side), A335273 (even side).

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][1]);} \\ Michel Marcus, May 28 2020

a(n) = A335034(3n-2).
a(n) = A335035(n) - A335347(n) - A335348(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; if c is the smallest odd side, then:
1st class of triangles: c = u^2+v^2 with u/v > 3 and 5 doesn't divide u-3v,
2nd class of triangles: c = u^2+v^2 with 1 < u/v < 2 and 5 doesn't divide u-2v.

A335273 Even side of primitive triples for integer-sided triangles that have two perpendicular medians, the triples being ordered by increasing perimeter.

Original entry on oeis.org

22, 22, 38, 58, 58, 62, 82, 122, 118, 158, 178, 122, 142, 178, 262, 242, 302, 278, 218, 202, 358, 362, 262, 382, 418, 298, 478, 398, 398, 302, 482, 562, 502, 362, 458, 638, 418, 418, 638, 538, 622, 718, 422, 758, 662, 478, 698, 802, 902, 682, 542, 922, 662, 818

Offset: 1

Bernard Schott, Jun 09 2020

nonn

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).
For the corresponding primitive triples and miscellaneous properties, see A335034.
In each increasing triple (c,a,b), c is always the smallest odd side (A335036), but the even side a' can be either the middle side a (A335347) or the largest side b (A335348) (see formulas and examples for explanations).
The even side a' is not divisible by 4, and the odd prime factors of this even side term a' are all of the form 10*k +- 1.
The repetitions for 22, 58, 122,... correspond to even sides for triangles with distinct perimeters (see examples).
This sequence is not increasing: a(8) = 122 for triangle with perimeter = 266 and a(9) = 118 for triangle with perimeter = 352; hence the even side is not an increasing function of the perimeter of these triangles.

			The triples (13, 19, 22), and (17, 22, 31) correspond to triangles with respective perimeters equal to 54 and 70, so a(1) = a(2) = 22.
The triples (37, 58, 59) and (41, 58, 71) correspond to triangles with respective perimeters equal to 154 and 170, so a(4) = a(5) = 58.
-> 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) = 22 = a' = a.
(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 = a' = b.
-> 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 = a' = b.
(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) = 38 = a' = a.

Annales Concours Général, Sujet Concours Général 2007.
Maths Challenge, Perpendicular medians, Problem with picture.
Index to sequences related to Olympiads and other Mathematical competitions.

Cf. A335034 (primitive triples), A335035 (corresponding perimeters), A335036 (smallest side), A335347 (middle side), A335348 (largest side), this sequence (even side), A335418 (largest odd side).

mycmp(x, y) = {my(xp = vecsum(x), yp = vecsum(y)); if (xp!=yp, return (xp-yp)); return (x[1] - y[1]); }
triples(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); } \\ A335034
lista(nn) = my(w=triples(nn)); vector(#w, k, select(x->!(x%2), w[k])[1]); \\ Michel Marcus, Jun 10 2020

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').

More terms from Michel Marcus, Jun 10 2020

A335347 Middle side of primitive triples for integer-sided triangles that have two perpendicular medians, the triples being ordered by increasing perimeter.

Original entry on oeis.org

19, 22, 38, 58, 58, 62, 82, 79, 118, 121, 139, 122, 142, 178, 191, 229, 179, 269, 218, 202, 241, 251, 262, 341, 311, 298, 319, 398, 398, 302, 389, 319, 421, 362, 458, 401, 418, 418, 509, 538, 569, 491, 422, 479, 631, 478, 671, 589, 499

Offset: 1

Bernard Schott, Jun 03 2020

nonn

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 middle side a with c < a < b is not divisible by 3, 4, or 5, and the odd prime factors of this middle side term a are all of the form 10*k +- 1.
In each increasing triple (c,a,b), c is the smallest odd side (A335036), but the middle side a can be either the even side (A335273) or the largest odd side (see formulas and examples for explanations).
The consecutive repetitions for 58, 398, 418,... correspond to middle sides for triangles with distinct perimeters (see examples).
This sequence is not increasing: a(7) = 82 for triangle with perimeter = 252 and a(8) = 79 for triangle with perimeter = 266; hence the middle side is not an increasing function of the perimeter of these triangles.

			The triples (37, 58, 59) and (41, 58, 71) correspond to triangles with respective perimeters equal to 154 and 170, so a(4) = a(5) = 58.
--> 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) = 22 = a = a',
(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) = 139 = a = b'.
--> 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) = 19 = a = b'
(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) = 22 = a = a'.

Maths Challenge, Perpendicular medians, Problem with picture.

Cf. A335034 (primitive triples), A335035 (corresponding perimeters), A335036 (smallest side), A335348 (largest side), A335273 (even side).

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][2]); }  \\ Michel Marcus, Jun 03 2020

a(n) = A335034(3n-1).
a(n) = A335035(n) - A335036(n) - A335348(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').

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

Original entry on oeis.org

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

Bernard Schott, Jun 06 2020

nonn

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.

			-> 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'.

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

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).

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

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').

A235509 Decimal expansion of arccos(4/5).

Original entry on oeis.org

6, 4, 3, 5, 0, 1, 1, 0, 8, 7, 9, 3, 2, 8, 4, 3, 8, 6, 8, 0, 2, 8, 0, 9, 2, 2, 8, 7, 1, 7, 3, 2, 2, 6, 3, 8, 0, 4, 1, 5, 1, 0, 5, 9, 1, 1, 1, 5, 3, 1, 2, 3, 8, 2, 8, 6, 5, 6, 0, 6, 1, 1, 8, 7, 1, 3, 5, 1, 2, 4, 7, 4, 8, 1, 1, 6, 2, 1, 0, 8, 8, 7, 1, 2, 8, 1, 6, 8, 4, 4, 7, 0, 1, 2, 8, 2, 7, 4, 8, 8

Offset: 0

Jean-François Alcover, Jan 14 2014

Given a square ABCD, there is one point M equidistant from A, B and the middle of CD. The measure of the angle BAM is arccos(4/5) (or arcsec(5/4)). This angle is the smallest angle of the well-known (3, 4, 5) Pythagorean triangle.
Also the polar angle phi of the viewing cone that cuts out exactly 10% of the celestial sphere; phi = arccos(1-2f), where f is the cut-out fraction of the full solid angle. - Stanislav Sykora, Feb 14 2016
Given a triangle ABC whose medians drawn from A and B are perpendicular in centroid G, then angle C <= arccos(4/5) (see Maths Challenge link with figure and proof). - Bernard Schott, Mar 29 2023

			0.64350110879328438680280922871732263804151059111531238286560611871351...
In degrees: 36.869897645844...°

Jean-François Alcover, Figure showing square ABCD and angle BAM.
Maths Challenge, Perpendicular medians, Problem ID: 296, 10 Dec 2006.
Index entries for transcendental numbers

Cf. A195771, A335034.

RealDigits[ArcCos[4/5], 10, 100] // First

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```
asin(3/5) \\ Michel Marcus, Feb 07 2019
```

Cos(A235509) + cos(A195771) = 1.
Equals arcsin(3/5). - Michel Marcus, Feb 07 2019
Equals arctan(3/4). - Amiram Eldar, Jul 04 2023

A335418 Largest odd side of primitive triples for integer-sided triangles that have two perpendicular medians, the triples being ordered by increasing perimeter.

Original entry on oeis.org

19, 31, 41, 59, 71, 101, 109, 79, 149, 121, 139, 211, 209, 271, 191, 229, 179, 269, 341, 361, 241, 251, 419, 341, 311, 449, 319, 439, 451, 551, 389, 319, 421, 599, 541, 401, 659, 701, 509, 649, 569, 491, 781, 479, 631, 811, 671, 589, 499, 761, 929, 571, 859, 739

Offset: 1

Bernard Schott, Jun 11 2020

nonn

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).
For the corresponding primitive triples and miscellaneous properties, see A335034.
In each increasing triple (c,a,b), c is always the smallest odd side (A335036), but the largest odd side b' can be either the middle side a (A335347) or the largest side b (A335348) (see formulas and examples for explanations).
The largest odd side b' is not divisible by 3 or 5, and the odd prime factors of this odd side b' are all of the form 10*k +- 1.
The repetitions for 319, 341 ... correspond to largest odd sides for triangles with distinct perimeters (see examples).
This sequence is not increasing: a(7) = 109 for triangle with perimeter = 252 and a(8) = 79 for triangle with perimeter = 266; hence the largest odd side is not an increasing function of the perimeter of these triangles.

			The triples (257, 319, 478) and (289, 319, 562) correspond to triangles with respective perimeters equal to 1054 and 1170, so a(27) = a(32) = 319.
-> 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) = 139 = 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) = 19 = 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.

Annales Concours Général, Sujet Concours Général 2007.
Maths Challenge, Perpendicular medians, Problem with picture.
Index to sequences related to Olympiads and other Mathematical competitions.

Cf. A335034 (primitive triples), A335035 (corresponding perimeters), A335036 (smallest side), A335347 (middle side), A335348 (largest side), A335273 (even side), this sequence (largest odd side).

mycmp(x, y) = {my(xp = vecsum(x), yp = vecsum(y)); if (xp!=yp, return (xp-yp)); return (x[1] - y[1]); }
triples(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); } \\ A335034
lista(nn) = my(w=triples(nn)); vector(#w, k, vecmax(select(x->(x%2), w[k]))); \\ Michel Marcus, Jun 11 2020

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

More terms from Michel Marcus, Jun 11 2020

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A335893 Primitive triples for integer-sided triangles whose angles A < B < C are in arithmetic progression.

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

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A335036 Smallest side c of the primitive triples (c,a,b) for integer triangles that have two perpendicular medians, ordered by increasing perimeter.

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A335273 Even side of primitive triples for integer-sided triangles that have two perpendicular medians, the triples being ordered by increasing perimeter.

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A335347 Middle side of primitive triples for integer-sided triangles that have two perpendicular medians, the triples being ordered by increasing perimeter.

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

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A235509 Decimal expansion of arccos(4/5).

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