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

Author

Bernard Schott, Jun 29 2020

Keywords

• nonn
• tabf

Comments

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

Examples

			(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.
		

References

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

Links

• 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.

Crossrefs

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).

Programs

Maple

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;

PARI

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