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A353618
Three-column array giving list of primitive triples for integer-sided triangles whose angle B = 3*C.
4
3, 10, 8, 35, 48, 27, 119, 132, 64, 112, 195, 125, 279, 280, 125, 20, 357, 343, 253, 504, 343, 539, 510, 216, 552, 665, 343, 91, 792, 729, 923, 840, 343, 533, 840, 512, 476, 1035, 729, 1455, 1288, 512, 224, 1485, 1331, 1504, 1575, 729, 17, 1740, 1728, 799
OFFSET
1,1
COMMENTS
This sequence is inspired by the 1st problem proposed during the 46th Czech and Slovak Mathematical Olympiad 1997 (see link).
The triples (a, b, c) are displayed in increasing order of side b, and if sides b coincide then in increasing order of the side c.
If in triangle ABC, B = 3*C, then the corresponding metric relations between sides are c*a^2= (b-c)^2 * (b+c). <===> a/(b-c) = sqrt(1+b/c).
This metric relation is equivalent to a = m(m^2-2k^2), b = k(m^2-k^2), c = k^3, gcd(k,m) = 1 and sqrt(2) * k < m < 2*k; hence every c is a cube number and always c < b.
When A <> 3*Pi/7 and A <> Pi/5, table below shows there exist these 3 possible configurations: c < b < a; c < a < b and a < c < b:
----------------------------------------------------------------------------
| A | Pi | decr. | 3*Pi/7 | decr. | Pi/5 | decr. | 0 |
---------------------------------------------------------------------------
| B | 0 | incr. | 3*Pi/7 | incr. | 3*Pi/5 | incr. | 3*Pi/4 |
----------------------------------------------------------------------------
| C | 0 | incr. | Pi/7 | incr. | Pi/5 | incr. | Pi/4 |
----------------------------------------------------------------------------
| < | No | c < b < a | c < b=a | c < a < b | c=a < b | a < c < b | No |
----------------------------------------------------------------------------
where 'No' means there is no such corresponding triangle.
If (A,B,C) = (3*Pi/7,3*Pi/7,Pi/7) then a = b with c = 2*a*cos(Pi/7), so isosceles ABC is not an integer-sided triangle.
If (A,B,C) = (Pi/5,3*Pi/5,Pi/5) then a = c with b = a*(1+sqrt(5))/2, so ABC is not an integer-sided triangle.
LINKS
The IMO Compendium, Problem 1, 46th Czech and Slovak Mathematical Olympiad 1997.
Mathematical Reflections, Solution to problem O467, Issue 5, 2018, p. 26.
EXAMPLE
The table begins:
3, 10, 8;
35, 48, 27;
119, 132, 64;
112, 195, 125;
279, 280, 125;
20, 357, 343;
253, 504, 343,
539, 510, 216;
................
The smallest such triangle is (3,10,8), it is of type a < c < b with 3/(10-8) = sqrt(1+10/8) = 3/2.
The 2nd triple (35, 48, 27) is of type c < a < b with 35/(48-27) = sqrt(1+48/27) = 5/3.
The 8th triple (539, 510, 216) is the first of type c < b < a with 539/(510-216) = sqrt(1+510/216) = 11/6.
MAPLE
for b from 1 to 2500 do
for q from 2 to floor((b-1)^(1/3)) do
a := (b-q^3) * sqrt(1+b/q^3);
if a= floor(a) and q^3 < b and igcd(a, b, q)=1 and (b-q^3) < a and a < b+q^3 then print(a, b, q^3); end if;
end do;
end do;
PROG
(PARI) lista(nn) = {for (b = 1, nn, for (q = 2, sqrtnint(b-1, 3), if (issquare(z=1+b/q^3), a = (b-q^3) * sqrtint(numerator(z))/sqrtint(denominator(z)); if ((q^3 < b) && (gcd([a, b, q]) == 1) && ((b-q^3) < a) && (a < b+q^3), print1([a, b, q^3], ", ")); ); ); ); } \\ Michel Marcus, May 11 2022
(Python)
from math import gcd
from itertools import count, islice
from sympy import integer_nthroot
def A353618_gen(): # generator of terms
for b in count(1):
q, c = 2, 8
while c < b:
d = (b-c)**2*(b+c)
s, t = divmod(d, c)
if t == 0:
a, r = integer_nthroot(s, 2)
if r and b-c < a < b+c and gcd(a, b, q) == 1:
yield from (a, b, c)
c += q*(3*q+3)+1
q += 1
A353618_list = list(islice(A353618_gen(), 30)) # Chai Wah Wu, May 14 2022
CROSSREFS
Cf. A335893 (A < B < C are in arithmetic progression), A343063 (B = 2*C).
Sequence in context: A280461 A373128 A222345 * A202339 A234642 A038228
KEYWORD
nonn,tabf
AUTHOR
Bernard Schott, Apr 30 2022
STATUS
approved