Hello Ed!
Alan Lemm wondered whether every integer-sided triangle, with one angle = k
times another angle, always has one side which is a power of k. Nice
problem - I thought I'd write and show you an easy way of seeing that this
is always the case.
Define the polynomials:
p_0 = 1
p_1 = x
p_2 = x^2 - 1
p_3 = x^3 - 2x
p_{n+1} = x.p_n - p_{n-1} (hope this makes sense)
Then a triangle with sides
( 1 , p_{n-1} , p_n )
has angles t and nt, where x = 2cos(t).
[To work this out, I applied the Sine Rule to all 3 angles, and used trig
identities]
These polynomials are pretty much the Chebyshev polynomials (the indices are
shifted by 1, and x is scaled by a factor of 2). Doing it this way makes
the number-theory bit easier.
Suppose we have an integer-sided triangle with angles t and nt. Then the
Cosine Rule tells us that cos(t) is a rational number, and so x = 2cos(t) is
also rational.
Write x = r / s in lowest terms (r and s have no common factor). The sides
of the triangle are proportional to ( 1, p_{n-1}(x), p_n(x) ) so we just
need to multiply by some factor to clear out the denominators.
Now each p_n is of degree n, with integer coefficients, and with the highest
coefficient equal to 1 (easy proof by induction). Therefore, multiplying by
s^n will be enough to clear out the fractions.
This gives us the triple ( s^n , s^n.p_{n-1}(r/s), s^n p_n(r/s) ) of
integers, which is proportional to the sides of the triangle we started
with. We can already see that the first number is a power of n, which is
what we want. The crucial question is: are there any common factors that we
might be able to divide by?
In fact, the first and third numbers of the triple are coprime: The third
number is equal to
r^n + s.(some integer).
So, for any prime factor p of s, the third number fails to be a multiple of
p. This shows that the first and third numbers are coprime, and so there
are no common factors we could divide by. This triple of numbers must
therefore be the same as the sides of the original triangle we started with.
So, provided I haven't made any mistakes, /every/ integer-sided triangle
with angles t and nt must have a side that is a perfect power of n (in fact,
the side opposite the angle t). This also gives a general formula for
churning out lots of these triples.
I don't normally feel motivated to actually write to authors of Web-sites,
but I really enjoy thinking about your questions, and when I came up with
this solution, I thought I should share it with you. Cheers for the
excellent site!
All the best,
Lambros Lambrou
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I seem to remember starting up this question of integer triangles with
integer angle ratios a long time ago.
Yes, you can parameterize all solutions, giving one side as a power of n.
Law of Sines gives: b/a = sin(NA)/sin(A) and c/a = sin((N+1)A)/sin(A)
Each of these are polynomials of 2cos(A), with power N-1 and N respectively.
Call them P() and Q().
By the Law of Cosines, we know that 2cos(A) is rational, so let it be = p/q.
[a:b:c] = [1 : P(p/q) : Q(p/q)] = [q^n : P'(p,q) : Q'(p,q)]
Examples:
sin(2A)/sin(A) = 2cos(A)
sin(3A)/sin(A) = 4cos(A)^2-1
sin(4A)/sin(A) = 8cos(A)^3 - 4cos(A)
sin(5A)/sin(A) = 16cos(A)^4 - 12cos(A)^2 + 1
sin(6A)/sin(A) = (4cos(A)^2-1)((4cos(A)^2-3)(2cos(A))
sin(7A)/sin(A) = 4cos(A)^2(4cos(A)^2-2)((4cos(A)^2-3) - 1
Giving the following parameterizations for N=2 through 6:
2x: q^2, pq, p^2-q^2
3x: q^3, q(p^2-q^2), p(p^2-2q^2)
4x: q^4, pq(p^2-2), p^4-3p^2q^2+q^4
5x: q^5, q(p^4-3p^2q^2+q^4), p(p^2-q^2)(p^2-3q^2)
6x: q^6, pq(p^2-q^2)(p^2-3q^2), p^2(p^2-2q^2)(p^2-3q^2) - q^6
To find the "smallest" solution, find the lowest p and q such that:
2> p/q > 2cos(pi/(N+1)), or the lowest integer q>1/2cos(pi/(N+1)).
2: (3,2) gives (4,6,5)
3: (3,2) gives (8,10,3)
4: (5,3) gives (81, 105, 31)
5: (7,4) gives (1024, 1220, 231)
6: (11,6) gives (46656, 72930, 30421)
A nice variation is to also require that the third angle be obtuse!
This translates into finding the lowest integer q>1/2cos(pi/2(N+1)).
2: (7,4) gives (16,28,33)
3: (13,7) gives (343, 840, 923)
4: (21,11) gives (14641,45969,49039)
5: (29,15) gives (759375,2853465,2965424)
6: (39,20) gives (64000000,280675980,288021761)
etc....
Regards,
Nick
I just stumbled across your page in which Alan Lemm asks whether, in an
integer-sided triangle in which one angle is an integer multiple of
another, one side must be a power of an integer. The answer is yes, if you
consider only primitive triangles (those with sides that have no common
divisor greater than 1).
In a paper in The Fibonacci Quarterly, vol 29, Feb 1991, pp 3-6, Joe
Carroll and I proved that primitive integer triangles with two sides in the
ratio j:k must have sides equal to
s^k*P[j](r,s) and s^j*P[k](r,s) and P[j+k](r,s)
where r and s are suitable integers and P[1](r,s)=1 and, inductively,
P[n+1](r,s)=r*P[n](r,s)-s^2*P[n-1](r,s).
If you take k=1 you get that the sides must be
s*P[j](r,s) and s^j and P[j+1](r,s).
Ken Yanosko
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I found the page at:
www.mathpuzzle.com/Chebychev.html
I should clarify that my discovery involved reduced solutions, meaning that the three sides had 1 as the greatest common factor. Obviously, if (4,5,6) is a solution for k=2, then so is (8,10,12), but the second case does not have a square number for a side.
Alan Lemm
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