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Bill Daly
The problem of constructing integer-sided triangles with one angle equal to k times another angle can be solved completely as follows.
The Chebyshev polynomials of the second kind are defined by the recursion:
U[0] = 1
U[1] = 2x
U[k+2] = 2xU[k+1] - U[k]
These have the geometric interpretation that if x = cos(t), then U[k-1] = sin(kt)/sin(t).

If the sides of the triangle (a,b,c) = (1,U[k-1],U[k]), then the angle opposite b will be k times the angle opposite a, a fact easily derived from the Law of Sines.

If we let x be any rational number satisfying cos(Pi/(k+1)) < x < 1, then the sides of the triangle (1,U[k-1],U[k]) will be rational numbers satisfying the triangle inequalities. If the denominator of 2x is d, then (d^k,U[k-1]*d^k,U[k]*d^k) will be an integer-sided triangle. The smallest integer-sided triangle for a given k will be that corresponding to x = 1-1/2n, where n is the smallest integer such that x > cos(Pi/(k+1)). For k = 2 to 7, these are
k      x       a       b       c
2    3/4       4       6       5
3    3/4       8      10       3
4    5/6      81     105      31
5    7/8    1024    1220     231
6  11/12   46656   72930   30421
7  13/14  823543 1024303  220597

Suppose the measure of the smaller of the angle pair is A.  Then the three angles are A, B=2A, and C=180-3A.  (Using degrees because I feel like it at the moment).  As the page says, cos A and cos B must be rational numbers (this will imply that cos C is rational).

A few facts, randomly arranged: cos B = cos(2A) = 2cos2(A)-1  thus, cos B is guaranteed to be rational if cos A is. For cos(B) to be >0, cos(A) must be > 1/sqrt(2).  (i.e., A must be < 45 degrees -- this is obvious).  The simplest fraction greater than 1/sqrt(2) (== .707) is 3/4 = .75; this would imply that cos B = 1/8.

So we have: cos A = 3/4, cos B = 1/8 therefore
sin A = sqrt(1-(3/4)^2) = sqrt(7/16) = sqrt(7)/4 and
sin B = sqrt(1/(1/8)^2) = sqrt(63/64) = 3 sqrt(7)/8
cos C = cos(180-(A+B)) = -cos(A+B) = sin A sin B - cos A cos B = 21/32-3/32 = 18/32 = 9/16 thus
sin C = sqrt(1-(9/16)^2) = sqrt(175/256) = 5 sqrt(7)/16

Now we start kicking in with the law of sines:
b/a = sin(B)/sin(A); b/a = (3/8)/(1/4) = 3/2  likewise,
c/a = sin(C)/sin(A) = (5/16)/(1/4) = 5/4
So if we let a=4, we get b=6 and c=5, and the 4-5-6 triangle is a solution.

But why did we say cos(B) had to be > 0?  If we take cos(A) = 1/2, then A will be 60 degrees and B will be 120 degrees, so there's no
triangle.  This says that A _must_ be less than 60 degrees (i.e., cos(A) > 1/2), but we could try cos(A) = 2/3:

cos(B) = 2cos^2(A)-1 = -1/9
sin(A) = sqrt(1-(2/3)^2) = sqrt(5/9) = sqrt(5)/3
sin(B) = sqrt(1-(1/9)^2) = sqrt(80/81) = 4 sqrt(5)/9
cos(C) = sin(A)sin(B)-cos(A)cos(B) = 20/27+2/27 = 22/27
sin(C) = sqrt(1-(22/27)^2) = sqrt(245)/27 = 7 sqrt(5)/27

again, kick in with the law of sines:
b/a = sin(B)/sin(A) = (4/9)/(1/3) = 4/3
c/a = sin(C)/sin(A) = (7/27)/(1/3) = 7/9

letting a=9 gives us b=12 and c=7, so a 7-9-12 triangle is a second (not-quite-minimal) solution.  A very cute problem...

Just for laughs, let's try and solve the 3x problem.  Again, assume that cos(B) > 0 -- this should insure that the triangle is acute and allow for a little more 'minimalness'.  This implies that B = 3A is less than 90 degrees, or that A < 30; in other words, cos(A) is greater than sqrt(3)/2 ~= .866.  The simplest value possible here is cos(A) = 7/8 = .875; then
cos(B) = cos(A)*(4cos^2(A)-3) = 7/8*(49/16-3) = 7/128 and
cos(C) = -cos(4A) = -cos(2(2A));
cos(2A) = 2cos^2(A)-1 = (49/32)-1 = 17/32 and so
cos(4A) = 2cos^2(2A)-1 = 2*(17/32)^2-1 = 289/512-1 = -223/512, implying
cos(C) = 223/512.  Plugging through:
sin(A) = sqrt(1-(7/8)^2) = sqrt(15)/8
sin(B) = sqrt(1-(7/128)^2) = 33*sqrt(15)/128
sin(C) = sqrt(1-(223/512)^2) = 119*sqrt(15)/512

which implies b/a = 33/16, c/a = 119/64, and a triangle of 64-119-132.

Plugging through the same arithmetic, no longer assuming cos(B)>0 but instead just assuming that A+B = 4A is less than 180 degrees (i.e., A < 45, cos(A) > .707), we may as well try the cos(A) = 3/4 solution again:
cos(B) = 3/4*(9/4-3) = -9/16
cos(2A) = 1/8
cos(4A) = 1/32-1 = -31/32, so cos(C) = 31/32
sin(A) = sqrt(7)/4
sin(B) = sqrt(1-(9/16)^2) = 5 sqrt(7)/16
sin(C) = sqrt(1-(31/32)^2) = 3 sqrt(7)/32  so
b/a = 5/4, c/a = 3/8, and we get a 3-8-10 triangle as a solution.

It's obvious that the technique can be extended to find some integer solution easily enough, but finding a minimal integer solution strikes me as being a much trickier problem.

A little bit more basework for the general problem: firstly,  Chebyshev polynomials, cos(nA) = T_n(cos(A)).
T_2(x) = 2x2-1
T_3(x) = 4x3-3x,
T_4(x) = 8x4-8x2+1  and in general
T_n(x) = 2xT_{n-1}(x)-T_{n-2}(x).

Assume that our angle ratio is n.  Then we need to have A+B less than 180 degrees; this implies that (n+1) A < pi.  Within this region, the 'simplest' fraction (the one likely to produce the smallest sides) is the one with the smallest even denominator (the denominator being even means that the factor of 2 in the Chebyshev recurrence allows for some simplification).  So, a quick table of these angles, minimum values for cos A, and the 'simplest' fraction in that range:

n        cos(pi/(n+1))    simplest in region
4        .8090            5/6
5        .8660            7/8
6        .9010            11/12
7        .9239            13/14
The executive summary (B=nA, C=pi-(n+1)A) :
n    cos(A)    cos(B)    cos(C)    sin(A)    sin(B)    sin(C)    triangle

4    5/6       -113/     475/      1/6       35/       31/        (31,
162      486       *s(11)    162*s(11) 486*s(11)   81,
105)

5    7/8       -1673/    8143/     1/8       305/      231/       (231,
2048     8192      *s(15)    2048*s(15)8192*s(15)  1024,
1220)

6    11/12    -72863/  -540529/    1/12    12155/      30421/     (30421,
93312    559872     *s(23)  93312*s(23)559872*s(23) 46656,
72930)

7    13/14  -1461083/ -11472481/   3/14   438987/     661791/     (220597,
1647086   11529602    *s(3) 1647086*s(3)11529602*s(3) 823543,
1024303)

While there are occasional cancellations (e.g., in the n=7 case), in general the denominator of cos(B) approximately determines the size of the solution. Note that for a given (large) n, we'll have cos(pi/(n+1)) ~= 1-pi^2/(2(n+1)^2) ~= 1-pi^2/(2n^2); this means that the 'simplest' fraction larger than cos(pi/(n+1)) will have denominator ~2n^2/pi^2.  Because the Chebyshev polynomial of degree n (i.e., cos(B)) has leading term (2x)^n, this implies that the denominator of cos(B) will be ~(n^2/pi^2)^n, or (n/pi)^(2n), a superexponential growth with n.

One interesting random note, BTW: the areas of the minimal 2x triangle (4, 5, 6) and the minimal 3x triangle (3, 8, 10) are identical; both are 15*sqrt(7)/4.  This can't hold true much further, though; a triangle with integer sides a<b<c has area Omega(c) (in fact, I don't think it can get smaller than approximately c/2) so the areas grow at least as quickly as the largest side, i.e. superexponentially with the ratio of angles.  The minimal 4x triangle (31, 81, 105) has area 1085*sqrt(11)/4; the minimal 5x triangle (231, 1024, 1220) has area 70455*sqrt(15)/4; etc.

Hopefully all of the above are right; while I had Windows Calculator handy to assist me, all of the calculations were done pencil-and-paper.  Everything matched up okay (e.g., the radicals were all correct) so I have some confidence in the solutions, but I'm not 100% sure.  Let me know if there's something I missed...