Bertie

Successive iterates are 1, 2/3, 3/4, 8/11, 11/15, 30/41, 41/56, 112/153, 153/209, 418/571, 571/780 ...

They are converging to sqrt(3) - 1 = 0.7320508... .

The last one that I've listed, 571/780 = 0.732051 6dp.

Put the expression equal to N (an integer) and square it.

That gets you N^2 =  t- N  or, t = N^2 + N.

The largest integer value for N that can be substituted so that t is still a 4 digit numbe is 99 and that gets

you t = 9900.

Put the continued fraction equal to x, and make the substitution further into the fraction to get

$$x=\frac{1}{1+\frac{1}{2+x}}}$$  

Tidying that up gets you quadratic $$x^{2}+2x-2=0,$$  from which $$x=\sqrt{3}-1.$$

Let xy=k, so that y = k/x.

Substitute this into the first equation to get   $$x^{4}-20x^{2}-7k=0.$$

Solve this (using the usual formula) as a quadratic in x squared to get $$x^{2}=10\pm\sqrt{100+7k}$$ .

The original equations are symmetric in x and y so repeating the procedure will produce an identical result for y.

Since x and y are different, one will have the positive sign in the middle, the other the negative sign.

Multiplying the two results produces (difference between two squares)

$$x^{2}y^{2}=100-(100+7k)=-7xy.$$

Since $$xy\ne 0,$$  it follows that xy = -7.

Write tan4x as sin4x/cos4x and use the Maclaurin series for these together with the series for exp(x). An x cancels top and bottom.

Label the four small angles at C as stated earlier, - use the sine rule, a/sinA=b/sinB etc, (remember that sin(90-P)=cosP) in the triangles ACD and ACB, - divide one equation by the other to remove the common sides (AC  and AD, AB(=2AD)), - cross multiply, - then use first the identity for sin(2P), and then the identity for sin(P+Q). That gets you to the equation I stated in my earlier post.

Post again if you have any problems.

The only method that I can see uses trig.

I called each of the four small angles at C, $$\alpha$$ , and after five (easy) lines arrived at

$$\sin2\alpha\cos4\alpha=0,$$

implying that $$4\alpha=90$$ , as required.

Hi Melody

Your second category, two letters the same, covers the possibility that two letters are the same and that the other two letters are different and different from each other. In other words, it doesn't allow for the possibility that your four letters consist of just two letters forming two pairs. This produces another 1950 combinations. You have to be quite careful as to how you calculate this, it is potentially confusing and it's very easy to overcount.

Hi Melody

Your missing combinations are in your 'two the same' section. You also have to include contributions from two the same and the other two the same as well, (and you have to be careful as to the calculation of that number).

Cosine rule,

a^2 = b^2 + c^2 - 2bc*cos(A).