Tangents in Harmonic Progression Here is a problem I am working on:
In triangle ABC, $\tan(A)$, $\tan(B)$, and $\tan(C)$ form a harmonic progression. Also, $BC=189$ and $AB=459$. Find AC.
My progress so far:
Because tanA, tanB, and tanC are in harmonic progression, $\tan(B)=\frac{(2\tan(A)\cdot\tan(C))}{(\tan(A)+\tan(C))}$. Also, we can compute AC using law of Cosines if we can find $\cos(B)$.
However, I am stuck from here. Any tips?

We know that $\cot C,\cot B,\cot A$ are in arithmetic progression, so $\tan A,\tan B,\tan C$ are $\frac{1}{x+y},\frac{1}{x},\frac{1}{x-y}$ and since $\tan A+\tan B+\tan C=\tan A\tan B\tan C$ must hold, we must have
$$ 3x^2-y^2 = 1,$$
so:
$$\cot A = x+\sqrt{3x^2-1},\qquad\cot B = x,\qquad \cot C=x-\sqrt{3x^2-1}$$
and since $\frac{1}{\sin^2 z}=\cot^2 z+1$ we have:
$$\frac{1}{\sin^2 A}=4x^2+2x\sqrt{3x^2-1},\qquad \frac{1}{\sin^2 B}=x^2+1,\qquad \frac{1}{\sin^2 C}=4x^2-2x\sqrt{3x^2-1},$$
but since we know the ratio between $BC$ and $AB$ we also know the ratio between $\sin^2 A$ and $\sin^2 C$. From that, we can compute $x$, then $\sin B$, then $AC$.
