Complicated Tangency of a system of Circles!

This is rather a tedious solution, hope to find some nicer one: Joining the centres of each of the small circles with the centre of the big circle we form 9 angles around the centre of the big circle. 8 of these angles are equal to, say, $\theta$ , and the one inside the triangle made by the distance wanted ( $d$ ) should be $360˚-8\theta$ . Since all $\theta$ 's are inside triangles of sides 3/2, 3/2 and 1, we can use the cosine theorem to find $\theta$ to satisfy

$1^2=\left(\frac{3}{2}\right)^2+\left(\frac{3}{2}\right)^2-2\left(\frac{3}{2}\right)\left(\frac{3}{2}\right)\cos\theta$ $\Longrightarrow \cos\theta=\frac{7}{9}$

Since the remaining angle is inside a triangle of sides 3/2, 3/2 and ( $d$ ), we can use the cosine theorem again to write

$d^2=\left(\frac{3}{2}\right)^2+\left(\frac{3}{2}\right)^2-2\left(\frac{3}{2}\right)\left(\frac{3}{2}\right)\cos(360˚-8\theta)$ $\Longrightarrow d^2=\frac{9}{2}(1-\cos8\theta)$

Now, applying the double-angle formula for cosine three times we get

$\cos8\theta=128\cos^8(\theta) - 256\cos^6(\theta) + 160\cos^4(\theta) - 32\cos^2(\theta) + 1.$

Substituting this and simplifying, we get

$d=\frac{1904\sqrt{2}}{2187}$

The desired distance BC can be found using the triangle $\Delta OMC, OC = \frac{3}{2}$ , and $MC = OC \sin4\theta$ $\cos \theta = \frac{\left( \frac{3}{2}\right) ^2 + \left( \frac{3}{2}\right) ^2 - 1^2}{2 \left( \frac{3}{2}\right) \left( \frac{3}{2}\right) } = \frac{7}{9}; \sin \theta = \frac{4 \sqrt{2}}{9}$ $\sin 2 \theta = 2 \sin \theta \cos \theta = \frac {56 \sqrt{2}}{81}; \cos 2\theta = \frac{17}{81}$ $\sin 4 \theta = 2 \sin 2\theta \cos 2\theta = \frac{1904 \sqrt{2}}{81^2}$ $BC = 2 MC = 2 \left( \frac{3}{2} \right) \sin 4 \theta = \frac{1902 \sqrt{2}}{2187}$