4. In triangle $ABC$, the angle at vertex $A$ is twice the angle at vertex $C$. A tangent line $\ell$ to the circumcircle $\Omega$ of triangle $ABC$ is drawn through vertex $B$. The distances from points $A$ and $C$ to this tangent line are 4 and 9, respectively.
a) Find the distance from point $A$ to the line $BC$.

b) Find the radius of the circumcircle $\Omega$ and the length of side $AB$.
Answer. a) $5 ;$ b) $R=\frac{32}{7} ; A B=\frac{16}{\sqrt{7}}$.

Solution. Let $\angle A C B=\gamma$, then $\angle B A C=2 \gamma$. By the generalized sine theorem, we find that $A B=2 R \sin \gamma, B C=2 R \sin 2 \gamma$. Note that $\gamma<90^{\circ}$ (otherwise the sum of the angles in the triangle would be greater than $180^{\circ}$ ).

By the theorem on the angle between a tangent and a chord, the angle between the tangent and $A B$ is $\gamma$, and the angle between the tangent and $B C$ is $2 \gamma$. Therefore, the distance $d_{A}$ from point $A$ to the tangent is $A B \sin \gamma=$ $2 R \sin ^{2} \gamma$. Similarly, the distance $d_{C}$ from point $C$ to the tangent is $B C \sin 2 \gamma=2 R \sin ^{2} 2 \gamma$.

Thus, $\frac{9}{4}=\frac{d_{C}}{d_{A}}=4 \cos ^{2} \gamma$, from which $\cos ^{2} \gamma=\frac{9}{16}$. Then $\sin ^{2} \gamma=\frac{7}{16}, R=\frac{d_{A}}{2 \sin ^{2} \gamma}=\frac{32}{7}$.

Therefore, $A B=2 R \sin \gamma=\frac{16}{\sqrt{7}}$.

The distance from point $A$ to line $B C$ is

$$
A B \cdot \sin \angle A B C=A B \sin 3 \gamma=A B\left(3 \sin \gamma-4 \sin ^{3} \gamma\right)=5
$$