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 5 and 12, 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 $BC$.
Answer. a) 7; b) $R=\frac{25}{4}; BC=5\sqrt{6}$.

Solution. Let $\angle ACB = \gamma$, then $\angle BAC = 2\gamma$. By the generalized sine theorem, we find that $AB = 2R \sin \gamma, BC = 2R \sin 2\gamma$. 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 $AB$ is $\gamma$, and the angle between the tangent and $BC$ is $2\gamma$. Therefore, the distance $d_A$ from point $A$ to the tangent is $AB \sin \gamma = 2R \sin^2 \gamma$. Similarly, the distance $d_C$ from point $C$ to the tangent is $BC \sin 2\gamma = 2R \sin^2 2\gamma$.

Thus, $\frac{12}{5} = \frac{d_C}{d_A} = 4 \cos^2 \gamma$, from which $\cos^2 \gamma = \frac{3}{5}$. Then $\sin^2 \gamma = \frac{2}{5}, R = \frac{d_A}{2 \sin^2 \gamma} = \frac{25}{4}$.

Therefore, $AB = 2R \sin \gamma = \frac{16}{\sqrt{7}}, BC = 4R \sin \gamma \cos \gamma = 5\sqrt{6}$.

The distance from point $A$ to line $BC$ is

$$
AB \cdot \sin \angle ABC = AB \sin 3\gamma = AB \left(3 \sin \gamma - 4 \sin^3 \gamma\right) = 7
$$