[ Touching Circles $\quad$ [

Two circles touch each other externally at point $A$, and a third circle at points $B$ and $C$. The extension of chord $A B$ of the first circle intersects the second circle at point $D$, the extension of chord $A C$ intersects the first circle at point $E$, and the extensions of chords $B E$ and $C D$ intersect the third circle at points $F$ and $G$ respectively. Find $B C$, if $B F=12$ and $B G=15$.

#
Let $S 1, S 2$ and $S 3$ be the first, second, and third circles, respectively. Draw the common tangents $l_{a}, l_{b}, l_{c}$ through points $A, B$ and $C$ to the circles $S 1$ and $S 2, S 1$ and $S 3, S 2$ and $S 3$ respectively. Then the tangents $l_{a}$ and $l_{b}$ form equal angles with the chord $A B$. Denote these angles by $\gamma$. Similarly, the equal angles formed by the tangents $l_{a}$ and $l_{c}$ with the chord $A C$ are denoted by $\beta$, and the equal angles formed by the tangents $l_{b}$ and $l_{c}$ with the chord $B C$ are denoted by $\alpha$. The sum $2 \alpha+2 \beta+2 \gamma$ is the sum of the angles of triangle $A B C$, so $\alpha+\beta+\gamma=90^{\circ}$. On the tangent $I_{a}$, mark point $P$ inside angle $D A E$ and point $Q$ inside angle $B A C$. From the theorem about the angle between a tangent and a chord, it follows that

$$
\angle B E C=\angle B E A=\angle B A Q=\angle P A D=\angle A C D=\angle E C D
$$

Thus, $B E \| C D$, and since

$$
\angle B C D=\angle A C B+\angle A C D=\alpha+\beta+\gamma=90^{\circ}
$$

then $\angle B C G=90^{\circ}$, so quadrilateral $B C G F$ is a rectangle. Therefore,

$$
B C=\sqrt{B G^{2}-C G^{2}}=\sqrt{B G^{2}-B F^{2}}=\sqrt{15^{2}-12^{2^{2}}}=9
$$

## Answer

9.