Asia Pacific Mathematics Olympiad (APMO)

Let $S$ be the tangent point of the circles $O$ and $O_1$ and let $T$ be the intersection point, different from $S$, of the circle $O$ and the line $SP$. Let $X$ be the tangent point of $\ell$ to $O_1$ and let $M$ be the midpoint of the line segment $XP$. Since $\angle TBP=\angle ASP$, the triangle $TBP$ is similar to the triangle $ASP$. Therefore, $$\frac{PT}{PB}=\frac{PA}{PS}$$ Since the line $\ell$ is tangent to the circle $O_1$ at $X$, we have $$\angle SPX=90^{\circ }-\angle XSP=90^{\circ }-\angle PAM=\angle PAM$$ which implies that the triangle $PAM$ is similar to the triangle $SPX$. Consequently, $$\frac{XS}{XP}=\frac{MP}{MA}=\frac{XP}{2MA}\text{and}\frac{XP}{PS}=\frac{MA}{AP}$$ From this and the above observation follows $$\begin{array}{c}\frac{XS}{XP}\cdot \frac{PT}{PB}=\frac{XP}{2MA}\cdot \frac{PA}{PS}=\frac{XP}{2MA}\cdot \frac{MA}{XP}=\frac{1}{2}.\end{array}\text{\hspace{1em}(1)}$$ Let $A^{\prime }$ be the intersection point of the circle $O$ and the perpendicular bisector of the chord $BC$ such that $A$, $A^{\prime }$ are on the same side of the line $BC$, and $N$ be the intersection point of the lines $A^{\prime }Q$ and $CT$. Since $$\angle NCQ=\angle TCB=\angle TCA=\angle TBA=\angle TBP$$ and $$\angle C{A}^{\prime }Q=\frac{\angle CAB}{2}=\frac{\angle XAP}{2}=\angle PAM=\angle SPX,$$ the triangle $NCQ$ is similar to the triangle $TBP$ and the triangle $C{A}^{\prime }Q$ is similar to the triangle $SPX$. Therefore $$\frac{QN}{QC}=\frac{PT}{PB}\text{and}\frac{QC}{Q{A}^{\prime }}=\frac{XS}{XP}$$ and hence $Q{A}^{\prime }=2QN$ by (1). This implies that $N$ is the midpoint of the line segment $Q{A}^{\prime }$. Let the circle $O_2$ touch the line segment $AC$ at $Y$. Since $$\angle ACN=\angle ACT=\angle BCT=\angle QCN$$ and $|CY|=|CQ|$, the triangles $YCN$ and $QCN$ are congruent and hence $NY\perp AC$ and $NY=NQ=N{A}^{\prime }$. Therefore, $N$ is the center of the circle $O_2$, which completes the proof.