China National Team Selection Test

As shown in Fig. 1, draw the circumcircles of $△ABC$, $△ABP$ and $△ACP$, respectively. Let the extension of $AP$ meet $\odot O$ at $D$, join $BD$, and draw the line tangent to $\odot O$ at $A$, intersecting $\odot O_1$ and $\odot O_2$ at $E$ and $F$ respectively.

It is clear that $△AMC\sim △ABD$, hence $$\frac{AB}{BD}=\frac{AM}{MC}.$$ Since $△EAB\sim △PDB$, we have $\frac{AB}{BD}=\frac{AE}{PD}$. Consequently, $\frac{AM}{MC}=\frac{AE}{PD}$, i.e. $$AE=\frac{AM\times PD}{MC},$$ and, similarly, $$AF=\frac{AM\times PD}{MB}.$$ Fig. 1

It follows that $$AE=AF.①$$

Draw the perpendicular lines $O_1{E}^{\prime }\perp AE$ with foot $E^{\prime }$, and $O_2{F}^{\prime }\perp AF$ with foot $F^{\prime }$. Since $E^{\prime }$, $F^{\prime }$ are the midpoints of $AE$, $AF$ respectively, it follows from ① that $A$ is the midpoint of $E^{\prime }{F}^{\prime }$. In the right-angled trapezoid $O_1{E}^{\prime }{F}^{\prime }{O}_2$, $AO$ is the extension of the median, and hence it bisects the segment $O_1{O}_2$.

Solution 2:

As shown in Fig. 2, draw segments $A{O}_1$, $O{O}_1$, $A{O}_2$, $O{O}_2$. Denote by $Q$ the intersection of $AO$ and $O_1{O}_2$. Then Fig. 2 $$\frac{O_{1}Q}{Q{O}_2}=\frac{S_{△O{O}_1}}{S_{△O{O}_2}}=\frac{AB\times O{O}_1}{AC\times O{O}_2},$$ where $\frac{AB}{AC}=\frac{\sin \angle ACB}{\sin \angle ABC}$. Since $\angle O{O}_{1}Q=\angle BAP=\angle CAM$, and $\angle O{O}_{2}Q=\angle CAP=\angle BAM$, it follows that $$\begin{array}{rl}\frac{O{O}_1}{O{O}_2}& =\frac{O{O}_1}{OQ}\times \frac{OQ}{O{O}_2}\\ & =\frac{\sin \angle OQ{O}_1}{\sin \angle O{O}_{1}Q}\times \frac{\sin \angle O{O}_{2}Q}{\sin \angle OQ{O}_2}\\ & =\frac{\sin \angle O{O}_{2}Q}{\sin \angle O{O}_{1}Q}=\frac{\sin \angle BAM}{\sin \angle CAM},\end{array}$$ and thus $$\frac{O_{1}Q}{Q{O}_2}=\frac{\sin \angle ACM}{\sin \angle CAM}\times \frac{\sin \angle BAM}{\sin \angle ABM}=\frac{AM}{CM}\times \frac{BM}{AM}=\frac{BM}{CM}.$$ Note that $M$ is the midpoint of $BC$, and therefore $O_{1}Q=Q{O}_2$, i.e. line $AO$ bisects segment $O_1{O}_2$.