The 35th Japanese Mathematical Olympiad

Let $Q^{\prime }$ be the intersection of the perpendicular bisector of $AB$ with line $BC$. Then, we have $$\angle A{Q}^{\prime }{O}_1=\angle B{Q}^{\prime }{O}_1=90^{\circ }-\angle ABC.$$ Moreover, since $O_1$ and $O_2$ lie on the perpendicular bisector of segment $AO$ and $O_2$ is the circumcenter of triangle $ACO$, we have $$\angle A{O}_2{O}_1=\frac{1}{2}\angle A{O}_{2}O=\angle ACO=\frac{1}{2}(180^{\circ }-\angle AOC)=90^{\circ }-\angle ABC.$$ Hence, the four points $A,O_1,O_2,Q^{\prime }$ are concyclic. Similarly, let $P^{\prime }$ be the intersection of the perpendicular bisector of $AC$ with $BC$. Then, $A,O_1,O_2,P^{\prime }$ are concyclic. Since triangle $ABC$

is acute, $O$ does not lie on line $BC$ and $P^{\prime }$ and $Q^{\prime }$ are distinct. Hence \{$P,Q$\} = \{$P^{\prime },Q^{\prime }$\}, and $O_3$ is the circumcenter of triangle $O{P}^{\prime }{Q}^{\prime }$. Now $O_1$, $O$ and $Q^{\prime }$ are collinear and $O_2$, $O$ and $P^{\prime }$ are collinear. Let $D$ be the intersection of $O_1{O}_2$ and $O{O}_3$. Then, we have $$\begin{array}{rl}\angle O_{2}D{O}_3& =\angle O_{2}O{O}_3-\angle D{O}_{2}O\\ & =(180^{\circ }-\angle P^{\prime }O{O}_3)-\angle O_1{O}_2{P}^{\prime }\\ & =180^{\circ }-\frac{1}{2}(180^{\circ }-\angle O{O}_3{P}^{\prime })-\angle O_1{O}_2{P}^{\prime }\\ & =180^{\circ }-(90^{\circ }-\angle O{Q}^{\prime }{P}^{\prime })-\angle O_1{Q}^{\prime }{P}^{\prime }\\ & =90^{\circ },\end{array}$$ so lines $O_1{O}_2$ and $O{O}_3$ meet at right angles. On the other hand, $O_1{O}_2$ is the perpendicular bisector of $AO$, so both $A$ and $O_3$ lie on the perpendicular to line $O_1{O}_2$ through $O$. Therefore $A$, $O$, $O_3$ are collinear, as claimed.