Japanese Mathematical Olympiad

Let $Q'$ be the intersection of the perpendicular bisector of $AB$ with line $BC$. Then, we have
$$\angle AQ'O_1 = \angle BQ'O_1 = 90^\circ - \angle ABC.$$
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
$$\angle AO_2O_1 = \frac{1}{2}\angle AO_2O = \angle ACO = \frac{1}{2}(180^\circ - \angle AOC) = 90^\circ - \angle ABC.$$
Hence, the four points $A, O_1, O_2, Q'$ are concyclic. Similarly, let $P'$ be the intersection of the perpendicular bisector of $AC$ with $BC$. Then, $A, O_1, O_2, P'$ are concyclic. Since triangle $ABC$

is acute, $O$ does not lie on line $BC$ and $P'$ and $Q'$ are distinct. Hence \{$P, Q$\} = \{$P', Q'$\}, and $O_3$ is the circumcenter of triangle $OP'Q'$. Now $O_1$, $O$ and $Q'$ are collinear and $O_2$, $O$ and $P'$ are collinear. Let $D$ be the intersection of $O_1O_2$ and $OO_3$. Then, we have
$$\begin{align*} \angle O_2 DO_3 &= \angle O_2 OO_3 - \angle DO_2 O \\ &= (180^\circ - \angle P' OO_3) - \angle O_1 O_2 P' \\ &= 180^\circ - \frac{1}{2}(180^\circ - \angle OO_3 P') - \angle O_1 O_2 P' \\ &= 180^\circ - (90^\circ - \angle OQ' P') - \angle O_1 Q' P' \\ &= 90^\circ, \end{align*}$$
so lines $O_1O_2$ and $OO_3$ meet at right angles. On the other hand, $O_1O_2$ is the perpendicular bisector of $AO$, so both $A$ and $O_3$ lie on the perpendicular to line $O_1O_2$ through $O$. Therefore $A$, $O$, $O_3$ are collinear, as claimed.