Japan Mathematical Olympiad

If $AB = AC$, then both of the points $P, Q$ lie on the bisector of the angle $\angle BAC$, and we get $\angle PAO = \angle QAO = 0^\circ$; so the assertion is valid in this case.

Let us assume in the sequel that $AB \ne AC$. Let $\gamma$ be the circle with $O$ as its center and tangent to the line $BC$ at the point $P$ and tangent to the circle $\Gamma$ at the point $Q$. We also denote by $P'$ ($M$, respectively) the point of intersection of the perpendicular bisector of the line segment $BC$ and the circle $\Gamma$ and lying on the same side (opposite side, respectively) as the point $A$ with respect to the line $BC$. Then, since the arc $BM$ has the same length as the arc $MC$, we have $\angle BP'M = \angle CP'M$. By the assumption for the problem, we also have $\angle BAO = \angle CAO$, and therefore, the arcs of the circle $\Gamma$ which subtend these angles must have the same length. This means that the three points $A, O$ and $M$ must be collinear. We also see that if we denote by $O'$ the center of the circle $\Gamma$, then the three points $Q, O$ and $O'$ are also collinear, since the circles $\Gamma$ and $\gamma$ are tangent at the point $Q$.

Next, we show that the three points $P', P$ and $Q$ are also collinear. Let us denote by $R$ the point of intersection of the line segment $P'Q$ and the circle $\gamma$. Then we see that
$$\angle MP'R = \angle O'P'Q = \angle O'QR = \angle ORQ.$$
Therefore, the line segment $OR$ must be parallel to the line segment $MP'$, and since $MP' \perp BC$, we must have $OR \perp BC$. This means that the point $R$ on $\gamma$ must coincide with the point $P$ of the tangency of $\gamma$ with $BC$. Thus, we see that the three points $P', P$ and $Q$ are collinear. Furthermore, we see that $\angle QP'M = \angle PQO = \angle QPO$ hold. We also have $\angle QAO = \angle QP'M$, as they are subtended by the same arc $QM$ of the circle $\Gamma$, and therefore, $\angle QAO = \angle QPO$. It then follows

that the four points A, P, O, Q must lie on the same circle, which in turn implies that $\angle PQO = \angle PAO$, as these two angles are subtended by the same arc PO of that circle. As $\angle PQO = \angle QPO = \angle QAO$, as we saw above, we finally obtain that $\angle PAO = \angle QAO$, proving the assertion of the problem.