, 2020

From $\angle RP_1S = 120^\circ = 180^\circ - \angle SAR$ we know $A, S, P_1, R$ lie on a common circle $\Gamma_1$. From $\angle RBA = \angle SCA$ we know $\overline{AR} = \overline{SB}$, and similarly $\overline{RC} = \overline{AS}$, so it is easy to see that $\overline{OR} = \overline{OS}$. Therefore, since $AO$ is the internal angle bisector of $\angle RAS$, we obtain $O \in \Gamma_1$.

Let $M$ be the midpoint of $\overline{BC}$. Clearly,
$$\frac{AB}{BM} = \frac{AC}{CM} = \frac{AO}{OM} = 2,$$
so $\odot(BOC)$ is the $A, M$-Apollonius circle, hence $P_1O, P_2O$ bisect $\angle AP_1M, \angle AP_2M$ respectively.

Let $V_1$ be the intersection point of $\Gamma$ and $\odot(ABC)$ other than $A$. Notice that
$$\frac{RV_1}{V_1S} = \frac{CR}{SB} = \frac{AS}{RA},$$
it is easy to obtain that $ARSV_1$ is an isosceles trapezoid with $AV_1 \parallel RS$.

Since $\triangle OAV_1$ is an isosceles triangle, we know
$$\angle MP_1O = \angle OP_1A = \angle OV_1A = \angle V_1AO = 180^\circ - \angle V_1PO,$$
therefore $M, P_1, V_1$ are collinear.

By Menelaus' theorem,
$$\frac{RQ_1}{Q_1S} = \frac{RP_1}{P_1B} \cdot \frac{BA}{AS} = \frac{RP_1}{P_1B} \cdot \frac{CB}{CR} = \frac{\sin \angle ACS}{\sin \angle SCB} = \frac{AS}{SB} = \frac{RV_1}{V_1S},$$
so $V_1Q_1$ is the internal angle bisector of $\angle RV_1S$, or $V_1, Q_1, O$ are collinear.

Defining $V_2$ analogously, we likewise have $M, P_2, V_2$ and $V_2, Q_2, O$ collinear respectively.

Finally, combining the above results,
$$\begin{align*} 2 \angle Q_2UQ_1 + \angle Q_2OQ_1 &= 2 (360^\circ - \angle Q_1UO - \angle OUQ_2) + (\angle AOV_1 - \angle AOV_2) \\ &= 720^\circ - 2 (180^\circ - \angle OP_1A) - 2 \angle OP_2A + \angle AP_1V_1 - \angle AP_2V_2 \\ &= 360^\circ + (\angle MP_1A + \angle AP_1V_1) - (\angle MP_2A + \angle AP_2V_2) \\ &= 360^\circ + 180^\circ - 180^\circ = 360^\circ. \end{align*}$$