Lemma: Let $A$ , $B$ , $C$ , and $D$ be points on a circle, and let $\overline{AB}$ and $\overline{CD}$ intersect at $X$ . Then, $\tfrac{CX}{XD}=\tfrac{CA}{AD} \cdot \tfrac{CB}{BD}$ .

Proof: We have
$\frac{CX}{XD}=\frac{[ABC]}{[ABD]}=\frac{\frac{1}{2} CA \cdot CB \cdot \sin \angle ACB}{\frac{1}{2} DA \cdot DB \cdot \sin \angle ADB}=\frac{CA}{AD} \cdot \frac{CB}{BD},$ as desired. $\square$

Let $\overline{AQ}$ and $\overline{BC}$ intersect at $Z$ . Notice that $QBX \sim QCY$ , so $\tfrac{QB}{QC}=\tfrac{BX}{CY}$ . By the lemma, we have
$\frac{ZB}{ZC}=\frac{BA}{AC} \cdot \frac{BQ}{QC}=\frac{AB}{AC} \cdot \frac{BX}{CY}=\frac{AB}{AC} \cdot \frac{2BO_1 \cdot \cos \angle O_1BA}{2CO_2 \cdot \cos \angle O_2CA}=\frac{AB}{AC} \cdot \frac{BO_1}{CO_2} \cdot \frac{\sin \angle ABC}{\sin \angle ACB}=\frac{BO_1}{CO_2}.$ Thus, we have $ZBO_1 \sim ZCO_2$ , so $Z$ , $O_1$ , and $O_2$ are collinear, as desired. $\square$

Remark: This problem has 3 main components: the circles and intersections, the definition of $Q$ , and the definition of $Z$ . These can be dealt with mostly independently of each other: using the lemma, length relations involving $Z$ decompose to those involving $Q$ , which decompose by spiral similarity to those involving $X$ and $Y$ , which are easily calculated.