A quick angle chase reveals that $CA_1QA_2$ is cyclic, meaning that we can forget about the $B_i$'s: we're looking for the locus of the circumcenter of $A_1A_2Q$. Since $\angle PA_iQ$ are constant angles, it means that the triangles $A_1A_2Q$ are all similar, and this means that the triangles $A_1QO$ are all similar, where $O$ is the circumcenter of $A_1A_2Q$. This means that the locus of $O$ is the image of $S_1$ through the spiral similarity of center $Q$ which turns $A_1$ into $O$. In other words, this locus is a circle, Q.E.D. Comment: it can be shown fairly easily now, by choosing particular positions of $A_1$, the the circle is, in fact, the circumcircle of $QO_1O_2$, where $O_i$ is the center of $S_i$.

If we regard $ A_1A_2PB_1B_2C$ as a complete quadrilateral the problem becomes trivial once we know that the circumcenters of the $ 4$ "small" triangles of the complete quadrilateral and the miquel point of this complete quadrilateral (which in this case is point $ Q$) lie on a circle. Is this a well-known fact? If so, where can I find a proof to this fact?

Hmm. This problem did not involve the introduction of a new point other than the center of the indicated circle. Solution Let the centers of $S_1$ and $S_2$ be $O_1$ and $O_2$. First note the following. \[\angle{CA_1 Q} + \angle{CA_2 Q} = \angle{QPB_2} + \angle{CA_2 A_1} +\angle{PA_2 Q} = \angle{PQB_2}+\angle{PB_2 Q} +\angle{QPB_2}=180\] Therefore $QA_1 C A_2$ is cyclic. Let $\gamma$ and $\omega$ denote the perpendicular bisectors of $A_1 Q$ and $B_2 Q$ and let $X=\omega \cap \gamma$. Note that $X$ is the circumcenter of $QA_1 C A_2$. Since $O_1 \in \gamma$ and $O_2 \in \omega$, we have that $\angle{O_1 X O_2}=180-\angle{A_1 Q A_2}$. However, since $Q$ is the center of the spiral similarity mapping $S_1$ to $S_2$, it follows that $\angle{O_1 Q O_2}=\angle{A_1 Q A_2}$. This yields that $O_1 X O_2 Q$ is cyclic and since $O_1$, $O_2$ and $Q$ are fixed, that $X$ always lies on one circle.

Actually, a similar problem to this has appeared in two different UK texts, however they were both published after 2002, so it's OK! It also reminds of this years British MO round 1 Q5, which is of the same essence, posted here. On these grounds I'm guessing this was proposed by the UK

$O_1,O_2,O_3$ are the centres of $S_1,S_2$ and $\odot(A_1QA_2)$. As previously shown, we only need to study the locus of $O_3$. $O_1O_3$ and $O_2O_3$ are perpendicular to $A_1Q$ and $A_2Q$ respectively as by definition all of $\triangle A_2QO_2,\triangle A_2QO_3,\triangle A_1QO_1,\triangle A_2QO_3$ are isosceles. Let the midpoints of $QA_1,QA_2$ be $M_1,M_2$ respectively. To show that the quadrilateral $QO_1O_2O_3$ is cyclic it suffices to prove $\angle O_2QM_2=\angle O_1QM_1$ as then it would follow $\angle QO_1M_1=\angle QO_2M_2$ if we recall the right angles $\angle M_1,\angle M_2$. There is a fairly quick angle chase:

But a more appealing argument is that no matter where we place $A_1$ on $S_1$, $\triangle A_1A_2Q$ has fixed angles. So if we consider the diagram, but with another $A_n$ diametrically opposite $Q$ on $S_1$ then easily $\angle A_nQA_{n+1}=\angle A_1QA_2\implies\angle O_1QO_2=\angle A_1QA_2$, since $O_1,O_2$ lie on $QA_n,QA_{n+1}$ respectively. So this rotation must leave an equal against the common $\angle A_1QA_2$, i.e. $\angle O_1QM_1=\angle M_2QO_2$. Then the locus of $O_3$ is the circumcentre of $O_1QO_2$.

Do I feel guilty reviving a 7 year old thread? No.

My solution: Let's show that $O_1,Q,O_2,O$ are concyclic,where $O_1,O_2,O$ be the centers of $S_1,S_2,(A_1A_2C),$ respectively. Let $S=OO_1\cap A_1Q$ and $T=OO_2\cap A_2Q.$ We know $\angle CA_1Q=180-\angle B_1PQ=\angle QPB_2=\angle QA_2B_2=180-\angle QA_2C\rightarrow A_1,Q,A_2,C$ are cyclic. Then $OS\perp AQ,OT\perp QA_2\to S,Q,T,O$ are cyclic. $Claim:$ $\angle A_1QA_2=\angle O_1QO_2.$ $Proof:$ We have $\angle QPA_2=180-\angle QPA_1=180-\frac{\angle A_1O_1Q}{2}=90+\angle O_1QA_1.$ Also we know $\angle QPA_2=180-\frac{\angle QO_2A_2}{2}=90+\angle O_2QA_2,$ then we know $\angle O_2QA_2=\angle O_1QA_1\to \angle O_1QO_2=\angle O_1QA_1+\angle A_1QO_2=\angle O_2QA_2+\angle A_1QO_2=\angle A_1QA_1.$ As desired. Also we know $$\angle O_1OO_2=\angle SOT=180-\angle SQT=180-\angle A_1QA_2=^{\text{Claim}}180-\angle O_1QO_2\implies O_1,QO_2,O$$are concyclic.

Lemma: This problem is trivial. Proof : For any $B$, by a trivial angle chase, we have $\angle A_1CA_2 = \angle A_1QA_2$, so we put the $B$ aside the picture, and focus on the locus of the circumcentre of $\Delta A_1QA_2$, as $A_1$ varies. We claim the locus is on $\omega_{O_1QO_2}$. Indeed, inverting around $Q$, the problem becomes trivial, because taking a homothety of factor $.5$ w.r.t $Q$ maps $O_1^{*} O_2^{*}$ to the $Q$-simson line of $\omega_{A_1^{*}P^{*}A_2^{*}}$, hence the desired result. As a corollary, we have: Lemma : 10 year or more old IMOSL problems are trivial. (Sketch of proof is trivial, just also note that JMO P1 was 1998 G8, then trivial by induction)

Let the centers of $S_1, S_2$ be $O_1, O_2$ resp. Some angle chasing shows that $\angle A_1QA_2=\angle A_1QP+\angle PQA_2=\angle AB_1B_2+\angle CB_2B_1=180-\angle A_1CA_2$ $\Longrightarrow A_1CA_2Q$ is cyclic. Thus the problem is equivalent to find the locus point of the circumcenter of $\triangle QA_1A_2$ which doesn't depend on $B_1$ so we can ignore $B_1$ in the problem. We will prove that $O\in (O_1QO_2)$ which is a fixed circle where $O$ is the circumcenter of $\triangle QA_1A_2$. We have $(A_1CA_2Q)\cap S_2=A_2Q$, $(A_1CA_2Q)\cap S_1=A_1Q$ $\Longrightarrow OO_2\perp A_2Q,\ OO_1\perp A_1Q\Longrightarrow \angle O_1OO_2=180 -\angle A_1QA_2$. But $\angle A_1QO_1=\frac{180-2\angle A_2PQ}{2}=90-\angle A_2PQ=\angle A _2QO_2\Longrightarrow \angle A_1QA_2=\angle O_1QO_2$. $\Longrightarrow \angle O_1OO_2=180 -\angle O_1QO_2\Longrightarrow O\in (O_1QO_2)$ which is a fixed circle as desired.

Adding an Inversive Solution. ISL 2002 G4 wrote: Circles $S_1$ and $S_2$ intersect at points $P$ and $Q$. Distinct points $A_1$ and $B_1$ (not at $P$ or $Q$) are selected on $S_1$. The lines $A_1P$ and $B_1P$ meet $S_2$ again at $A_2$ and $B_2$ respectively, and the lines $A_1B_1$ and $A_2B_2$ meet at $C$. Prove that, as $A_1$ and $B_1$ vary, the circumcentres of triangles $A_1A_2C$ all lie on one fixed circle. Note that $$\angle AQA_2=\angle AQP+\angle PQA_2=\angle AC_1B_2+\angle B_1B_2C=180^\circ-\angle A_1QA_2\implies A,Q,A_2,C\text{ are concyclic.}$$So the Circumcenter of $\triangle A_1A_2C$ is same as the Circumcenter of $\triangle A_1QA_2$. So now we can safely delete the points $C,B_1,B_2$ from the Diagram. So, the problem can now be restated as Reduced Problem wrote: $\omega_1$ and $\omega_2$ are two circles and $\omega_1\cap\omega_2=\{P,Q\}$. A line through $P$ intersects $\omega_1$ and $\omega_2$ at $A_1,A_2$ respectively. Then Prove that the locus of the Circumcenters of such triangles $A_1QA_2$ is a circle. For this Invert around $Q$ and let this map be denoted as $\Psi$. So, $\Psi:P\leftrightarrow \omega_1'\cap\omega_2'=P'$ where $\omega_1'$ and $\omega_2'$ are the Inverted Image of $\omega_1$ and $\omega_2$ respectively during this transformation $\Psi$. We have to prove that the circles passing through $P',Q$ intersecting $\omega_1'$ and $\omega_2'$ at points $\{A_1',A_2'\}$ respectively, then the reflections of $Q$ on $A_1'A_2'$ are collinear. Drop Perpendiculars from $Q$ onto $\omega_1'$ and $\omega_2'$, So by Simson Line we get that the Projections of $Q$ on such lines $A_1'A_2'$ are collinear, so by a Homothety at $Q$ with a scale factor of $2$ we get that the reflections of $Q$ of such lines $A_1A_2$ are also collinear, and the reflections of $Q$ on such lines $A_1A_2$ are the points where the circumcenters of such triangles $QA_1A_2$ map! Hence Inverting back we get that the Locus of the Circumcenters of such triangles $QA_1A_2$ which are the circumcenters of $\triangle A_1A_2C$ will lie on a circle. $\blacksquare$

What troll! orl wrote: Circles $S_1$ and $S_2$ intersect at points $P$ and $Q$. Distinct points $A_1$ and $B_1$ (not at $P$ or $Q$) are selected on $S_1$. The lines $A_1P$ and $B_1P$ meet $S_2$ again at $A_2$ and $B_2$ respectively, and the lines $A_1B_1$ and $A_2B_2$ meet at $C$. Prove that, as $A_1$ and $B_1$ vary, the circumcentres of triangles $A_1A_2C$ all lie on one fixed circle. Let $O_1,O_2$ be center of $S_1,S_2$ respectively.We have $Q$ is the miquel point of complete quadrilateral $PB_2CA_1A_2B_1$.Hence center of $\odot{A_1A_2C}$ lies on the miquel circle $\odot{QO_1O_2}$ which is fixed.$\square$

Easy G4. Notice that $\angle A_1QA_2=\angle A_1B_1B_2+\angle CB_2B_1=180^{\circ}_\angle C,$ so $A_1CA_2Q$ is cyclic and the problem reduces to showing that the circumcenter of $(A_1A_2Q)$ lie on a fixed circle (the $B$'s were just a distraction!). We finish with Claim: Denote the center of $S_1$ by $O_1$ and that of $S_2$ by $O_2$. Then the desired circle is $(O_1QO_2)$. Proof: Denote the midpoints of $A_1Q$ and $A_2Q$ by $M_1, M_2,$ respectively. Then the circumcenter of $\triangle A_1A_2Q$ is the intersection of the perpendicular at $M_1$ and $M_2$ to $A_1Q$ and $A_2Q$. Notice that $\angle AO_1Q=2\angle A_1PQ=180^{\circ}-\angle B_2PQ=\angle QO_2A_2,$ and this easily implies the result. We are done.

Obviously $Q$ is the Miquel point of $A_1CB_2P,$ and so circumcenter of $\triangle A_1A_2C$ lie on circle passing through $Q$ and centers of $S_1,S_2,$ which is fixed.

Let $O,O_1,O_2$ be the centers of $S_1,S_2,(A_1A_2C),$ respectively. Notice that $Q$ is the Miquel point of $A_1B_1CA_2B_2P$ so $OO_1QO_2$ is cyclic. $\square$

No one: EGMO Ch. 11: Let $O_1,O_2$ be centers of $S_1,S_2.$ Let $O$ be center of $(A_1CA_2).$ We claim that $O_1OO_2Q$ is cyclic. First, we claim that $CA_1QA_2$ is cyclic. Note that $\angle B_1A_1Q=\angle B_1PQ=\angle CA_2Q.$ This implies the claim. Thus, $O$ is the center of $(QA_1A_2).$ Note that $OO_1$ is the perpendicular bisector of $A_1Q$ and $OO_2$ the perpendicular bisector of $A_2Q.$ Thus, $\angle OO_2Q=\frac{1}{2}\overset{\huge\frown}{QPA_2}=180^\circ-\angle QPA_2$ and $\angle OO_1Q=\frac{1}{2}\overset{\huge\frown}{QPA_1}=180^\circ-\angle A_1PQ.$ These sum to $180^\circ$ so we are done.

Wut Observe that $Q$ is miquel point of quad $A_1PB_2C$ which means the circumcenter of $\triangle A_1A_2C$ belongs to $\odot(Q0_1O_2)$ where $O_1$ and $O_2$ are the centres of $S_1$ and $S_2$ which are obviously fixed $\blacksquare$

Claim: We can simply consider the circumcenter of $A_1QA_2$ Proof: Observe that $Q$ is the miquel point of quadrilateral $A_1PB_2C$ Therefore, $Q$ lies on $(A_1A_2C)$ As such, the circumcenter of $(A_1A_2C)$= the circumcenter of $(A_1QA_2)$ Rephrased problem: the circumcenters of $(A_1A_2Q)$ lie on a circle as $A_1A_2$ varies Claim: $\angle A_1QA_2$, $\angle A_2A_1Q$, $\angle A_1A_2Q$ remain invariant over all $A_1$,$A_2$ Proof: Fix one $A_1$,$A_2$, Now, suppose we take another pair, say $B_1,B_2$ Note that $Q$ is the center of the spiral sending $A_1 \rightarrow A_2$, $B_1 \rightarrow B_2$ Which implies that $\triangle A_1QA_2 \sim \triangle B_1QB_2$ Claim: If $O$ is the circumcenter of $(A_1QA_2)$, $\triangle A_1QO$ remains similar as $A_1$ varies Proof: Note $Q$ is the spiral center sending all $A_1 \rightarrow O$ As such, the possible positions of $O$ are simply a fixed scaling and translation from every possible position of $A_1$ with respect to the spiral center $Q$ This means that all possible positions of $O$ must cover a circle

Viewing $CA_1PB_1$ as a complete quadrilateral, $CA_1QA_2$ is cyclic. Therefore, we can delete $C,B_1,B_2$ from the diagram and focus on the center of $(A_1A_2Q)$ as $A_1$ varies around $S_1$. As $A_1$ varies around $S_1$, $\triangle QA_1A_2$ is directly similar to a fixed triangle for spiral similarity reasons, so $\triangle QA_1O$ is directly similar to a fixed triangle as well, where $O$ is the (variable) circumcenter of $\triangle QA_1A_2$. Therefore, it suffices to prove the following "independent" lemma. Lemma: Let $P$ be a point and suppose that $\triangle PAB$ is directly similar to a fixed triangle as $A$ varies around a circle. Then $B$ varies around a circle too. Proof: We use complex numbers. WLOG let $p=0$. Then there exists some fixed complex number $z \neq 0$ such that $b=za$ always holds, and the rest is clear. $\blacksquare$

wow how did I not see the Miquel point We first do some angle chasing: \begin{align*} \angle A_1QA_2 &= \angle A_1QP + \angle PQB_2 + \angle B_2QA_2 \\ &= \angle A_1B_1P + \angle PA_2B_2 + \angle B_2PA_2 \\ &= \angle A_1B_1P + \angle A_1PB_1 + \angle A_1A_2C \\ &= \angle CA_1A_2 + \angle A_1A_2C \\ \angle A_1QA_2 &= 180^\circ - \angle A_1CA_2 \end{align*}From here, we see that $CA_1QA_2$ is a cyclic quadrilateral. Obviously, the circumcenter of $CA_1A_2$ is the same as the circumcenter of $A_1QA_2$, a point we will label as $O$. The centers of $S_1$ and $S_2$ will be called $O_1$ and $O_2$, respectively. Now, we claim that $OO_1QO_2$ is a cyclic quadrilateral. In order to prove this, construct the perpendicular bisectors to $A_1Q$ and $A_2Q$. Denote the midpoints of $A_1Q$ and $A_2Q$ as $M$ and $N$, respectively. Visibly, $OMQN$ is a cyclic quadrilateral, which means that $\angle MQN + \angle MON$. It is easy to see that the perpendicular bisector of $A_1Q$ passes through both $O_1$ and $O$, while the perpendicular bisector of $A_2Q$ passes through both $O_2$ and $O$. As a result, $\angle MON = \angle O_1OO_2$. We finish with a short angle chase: \begin{align*} 180^\circ &= \angle MQN + \angle MON \\ &= \angle A_1QA_2 + \angle O_1OO_2 \\ 180^\circ &= \angle O_1QO_2 + O_1OO_2 \end{align*}Thus, $OO_1QO_2$ is a cyclic quadrilateral, and $O$ always lies on $(O_1QO_2)$, a fixed circle. $\blacksquare$

Let $O$ be the circumcenter of $\triangle A_1A_2C$. Since $Q$ is the Miquel point, we have thta $Q$ is concyclic with the centers of $(A_1PB_1)$, $(A_1A_2C)$, $(B_2B_1C)$, and $(B_2PA_2)$. Since three of these points are $O_1$, $O_2$, and $O$, $O$ always lies on $(QO_1O_2)$, as desired.

Forgot to post this from WOOT oops Note $Q$ is the Miquel point of $A_1 B_1 A_2 B_2$. It is a well known Miquel point property that $Q$, $O(A_1 B_1 P)$, $O(A_2 B_2 P)$, $O(A_1 A_2 C)$, and $O(B_1 B_2 C)$ are concyclic. As the first three are fixed, the locus of $O(A_1 A_2 C)$ lies on a fixed circle. $\blacksquare$

By Miquel, $(A_1 A_2 C)$ also passes through $Q,$ so we just need to show that as $A_1$ varies, the circumcenter of $\triangle A_1 A_2 Q$ lies on a fixed circle. Invert at $Q$ with arbitrary radius; the problem reduces to the following: Let $Q$ be a fixed point and $\ell_1, \ell_2$ two fixed lines in the plane which intersect at $P.$ An arbitrary circle passing through $P,Q$ intersects $\ell_1, \ell_2$ at $A_1, A_2,$ respectively. Prove that as the circle varies, the reflection $Q'$ of $Q$ over $A_1 A_2$ lies on a fixed line. Indeed, if we reflect $Q$ over $\ell_1, \ell_2$ to obtain $Q_1, Q_2,$ the line $Q_1 Q_2$ is fixed and passes through $Q'$ by Simson, so we are done. Remark: Interesting how I didn't use any complicated Miquel properties.