Let $O_1, O_2$ be the centers of the 2 circles and let the points B, C move on the circles $(O_1), (O_2)$, recpectively. Let A' be the other intersection of the 2 circles different from the point A. Since the angles $\angle BO_1A = \angle CO_2A$ are equal, the isosceles triangles $\triangle AO_1B \sim \triangle AO_2C$ are similar. Consequently, the angles $\angle BAO_1 = \angle CAO_2$ are equal and the angles $\angle BAC = BAO_1 + \angle O_1AO_2 + \angle O_2AC =$ $= BAO_1 + \angle O_1AO_2 - \angle CAO_2 = \angle O_1AO_2$ are also equal. The angles $\angle A'BA = \frac{\angle A'O_1A}{2} = \angle O_2O_1A$ are equal and the angles $\angle A'CA = \frac{\angle A'O_2A}{2} = \angle O_1O_2A$ are also equal. As a result, the point A' is on the side BC of the triangle $\triangle ABC$, which is always similar to the fixed triangle $\triangle O_1AO_2$. Lemma (Yaglom's Geometric Transformations II, pp 68-69). Assume that a figure F' (the triangle $\triangle ABC$) moves in such a way that it remains at all times similar to a fixed figure F (the triangle $\triangle O_1AO_2$) and in addition, some point of the figure F' does not move at all (the point A). If a point B of the figure F' describes a curve $\Gamma$ (the circle $(O_1)$), then any other point M of the figure F' describes a curve $\Gamma'$ similar to the curve $\Gamma$. Consequently, the midpoint M of the segment BC decribes a circle (Q). When the points B, C, become identical with the point A, the midpoint M of BC also becomes identical with A. Hence, the circle (Q) passes through the point A. The line BC always passes through A'. Just after the start of motion counter-clockwise, the points B, C, A' follow on the line BC in this order. The points B, C both pass through the point A' one after another and just before the end of one period of motion, the points A', B, C follow on the line BC In this order. From the continuity principle, the midpoint M of BC also passes through the point A' and consequently, the circle (Q), the locus of the midpoints M, also passes through the point A'. Thus the circle (Q) belongs to the same pencil with the circles $(O_1), (O_2)$ and it is centered on their center line $O_1O_2$. Let p be the perpendicular bisector of the segment BC intersectong the circle (Q) at a point P different from the point M. The point A' and the circle (Q) passing through this point are fixed and the angle $\angle A'MP = 90^\circ$ is right. Therefore, A'P is a fixed diameter of the circle (Q), which means that the point P is also fixed. Q.E.D. When the points B, C are both diametrally opposite to the point A on their respective circles, the segment $O_1O_2$ is the midline of the triangle $\triangle ABC$, i.e., $BC \parallel O_1O_2$, $BC \perp AA'$. The cyclic quadrilateral AA'MP inscribed in the circle (Q) is then a rectangle. Its circumcenter Q is the intersection of its diagonals AM, A'P intersecting at their midpoint and the midpoint of AM is identical with the midpoint of $O_1O_2$. Thus the fixed point P is the reflection of the intersection A' of the circles $(O_1), (O_2)$ in the midpoint Q of the segment $O_1O_2$.

yetti wrote: $\angle BAC = BAO_1 + \angle O_1AO_2 - \angle O_2AC = \angle O_1AO_2$ . I'm sorry, and maybe I mistake something. but why is this true? $O,A,C$ can not be on a line.

By the problem conditions, the angles $\angle AO_1B = \angle AO_2C$ are equal, because the points A, B travel on their respective circles with constant velocity and the same period, hence, with the same angular velocity. The triangles $\triangle s AO_1B, \triangle AO_2C$ are both isosceles, hence all their angles are equal and $\angle BAO_1 = \angle CAO_2$. The angle $\angle BAC$ can be expressed as a sum of directed angles $\angle BAC = \angle BAO_1 + \am\angle O_1AO_2 + \angle O_2AC =$ $= \angle BAO_1 + \angle O_1AO_2 - \angle CAO_2 = \angle O_1AO_2$ The equation you refer to has a wrong sign (tx, fixed).

ok,thanks . I haven't notice the condition which is "each travelling along its own circle in the same sense." but if "each travelling along its own circle in the different sense" can the conclusion also be true?

The conclusion is true even if the 2 points B, C travel on the circles $(O_1), (O_2)$ in the opposite directions. The fixed point of the perpendicular bisector of the segment BC is then the reflection of the intersection A in the midpoint Q of the segment $O_1O_2$, but I do not have a proof yet.

yetti wrote: The conclusion is true even if the 2 points B, C travel on the circles $(O_1), (O_2)$ in the opposite directions. The fixed point of the perpendicular bisector of the segment BC is then the reflection of the intersection A in the midpoint Q of the segment $O_1O_2$, but I do not have a proof yet. sorry.I don't know which is the fixed point you wrote because my weak english. and I guess the fixed point is let $AO_1$ intersect circle $(O_1)$ at $M$,$AO_2$ intersect circle $(O_2)$ at $N$, and I think the fixed point is the midpoint of $MN$. but no proof,either? is your fixed point the same as me ?

zhaobin wrote: ... is your fixed point the same as me ? Certainly. With your notation, the points M, N are diametrally opposite to the point A on their respective circles $(O_1), (O_2)$, so that $O_1O_2$ is the midline of the triangle $\triangle AMN$ parallel to the side MN. If P is the midpoint of MN, AP is the A-median of the triangle $\triangle AMN$, which cuts the midline segment $O_1O_2$ at its midpoint Q and AQ = PQ, so that P is a reflection of A in Q. Let $B_0, C_0$ be the points on the circles $(O_1), (O_2)$ diametrally opposite to their intersection A and let A' be the other intersection of these 2 circles. Let B, C be the points on the circles $(O_1), (O_2)$ traveling in the same sense counter-clockwise. Denote $\alpha = \angle BAC = \angle B_0AC_0 = \angle O_1AO_2$, $\beta = \angle ABC = \angle AB_0C_0 = \angle AO_1O_2$, $\gamma = \angle ACB = \angle AC_0B_0 = \angle AO_2O_1$ the angles of the similar triangles $\triangle ABC \sim \triangle AB_0C_0 \sim \triangle AO_1O_2$. We know that $A' \equiv (O_1) \cap (O_2) \equiv B_0C_0 \cap BC$. Furthermore, denote $\phi = \angle BAO_1 = \angle CAO_2$ the base angles of the similar isosceles triangles $\triangle AO_1B \sim \triangle AO_2C$ and $k = \frac{BA}{O_1A} = \frac{CA}{O_2A}$ the ratio of their base to the shoulder side. Let $M_0$ be the midpoint of $B_0C_0$. Since $O_1O_2$ is a midline of the triangle $\triangle AB_0C_0$, the point $M_0$ is a reflection of the point A in the midpoint Q of the segment $O_1O_2$. Let C' be a point traveling in the opposite sense on the circle $(O_2)$ than the the point B on the circle $(O_1)$. The point C' is obviously a reflection of the point C in the diameter $AC_0$ of the circle $(O_2)$. We will show that the triangles $\triangle C'BC \sim \triangle C'M_0C_0$ are similar. The angle $\angle BCC' \equiv \angle A'CC'$ and the angle $\angle M_0C_0C' \equiv \angle A'C_0C'$ spanning the same arc A'C' of the circle $(O_2)$ are equal. The isosceles triangles $\triangle ACC' \sim \triangle O_2C_0C'$ have the angles at the vertices $A, O_2$ equal, hence, they are similar and $\frac{CC'}{C_0C'} = \frac{CA}{O_2C_0} = \frac{CA}{O_2A} = k$. From similarity of the triangle $\triangle s ABC \sim \triangle AO_1O_2$ with similarity coefficient $\frac{BC}{O_1O_2} = \frac{CA}{O_2A} = k$ and from the parallelogram $O_1M_0C_0O_2$, we have $\frac{BC}{M_0C_0} = \frac{BC}{O_1O_2} = k$. Thus the similarity of the triangles $\triangle C'BC \sim \triangle C'M_0C_0$ by SAS is proven. But then the angles $\angle C'BA' \equiv \angle C'BC = \angle C'M_0C_0 \equiv \angle C'M_0A'$ are equal, which means that the quadrilateral $BC'A'M_0$ is always cyclic, regardles of the position of the point B on the circle $(O_1)$ and the corresponding position of the point C' on the circle $(O_2)$. From the quadrilateral ABA'C (concave in the figure below), the angle $\angle BA'C' = \angle ABA' + \angle AC'A' + \angle BAC' =$ $= \angle AB_0A' + \angle AC_0A' + \angle B_0AC_0 - 2 \phi = \beta + \gamma + \alpha - 2 \phi = 180^\circ - 2 \phi$. The angles $\angle BM_0C' = \angle BA'C' = 180^\circ - 2 \phi$ are equal, because the quadrilateral $BC'A'M_0$ is cyclic. For the same reason, the angles $\angle BC'M_0 = \angle BA'M_0$ are also equal. But the angle $\angle BC'M_0 = \angle CA'C_0 = \angle CAC_0 \equiv \angle O_2AC_0 = \phi$. As a result, the remaining angle of the triangle $\triangle BCM_0$ is equal to $\angle C'BM_0 = 180^\circ - (\angle BM_0C' + \angle BC'M_0) = 180^\circ - (180^\circ - 2 \phi + \phi) = \phi$ This means that the triangle $\triangle BC'M_0$ is always isosceles, i.e., $BM_0 = C'M_0$, regardles of the position of the point B on the circle $(O_1)$ and the corresponding position of the point C' on the circle $(O_2)$. The point $M_0$ is fixed and the perpendicular bisector of the segment BC' always passes through this point. The locus of midpoints N of the segments BC' is an ellipse centered at the midpoint Q of the seqment $O_1O_2$ and passing through the points $A, M_0$ (no proof).

As above, $O_1$ and $O_2$ are the centers of the 2 circles, $A$ and $A'$ being their common points; also, as already shown, $B, C, A'$ are collinear. When $BC$ becomes parallel to $O_1O_2$, their position become $D$, respectively $E$, diametrically opposite to $A$ on the 2 circles; let’s call $F$ the midpoint of $DE$. It’s easy to see that always $\Delta ABC \sim \Delta AO_1O_2$, having $\hat B=\hat O_1$ and $\hat C=\hat O_2$, so always $\widehat{AQC}=\widehat{AOO_2}=\widehat{AFA'}$ ($O$-midpoint of $O_1O_2$). Now let’s take $P'$ the symmetrical of $A$ w.r.t perpendicular bisector of $O_1O_2$, and we claim that this is the required fixed point. $AP'O_1O_2$ is isosceles trapezoid, hence $O_1P'=O_2A=O_2E \ (\ 1\ )$ and $O_2P'=O_1A=O_1D \ (\ 2\ )$ and $\widehat{P'O_1A}=\widehat {P'O_2A}$, resulting that $\Delta DO_1P'=\Delta P'O_2E$, i.e. $P'D=P'E$, hence $P'F \perp DE$ or $AP'FA'$ is a rectangle. Let’s call $M$ the intersection of $BO_1$ and $O_2C$; as $\widehat{AO_1B}=\widehat{AO_2C}$, $AO_1MO_2$ is cyclic, or $\widehat {BO_1P'}=\widehat{P'O_2C}$; with (1) and (2) we get $\Delta BO'P'=\Delta P'O_2C$, hence $P'B=P'C$, and $P'$ is always on the perpendicular bisector of $BC$, and $P'\equiv P$, done. Notes: 1) There is not necessary for $DE$! We get easily the result from the other triangles and parallelogram. 2) When the points move in opposite direction, the result is even more obvious (of course, once you know the required position). Best regards, sunken rock

Let $O1$ and $O2$ be the centers of the two circles in the problem, denoted Γ and Φ. Using $O1$ as the center draw another circle, denoted Π, that has the same radius as the radius of Φ. Now use $O2$ as the center draw another circle, denoted Δ, that has the same radius as the radius of Γ. There are now two sets of concentric circles. Now let Π intersect Δ at P with $P$ closer to $A$ than to $B$. This is the point $P$ in the problem where the two points $E$ and $F$ are equidistant from. Indeed, with $PE = PF$, it’s easily seen that ∠$AO1E$ = ∠$AO2F$ = α.

Let $X,Y$ be the two points, $\omega_X$, and $\omega_Y$ be the corresponding circles, and $B$ be the other intersection. It is simple to check that $BXY$ is always a line. Now, invert at $B$ with radius $1$ so that $\omega_X$ and $\omega_Y$ map to lines $\ell_X$ and $\ell_Y$ through $A'$, and let $C$ be the point on $BX'Y'$ such that $(B,C;X',Y')=-1$. By well known facts about projective, $A'C$ is a fixed line - let $T$ be the foot from $B$ to this line. Now, if we invert back, $\angle BC'T'=90^{\circ}$, and $C'$ is the midpoint of $XY$, so we are done as $T'$ is the fixed point. Edit: Holy I'm bad at math, here's a quicker sol: Let $X,Y$ be the two points, $\omega_X$, and $\omega_Y$ be the corresponding circles, and $B$ be the other intersection. It is simple to check that $BXY$ is always a line. Now, if $M$ is the midpoint of $XY$, since the ratios of the powers of $M$ to $\omega_X$ to $\omega_Y$ is $-1$, by the forgotten coaxiality lemma, $M$ lies on a fixed circle through $A$ and $B$, and now the fixed point is just the $B$-antipode.

Solution from Twitch Solves ISL: Let $B$ and $C$ be the antipodes of $A$ on the two circles and let $D$ be the foot of the altitude from $A$, which is the other intersection point of the two circles. Also, let $M$ be the midpoint of $\overline{BC}$, and construct rectangle $ADMZ$. Our claim is that $Z$ is the fixed point. We let $X$ and $Y$ be the two points; by the condition the angles are the same. So we have a spiral similarity \[ \triangle AXY \sim \triangle ABC. \][asy][asy]size(10cm); pair A = dir(110); pair B = dir(210); pair C = dir(330); pair D = foot(A, B, C); pair M = midpoint(B--C); pair Z = A+M-D; pair X = midpoint(A--B)+dir(200)*abs(A-B)/2; pair Y = -D+2*foot(midpoint(A--C), X, D); draw(circumcircle(A, B, D), blue); draw(circumcircle(A, C, D), blue); draw(circumcircle(A, D, M), dashed+deepgreen); pair N = midpoint(X--Y); filldraw(A--X--Y--cycle, invisible, purple); filldraw(A--B--C--cycle, invisible, blue); draw(A--D, blue); draw(A--Z--M, deepgreen); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$D$", D, dir(D)); dot("$M$", M, dir(M)); dot("$Z$", Z, dir(Z)); dot("$X$", X, dir(X)); dot("$Y$", Y, dir(Y)); dot("$N$", N, dir(N)); /* TSQ Source: A = dir 110 B = dir 210 C = dir 330 D = foot A B C M = midpoint B--C Z = A+M-D X = midpoint(A--B)+dir(200)*abs(A-B)/2 Y = -D+2*foot midpoint A--C X D circumcircle A B D blue circumcircle A C D blue circumcircle A D M dashed deepgreen N = midpoint X--Y A--X--Y--cycle 0.1 red / purple A--B--C--cycle 0.1 yellow / blue A--D blue A--Z--M deepgreen */ [/asy][/asy] Now let $N$ be the midpoint of $\overline{XY}$. By spiral similarity, since $N$ maps to $M$, it follows $M$, $N$, $A$, and $D$ are cyclic too. So actually $N$ lies on the circumcircle of rectangle $ADMZ$, meaning $\overline{ZN} \perp \overline{XY}$, hence $ZX = ZY$ as needed. Remark: The special point $Z$ can be identified by selecting the special case $X \to A$, $Y \to A$ and $X=B$, $Y=C$.

Let the points on the circles be $E$ and $F$. Then, let $B$ the the other intersection of the two circles, and $C$ and $D$ be the centers of the circles containing $E$ and $F$, respectively. WLOG, let us name $E$ and $F$ such that $CA>DA$. Let $O$ be the midpoint of $CD$, and the antipode of $B$ with respect to the circle centered at $O$ with radius $OA$ be $P$. Let $M$ be where this such circle hits $EF$. We first note that $$\angle ADF = 2\angle ABF = 2 \angle APM = \angle AOM$$and also $$2 \angle APM = 360^{\circ} - 2\angle MBA = 360^{\circ} - 2\angle EBA = \angle ACE$$meaning that since $AD=DF$, $AO=OM$, and $AC=CE$, then SAS similarity gets us that $\triangle ADF \sim \triangle AOM \sim \triangle ACE$. This means we have $\frac{AO}{AM} = \frac{AC}{AE}$ so $\frac{AO}{AC} = \frac{AM}{AE}$ and because $\angle MAE = \angle OAC$, then $\triangle AOC \sim \triangle AME$ and similarly, we also have $\triangle AOD \sim \triangle AMF$. This means that $$\frac{EM}{MA} = \frac{OC}{OA} = \frac{DO}{OA} = \frac{FM}{MA} \to EM=FM$$meaning that $M$ is the midpoint of $EF$. Now, since $P$ is the antipode of $B$, then $\angle BMP = 90^{\circ} = \angle EMP = \angle FMP$ so $EP=FP$ for all $E$ and $F$ for some fixed $P$ and we are done.

Let $P_1, P_2$ be the moving points (pun intended). Denote by $M$ the midpoint of $P_1P_2$. Claim. The second intersection of the circles $B$ lies on $P_1P_2$. Denote by $P_1', P_2'$ the $P_1, P_2$-antipodes. Evidently a spiral similarity at $A$ still sends $\overline{P_1P_1'} \to \overline{P_2P_2'}$. Thus we are done by spiral similarity lemma. $\blacksquare$ Claim. $M$ moves along a fixed circle through $A, B$. Observe that $\angle AMB$ is constant because a spiral similarity at $A$ takes $\overline{O_1O_2} \to \overline{P_1P_2}$, and hence $$\angle AMP_1 = \angle ANO_1$$which is constant where $N$ is the midpoint of $\overline{O_1O_2}$. $\blacksquare$ Therefore the $B$-antipode in the fixed circle $(AMB)$ will always satisfy $\angle PMB = 90^{\circ}$, and thus lies on the perpendicular bisector.

Let \(O_1\) and \(O_2\) be the centers of the two circles, and let \(P\) be the point such that \(PAO_1O_2\) is an isosceles trapezoid with \(\overline{PA}\parallel\overline{O_1O_2}\). I claim \(P\) is the fixed point equidistant from the two particles. Let the particle on \((O_1)\) be \(C\) and let the particle on \((O_2)\) be \(D\). Then \(\triangle AO_1C\sim\triangle AO_2D\), so since spiral similarites come in pairs, we have \(\triangle ACD\sim\triangle AO_1O_2\). Now let \(P'\) be the point such that \(\triangle ACD\cup P'\sim\triangle AO_1O_2\cup P\). Since \(\triangle AO_1C\sim\triangle APP'\), we have \(PA=PP'\). But \(P'ACD\) is an isosceles trapezoid, so the perpendicular bisectors of \(\overline{CD}\) and \(\overline{AP'}\) coincide. Thus \(PC=PD\), as needed.

There is no need to identify the point $P$, we can just prove its existence as follows: Let $\omega_1,\omega_2$ be the two circles with centers $O_1,O_2$ respectively and $B = \omega_1 \cap \omega_2 \ne A$. Let $X,Y$ be the points which move and $Z$ be the midpoint of segment $XY$ ($Z$ is also moving). Claim: As $X,Y,Z$ move, $B$ always lies on line $XY$ and there is a fixed circle $\Gamma$ which passes through $A,B$ and $Z$. Proof: If we consider two such positions $(X_1,Y_1,Z_1)$ and $(X_2,Y_2,Z_2)$. It suffices to show $B = \overline{X_1Y_1} \cap \overline{X_2Y_2}$ and points $A,B,Z_1,Z_2$ are concylic. Now it is not hard to note by $SAS$ criterion that $$\triangle AX_1Y_1 \sim \triangle AO_1O_2 \sim \triangle AX_2Y_2$$So $A$ is the center of spiral sim $\tau$ sending $\overline{X_1Y_1}$ to $\overline{X_2Y_2}$, which implies $$\overline{X_1Y_1} \cap \overline{X_2Y_2} = \odot(AX_1X_2) \cap \overline{AX_2Y_2} \ne A = \omega_1 \cap \omega_2 \ne A = B$$As $\tau$ sends $Z_1$ to $Z_2$, so it follows that points $A,B,Z_1,Z_2$ are concyclic. This proves our claim. $\square$ Now just note that $P$ as the antipode of $B$ wrt $\Gamma$ works. This completes the proof of the problem. $\blacksquare$

Let $X$ and $Y$ be the particles and $O_1$ and $O_2$ be the circumcenters. First, we claim that $X$, $Y$, and $B$ are always colinear. Indeed, notice that \[ \angle ABX = \dfrac{1}{2} \angle AO_1X = \dfrac{1}{2} \angle AO_2Y = 180^{\circ}-\angle ABY \]as desired. Thus, by spiral similarity, $\triangle XYA$ remains similar to every other instance of $\triangle XYA$. If $M$ is the midpoint of $XY$ then $M$ moves along a circle passing through $A$ and $B$ since $\angle AMX$ and $\angle AMY$ are fixed. So, the perpendicular bisector of $XY$ always passes through the antipode of $B$ in $(ABM)$ which we just established as fixed. $\blacksquare$

Let $X$ and $Y$ be the moving points on $\omega_1$ and $\omega_2$, respectively. Let $O_1$ and $O_2$ be the centers of $\omega_1$ and $\omega_2$, respectively and let $A_1$ and $A_2$ be the $A$ antipode with respect to $\omega_1$ and $\omega_2$, respectively. Let $B$ be the other intersection of $\omega_1$ and $\omega_2$. Claim: $\overline{XY}$ passes through $B$. Proof. Since $X$ and $Y$ have equal angular velocities, \[\angle AA_1X=\tfrac{1}{2}\angle AO_1X=\tfrac{1}{2}\angle AO_2Y=\angle AA_2Y\]so $A$ is the center of the spiral similarity $\overline{A_1X}\mapsto\overline{A_2Y}$. Hence, $A$ is the Miquel point of $A_1XYA_2$ so $\overline{XY}\cap\overline{A_1A_2}$ lies on $(AA_1X)$ and $(AA_2Y)$. $\blacksquare$ Let $M$ and $O$ be the midpoints of $\overline{XY}$ and $\overline{O_1O_2}$, respectively. Claim: $M$ lies on $\gamma$, the circle with center $O$ passing through $A$ and $B$. Proof. Notice $A$ is the center of the spiral similarity $\overline{OO_1}\mapsto\overline{XM}$ so \[\angle AMX=\angle AOO_1=\tfrac{1}{2}\angle AOB\]and $O$ lies on the perpendicular bisector of $\overline{AB}$. $\blacksquare$ Since the perpendicular bisector of $\overline{XY}$ is the line perpendicular to $\overline{BM}$ that passes through $M$, it must always pass through the antipode of $B$ with respect to $\gamma$. Since $\gamma$ and $B$ are fixed we are done. $\square$

Denote the two circles as $\omega_1, \omega_2$. Let $P_i$ and $O_i$ denote the moving point and center of $\omega_i$. Additionally, let $B \neq A$ be the second intersection of the two circles. Claim: $B$ lies on $P_1P_2$ Proof: Let $P_1B$ intersect $\omega_2$ again at $P_2'$. Since $\measuredangle ABP_2' = \measuredangle ABP_1$, we can see that the directed arcs $\widehat{AP_1}$ of $\omega_1$ and $\widehat{AP_2'}$ of $\omega_2$ have equal measures, which implies that $P_2' = P_2$, as desired. Now, let $M$ be the midpoint of $\overline{P_1P_2}$ and $O$ be the midpoint of $\overline{O_1O_2}$. Claim: $M$ always lies on the circle centered at $O$ through $A$ and $B$ Proof: Note that since $\triangle AO_1P_1 \overset{+}{\sim} \triangle AO_2P_2$, $A$ is the center of the spiral similarity sending $\overline{O_1P_1}$ to $\overline{O_2P_2}$. Hence $A$ is also the center of the spiral similarity which takes $\overline{O_1O_2}$ to $\overline{P_1P_2}$, which also sends $O$ to $M$. Thus $\triangle AOM$ is similar to both $\triangle AO_1P_1$ and $\triangle AO_2P_2$, so $OM = OA = OB$, which proves the claim. So, if we let $B'$ be the antipode of $B$ wrt the circle from the last claim (which is fixed), then we always have $\overline{B'M} \perp \overline{P_1P_2}$, and thus $B'P_1 = B'P_2$, so we are done.

Suppose the second intersection of the two given circles is $B$. Denote segments $P_1Q_1$ and $P_2Q_2$, with the intersection of their perpendicular bisectors as $X$, and $P_1P_2 \cap Q_1Q_2 = C$. Let $O_P$, $O_Q$, $O_1$, and $O_2$ be the circumcenters of $\triangle AP_1P_2$, $\triangle AQ_1Q_2$, $\triangle CP_1Q_1$, and $\triangle CP_2Q_2$, respectively. By Miquel point properties, it is well known that $AO_1O_2O_PO_Q$ is cyclic (and fixed), so proving that $X$ also lies on the circle suffices. It's easy to show $B$ will lie on $P_1Q_1$ and $P_2Q_2$ through the angle condition implied by the problem. Spiral similarity at $A$ gives \[\triangle AO_1O_2 \sim \triangle AP_1P_2 \sim \triangle AQ_1Q_2\]\[\implies \angle O_1AO_2 = \angle Q_1AQ_2 = \angle Q_1BQ_2 = \angle O_1XO_2.\] Thus $X$ lies on the fixed circle as a fixed point, completing the proof. $\blacksquare$ [asy][asy] size(250); defaultpen(linewidth(.4)); pair A,B,C,P1,P2,Q1,Q2,O1,O2,OP,OQ,X; A = (0,24); B = (0,-24); P1 = 40dir(200)-32; P2 = 40dir(260)-32; Q1 = 51dir(315)+45; Q2 = 51dir(15)+45; OP = (-32,0); OQ = (45,0); C = extension(P1,P2,Q1,Q2); O1 = circumcenter(C,P1,Q1); O2 = circumcenter(C,P2,Q2); X = extension(O1,foot(O1,P1,Q1),O2,foot(O2,P2,Q2)); filldraw(A--P1--P2--cycle^^A--O1--O2--cycle^^A--Q1--Q2--cycle, palegreen); draw(A--P1--C--Q2--cycle^^Q1--P2--A--Q1--P1--P2--Q2^^A--O1--O2--cycle); draw(circle(OP,40)^^circle(OQ,51)); draw(circumcircle(A,O1,OQ),blue+linewidth(.6)+dashed); draw(O1--X--O2, blue+linewidth(.6)); dot("$A$", A, dir(90)); dot("$B$", B, dir(135)); dot("$C$", C, dir(270)); dot("$P_1$", P1, dir(180)); dot("$P_2$", P2, dir(225)); dot("$Q_1$", Q1, dir(0)); dot("$Q_2$", Q2, dir(20)); dot("$O_1$", O1, dir(270)); dot("$O_2$", O2, dir(0)); dot("$X$", X, dir(90)); dot(OP); dot(OQ); [/asy][/asy]

I used Evan's diagram for this problem (cannot secretly draw diagrams in class): Let the second intersection of the two circles be $D,$ and draw the line perpendicular to $AD$ passing through $D$ such that it intersects the circle on the left at $B$ and the other circle at $C.$ Notice $AB$ and $AC$ are both diameters so when one point is at $B$ the other is at $C.$ Let $M$ be the midpoint of $BC$ and $Z \neq M$ the point on $(AMD)$ such that $ZM \perp BC.$ We claim $Z$ is the point we are looking for. Proof: Draw another point $X$ on the left circle such that when the point is at $X$ the other point is at $Y$ on the right circle. Observe there is a spiral similarity centered at $A$ between $XB$ and $YC$ so there is also one between $XY$ and $BC.$ Thus $XY$ passes through $D,$ the second intersection of the two circles. By Spiral Similarity $\angle AND=\angle AMC=180^\circ-\angle AMD$ so $N$ lies on $(AMDZ).$ As $\angle ZMD=90^\circ$ we know $\angle ZND=180=90=90^\circ$ so $ZN$ is the perpendicular bisector and $ZX=ZY.$ For uniqueness, observe two different perpendicular bisectors intersect at exactly one point, so $Z$ is unique.

This is also a quick bash! Toss the figure in complex plane! Let the center of circles be $O_1 (x_1)$ and $O_2 (x_2) $ and their point of intersection be $P(ai)$ After one of the particles has turned through angle $\theta$ other particle must also have turned through the same angle Thus, $A(z_1) \equiv ((ai-x_1)(\cos{\theta} +i\sin{\theta})+x_i) $ and $B(z_2) \equiv ((ai-x_2)(\cos{\theta} +i\sin{\theta})+x_2)$ Consider $z=x_1+x_2+ai$ t is easy to check that $|z-z_1|^2=|z-z_2|^2$ $= x_1^2+x_2^2+2a^2+2x_1x_2\cos{\theta}+2x_2a\sin{\theta}+2x_1a\sin{\theta}-2a^2\cos{\theta}$

Draw a few good diagrams, find the correct point, and it's school level geo

XOOKS Let the points be $X,Y$ on circles centred at $O_1, O_2$. now xoink $\angle AO_1X=\angle AO_2Y$ we skibidiconstruct $P$ as the point such $AO_1O_2P$ is an isosceles trapezium. we are done because $O_1P=AO_2=O_2Y$, $XO_1=O_1A=O_2P$. check that $\angle XO_1P=\angle YO_2P$ because they are differences of equal angles (bowtie + our previous angle inequality) so done by SAS congruence. ($PX=PY$)

I'll present two solutions; the first shows a succinct way to prove the main claim by complex numbers, and the second finishes the problem via a pure length chase.