I still have to write up a complex numbers approach for this, but in the interim... Let $O$ be the center of $\omega$, and let $M = \overline{PQ} \cap \overline{AB}$ and $N = \overline{PQ} \cap \overline{AC}$ be the midpoints of $\overline{AB}$ and $\overline{AC}$ respectively. Refer to the diagram below. [asy][asy] size(9cm); pair A, B, C, O, M, N, P, Q, E1, F1, E2, F2, O1, O2; A = dir(105); B = dir(190); C = dir(350); O = origin; M = (A + B)/2; N = (A + C)/2; P = point(unitcircle, intersections(unitcircle, M, N)[1]); Q = point(unitcircle, intersections(unitcircle, M, N)[0]); E1 = extension(A, C, P, rotate(90, P) * origin); F1 = extension(A, B, P, rotate(90, P) * origin); E2 = extension(A, C, Q, rotate(90, Q) * origin); F2 = extension(A, B, Q, rotate(90, Q) * origin); O1 = circumcenter(A, E1, F1); O2 = circumcenter(A, E2, F2); fill(O--P--E1--cycle^^O--Q--E2--cycle, lightblue + lightred + gray(0.5)); draw(O--P^^O--Q^^O--M^^O--N, gray(0.5)); draw(E1--F1^^E2--F2); draw(arc(O1, F1, E1)^^arc(O2, F2, E2), dashed); draw(unitcircle); draw(A--B--C--cycle); draw(P--Q); dot(A^^B^^C^^O^^M^^N^^P^^Q^^E1^^F1^^E2^^F2); label("$A$", A, dir(60)); label("$B$", B, dir(B)); label("$C$", C, dir(C)); label("$O$", O, dir(200)); label("$M$", M, dir(120)); label("$N$", N, dir(60)); label("$P$", P, dir(P)); label("$Q$", Q, dir(Q)); label("$E_1$", E1, dir(P)); label("$F_1$", F1, dir(P)); label("$E_2$", E2, dir(Q)); label("$F_2$", F2, dir(Q)); [/asy][/asy] The main idea is to prove two key claims involving $O$, which imply the result: quadrilaterals $AOE_1F_1$ and $AOE_2F_2$ are cyclic (giving the radical axis is $\overline{AO}$), $\triangle OE_1F_1 \cong \triangle OE_2F_2$ (giving the congruence of the circles). We first note that (i) and (ii) are equivalent. Indeed, because $OP = OQ$, (ii) is equivalent to just the similarity $\triangle OE_1F_1 \sim \triangle OE_2F_2$, and then by the spiral similarity lemma (or even just angle chasing) we have (i) $\iff$ (ii). We now present four proofs, two of (i) and two of (ii). Proof of (i) by angle chasing Note that \[ \measuredangle F_2E_2O = \measuredangle QE_2O = \measuredangle QNO = \measuredangle MNO = \measuredangle MAO = \measuredangle F_2AO \]and hence $E_2OAF_2$ is cyclic. Similarly, $E_1OAF_1$ is cyclic. Proof of (i) by Simson lines Since $P$, $M$, $N$ are collinear, we see that $\overline{PMN}$ is the Simson line of $O$ with respect to $\triangle AE_1F_1$. Proof of (ii) by butterfly theorem By butterfly theorem on the three chords $\overline{AC}$, $\overline{PQ}$, $\overline{PQ}$, it follows that $E_1N = NE_2$. Thus \[ E_1P = \sqrt{E_1A \cdot E_1C} = \sqrt{E_2A \cdot E_2C} = E_2P. \]But also $OP = OQ$ and hence $\triangle OPE_1 \cong \triangle OQE_2$. Similarly for the other pair. Proof of (ii) by projective geometry Let $T = \overline{PP} \cap \overline{QQ}$. Let $S$ be on $\overline{PQ}$ with $\overline{ST} \parallel \overline{AC}$; then $\overline{TS} \perp \overline{ON}$, and it follows $\overline{ST}$ is the polar of $N$ (it passes through $T$ by La Hire). Now, \[ -1 = (PQ;NT) \overset{T}{=} (E_1E_2;N\infty) \]with $\infty = \overline{AC} \cap \overline{ST}$ the point at infinity. Hence $E_1 N = N E_2$ and we can proceed as in the previous solution. Remark: The assumption that $\triangle ABC$ is acute is not necessary; it is only present to ensure that $P$ lies on segment $E_1F_1$ and $Q$ lies on segment $E_2F_2$, which may be helpful for contestants. The argument presented above is valid in all configurations. When one of $\angle B$ and $\angle C$ is a right angle, some of the points $E_1$, $F_1$, $E_2$, $F_2$ lie at infinity; when one of them is obtuse, both $P$ and $Q$ lie outside segments $E_1F_1$ and $E_2F_2$ respectively.

Due to the $A-$midline of $ABC$ being the Simson line of $O$ with respect to $AE_1F_1$, we have that $OEA_1F_1$ is a cyclic quadrilateral. So we have $\angle OE_1F_1 = 90^\circ - \angle C, \angle OF_1E_1 = 90^\circ - \angle B$. So since the altitudes from $O$ to the tangents at $P, Q$ are equal, triangles $OE_1F_1$ and $OE_2F_2$ are congruent. Hence their circumcircles are congruent, proving the first part. Also since the radical axis is $AO$, the line joining the centers is perpendicular to it, giving (ii).

v_Enhance wrote: Let $ABC$ be an acute triangle with circumcircle $\omega$, and let $H$ be the foot of the altitude from $A$ to $\overline{BC}$. Let $P$ and $Q$ be the points on $\omega$ with $PA = PH$ and $QA = QH$. The tangent to $\omega$ at $P$ intersects lines $AC$ and $AB$ at $E_1$ and $F_1$ respectively; the tangent to $\omega$ at $Q$ intersects lines $AC$ and $AB$ at $E_2$ and $F_2$ respectively. Show that the circumcircles of $\triangle AE_1F_1$ and $\triangle AE_2F_2$ are congruent, and the line through their centers is parallel to the tangent to $\omega$ at $A$. Ankan Bhattacharya and Evan Chen Let $PQ$ intersect $AB$, $AC$ at $M$, $N$ respectively. Since $PQ$ is a perpendicular bisector of $AH$, and $AH$ is perpendicular to $BC$, $PQ$ must be the $A$-midline of $ABC$. Thus, $M$ and $N$ are the midpoints of $AB$ and $AC$ respectively. Since $OP \perp E_1F_1$, $OM \perp AF_1$, $ON \perp AE_1$, so $O$ lies on the circumcircle of $AE_1F_1$ by converse of Simson Line. Similarly, $O$ lies on circumcircle of $AE_2F_2$. Since the line through the centre of these circles is perpendicular to their radical axis, which is $OA$, we have the line through their centres is parallel to the tangent of $\omega$ at $A$, proving the second part. Angle chasing in the cyclic quadrilaterals, we get $\angle OE_1F_1 = \angle OAF_1 = \angle OAB$, and $\angle OE_2F_2 = \angle OAB$, so $\angle OE_1F_1 = \angle OE_2F_2$. Similarly, $\angle OF_1E_1 = \angle OF_2E_2$, so we have $OE_1F_1 \sim OE_2F_2$. However, $OP$ and $OQ$ are their respective altitudes from $O$ and are equal in length by radius of circle. Therefore, $OE_1F_1 \cong OE_2F_2$ so their circumcircles are congruent, proving the first part.

Let $P_1$ be the midpoint of $AB$ and $Q_1$ the midpoint on $AC$.Points $P_1$ and $Q_1$ are on $PQ$. We have $OQ_1 \perp AE_2$ so $OQ_1QE_2$ is cyclic.Also $OP_1PF_1$ is cyclic.After some angle chasing $\angle OE_2Q = \angle OQ_1P = \angle OAP_1 $ ( $AP_1OQ_1$ is cyclic) , we obtain $AF_2E_2O$ cyclic .Also $AOF_1E_1$ is cyclic . Angle chasing $\angle AE_2O= \angle OQP = \angle OPQ = \ AF_1O $ . Notice that circle $(AE_1F_1)$ is circle $(AOF_1)$ (with $R_1$ radius )so after law of sines $\frac {AO}{sin \angle AF_1O}=2*R_1$ and circle $(AE_2F_2)$ is circle $(AOE_2)$ (with $R_2$ radius )so after law of sines $\frac {AO}{sin \angle AF_2O}=2*R_2$ , then we have $R_1=R_2$. For the second part the centre of circle $(AE_1F_1)$ is on the perpendicular bisector of $AO$ and the centre of circle $(AE_2F_2)$ is on the perpendicular bisector of $AO$ therefore the line through their centers is perpendicular to $AO$ so is parallel to the tangent to $\omega$ at $A$

Let $O,O_1,O_2$ be the centers of $\omega,\odot (AE_1F_1), \odot (AE_2F_2)$ and $M,N$ be the midpoints of $\overline{AB},\overline{AC}.$ Easily see that $P,M,N,Q$ are collinear. Considering the power of $E_2$ with respect to $\omega$ and the circle with diameter $\overline{AC},$ we see that $E_2Q^2=E_2A\cdot E_2C=E_2N^2-\left(\frac{b}{2}\right)^2.$ Moreover, by Law of Sines in $\triangle E_2QN,$ we see that \[\frac{E_2Q}{E_2N}=\frac{\sin\angle E_2NQ}{\sin\angle E_2QN}=\frac{\sin\angle C}{\sin\angle PAQ}=\frac{c}{PQ}.\] Substituting, we find $E_2Q=\frac{bc}{\sqrt{2(PQ^2-b^2)}}.$ By symmetry, we get $F_1Q=E_2Q,F_2P=E_1P=\frac{bc}{\sqrt{2(PQ^2-c^2)}},$ which shows $E_1F_1=E_2F_2.$ Since $\angle E_1AF_1=\angle E_2AF_2,$ Law of Sines shows that the circumradii of the two triangles are indeed equal. The second part is also easy. $E_1P=F_2Q$ implies that $E_1,F_2$ have the same power WRT $\omega,$ so $OE_1=OF_2.$ Similarly $OE_2=OF_1,$ so $O$ is the spiral center that sends $E_1F_1\to F_2E_2.$ Thus, $O$ lies on both $\odot (AE_1F_1)$ and $\odot (AE_2F_2),$ so $O_1O_2\perp AO \perp \text{tangent at } A$ and we are done.

This problem is so amazing yet so simple. Drawing the diagram itself using compass and ruler is a pain though. Let $O$ be the centre of $\omega$ and $M$ and $N$ the mid points of $AB$ and $AC$, respectively. Claim 1: $O = \odot(AE_1F_1) \cap \odot(AE_2F_2)$. Proof: Note that $M, N, P$ are collinear and $OP\perp E_1F_1, OM\perp AF_1$ and $ON \perp AE_1$. Thus, $MNP$ is the Simson Line of $O$ w.r.t. $\odot(AE_1F_1)$, therefore, $O\in \odot(AE_1F_1)$. Similarly, $O\in \odot(AE_2F_2)$. Claim 2: $\triangle OE_1F_1\cong \triangle OE_2F_2$ Proof: Note that by spiral similarity $\triangle OE_1F_1 \sim \triangle OE_2F_2$. Also, note that $OP = OQ$. Since, the two triangles are similar and their analogous altitudes are equal, the two triangles are congruent. Main Proof: Note that, since, $\triangle OE_1F_1\cong \triangle OE_2F_2$, their circumradii are equal. Note that $AO$ is the radical axis of $\odot(PE_1F_1)$ and $\odot(QE_2F_2)$, thus, it is perpendicular to the line joing the centres of the two circles. Also, $AO \perp \text{Tangent to }\omega\text{ at }A$. Thus, the two lines are parallel.

A few people have asked me how I created the problem, so I will offer some comments. Some accounts may be omitted, exaggerated, or wrong. I was originally trying to write an AMC/AIME-level problem involving an equilateral triangle in some way. Eventually, I decided on the following. v0 wrote: Let $ABC$ be an equilateral triangle with side 2 inscribed in circle $\omega$, and let $P$ be a point on small arc $AB$ of its circumcircle. The tangent line to $\omega$ at $P$ intersects lines $AC$ and $AB$ at $E$ and $F$. If $PE = PF$, find $EF$.

Of course, this is equivalent to $P$ being on the $A$-midline. At this point, it was natural to conjecture that we can relax the condition on $\triangle ABC$ to only $AB = AC$, and it turns out this property still holds. So we get the following problem. v1 wrote: Let $ABC$ be an isosceles triangle with $AB = AC$, and let $M$ be the midpoint of $\overline{BC}$. Let $P$ be a point on the circumcircle with $PA = PM$. The tangent to the circumcircle at $P$ intersects lines $AC$ and $AB$ at $E$ and $F$, respectively. Show that $PE = PF$. After some time, I was able to find the following solution (hidden for length).

I was happy with this solution; I felt like I was doing nontrivial geometry, so I showed the problem to Evan. While he believed it was nice, he was worried that people would destroy the problem by coordinates. So I suggested the following variant. v2 wrote: Let $ABC$ be an isosceles triangle with $AB = AC$ and circumcircle $\omega$, and let $M$ be the midpoint of $\overline{BC}$. Let $P$ be a point on $\omega$ with $PA = PM$. The tangent to $\omega$ at $P$ intersects lines $AC$ and $AB$ at $E$ and $F$, respectively. Show that the circumcircle of $\triangle AEF$ passes through the center of $\omega$. To solve this after $PE = PF$, note that it follows that the center $O$ of $\omega$ is equidistant from $E$ and $F$, and $\overline{AO}$ externally bisects $\angle EAF$, so $O$ is the arc midpoint. Both of us were happy about this change, and we moved on to bigger and better things in life. Well, that's not really true. Over a month later, I returned to the problem above (v2), and noticed it was true for any triangle $ABC$. Oops. v3 wrote: Let $ABC$ be a triangle with circumcircle $\omega$, and let $H$ be the foot of the altitude from $A$ to $\overline{BC}$. Let $P$ be a point on $\omega$ with $PA = PH$. The tangent to $\omega$ at $P$ intersects lines $AC$ and $AB$ at $E$ and $F$, respectively. Show that the circumcircle of $\triangle AEF$ passes through the center of $\omega$. After trying the generalized problem for a while, I finally had the idea to add in the midpoints of $\overline{AB}$ and $\overline{AC}$, after which the problem is just angle chasing with cyclic quads. (Yep, I didn't notice the Simson line.) Welp. I considered removing the problem because I thought the angle chase solution was too easy, but was persuaded otherwise by others. Nevertheless I still felt that the problem has no content, so I searched for more properties of this diagram. Now I decided to add in both intersections $P$ and $Q$ of the $A$-midline with $\omega$, and all four points $E_1$, $F_1$, $E_2$, $F_2$. Then $\triangle OE_1F_1 \sim \triangle OE_2F_2$ by spiral sim lemma. However, the spiral sim lemma also applies to smaller triangles: indeed, we obtain $\triangle OPE_1 \sim \triangle OQE_2$ and $\triangle OPF_1 \sim \triangle OQF_2$. But $OP = OQ$, so in fact $E_1F_1 = E_2F_2$. So now I had something like this. v4 wrote: Let $ABC$ be an acute triangle with circumcircle $\omega$, and let $H$ be the foot of the altitude from $A$ to $\overline{BC}$. Let $P$ and $Q$ be the points on $\omega$ with $PA = PH$ and $QA = QH$. The tangent to $\omega$ at $P$ intersects lines $AC$ and $AB$ at $E_1$ and $F_1$ respectively; the tangent to $\omega$ at $Q$ intersects lines $AC$ and $AB$ at $E_2$ and $F_2$ respectively. Show that $E_1F_1 = E_2F_2$. To make the problem more substantial, Evan suggested to combine the two parts and reworded the problem: v5 wrote: Let $ABC$ be an acute triangle with circumcircle $\omega$, and let $H$ be the foot of the altitude from $A$ to $\overline{BC}$. Let $P$ and $Q$ be the points on $\omega$ with $PA = PH$ and $QA = QH$. The tangent to $\omega$ at $P$ intersects lines $AC$ and $AB$ at $E_1$ and $F_1$ respectively; the tangent to $\omega$ at $Q$ intersects lines $AC$ and $AB$ at $E_2$ and $F_2$ respectively. Show that the circumcircles of $\triangle AE_1F_1$ and $\triangle AE_2F_2$ are congruent, and the line through their centers is parallel to the tangent to $\omega$ at $A$. However many of the testsolvers thought this problem was too hard for 1/4/7, so we bumped it up to a possible candidate for 2/5/8. Then this problem ended up being selected. The end.

It is easy to show that $E_1 F_1=PH\cdot QH (\frac {1}{BH}+\frac {1}{CH})=E_2 F_2$ For second part if $O,O_1,O_2$ are circumcenters of $(ABC),(AE_1 F_1),(AE_2 F_2) $ then we can easily show that triangles $AOO_1$ and $AOO_2$ are congruent ($AO $ is common ,according to first part $AO_1=AO_2$ and $\angle OAO_1=\angle PBQ=\angle OAO_2$ )So $OO_1=OO_2$ $\implies$ $O $ lies on radical axis of $(AE_1 F_1) $ and $(AE_2 F_2) $.

It is true for all \(PQ||BC.\) And thus we don't need circumcenter \(O.\)

Since several people already presented solutions, I guess I will be lazy.

I'm back baby!

Let $M$ and $N$ be the midpoints of $\overline{AB}$ and $\overline{AC}.$ Note that $M, N \in PQ.$ It follows that $\overline{MNP}$ is the Simson line of $O$ WRT $\triangle AE_1F_1$, so $O \in \odot(AE_1F_1).$ Similarly, $O \in \odot(AE_2F_2).$ Therefore, $\odot(AE_1F_1)$ and $\odot(AE_2F_2)$ have radical axis $AO$, whence the second part follows. For the first part, by Desargues' involution theorem for quadrilateral $PPQQ$ cut by $AC$, there is an involution fixing $N$ and swapping $(A, C), (E_1, E_2).$ This involution must be reflection in $N.$ It follows from symmetry in $N$ that $\angle AE_1O = \angle AE_2O.$ Therefore by the law of sines, $\triangle AE_1O$ and $\triangle AE_2O$ have equal circumradii, hence congruent circumcircles.

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label("$F_{1}$", (-14.446714545608417,-11.883167845590842), NE * labelscalefactor); dot((6.017017631714434,-7.24778058269769),linewidth(4pt) + dotstyle); label("$E_{2}$", (6.1993931400133215,-6.861141651790959), NE * labelscalefactor); dot((0.22577273433484024,13.389266114459943),linewidth(4pt) + dotstyle); label("$F_{2}$", (0.4164538865467904,13.78496603383078), NE * labelscalefactor); dot((-6.27305716110433,0.6510061141266021),linewidth(4pt) + dotstyle); label("$M$", (-6.07667088927528,1.0523541687421896), NE * labelscalefactor); dot((-0.2430571611043312,0.5810061141266015),linewidth(4pt) + dotstyle); label("$N$", (-0.04009394925319893,1.0016266314310796), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Note that $PQ$ is the perpendicular bisector of $AH$, so a homothety at $A$ with scale factor $2$ shows that it is the same line as $MN$. Now, note that $ONE_1P$ is cyclic because $\angle ONE_1=\angle OPE_1=\pi/2$, and $ONAM$ is cyclic with diameter $OA$. Therefore, the center of the spiral similarity that sends $AM\to E_1P$ is $(AMN)\cap (E_1PN)=O$. Therefore, $\angle OE_1P=\angle OAM$, and similarly we can show that $\angle OF_1P=\angle OAN$. Thus, \[\angle E_1OF_1=\pi-\angle OE_2P-\angle OF_1P=\pi-\angle A=\angle E_1AF_1,\]so $(AE_1F_1)=(OE_1F_1)$. By the cyclic quads $E_1NOP$ and $F_2ONQ$, we have that \[\angle OE_1E_2=\angle OPQ=\angle OQP=\angle OE_2E_1,\]so $OE_1=OE_2$. Similarly, $OF_1=OF_2$, and $\angle E_1OF_1=\pi-A=\angle E_2OF_2$, so $\triangle OE_1F_1\cong\triangle OE_2F_2$. Therefore, the circumcircles are also congruent. Note that the line connecting their centers is perpendicular to their radical axis which is $OA$, which is perpendicular to the tangent at $A$. Therefore, the line connecting the centers is perpendicular to the tangent at $A$, as desired. $\blacksquare$

this seems like a pretty silly problem Construct $P'$ on $(ABC)$ such that $PP' \parallel AB$. It is easy to see that $QABP'$ is a harmonic quadrilateral, so $E_2P'$ is tangent to $(ABC)$; thus it follows by symmetry that $PF_1 = P'E_2 = QE_2$. Therefore similarly $PE_1 = QF_2$, so $E_1F_1 = E_2F_2$ and it follows by the Extended Law of Sines that $(AE_1F_1)$ and $(AE_2F_2)$ have the same circumradius, so we may conclude for the first part. As for the second part: observe now that if $O$ is the circumcenter of $ABC$, then $OE_1 = OF_2$ and $OE_2 = OF_1$, and $\angle POF_1 = \angle{QOE_2}$ and $\angle POE_1 = \angle{QOF_2}$; thus triangles $E_1OF_1$ and $F_2OE_2$ are congruent. Therefore by the spiral similarity lemma applied to $E_1F_2$ and $E_2F_1$ we know that $(E_1AF_1)$ and $(E_2AF_2)$ meet at $O$ and we are done.

RC. wrote: It is true for all \(PQ||BC.\) And thus we don't need circumcenter \(O.\) jeff10 wrote: As mentioned above, it is worth noting that we only needed $PQ$ to be parallel to $BC$. On geogebra it looks like it needs to be midline.

Let $B'$ and $C'$ be the midpoints of $AB$ and $AC$, and let the tangents at $P$ and $Q$ meet at $X$. Also, call the circumcenter of $ABC$ to be $O$. Claim 1. $OE_1=OE_2$ and $OF_1=OF_2$. Proof. $OPE_1C'$ is cylic $\Rightarrow \angle OE_1C'=\angle OPC'=\angle OPQ$. Similarly, $OC'QE_2$ is cyclic$\Rightarrow \angle OE_2C'=\angle OQC'=\angle OCP$. So $\angle OE_1E_2=\angle OE_2E_1$ and so $OE_1=OE_2$. Similarly $OF_1=OF_2.\qquad{\blacksquare}$ Claim 2. $OE_1XE_2$ and $OF_1XF_2$ are cyclic. Proof. We have $\angle E_1OC'=\angle E_1PQ=\angle PBQ$ and $\angle E_2OC'=180^{\circ}-\angle E_2QC'=\angle PBQ$. So $$\angle E_1OE_2=2\angle PBQ=\angle POQ=180^{\circ}-\angle PXQ=180^{\circ}-\angle E_1XE_2,$$proving the claim.$\qquad\blacksquare$ We used the fact in Claim 2 that $\angle E_1OE_2=2\angle PBQ$. Similarly, we have $\angle F_1OF_2=2\angle PBQ$, so $$\angle E_1OE_2=\angle F_1OF_2\Rightarrow \angle E_1OF_1=\angle E_2OF_2.$$Since $OE_1=OE_2$ and $OF_1=OF_2$, we conclude that $\Delta OE_1F_1\cong\Delta OF_2E_2$, and so the circumcircles of $\Delta AE_1F_1$ and $\Delta AE_2F_2$ are congruent. This proves part (i). Now to prove part (ii). The second claim shows that $O$ is the center of spiral similarity taking $E_1F_1$ to $E_2F_2$. So circles $(AE_2F_1)$ and $(AE_2F_2)$ intersect at $A$ as well as at $O$. So $AO$ is the radical axis of circles $(AE_1F_1)$ and $(AE_2F_2)$. So the line joining their centers are perpendicular to $AO$, and so it is parallel to the tangent to $\omega$ at $A$.

Just angle-chasing let $X = AB\cap PQ $ and $O$ is the circumcenter of $(ABC)$ Claim(1): $OXPF_1$ is cyclic obviously $X$ is the midpoint of $AB$ thus $\angle OXF_1=90=OPF_1$ so $\angle OF_1A=OPQ=QPO=OE_2A$ which means that $R_1 =R_2$ Claim(2):$AOF_1E_1 , AOE_2F_2$ are cyclics $\angle OF_1A=\angle OF_1P= \angle OXC = \angle 90 -\angle BAC=\angle OAE_1$ thus $O_1O_2 $ is perpendicular to $AO $

Proof of part (a) Let $M,N$ be the midpoints of $AB$ and $AC$ and $PQ$ is the $A$ midline of $\triangle ABC$. So by degenerate butterfly on chords $AB$ and $AC$, we have $M$ and $N$ is the midpoint of $AB$ and $AC$ respectively. Now reflect $F_2$ over $N$ to get $X$ such that $F_2E_1XE_2$ is a parallelogram. Now we have $F_1X \parallel BC$, so $\angle E_1F_1X=\angle E_1PQ=\angle F_2QP=\angle (F_2E_2,BC)=\angle (E_1X,BC)=\angle E_1XF_1$, which implies, $E_1F_1=E_1X=E_2F_2$ and we are done. Proof of part (b) We have $90^\circ -\angle C=\angle ONM=\angle OE_2Q=180^\circ -\angle OAF_2$ So, $O,A,F_2,E_2$ are concyclic and similarly $O,A,F_1,E_1$ are concyclic, which implies $OA\perp O_1O_2$ and we are done.

This was also proposed as 2019 IITB Mathathon Round 4 p2

Let $E$ and $F$ be the midpoints of $\overline{AB}$ and $\overline{AC},$ respectively. Claim: $O,$ the center of $(ABC),$ lies on $(AE_1F_1).$ Proof. Notice $E,P,$ and $Q,$ the perpendiculars from $O$ to the sides of $\triangle AE_1F_1,$ are collinear. Simson line finishes. $\blacksquare$ Then, $$\measuredangle QE_2O=\measuredangle FAO=\measuredangle F_1AO=\measuredangle F_1E_1O,$$$OP=OQ,$ and $\angle OPE_1=\angle OQE_2=90$ so $\triangle OQE_1\cong\triangle OQE_2.$ Similarly, $\triangle OPF_1\cong\triangle OQF_2$ and so $\triangle OE_1F_1\cong\triangle OE_2F_2.$ Since $(OE_1F_1)=(AE_1F_1)$ and $(OE_2F_2)=(AE_2F_2),$ we are done with the first part of the problem. Notice $\overline{AO}$ is the radical axis of $(AE_1F_1)$ and $(AE_2F_2)$ so $\overline{AO}\perp\overline{O_1O_2}.$ But the tangent to $(ABC)$ at $A$ is also perpendicular to $\overline{AO}$ which suffices. $\square$

Let $E$ and $F$ be the midpoints of $AB$ and $AC$, respectively, and let $O$ be the circumcenter of $\triangle ABC$. Since $EF$ is the perpendicular bisector of $AH$, $P$ and $Q$ are the intersections of $EF$ with $\omega$. Since $\angle OPF_1 = \angle OEF_1 = 90^\circ$, $(OEPF_1)$ is cyclic, and since $\angle OPE_1 = \angle OFE_1 = 90^\circ$, $(OPE_1F)$ is cyclic. Similarly, $(OFQE_2)$ and $(OQF_2E)$ are cyclic. Now \[\angle AF_1O = \angle EF_1O = \angle EPO = \angle FPO = \angle FE_1O = \angle AE_1O,\]so $(AE_1F_1O)$ is cyclic, and similarly $(AE_2F_2O)$ is cyclic. Furthermore, \[\angle OE_1E_2 = \angle OE_1F = \angle OPF = \angle OQF = \angle OE_2F = \angle OE_2E_1,\]so $OE_1 = OE_2$, and similarly $OF_1 = OF_2$. Then $OP = OQ$, $OE_1 = OE_2$, and $\angle OPE_1 = \angle OQE_2 = 90^\circ$ implies $\triangle OPE_1 \cong \triangle OQE_2$, and similarly $\triangle OPF_1 \cong \triangle OQF_2$. Thus $E_1P = E_2Q$ and $F_1P = F_2Q$; adding the two together gives $E_1F_1 = E_2F_2$. Let $R_1$ and $R_2$ be the radii of $(AE_1F_1O)$ and $(AE_2F_2O)$, respectively, and $O_1$ and $O_2$ be their centers. By the Extended Law of Sines, \[2R_1 = \frac{E_1F_1}{\sin \angle E_1AF_1} = \frac{E_2F_2}{\sin \angle E_2AF_2} = 2R_2,\]so $R_1 = R_2$, giving that the two circumcircles are congruent, as desired. Let $\ell$ be the tangent to $\omega$ at $A$; then $O_1O_2 \perp AO \perp \ell$ gives $O_1O_2 \parallel \ell$, as desired.

Let $\overline{PQ} \cap \overline{AB} = M$ and $\overline{PQ} \cap \overline{AC} = N$, and it is easy to see that $M,N$ are the midpoints of $\overline{AB}, \overline{AC}$ respectively. We also let $O$ be the circumcenter of $ABC$. The main claim of the problem is the following: Claim: $AOE_1F_1$ is cyclic; similarly, $AOF_2E_2$ is cyclic. Proof. We prove that $AOE_2F_2$ is cyclic (the other proof is similar.) Indeed, we have \[ \measuredangle F_2E_2O = \measuredangle QE_2O = \measuredangle QNO = \measuredangle MNO = \measuredangle MAO = \measuredangle F_2AO \]as desired. $\blacksquare$ Now, \[\measuredangle OE_1A = \measuredangle OE_1M = \measuredangle OPM = \measuredangle NQO = \measuredangle NF_2O = \measuredangle AF_2O = \measuredangle AE_2O \]\[\implies OE_1 = OE_2 \implies \text{Pow}_{(ABC)}(E_1) = \text{Pow}_{(ABC)}(E_2) \implies E_1P = E_2Q.\]In a similar fashion, we have $F_1P = F_2Q \implies E_1F_1 = E_2F_2$, which when coupled with $\measuredangle F_1AE_1 = \measuredangle F_2AE_2$ implies $(AE_1F_1)$ and $(AE_2F_2)$ are congruent, which solves the first part. For the second part, note that $AO$ is the radical axis of $(AE_1F_1)$ and $(AE_2F_2)$, hence $AO$ is perpendicular to the line connecting the centers of the two circles, but $AO \perp AA$, which is sufficient to finish. $\square$ (note that the point H is useless for this proof, and can be thrown away pretty quickly)

$\text{the critical observation is:}\; \triangle E_1AP\sim \triangle QCH\; \text{and}\; \triangle E_2QA\sim \triangle PHC$

Let $O$ be the center of $\omega$. The key claim is the following. Claim: $AE_1F_1O$ and $AE_2F_2O$ are cyclic. Proof: Let $M$ and $N$ be the midpoints of $\overline{AB}$ and $\overline{AC}$ respectively, so $M,N,P,Q$ are collinear. Then the feet of the perpendiculars from $O$ to the sides to $\triangle AE_1F_1$ and $\triangle AE_2F_2$ are $P,M,N$ and $Q,M,N$ respectively, so $O$ lies on $(AE_1F_1)$ and $(AE_2F_2)$ by Simson lines. $\blacksquare$ This immediately yields that the line through the centers of $(AE_1F_1O)$ and $(AE_2F_2O)$ is parallel to the tangent at $A$, since both are perpendicular to $\overline{AO}$. Furthermore, due to spiral similarity, the claim implies that $\triangle OE_1F_1 \sim \triangle OE_2F_2$. It suffices to show that this similarity is actually a congruence, which follows because the $O$-altitudes of the respective triangles are $\overline{OP}$ and $\overline{OQ}$, which are also both radii of $\omega$ and therefore have equal length. $\blacksquare$

Make the config s.t. Q is to the right of P. We'll show that O is the second intersection of the circles, and that the circles are congruent, from which the second part follows by radax (where $O_1,O_2$ are the circumcenters of $(AE_1F_1)$ etc.). Note that PQ is just the midline of ABC, so if M,N are the midpoints of AB,AC, then $ONE_2=90=OQE_2$, so $F_2E_2O=180-ONQ=ONM=OAM=180-OAF_2$, and analogous results makes cyclic quads. To finish, $OE_1F_1=OAB=90-ACB=90-ANM=ONM=180-ONQ=OE_2F_2$ means spiral sim taking $E_1P\to E_2Q$ is a rotation, since OP=OQ, so $OE_1F_1\cong OE_2F_2$, so the circles are congruent, as desired.

Let $K$ be where $\overline{PQ}$ hits $\overline{AB}$. Let $\angle PAB = \alpha$, so $\angle PBA = \angle C - \alpha$. Now \[\angle BPK = 180 - \angle PBA - \angle PBK = 180 - (\angle C - \alpha) - \angle B = \angle A + \alpha\]Notice since $K$ is the midpoint of $AB$, $\overline{OK} \perp \overline{AB}$, so $PKOF_1$ is cyclic. Now \[\angle POK = \angle PF_1K = 180 - \angle F_1PB - \angle F_1BP = 180 - \alpha - (180-\angle C+\alpha) = \angle C - 2\alpha\]\[\implies \angle KPO = 180 - (90+\angle B) - (\angle C - 2\alpha) = 2\alpha + \angle A - 90 \implies \angle KF_1O = 2\alpha + \angle A - 90\]Now a quick check yields $\angle E_1F_1O + \angle E_1AO = 180$, since $\angle E_1AO = 90 + \angle B$. Thus $E_1AOF_1$ cyclic. Lastly, \[\angle AQP = \angle PBA = \angle C - \alpha \implies \angle QAC = \alpha\]Repeating the answer chase gives $F_2AOE_2$ cyclic. Now the centers of these two circles are perpendicular to $OA$ (the radical axis), so this proves the second condition. Now notice $\angle E_1OF_1 = \angle E_1AF_1 = \angle E_2AF_2 = \angle E_2OF_2$. Furthermore, since $OP = OQ = $ radius of $(ABC)$, this implies $\Delta E_1OF_1 \cong \Delta E_2OF_2$, proving the first condition.

i dont like this very much WLOG $P$ is on minor arc $AB$ and $Q$ on minor arc $AC$. Both parts are killed by the following claim. Claim: $(AE_1F_1)$ and $(AE_2F_2)$ pass through $O$, the circumcenter of $ABC$. Proof. Add $M$ and $N$, the midpoints of $AC$ and $AB$ so $M,N$ lie on line $PQ$. Then, $ANOM$ is cleary cyclic. By right angles, $F_2NOQ$ is cyclic. Thus \[\measuredangle OF_2E_2=\measuredangle OF_2Q=\measuredangle ONQ=\measuredangle ONM=\measuredangle OAM=\measuredangle OAE_2\]and the claim is proved. $\blacksquare$ Then, the circumradii are equal since \[\angle AF_2O=\angle NF_2O=\angle NQO=\angle PQO=\angle QPO=\angle MPO=\angle ME_1O=\angle AE_1O\]. The other part follows from $AO$ being the radical axis of the two circles.

Consider the following claim: Claim: $E_2A = E_1C$ and $F_2B = F_1A$. Proof. We only need to prove that $E_2A = E_1C$ since the other one is exactly same. Assume $X$ be the reflection of $E_1$ wrt the midpoint of $AC$ and let the tangent from $X$ touches $\omega$ at $Y$. Assume $Y, P$ lie on different sides of $AC$. Then we'll prove that $PY$ bisects $AC$. By applying law of sine few times, we get \begin{align*} \frac{\sin{\angle AYP}}{\sin{\angle CYP}} = \frac{\sin{\angle PCE_1}}{\sin{\angle CPE_1}} = \frac{PE_1}{CE_1} = \frac{XY}{XA} = \frac{CY}{AY}. \end{align*} Hence $PY$ bisects $AC$. Conversely, since $PQ$ bisects $AC$, thus we get $E_2A = E_1C$ and $E_2Q = E_1P$. $\blacksquare$ Since $E_2Q = E_1P$ and $F_2Q = F_1P$, thus we have $E_1F_1 = E_2F_2$. Now we'll prove that $A-$tangent of $\omega$ is perpendicular to the radical axis of $(AE_1F_1), (AE_2F_2)$. Let $O$ be the second intersection of $(AE_1F_1), (AE_2F_2)$. Then $O$ is the center of the spiral similarity sending $E_1F_1$ to $E_2F_2$. Since $E_1F_1 = E_2F_2$, therefore we get $\triangle E_1F_1O = \triangle E_2F_2O$. Hence $O$ is the intersection of the perpendicular bisector of $F_1F_2$ and $E_1E_2$, in other word, $O$ is the center of $\omega$. Thus $AO$ is the radical axis of $(AE_1F_1), (AE_2F_2)$ and since $AO$ is perpendicular to $A$-tangent, so we're done. $\blacksquare$

only requires three constructions, interesting! Let $O$ be the center of $\omega$, and let $M$ and $N$ be the midpoints of $AB$ and $AC$. Notice that $F_1PMO$ is cyclic with diameter $F_1O$. Similarly, $F_2QOM$ is cyclic with diameter $F_2O$. As a result, $M\in PQ$ implies that $O$ is the center of spiral similarity $F_1P\to F_2Q$. From $OP=OQ$ we then get $F_1P=F_2Q$. Similarly we may write $E_1P=E_2Q$. Hence $O$ sends $F_1PE_1\to F_2QE_2$. Note the following: $A=E_1E_2\cap F_1F_2$, hence $O=(AE_1F_1)\cap (AE_2F_2)$, proving the second part of the question. $\triangle OE_1F_1\cong \triangle OE_2F_2$, proving the first part of the question. Done!

LOL WHAT IS THIS PROBLEM Let $M$ and $N$ be the midpoints of $\overline{AC}$ and $\overline{AB}$, respectively, and let $T$ be the intersection of the tangents to $\omega$ through $P$ and $Q$. Since $M$, $N$ and $Q$ are the feet from $O$ to the sides of $\triangle E_2 A F_2$, it follows from the converse of Simson's theorem that $O$ lies on $(AE_2 F_2)$; analogous reasoning shows that $O$ lies $(AE_1 F_1)$. Therefore the circumcenters of $\triangle AE_1 F_1$ and $\triangle AE_2 F_2$ lie on the perpendicular bisector of $\overline{AO}$; this proves the second part of the problem. Since $M$, $P$ and $Q$ are the feet from $O$ to the sides of $\triangle TE_1 E_2$, it follows from the converse of Simson's theorem that $O$ lies on $(TE_1 E_2)$. Since $O$ also lies on the angle bisector of $\angle E_1 T E_2$, it follows that $OE_1 = OE_2$. Therefore, the radii of $(AE_1F_1O)$ and $(AE_2F_2O)$ are equal (since they share arc $AO$ and $\angle AE_1O = \angle AE_2O$.)

Nice problem! I have no clue why I inverted oops. Let the tangents at $A$ and $C$ to $\omega$ intersect at $B'$; define point $C'$ similarly. Also let $M, N$ be the midpoints of $AB, AC,$ respectively. Observe that $P, M, N, Q$ are collinear since they all lie on the perpendicular bisector of $AH.$ By inverting this common line about $\omega,$ we discover that $B', P, O, Q, C'$ are concyclic. In particular, $B', P, O, C'$ are concyclic, so by the Simson line from $O$ to $\triangle B'PC',$ we get that $A$ is collinear with the foot of the altitude from $O$ to $B'P$ and the foot of the altitude from $O$ to $C'P.$ The former point lies on both $(AOC)$ and $(OP)$, so its inverse is just $E_1.$ Similarly, the inverse of the latter point is $F_1.$ Therefore, $A, E_1, F_1, O$ are concyclic. Similarly, $A, E_2, F_2, O$ are concyclic. Hence $(AE_1 F_1)$ and $(AE_2 F_2)$ have radical axis $AO,$ meaning that the line through their centers is perpendicular to $AO$ and thus parallel to the tangent line from $A$ to $\omega.$ This solves the second part of the problem. To solve the first part of the problem, with our two circles $(AE_1 F_1 O)$ and $(AE_2 F_2 O),$ we discover that $O$ is the center of a spiral similarity sending $E_1 F_1$ to $E_2 F_2.$ In particular, it sends the foot of the altitude from $O$ to $E_1 F_1$ to the foot of the altitude from $O$ to $E_2 F_2.$ These points are just $P$ and $Q,$ respectively. However, since $P$ and $Q$ lie on $\omega,$ we have that $OP = OQ.$ Therefore, the spiral similarity described above is in fact a rotation, implying that $(AE_1 F_1 O)$ and $(AE_2 F_2 O)$ are congruent. This solves the first part of the problem, and we are done.