This problem was proposed by my friend Mykhailo Shtandenko from Ukraine (who might be the youngest IMO author in history — he turns 18 on 25th of July)

OHMYGOD, This is my problem. I'm quite surprised to see it on IMO. I even doubted that this problem will be on the Shortlist. The story of how I came up with it is quite lengthy, and I don't remember all the details, but I got chain of problems, each time improving former version by doing inversion, watching some new stuff in construction in Geogebra, until finally, in few months came up with the final version of this problem. I actually sent slightly different proposal, and in that statement the fact of this problem was very unexpected for me, so I decided to send it on IMO. Thanks to Fedir Yudin, who first solved this problem and to Anton Trygub who first solved it with the nice solution which I present you here.

I am very happy that a problem by Mykhailo Shtandenko made it to IMO. He really deserved it with all the hard work he has put into problemsetting. I am sure that in next few years you will have many more chances to enjoy his problems on various international competitions, including IMO, RMM, and EGMO.

Oh, thanks, MS_Kekas, you are also a great proposer!

Will make a diagram later. Let $P$ be the point where $AD$ intersects $(BDC)$ for the second time. Then $\angle APC = \angle DBC = \angle FDA$. As $\angle FAD = \angle CAP$ it follows that $\triangle AFD \sim \triangle ACP$, thus $AD/AP = AF/AC$. Similarly $AE/AB = AD/AP$, so $AB \cdot AF = AC \cdot AE$ and $BCEF$ is cyclic. Now we prove that $(DEF)$ and $(BDC)$ are tangent. Let $AD$ intersect $BC$ at $K$, then \[\angle FED = \angle AED - \angle AEF = \angle ABP - \angle ABC = \angle PBC = \angle KDC\] Now $\angle FED + \angle DCB = \angle DKB$ and some more easy angle chasing shows $\angle DKB = \angle FDB$, so these circumcircles are indeed tangent. Let $EF$ and $BC$ intersect at $T$, then $T$ is the radical center of $(DEF)$, $(DBC)$ and $(BCEF)$, so $D$ lies on the circle $\Omega$ with center $T$ that is orthogonal to $(BCEF)$. To finish it is enough to prove that the second intersection of $(ADC)$ and $(XDE)$ also lies on $\Omega$. We will actually generalize this, and prove that it holds for any point $D$ on $\Omega$. Now we invert at $D$, with the inverted diagram looking as follows: IMO 3, inverted wrote: Let $BCEF$ be an isosceles trapezoid with $BC \parallel EF$ and let $\Omega$ be its axis of symmetry. Let $D$ be any point on the plane, and $A$ be the second intersection of $(BDF)$ and $(CDE)$. Let $X$ be a point on $(CDE)$ such that $BD/DC = BX/XC$. Then $AC$ and $EX$ intersect on $\Omega$. Now let $P = BF \cap CE$, then $A, D, P$ are collinear on account of $AD$ being the radical axis of $(BDF)$ and $(CDE)$, and $P$ lies on $\Omega$. Rewording the problem wrt the circle $(CDE)$ and relabeling points, we get the following: IMO 3, inverted and reworded wrote: Let $ABCDE$ be a cyclic pentagon, let $P = AB \cap CD$ and $Q = BD \cap CE$, and let $X$ be the reflection of $D$ wrt $PQ$. Then $AX/AD = EX/ED$. To finish, notice that the required statement is equivalent to $AX/XE = AD/DE$, which is equivalent to $X$ lying on the $D$-Apollonian circle of $\triangle ADE$. Let $DD \cap AE = R$, then by Pascal on $DDBAEC$ we get $P, Q, R$ collinear, implying that $RD = RX$, which finishes as $R$ is the center of the Apollonian circle.

mshtand1 wrote: OHMYGOD, This is my problem. I'm quite surprised to see it on IMO. I even doubted that this problem will be on the Shortlist. The story of how I came up with it is quite lengthy, and I don't remember all the details, but I got chain of problems, each time improving former version by doing inversion, watching some new stuff in construction in Geogebra, until finally, in few months came up with the final version of this problem. I actually sent slightly different proposal, and in that statement the fact of this problem was very unexpected for me, so I decided to send it on IMO. Thanks to Fedir Yudin, who first solved this problem and to Anton Trygub who first solved it with the nice solution which I present you here. Congratulations! I really enjoyed your problem. Here goes my solution: Let $D'$ be the isogonal conjugate of $D$ with respect to $ABC$. Since $\angle ADE=\angle DCB=\angle ACD'$, $ED'DC$ is cyclic and similarly $FD'DB$ is cyclic. From POP we get $AE\cdot AC=AD'\cdot AD=AF\cdot AB$ therefore $BFEC$ is cyclic, too. Let $AD$ intersects $BC$ and $EF$ at $R$ and $S$. Notice that $\angle EDC=\angle ADC-\angle ADE=\angle ADC-\angle DCB=\angle CRA\ (1)$. Hence $$\angle EFD+\angle DBC=\angle EFD+\angle ADF=\angle ASF=\angle CRA=\angle EDC$$so $\odot (EDF)$ and $\odot (BDC)$ are tangent. It yields that $EF$, $BC$ and the common internal tangent of $\odot (EDF)$ and $\odot (BDC)$ are concurrent. Let $K$ be the concurrency point. Similarly we can prove that $KD'$ is tangent to both $\odot (ED'F)$ and $\odot (BD'C)$, so $KD'^2=KB\cdot KC=KD^2$ or $KD'=KD$. Suppose that $\odot (EDX)$ and $\odot (ADC)$ intersects again at $P$. We just need to prove that $K$ is the circumcenter of $\odot (PD'D)$, since $O_1O_2$ is the perpendicular bisector of $PD$. It's sufficient to prove that $\angle D'PD=90^\circ-\angle D'DK$ or $\angle APD'=\angle APD+\angle D'DK-90^\circ$. Notice that $$\angle D'DK=\angle D'DC-\angle KDC=\angle D'DC-\angle DBC=\angle D'DC-\angle ABD'$$$$\Longrightarrow \angle APD+\angle D'DK=\angle ACD+\angle D'DC-\angle ABD'=180-\frac12 \angle A-\angle ABD'=\angle AD'B$$hence we need to prove that $\angle APD'=\angle AD'B-90^\circ=\angle AFD-90^\circ$. 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label("\contour{white}{$R$}", (-1.503333333333333,-3.46), NE * labelscalefactor); dot((0.778276801693389,0.08082997656047487),linewidth(4pt) + dotstyle);[/asy][/asy] Now perform an inversion centered at $A$ that fixed $\odot (BFEC)$. The images of the points are denoted by primes. It's obvious that $X'$ lies on $AC$ and $\angle X'FE=180^\circ-\angle C$. Also $P'$ lies on $D'E$ and $\odot (X'D'C)$. We want to prove that $\angle ADP'=\angle AFD-90^\circ$ or $P'D$ passes through the circumcenter of $\odot (AFD)$. We claim that $\triangle FX'P'\sim \triangle DRB$. Notice that $\angle FX'E=\angle C-\angle AEF=\angle C-\angle B$ and $\angle P'X'C=\angle ED'C\overset{(1)}{=}\angle CRA$, hence $\angle FX'P'=\angle CRA+\angle C-\angle B=\angle ARB$. So to prove our claim we need to show that $\frac{X'F}{X'P'}=\frac{RD}{RB}$. It's clear that $\triangle X'EP'\sim \triangle D'EC$ and $\triangle D'EC\sim \triangle RDC$, therefore $$\triangle X'EP'\sim \triangle RDC \Longrightarrow X'P'=X'E\cdot\frac{CR}{RD}.$$So $$\frac{X'F}{X'P'}=\frac{RD}{RB} \Longleftrightarrow \frac{X'F}{X'E}=\frac{RC}{RB}=\frac{AC}{AB}$$The last equality holds by law of sines, hence the claim is proved and it yields $\angle X'P'F=\angle DBC$. Let $J$ be the second intersection of $AC$ and $\odot (AFD)$. We have $\angle AJF=\angle ADF=\angle DBC=\angle X'P'F$ so $X'P'JF$ is cyclic. On the other hand, we proved that $\angle P'X'C=\angle CRA=\angle B+\frac12 \angle A$ and $\angle FX'E=\angle C-\angle B$, therefore $2\angle P'X'C+\angle FX'E=\angle A+\angle B+\angle C=180^\circ$. It means that $X'P'$ is the external angle bisector of $\angle FX'E$, hence $P'F=P'J$. Also it's clear that $DF=DJ$ so $P'D$ is the perpendicular bisector of $FJ$ and passes through the circumcenter of $\odot (AFD)$, as desired.

Михайло Штанденко -- well-done mate. Your problem is definitely the hardest from this IMO...

Thank you!

Here is a sketch of my solution: Begin by proving $BCEF$ cyclic (e.g. by introducing isogonal conjugate of $D$ as other solutions have). Let $EF$ meet $BC$ at $Y$, by angle chasing one can show $YD^2 = YB \cdot YC$. Notice $(BFY)$ and $AC$ are inverses wrt the circle $\omega$ centred at $Y$ passing through $D$. Let $(BFY)$ meet $(AC)$ at $(P, Q)$. Angle chasing gives $XB$ tangent to $(BFY)$. DIT applied to $FYBB$ gives that $(P, Q), (X, E), (A, C)$ are pairs of an involution on $AC$. Hence $\omega$, $(ADC)$ and $(DEX)$ are coaxial (since $D$ also lies on radical axis) so we are done.

Kagebaka wrote: Let $D$ be an interior point of the acute triangle $ABC$ with $AB > AC$ so that $\angle DAB = \angle CAD.$ The point $E$ on the segment $AC$ satisfies $\angle ADE =\angle BCD,$ the point $F$ on the segment $AB$ satisfies $\angle FDA =\angle DBC,$ and the point $X$ on the line $AC$ satisfies $CX = BX.$ Let $O_1$ and $O_2$ be the circumcenters of the triangles $ADC$ and $EXD,$ respectively. Prove that the lines $BC, EF,$ and $O_1O_2$ are concurrent. The solution proceeds in three steps. Step 1. Identifying the point of concurrence. Let $D^\ast$ be the isogonal conjugate of $D$ in $\triangle ABC$. Note that $D^\ast$ lies on $\odot(CDE)$ and $\odot(FBD)$, and $A$ lies on $\overline{DD^\ast}$ hence $BFEC$ is cyclic by the converse of radical axis theorem. Note also that $\angle BDF+\angle CDE=180^{\circ}$ hence the circles $\odot(BDC)$ and $\odot(EDF)$ are tangent at $D$, so combining them with $(BFEC)$ and applying radical axes, we get $\overline{BC}, \overline{EF},$ and the line through $D$ tangent to $\odot(BDC)$ concur (say at point $T$). Step 2. Reformulating line through centres. Let $Y=\odot(ADC) \cap \odot(EXD)$ with $Y \ne D$. It suffices to show that the perpendicular bisector of $\overline{DY}$ passes through $T$. Thus, we wish to understand point $Y$. Let $M$ be the midpoint of arc $\widehat{BAC}$. Notice that $$\angle AYX=\angle DYX-\angle DYA=\angle DEX-\angle DCA=\angle CDE=\angle ADC-\angle ADE=\angle(AD, BC)$$hence $AMXY$ cyclic. Also, $\angle CYD=\tfrac{1}{2}\angle A$; which characterises $Y$ completely. Note also that $\overline{CD}$ meets the perpendicular bisector $\ell$ of $\overline{BC}$ at $S$, then $DSMY$ is cyclic since $\angle DYM=\angle DYA+\angle AYM=\angle DCA+(90^{\circ}-\angle C)=90^{\circ}-\angle SCB$, as claimed. In order to show $TD=TY$, since $TD^2=TB \cdot TC$, it suffices to show that $Y$ lies on the $D$-Appolonian circle $\omega$ in $\triangle BDC$. Step 3. Changing reference to $\triangle BDC$ and moving points. We employ the method of moving points (or the rather cheekily named "MMP"). Fix $\triangle BDC$ and move point $M$ along $\ell$; let $Y_1=\odot(DSM) \cap \omega$ with $Y_1 \ne D$ and $Y_2$ be the point on $\omega$ with $\angle CY_2D=\angle (CM, \ell)$. So $M \mapsto Y_1$ and $M \mapsto Y_2$ are projective mappings, hence it suffices to check $Y_1=Y_2$ for three choices of $M$. When $M$ is the point at infinity on $\ell$, we get $Y_1, Y_2$ are both the second intersection of ray $\overrightarrow{DC}$ with $\omega$. When $M=\ell \cap \overline{DB}$, we see that $Y_1=Y_2=A$. Finally, when $M$ coincides with the reflection of $S$ in $\overline{BC}$, we get $Y_1, Y_2$ both as the reflection of $D$ in $\overline{BC}$. Thus, we have solved the problem. Remark. The strategy for the final moving points steps after finding the right "base-change" is inspired from this solution to IMO 2018 P6

Solution finished by Li4 and me!! I think it is a medium 3,6 level problem. (~MOHS 40) First we prove that $B$, $C$, $E$, $F$ are concyclic. Let $D^*$ be the isogonal conjugate of $D$ w.r.t $\triangle ABC$, which lies on $AD$. Since $\measuredangle BDC + \measuredangle EDF = 180^{\circ}$, $D$ also has an isogonal conjugate w.r.t the complete quadrilateral $\mathcal{Q}\{BC, CA, AB, EF\}$, which is exactly $D^*$. Therefore, $\measuredangle ADF = \measuredangle CBD = \measuredangle D^*BF$, which implies that $B$, $D^*$, $D$, $F$ are concyclic. Similarly, $C$, $D^*$, $D$, $E$ are concyclic. Hence $$ AF \cdot AB = AD \cdot AD^* = AE \cdot AC, $$implying that $B$, $C$, $E$, $F$ are concyclic. Let $Y = AD \cap EF$, and $Z = AD \cap BC$. Then from $\triangle ABC \stackrel{-}{\sim} \triangle AEF$ we have $$ \triangle EYD \stackrel{+}{\sim} \triangle DZC \quad \text{and} \quad\triangle FYD \stackrel{+}{\sim} \triangle DZB. $$This means $$\measuredangle CDE = \measuredangle CZD = \measuredangle CBD + \measuredangle BDZ = \measuredangle CBD + \measuredangle DFE,$$implying that $\odot(BDC)$ is tangent to $\odot(EDF)$. By radical axis theorem we have $SD$ is the common tangent line of $\odot(BDC)$ and $\odot(EDF)$.

This is probably the only problem I like in Day 1, and I think this geo is pretty neat! Lemma 1. $BCEF$ is concyclic.

Now we invert about $D$ and label the image of a point $T$ as $T'$. Lemma 2. $B'C'E'F'$ is a trapezoid with $B'F'=C'E'$ and $B'C'\parallel E'F'$.

Lemma 3. Let $M$ be the Mique point of $(AB, BC, CA, EF)$. Then $BEMX$ are concyclic.

Now we can put everything together. Assume that $BC$ intersects $EF$ at $P$. Then $P'$ is the intersection of $DB'C'$ and $DE'F'$, showing that $P'$ is the reflection of $D$ w.r.t. the common perpendicular bisector $\ell$ of $EF$ and $BC$. Also $M'$ is the intersection of $A'B'C'$ and $A'E'F'$, showing that $M'$ is the reflection of $A'$ w.r.t. $\ell$. The statement is equivalent to showing that $DP'O_1'O_2'$ are concyclic. It is well-known that $O_1'$ is the reflection of $D$ w.r.t. $AC$ and $O_2'$ is the reflection of $D$ w.r.t. $EX$. Thus it is equivalent to showing that $\ell, AC, EX$ are concurrent. However, since $B'M'E'X'$, $B'M'A'C'$ and $A'C'E'X'$ are concyclic, their radical axes are concurrent. They are exactly $\ell, A'C'$ and $E'X'$, as desired.

Here is K2I 's Solution : Part 1 : $B,C,E,F$ cyclic pf) Let the mid point of arc $(BC)$ of $(ABC)$ is $M$ noticing that radical axis of $(DFB), (DEC)$ is $DM$ , so $AB*AF=AC*AE$ Part 2 : $PD^2=PC*PB$ Let $EF$ meets $BC$ at $P$. pf) By simple angle chase, $DE, DB$ is isogonal w.r.t $\angle FDC$ So, by Isogonality lemma on $(DF,DC)$ and $(DE,DB)$ $DA$ and $DP$ is isogonal w.r.t $\angle FDC$ so $\angle CDP=\angle FDA=\angle DBC$ Part 3 : Finish Let Miquel Point of $BCEF$ be $K$ (so, $K$ lies on circumcircle of $(AEF),(ABC),(BFP),(CEP)$) Since $BCEF$ is cyclic, well known that $K$ lies on $PA$ $BK$ meets $AC$, $(AEF)$ at $S,T$ respectively. $\angle KTE= \angle KAE = \angle KBC$. So $TE//BC$. also note that $\angle ATE= \angle AFE = \angle ACB= \angle XBC$. So $ \triangle ATE $ and $ \triangle XBC $ are homothetic, and the homothetic center is $S$. $SA:SX=SE:SC$. So $SA*SC=SE*SX...(1)$ . Let the circle centered at $P$ and radius $PD$ be $w$. Let $(ABC)$ meets $w$ at $U,V$. Easy to see that $U,V,S$ colinear (육점공선 또는 Pole & Polar), which gives $US*VS=AS*SC ... (2)$ In conclusion, powers of $S$ in three circle $(DXE), (DAC), w$ are all same by (1) and (2) and note that three circle all passes $D$. So three circles are coaxial. So centers of three circle, which is $O_1,O_2,P$are colliner. $Q.E.D$

Here's my solution which uses polar duality to show that $(DEF)$ and $(BCD)$ are tangent. Apart from that it is identical to juckter's solution though. Let $H$ be the intersection of $(BDF)$ with $AD.$ Since $\measuredangle HBA=\measuredangle HDF=\measuredangle CBD,$ it follows that $H$ is the isogonal conjugate of $D$ with respect to $ABC;$ thus, we have $\measuredangle EDH=\measuredangle DCB = \measuredangle ECH$ so $HECD$ is cyclic, and $BCEF$ is too by the converse of radical center. Next, take the dual at $D$ and denote the image of a figure $p$ by $p'.$ Then we immediately have \[\measuredangle b'a'c' = \measuredangle (C', B') = -\measuredangle BDC = \measuredangle DCB + \measuredangle CBD.\]However, we know that $\measuredangle (E', A') = \measuredangle EDA = \measuredangle DCB$ and similarly $\measuredangle (A', F') = \measuredangle CBD,$ so \[\measuredangle b'a'c' = \measuredangle (E', A') + \measuredangle (A', F') = \measuredangle b'(EF)'c'\]and thus $a'c'(EF)'b'$ is cyclic. Now let $T=EF\cap a;$ we then have \[\measuredangle TDB = \measuredangle (T', B') = \measuredangle (EF)'a'c' = \measuredangle (EF)'b'c' = \measuredangle (E', A') = \measuredangle DCB,\]so $TD$ is tangent to $(BCD),$ which by Power of a Point implies that $TD^2 = TB\cdot TC = TE\cdot EF,$ so $TD$ is also tangent to $(DEF).$ [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(15cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -12.561350000222955, xmax = 10.132480596525847, ymin = -4.01560611339061, ymax = 10.092394316050871; /* image dimensions */ /* draw figures */ draw(circle((-0.4271588523718617,0.7703217957553912), 2.902369505837254), linewidth(0.7)); draw(circle((-6.375544286809934,3.2789639515334055), 5.344850881161717), linewidth(0.7)); draw(circle((-4.255,-1.941807755344781), 4.965176017867541), linewidth(0.7)); draw(circle((-4.255,-11.417324790326727), 10.73455616989206), linewidth(0.7)); draw(circle((-2.6616158068456093,0.9827007538498959), 1.7674239406035193), linewidth(0.7)); draw(circle((-1.85,2.472346090809064), 0.8276539091909356), linewidth(0.7)); draw(circle((-4.255,-1.1141538461538465), 5.026806061259455), linewidth(0.7)); draw(circle((-6.517127061903484,1.2643223304684243), 4.16154513739736), linewidth(0.7)); draw(circle((-0.7346494013264171,-0.20871207599778657), 2.2242901642161015), linewidth(0.7)); draw((-1.85,3.3)--(-9.22,-1.9), linewidth(0.7)); draw((0.71,-1.9)--(-4.255,8.18515625), linewidth(0.7)); draw((-9.22,-1.9)--(-2.8868749533704117,-0.7703097146657338), linewidth(0.7)); draw((-2.8868749533704117,-0.7703097146657338)--(0.71,-1.9), linewidth(0.7)); draw((-2.8868749533704117,-0.7703097146657338)--(-2.62974968111699,2.749837402739709), linewidth(0.7)); draw((-2.8868749533704117,-0.7703097146657338)--(-1.19405786583458,1.9676175399764904), linewidth(0.7)); draw((-9.22,-1.9)--(5.904594458298186,-1.9), linewidth(0.7)); draw((-6.375544286809934,3.2789639515334055)--(5.904594458298186,-1.9), linewidth(0.7) + linetype("4 4")); draw((5.904594458298186,-1.9)--(-2.62974968111699,2.749837402739709), linewidth(0.7)); draw((-2.8868749533704117,-0.7703097146657338)--(5.904594458298186,-1.9), linewidth(0.7)); draw((-1.85,3.3)--(-2.8868749533704117,-0.7703097146657338), linewidth(0.7)); /* dots and labels */ dot((-1.85,3.3),dotstyle); label("$A$", (-1.9437189442369723,3.550614033934334), NE * labelscalefactor); dot((-9.22,-1.9),dotstyle); label("$B$", (-9.727017933300049,-2.654564578185156), NE * labelscalefactor); dot((0.71,-1.9),dotstyle); label("$C$", (0.8027749274745735,-2.610660012533367), NE * labelscalefactor); dot((-2.8868749533704117,-0.7703097146657338),dotstyle); label("$D$", (-3.1927936646626455,-1.5154898577594865), NE * labelscalefactor); dot((-1.19405786583458,1.9676175399764904),linewidth(4pt) + dotstyle); label("$E$", (-1.063183644680351,2.072493656990776), NE * labelscalefactor); dot((-2.62974968111699,2.749837402739709),linewidth(4pt) + dotstyle); label("$F$", (-3.0388890990108566,2.93595011480929), NE * labelscalefactor); dot((-4.255,8.18515625),linewidth(4pt) + dotstyle); label("$X$", (-4.195042661174634,8.306941979544792), NE * labelscalefactor); dot((-0.4271588523718617,0.7703217957553912),linewidth(4pt) + dotstyle); label("$O_{1}$", (-0.945121395552354,0.11142305787754121), NE * labelscalefactor); dot((-6.375544286809934,3.2789639515334055),linewidth(4pt) + dotstyle); label("$O_{2}$", (-6.9171257315855526,2.6749667074193978), NE * labelscalefactor); dot((5.904594458298186,-1.9),linewidth(4pt) + dotstyle); label("$T$", (6.034721135692205,-2.2301537768845305), NE * labelscalefactor); dot((-1.0407745976486813,3.6070848217345377),linewidth(4pt) + dotstyle); label("$K$", (-1.019279079028562,3.726232296541489), NE * labelscalefactor); dot((-2.355880114831864,1.3141397579687066),linewidth(4pt) + dotstyle); label("$H$", (-2.3486069172786545,0.8724355291752145), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy][/asy] Now suppose that $K$ is the second intersection of $(EDX)$ and $(ADC);$ it suffices to show that $TK=TD.$ Invert at $D$ with arbitrary radius $r$ and denote the image of a point $P$ by $P^*.$ By the inversion distance formula, we have \[\frac{B^*X^*}{C^*X^*}=\frac{\frac{r^2}{XD\cdot BD}\cdot BX}{\frac{r^2}{XD\cdot CD}\cdot CX}=\frac{CD}{BD} = \frac{\frac{r^2}{C^*D}}{\frac{r^2}{B^*D}} = \frac{B^*D}{C^*D}\]and \[\frac{KT}{DT}=\frac{\frac{r^2}{K^*D\cdot T^*D}\cdot K^*T^*}{\frac{r^2}{T^*D}} = \frac{K^*T^*}{K^*D}.\]Thus, omitting the $^*$s, we can reformulate the problem as follows: Quote: Let $BCEF, BCTD, FETD$ be three isosceles trapezoids in the plane. Let $A$ be the second intersection of $(DBF)$ and $(DCE),$ and let $X$ be the second intersection of the $D$-Apollonian Circle of $\triangle DBC$ with $(DCEA).$ If $K$ is the intersection of $XE$ and $AC,$ prove that $KD=KT.$ First, observe that the Apollonian Circle condition can be manipulated into $\tfrac{BX}{BD}=\tfrac{CX}{CD},$ meaning that $B$ lies on the $C$-Apollonian Circle of $\triangle CDX.$ Let $H=BF\cap CE$ and $G = XD\cap CC.$ Since $H$ is the radical center of $(BCEF), (DBFA), (DCEA)$ and therefore must lie on $AD$ (alternatively just notice that $H,A,D$ are collinear pre-inversion), Pascal's Theorem on $ADXECC$ implies that $G,H,K$ are collinear. This means that $GX\cdot GD=GC^2,$ so $G$ is the center of the $C$-Apollonian Circle of $\triangle CDX.$ Thus, $GB=GC,$ which finishes the problem as it implies that $HG$ is the perpendicular bisector of $BC$ and $DT$ as well, whence $KD=KT.$ $\blacksquare$ [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(12cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -8.19, xmax = 16.19, ymin = -8.64, ymax = 7.84; /* image dimensions */ /* draw figures */ draw(circle((0,-0.8078168044077136), 5.748582141678175), linewidth(0.7)); draw(circle((0,2.4204424778761053), 2.7184575958089634), linewidth(0.7)); draw(circle((0,-6.133266423357665), 6.602753634661262), linewidth(0.7)); draw(circle((2.6132511224036707,0.6859837683161479), 4.28448040064895), linewidth(0.7)); draw(circle((-6.389383232810618,2.8445165862484965), 5.398284858363812), linewidth(0.7)); draw((-1.65,0.26)--(0,-7.494908473527701), linewidth(0.7)); draw((0,-7.494908473527701)--(5.5,-2.48), linewidth(0.7)); draw((0,-7.494908473527701)--(-5.5,-2.48), linewidth(0.7)); draw((-5.5,-2.48)--(0,7.1416867469879435), linewidth(0.7)); draw((0,7.1416867469879435)--(5.5,-2.48), linewidth(0.7)); draw((5.5,-2.48)--(-5.5,-2.48), linewidth(0.7)); draw((-1.65,0.26)--(1.65,0.26), linewidth(0.7)); draw((-1.35,4.78)--(1.35,4.78), linewidth(0.7)); draw((1.35,4.78)--(-1.4541401382799275,-0.6605304856223962), linewidth(0.7)); draw((-0.9933060082157816,2.9988862667200404)--(5.5,-2.48), linewidth(0.7)); draw((0,7.1416867469879435)--(-1.65,0.26), linewidth(0.7)); draw((0,7.1416867469879435)--(0,-7.494908473527701), linewidth(0.7) + linetype("4 4")); /* dots and labels */ dot((5.5,-2.48),dotstyle); label("$C$", (5.89,-3), NE * labelscalefactor); dot((-5.5,-2.48),dotstyle); label("$B$", (-6.37,-3.05), NE * labelscalefactor); dot((-1.65,0.26),dotstyle); label("$D$", (-2.43,0.19), NE * labelscalefactor); dot((1.35,4.78),dotstyle); label("$E$", (1.43,4.98), NE * labelscalefactor); dot((-1.35,4.78),linewidth(4pt) + dotstyle); label("$F$", (-2.07,4.82), NE * labelscalefactor); dot((-0.9933060082157816,2.9988862667200404),linewidth(4pt) + dotstyle); label("$A$", (-1.65,3.02), NE * labelscalefactor); dot((1.65,0.26),linewidth(4pt) + dotstyle); label("$T$", (1.67,-0.46), NE * labelscalefactor); dot((0,7.1416867469879435),linewidth(4pt) + dotstyle); label("$H$", (0.09,7.3), NE * labelscalefactor); dot((0,-7.494908473527701),linewidth(4pt) + dotstyle); label("$G$", (0.09,-8.26), NE * labelscalefactor); dot((-1.4541401382799275,-0.6605304856223962),linewidth(4pt) + dotstyle); label("$X$", (-2.01,-1.28), NE * labelscalefactor); dot((0,2.160759950144464),linewidth(4pt) + dotstyle); label("$K$", (0.33,2.22), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy][/asy] Remarks: My first instinct for this problem was that taking the dual and/or inverting at $D$ would be helpful, and indeed, both transformations turned out to be useful. Taking the dual was especially important for motivational purposes as the equal angles from cyclic quadrilateral $a'c'(EF)'b'$ allowed me to notice the tangent circles immediately, although this did end up causing me to flounder on how $TB$ tangent to $(BCD)$ implied $BCEF$ cyclic for like 2 hours. I would say that probably the most difficult step by far was figuring out the Apollonian Circle stuff after inverting, and otherwise I think this solution flowed pretty naturally.

a nice configuration,also a medium hard problem for the P3 I proved the problem in the situation of $AB \le AC$ since the proof is not different i write my solution considering $AB \le AC$ For the first part of the proof refer to Figure 1 claim: $BCEF$ is cylic proof: extend $AD$ to intersect circle $BDC$ for the second time at $G$ and reflect $E,F$ to attain $E',F'$. trivially we have that $E'F'||BC$ so we have that $EF$ and $BC$ are antiparallel it implies that $BCEF$ is cyclic. claim: $\odot EFD$ is tangent to $\odot BDC$ proof: Let $EF$ intersect $BC$ at $T$. also note that $ BDG = EFD$ so circle $ EFD$ is tangent to circle $ BDC$. claim:$TD$ is tangent to $\odot EFD$ and $\odot BDC$ proof: by radical-axis theorem $TD$ is tangent to the circle $BDC$. [asy][asy] import graph; size(10cm); real labelscalefactor = 0.5; pen dps = linewidth(1) + fontsize(7); defaultpen(dps); real xmin = -4.293148965357704, xmax = 3.63258620525972, ymin = -2.5055781362405414, ymax = 1.5099999006260838; pen fffaev = rgb(1.,0.9803921568627451,0.8980392156862745); pen ffdxqq = rgb(1.,0.8431372549019608,0.); pair A = (-0.37714665198045905,0.9261535525494304), B = (-0.849056803289512,-0.5283015661417207), C = (0.8901303651910498,-0.4557059720530863), D = (-0.15654837692875345,-0.0878437357646118), F = (-0.5618383812607487,0.3569226718691055), G = (0.3140060935540723,-2.2507842980233113), T = (-2.9511406288887545,-0.6160448676730947); filldraw(A--B--C--cycle, fffaev, linewidth(0.) + ffdxqq); /* draw figures */ draw(circle((0.,0.), 1.), linewidth(0.4)); draw(A--D, linewidth(0.4)); draw(circle((0.013311101804950794,-0.3188967284880084), 0.8874281885143103), linewidth(0.4) + linetype("0 3 4 3")); draw(D--(-0.047279113289258365,0.5664606093166927), linewidth(0.4)); draw(D--F, linewidth(0.4)); draw(D--B, linewidth(0.4)); draw(C--D, linewidth(0.4)); draw(circle((0.04907771833694624,-1.1757647149454669), 1.1071829785523477), linewidth(0.4)); draw((-0.047279113289258365,0.5664606093166927)--F, linewidth(0.4)); draw(D--G, linewidth(0.4)); draw(G--C, linewidth(0.4)); draw(G--B, linewidth(0.4)); draw(D--(-0.5277682996998131,0.4619286870672731), linewidth(0.4)); draw(D--(0.027335751955441856,0.48509934353326345), linewidth(0.4)); draw((-0.5277682996998131,0.4619286870672731)--(0.027335751955441856,0.48509934353326345), linewidth(0.4)); draw(circle((-0.22217896613541185,0.2593928091965897), 0.35338448239305253), linewidth(0.4)); draw(T--F, linewidth(0.4)); draw(B--T, linewidth(0.4)); draw(T--D, linewidth(0.4)); dot(A,linewidth(3.pt)); label("$A$", (-0.4548697200949737,0.992479103512008), NE * labelscalefactor); dot(B,linewidth(3.pt)); label("$B$", (-0.9771823764415875,-0.5552914285976817), NE * labelscalefactor); dot(C,linewidth(3.pt)); label("$C$", (0.977896190433811,-0.5217484139699176), NE * labelscalefactor); dot(D,linewidth(3.pt)); label("$D$", (-0.22486047693316205,-0.22944500078511545), NE * labelscalefactor); dot((-0.047279113289258365,0.5664606093166927),linewidth(3.pt)); label("$E$", (-0.037977966864190145,0.6522570980018285), NE * labelscalefactor); dot(F,linewidth(3.pt)); label("$F$", (-0.6417522301639456,0.3982885586773284), NE * labelscalefactor); dot(G,linewidth(3.pt)); label("$G$", (0.30703589787852725,-2.3618223592644094), NE * labelscalefactor); dot((-0.5277682996998131,0.4619286870672731),linewidth(3.pt)); label("$E'$", (-0.6130010747687191,0.532460617188385), NE * labelscalefactor); dot((0.027335751955441856,0.48509934353326345),linewidth(3.pt)); label("$F'$", (0.09619409164686663,0.47016644716539446), NE * labelscalefactor); dot(T,linewidth(3.pt)); label("$T$", (-2.932260943316986,-0.5888344432254459), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); label("\color{red} Figure 1", (0,-2.4));[/asy][/asy] \[\rule{100mm}{.1pt}\]For the second part of the proof refer to the Figure 2. now we simply finish the problem: the circle centered at $T$ and radius $TD$ is called $\omega$. we have to prove $\odot ADC$,$\odot EXD$ and $\omega$ are coaxial. $S$ is the miquel point of $BCEF$. the inverse of line $AC$ with respect to $\omega$ is $\odot BSD$. let $BS \cap AC = N$. \[\angle ABX=\angle B-\angle C=\angle ETC \implies \text{BX is tangent to circle BFT}\]since \[\angle SEX=\angle STB=\angle SBX \implies \text{SEXB is cyclic}\] now by power of point we have: \[NE \times NX=NS \times NB=NA \times NC\]so we get that $N$ and $D$ have the same power of point wrt the three circles $\odot ADC$, $\odot EXD$ and $\omega$. $\blacksquare$ [asy][asy] import graph; size(17cm); real labelscalefactor = 0.5; pen dps = linewidth(0.7) + fontsize(9); defaultpen(dps); pen fffaev = rgb(1.,0.9803921568627451,0.8980392156862745); pen ffdxqq = rgb(1.,0.8431372549019608,0.); pen evefev = rgb(0.8980392156862745,0.9372549019607843,0.8980392156862745); pen qqwuqq = rgb(0.,0.39215686274509803,0.); pen fffsev = rgb(1.,0.9490196078431372,0.8980392156862745); pen ffxfqq = rgb(1.,0.4980392156862745,0.); pen qqffff = rgb(0.,1.,1.); pair A, B, C, D, E, F, G, M, T, W, X, O1, O2, I, P,L; A = dir(130); B = dir(210); C = dir(330); I = incenter(A, B, C); D = I * 1.2 - A *.2; G = extension(B, foot(D, I, B) * 2 - D, C, foot(D, I, C) *2- D); F = foot(2 * circumcenter (D, B, G) - B, A, B); E = foot(2 * circumcenter (D, G, C) - C, A, C); T = extension(E, F, B, C); M = foot(extension(E, B, F, C), A, T); W = extension(B, M, A, C); X = extension(A, C, origin, (B+C)/2); P = foot(circumcenter(D, B, M) * 2 - D, W, D); filldraw(A--B--C--cycle, fffaev, linewidth(0.4) + ffdxqq); draw(circle(T,distance(D,T)),blue); draw(circumcircle(A, D, C), blue); draw(circumcircle(M, D, B), blue); draw(circumcircle(A, B, C)); draw(circumcircle(D, E, X), blue); draw(circumcircle(E, F, B),heavygreen); draw(circumcircle(B,M,X), orange + opacity(0.5)); draw(T--B); draw(T--A); draw(A--D); draw(W--B); draw(W--A); draw(F--D--B); draw(E--D--C); draw(W--D,red+dashed); draw(T--E); draw(D--X); draw(circumcircle(A,E,F)); draw(B--X); dot("$A$", A, dir(A)*2*dir(340)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$T$", T, dir(T)); dot("$N$", W, dir(W)); dot("$D$", D, dir(270)*2); dot("$E$", E, dir(E)*2*dir(340)); dot("$F$", F, dir(F)); dot("$X$", X, dir(90)*2); dot("$S$", M, dir(M)); label("\color{red} Figure 2", (0,-1.4)); [/asy][/asy]

Solved with Albert Wang, Isaac Zhu, Kevin Wu, Nacho Cho. [asy][asy] size(18cm); pair A, B, C, D, E, F, G, M, T, W, X, O1, O2, I, P; A = dir(120); B = dir(210); C = dir(330); I = incenter(A, B, C); D = I * 1.2 - A *.2; G = extension(B, foot(D, I, B) * 2 - D, C, foot(D, I, C) *2- D); F = foot(2 * circumcenter (D, B, G) - B, A, B); E = foot(2 * circumcenter (D, G, C) - C, A, C); T = extension(E, F, B, C); M = foot(extension(E, B, F, C), A, T); W = extension(B, M, A, C); X = extension(A, C, origin, (B+C)/2); P = foot(circumcenter(D, B, M) * 2 - D, W, D); draw(circumcircle(A, D, C), cyan); draw(circumcircle(M, D, B), cyan); draw(circumcircle(E, D, C), purple); draw(circumcircle(F, D, B), purple); draw(circumcircle(A, B, C)); draw(circumcircle(D, E, X), cyan); draw(circumcircle(E, F, B)); draw(circumcircle(M,B,X), heavygreen); draw(A--B--C--cycle); draw(T--B); draw(T--A); draw(A--D); draw(W--B); draw(W--A); draw(F--D--B); draw(E--D--C); draw(W--D); draw(T--E); dot("$A$", A, dir(A)*2*dir(340)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$T$", T, dir(T)); dot("$W$", W, dir(W)); dot("$D$", D, dir(270)*2); dot("$E$", E, dir(E)*2*dir(340)); dot("$F$", F, dir(F)); dot("$X$", X, dir(90)*2); dot("$M$", M, dir(M)); dot("$P$", P, -2*dir(P)); [/asy][/asy] We assume $AB < AC$ since we are Americans here. Let $G$ be the isogonal conjugate of $D$ with respect to $\triangle ABC$. We know that $G$ lies on $AD$, and observe that \[\angle FBG = \angle DBC = \angle FDA = 180 - \angle FDG\]This proves that $(BFDG)$ is cyclic. Similarly, $(EDGC)$ is cyclic, so \[AF\cdot AB = AD\cdot AG = AE\cdot AC\]Therefore, $(BFEC)$ is cyclic. Let $T = FE\cap BC$. I claim that $(DBC)$ is tangent to $TD$. This is because, observe that \[\angle FDE = \angle DBC + \angle DCB = 180 - \angle BDC\]which means $C$ and $F$ are isogonal with respect to $\angle BDE$. Now, since $TB\cap AE$ and $TE\cap AB$ are isogonal with respect to $\angle BDE$, by the second isogonality lemma, $T$ and $A$ are isogonal with respect to $\angle BDE$. Finally, this means \[\angle TDB = \angle ADE = \angle DCA\]Therefore, $TD$ is tangent to $(BDC)$. Now, we have $TD^{2} = TB\cdot TC$. Define $M$ as the miquel point of $EFBC$. By miquel properties, we have $M\in (ABC)$. We have \[\angle BME = 180 - \angle TMB - \angle AME = 180 - \angle C - \angle C = 180 - 2\angle C = 180 - \angle AXB\]This shows that $(AXBM)$ is cyclic. Next, let $W = AC\cap MB$. We have \[WM\cdot WB = WA\cdot WC = WE\cdot WX\]by power of a point. This means that $W$ lies on the radical axis of $(DEX)$ and $(MDB)$, and $W$ lies on the radical axis of $(MDB)$ and $(ADC)$. Therefore, $WD$ is the radical axis of all three circles, and since all three circles pass through $D$, they therefore must pass through a second point. Let's call this point $P$. Finally, by inversion at $T$ with radius $\sqrt{TF\cdot TE}$, we get $(MDB)$ gets sent to $(ADC)$, so $(MDB)\cap (ADC)$ is fixed, which means $P$ is fixed. Therefore, $TP = TD$, and since $DP$ is the radical axis of $(DEX)$ and $(DAC)$, we have $O_{1}, O_{2}, T$ collinear.

Same groupsolve as above, another writeup:

Let $T=EF \cap BC, M=AT \cap (ABC), W=BM \cap AC, (DEX) \cap (ACD) = P$. We begin with three claims. Claim: $BCEF$ is cyclic.

Claim: $TD^2=TB \cdot TC$.

Claim: $MBEX$ is cyclic.

By the last claim, $W$ is the radical center of $(APDC), (ABC), (MBEX), (DPEX)$. In particular, $WDP$ are collinear and $MBDP$ is cyclic. Finally, invert at $T$ fixing $(ABC)$. This swaps $(MBDP)$ and $(ACDP)$, and hence fixes the intersections $D,P$. The fact that $TP=TD$ solves the problem.

I'd like to present my solution as well, even though some of the ideas are already present in the above solutions. I enjoyed the process of decripting the configurations this one involved $\textit{Step I.}$ We prove $BCEF$ is cyclic. Le the angle bisector of $\angle BAC$ meet $(ABC)$ again at $M$. Then $\angle DCM=\angle DCB+\angle BCM=\angle ADE+\frac 12 \angle BAC=\angle ADE+\angle DAE=\angle DEC$, so $MC$ is tangent to $(DEC)$. Similarly, $MB$ is tangent to $(DFB)$, which togrther with $MB=MC$ leads to $M$ being on the radical axis of the circles $(DFB)$ and $(DEC)$. Since $D$ lies on it too, we conclude that $AD$ is the radical axis of the $2$ circles, meaning that $AF\cdot AB=AE\cdot AC$ i.e. $BCEF$ is cyclic. $\textit{Step II.}$ We prove $(BDC)$ and $(EDF)$ are tangent to each other. In order to achieve that, we just have to prove that $\angle EDC=\angle EFD+\angle DBC$. However, if $H\in AD\cap BC$, then $\angle CDE=\angle DHC=\angle DBC+\angle BDH$ (since, in fact, $DE$ is tangent to $(DHC)$), so we just have to check that $\angle BDH=\angle DFE$. But $\angle DFE=\angle BFE-\angle BFD=180^o-\angle ACB-\angle MBD=180^o-\angle BMD-\angle MBD=\angle BDM$, done. $\textit{Step III.}$ We connect everything now. It follows that $BC$, $EF$ and the common tangent of the circles $(BDC)$ and $(EDF)$ are concurrent at a point $Z$. In order to finish the problem, if $Y$ is the second intersection of circles $(XDE)$ and $(ADC)$, we need to prove that $ZD=ZY$. In order to achieve this, it sounds reasonable to consider inverting around $Z$ with power $ZD^2$. This way, $A$ is sent to $A'$, the Miquel point of cyclic quad $BCEF$. Now we introduce point $X$ to the configuration: $\angle BA'E=\angle AA'E-\angle AA'B=180^o-\angle AFE-\angle ACB=180^o-2\angle ACB=\angle BXE$, so $BEA'X$ is cyclic. Now, the lines $XE, BA'$ and $DY$ are the pairwise radical axes of circles $(ABCA'),(XDEY),(BEA'X)$, thus they are concurrent, which by, say, Radical lemma, leads to the cyclicity of quad $BDA'Y$. Since the inversion at $Z$ sends $(ADCY)$ to $(BDA'Y)$, and since $Y$ is the second intersection point of these $2$, besides $D$, it follows that $ZY^2=ZD^2$ i.e. $ZD=ZY$, and so the problem is solved.

We denote the power of $\bullet$ wrt circle $\odot$ by $P(\bullet, \odot)$. Claim: Quadrilateral $BCEF$ is cyclic. Proof. Let $D'$ be the isogoanal conjugate of the point $D$. The angle condition implies quadrilateral $CEDD'$ and $BFDD'$ are cyclic. By power of point we have $AE\cdot AC=AD\cdot AD'=AF\cdot AB$. So $BCEF$ is cyclic. $\blacksquare$ Claim: Line $ZD$ is tangent to the circles $(BCD)$ and $(DEF)$ where $Z=EF\cap BC$. Proof. Let $\angle CAD=\angle BAD=\alpha$, $\angle BCD=\beta$, $\angle DBC=\gamma$, $\angle ACD=\phi$, $\angle ABD=\epsilon$. From $\triangle ABC$ we have $2\alpha+\beta+\gamma+\phi+\epsilon=180^\circ$. Let $\ell$ be a line tangent to $(BCD)$ and $K$ be a point on it in the same side of $AD$ as $C$ and $L=AD\cap BC$. From our labeling we have, \begin{align*} \angle AFE &= \beta + \phi, \qquad \angle BFD = \alpha + \gamma, \qquad \angle DFE = \alpha + \phi, \qquad \angle CDL = \alpha + \phi \end{align*}Now $\angle CDJ = 180^\circ - \gamma - \beta - (\alpha + \phi) = \alpha + \epsilon$. So $\angle DFE = \angle EDK = \alpha + \epsilon$, which menas $\ell$ is also tangent to $(DEF)$. Now by the radical center theorem we have $\ell$ passes through $Z$. $\blacksquare$ Let $M$ be the miquel point of the cyclic quadrilateral $BCEF$. From the miquel configuration we have $A$, $M$, $Z$ are collinear and $(AFEM)$, $(ZCEM)$ are cyclic. Claim: Points $B$, $X$, $M$, $E$ are cyclic. Proof. Notice that $\angle EMB = 180^\circ - \angle AMB -\angle EMZ = 180^\circ - 2\angle ACB = \angle EXB$. $\blacksquare$ Let $N$ be the other intersection of circles $(ACD)$, $(DEX)$ and let $R$ be the intersection of $AC$, $BM$. [asy][asy] import graph; size(14cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -11., xmax = 17.4, ymin = -13.5, ymax = 7.5; /* image dimensions */ pen ffxfqq = rgb(1.,0.4980392156862745,0.); pen qqwuqq = rgb(0.,0.39215686274509803,0.); pen ffdxqq = rgb(1.,0.8431372549019608,0.); pen qqffff = rgb(0.,1.,1.); /* draw figures */ draw((12.,-7.5)--(3.888164714306506,0.3692314047262663), linewidth(0.4) + red); draw((3.888164714306506,0.3692314047262663)--(0.5,-7.5), linewidth(0.4) + red); draw((2.731293854793663,-2.317675893626402)--(4.745944291861816,-3.928412178384129), linewidth(0.4)); draw((4.745944291861816,-3.928412178384129)--(5.4799348442300415,-1.1749330234631339), linewidth(0.4)); draw(circle((6.25,-6.904198198235134), 5.780785395340866), linewidth(0.4) + red); draw((6.25,5.854746416582486)--(12.,-7.5), linewidth(0.4)); draw((-9.73375741022764,-7.5)--(5.4799348442300415,-1.1749330234631339), linewidth(0.4) + red); draw((-9.73375741022764,-7.5)--(12.,-7.5), linewidth(0.4) + red); draw((-9.73375741022764,-7.5)--(4.745944291861816,-3.928412178384129), linewidth(0.4)); draw(circle((0.03259664736500382,-2.6347381854913), 4.887661855915456), linewidth(0.4) + ffxfqq); draw(circle((6.840565436013305,0.7567578002626845), 5.13208100738286), linewidth(0.4) + qqwuqq); draw(circle((13.092427672885751,3.8712206214193152), 11.423574599987449), linewidth(0.4) + ffxfqq); draw(circle((6.25,-5.311694240101265), 6.152331436033495), linewidth(0.4) + red); draw(circle((3.8881647143065052,-1.223272553407604), 1.5925039581338705), linewidth(0.4) + red); draw((3.888164714306506,0.3692314047262663)--(-9.73375741022764,-7.5), linewidth(0.4)); draw(circle((9.816894084966325,-0.5247258908263249), 7.3089260725747955), linewidth(0.4) + ffdxqq); draw((2.508611772870688,-0.42772083387717386)--(12.,-7.5), linewidth(0.4)); draw((1.83901321963071,1.9068581842023715)--(4.745944291861816,-3.928412178384129), linewidth(0.4)); draw((6.25,5.854746416582486)--(3.888164714306506,0.3692314047262663), linewidth(0.4)); draw((0.5,-7.5)--(4.745944291861816,-3.928412178384129), linewidth(0.4)); draw((4.745944291861816,-3.928412178384129)--(12.,-7.5), linewidth(0.4)); draw(shift((-9.73375741022764,-7.5))*xscale(14.913685022455285)*yscale(14.913685022455285)*arc((0,0),1,3.185775541948478,50.41705450522159), linewidth(0.4) + dotted + qqffff); /* dots and labels */ dot((12.,-7.5),linewidth(3.pt) + dotstyle); label("$B$", (12.471428644365055,-8.477712836766255), NE * labelscalefactor); dot((0.5,-7.5),linewidth(3.pt) + dotstyle); label("$C$", (-0.29392501614684113,-8.477712836766255), NE * labelscalefactor); dot((3.888164714306506,0.3692314047262663),linewidth(3.pt) + dotstyle); label("$A$", (3.426458804697535,1.1133683894471835), NE * labelscalefactor); dot((6.25,5.854746416582486),linewidth(3.pt) + dotstyle); label("$X$", (5.781564159177003,6.472086369929425), NE * labelscalefactor); dot((4.745944291861816,-3.928412178384129),linewidth(3.pt) + dotstyle); label("$D$", (4.655209424425953,-5.030384709194877), NE * labelscalefactor); dot((5.4799348442300415,-1.1749330234631339),linewidth(3.pt) + dotstyle); label("$F$", (5.645036312540512,-0.8662853867819248), NE * labelscalefactor); dot((2.731293854793663,-2.317675893626402),linewidth(3.pt) + dotstyle); label("$E$", (1.5833328751049083,-2.4363556231015626), NE * labelscalefactor); dot((-9.73375741022764,-7.5),linewidth(3.pt) + dotstyle); label("$Z$", (-10.772437245497516,-8.204657143493275), NE * labelscalefactor); dot((2.508611772870688,-0.42772083387717386),linewidth(3.pt) + dotstyle); label("$M$", (1.2078812968545585,-0.21777811525859622), NE * labelscalefactor); dot((1.83901321963071,1.9068581842023715),linewidth(3.pt) + dotstyle); label("$N$", (0.7982977569450858,2.478646855812086), NE * labelscalefactor); dot((3.2932895576506245,-1.0124045720848776),linewidth(3.pt) + dotstyle); label("$R$", (3.631250574652271,-1.1052091183957826), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Claim: Points $B$, $D$, $M$, $N$ are cyclic. Proof. By power of point we have \[ P(R, (ACD)) = RC \cdot RA = RM \cdot RB = RE \cdot RX = P(R, (DEX)) \]Menas $R$ lies on the radical axis of $(ACD)$ and $(DEX)$ and so $N$, $R$, $D$ are collinear. Also \[ RN \cdot RD = RA \cdot RC = RM \cdot RB \]So $BDMN$ is cyclic. $\blacksquare$ Notice that $(ACD)$, $(BDMN)$, $(DEX)$ are coaxial so their centers are collinear. Now we just need to prove the centers of $(ACD)$, $(BDMN)$ and $Z$ are collinear. To prove this, take a circle $\omega$ with radius $ZD$ centered at $Z$. Notice that by power of point $ZC \cdot ZB = ZD^2 = ZE \cdot ZF = ZM \cdot ZA$, which means inversion circle $\omega$ swaps $(ACD)$ and $(BDMN)$. So the centers of $(ACD)$ and $(BDMN)$ must have to be collinear with the center of inversion circle. As desired.

Posting for storage; note that inverted points retain their designations across inverted diagrams. Starting diagram Let $EF\cap BC=P$, $AD\cap BC=D_0$, $(XED)\cap (ADC)=\{ D,Q\}$ and $N\in (ADC)$ such that $DN\parallel BC$. Lemma 1: $PD$ is tangent to $(BDC)$ at $D$.

Lemma 2: $X,Q$ and $N$ are collinear. (The proof is just by angle chasing) Now, to finish the proof of the problem, we need to prove that $P$ lies on $O_1O_2$ or, equivalently, that $PD=PQ$. Note that because of Lemma 1, the circle with center $P$ and radius $PD$ is in fact the $D$-Apollonian circle of $\triangle BDC$, so we want to prove that $\frac{BQ}{QC}=\frac{BD}{DC}$. We accomplish this by way of two inversions. To start, invert about $D$ with arbitrary radius. First Inversion Placement of points: $D$ is on the interior of $\triangle ABC$, satisfying $\angle ABD=\angle DCA$ $X$ lies on $(ADC)$ such that $x\neq D$ and $\frac{BX}{XC}=\frac{BD}{DC}$ (i.e. $X$ lies on the $D$-Apollonian circle of $\triangle BDC$) $N$ is the intersection of $AC$ and the tangent at $D$ to $(BDC)$ $Q$ lies on $AC$ such that $D$, $X$, $N$ and $Q$ are concyclic Want to prove: $BQ=QC$ We now perform a second inversion, this time centered at $C$ and with arbitrary radius. Doubly-inverted Diagram Placement of points: $D$ is on the exterior of $\triangle ABC$, on the opposite side of $AB$ as $C$, satisfying $\angle BDC+\angle ACD=\angle BAC$ $N$ lies on $AC$ so that $(CDN)$ is tangent to $BD$ at the point $D$ $X$ lies on $AD$ such that $BD=BX$ $Q$ is the second intersection of the line $AC$ and $(DNX)$ Want to prove: $BQ=BC$ Now, let $(CDN)$ intersect $CB$ for the second time at $J$ and let $JN\cap AB=K$. We have that $$\angle CDN=180^{\circ}-\angle DNC-\angle NCD=180^{\circ}-\angle BDC-\angle ACD=180^{\circ}-\angle BAC$$so $$\angle CJK=\angle CJN=180^{\circ}-\angle BAC \rightarrow (ACJK)$$. Furthermore, $$\frac{BD}{BK}=\frac{BD^2}{BD\cdot BK}=\frac{BJ\cdot BC}{BD\cdot BK}=\frac{BA\cdot BK}{BD\cdot BK}=\frac{BA}{BD}$$and $\angle KBD=\angle ABD$, so $$\triangle ABD\sim \triangle DBK\rightarrow \angle DKB=\angle BDA\equiv\angle BDX=\angle DXB\rightarrow (DXKB)$$. Now, we have that $(DXKB)$ and $(DNXQ)$, so $AB\cdot AK=AX\cdot AD=AN\cdot AQ$, hence $(BKQN)$. This means that $\angle CQB\equiv \angle AQB=\angle NKA=180^{\circ}-\angle AJK=\angle JCA\equiv \angle BCQ$, i.e., $\angle CQB=\angle BCQ$, which immediately implies that $BC=BQ$, as desired.$\blacksquare$

Here's an isogonality and inversion only solution. Probably similar to something above, but haven't checked. Claim 1: $BCEF$ is cyclic. Proof: Consider the isogonal conjugate $D^*$ of $D$ w.r.t. $\triangle ABC$. By the angle conditions, we get that $BFD^*D$ and $CED^*D$ are cyclic. PoP at $A$ finishes. Let $R=EF\cap BC$, and $K$ be the second intersection of $(EDX)$ and $(CAD)$. Note that proving $RD=RK$ suffices. Also, let $RA\cap (ABC)=\left\{S,A\right\}$. Claim 2: $RD$ is tangent to $(BDC)$ and $(EFD)$, or the inversion $\mathcal{J}$ at $R$ with radius $RD$ fixes $(BFEC)$. Proof: Note that $DF$ and $DC$ are isogonal w.r.t. $\angle EDB$, as are $DB$ and $DE$ themselves. By the isogonality lemma, so are $DR$ and $DA$. The claim follows. To prove $RD=RK$, we'll show that $\mathcal{J}$ fixes $K$ by proving that $(SBD)$, $(ACD)$ and $(XED)$ are coaxial, because $(SBC)\stackrel{\mathcal{J}}{\longleftrightarrow}(ACD)$. Claim 3: $(SBD)$, $(ACD)$ and $(XED)$ are coaxial. Proof: We invert about $A$, fixing $(BCEF)$. Because of angle chasing, or noting that $S$ is the spiral center $FE\rightarrow BC$, $RBFS$ is cyclic. Hence, this inversion swaps $S\longleftrightarrow R$. After seeing where everything goes, it boils down to showing $RCX^*F$ is cyclic, where $X^*$ is the image of $X$. This follows by direct angle chasing.

Let $D'$ be the isogonal conjugate of $D$ with respect to $\triangle ABC$, let $P$ be the intersection of $EF$ and $BC$, let $Y$ and $Z$ be the intersection of $AD$ with $BC$ and $EF$, respectively. Let $S$ be the Miquel point of complete quadrilateral $BCEFAP$, and $G$ be the intersection of $BS$ and $AC$. $~$ We have $\angle ABD'=\angle CBD=\angle ADF$, so $BDD'F$ is cyclic. Similarly, $CDD'E$ is cyclic. We have \[AF\cdot AB=AD'\cdot AD=AE\cdot AC\]so $BCEF$ is cyclic. Thus, $AEZF\sim ABYC$. In particular, $\angle PYZ=\angle PZY$. We have \[\angle CDE=180^\circ-\angle CDY-\angle ADE=180^\circ-\angle CDY-\angle DCY=\angle DYC\]so let $P'$ be on the tangent to $(BDC)$ at $D$ on the same side as $E$ and $C$ as $AD$, then $\angle P'DC=\angle DBY$ and so \[\angle P'DE=\angle CDE-\angle P'DC = \angle DYC-\angle DBY = \angle BDY\]On the other hand, \[\angle DFZ=\angle DZE-\angle FDZ=\angle DYC-\angle DBY=\angle BDY\]so $P'D$ is tangent to $(DEF)$. Thus, by radical center theorem, $DP'$, $EF$, and $BC$ concur, at $P$. Since $BCEF$ is cyclic, $S$ lies on $AP$ and the following are concyclic: $ABCS$ $SAFE$ $SECP$ $SFBP$ We have \begin{align*} \angle SEX &= \angle SFA \\ &= \angle AFE - \angle SFE \\ &= \angle ACB - \angle SBC \\ &= \angle XBC - \angle SBC \\ &= SBX \end{align*}so $SBEX$ is cyclic. We have $PD^2 = PE\cdot PF = PC\cdot PB=PA\cdot PS$ so the inversion centered at $P$ fixing $D$ switches $A$ and $S$, $B$ and $C$, $E$ and $F$. Thus, $(ACD)$ is taken to $BDS$. We have \[GA\cdot GC = GS\cdot GB=GE\cdot GX\]so $G$ has equal power with respect to $(ACD)$, $(DEX)$, and $(BDS)$. Thus, they are coaxial, and so $P$ is equidistant from the two common intersection points. Thus, the centers of $(ACD)$ and $(DEX)$ are collinear with $P$, as desired.

Alright, another 45 MOHS... great. Why do I feel like I received the short end of 45 MOHS again and not actually solving the hard ones... But nonetheless one of my favorite problems, and wth SO MANY CIRCLESSSSS. What a journey it has been (2.5 hrs is not really considered a journey but wtv) Let $W=AD\cap BC,Y=AD\cap(BDC),Z=BC\cap EF$; note that $$AYC=DBC=ADF,AYB=DCB=ADE\implies\frac{AF}{AC}=\frac{AD}{AP}=\frac{AE}{AB},$$so BCEF is cyclic. Furthermore, let T be a point on the tangent to (DEF). Then $$DEF=AED-AEF=ABY-ABC=CBY=CDW\text{ and }FBD+FDB=AFD=180-FAD-ADF=180-BAW-DBW=ABD+DWB$$$$\implies DEF+DCB=CDW+DCB=DWB=FDB=FDT+TDB\Rightarrow TDB=DCB,$$so TD is also tangent to (CDB); in particular, the radcenter Z of (BDC),(DEF),(BCEF) means that the common tangent through D to (DEF) and (BDC) also passes through that point. Define M as the Miquel point of BCEF (so it lies on AZ, and ABCM is cyclic (!)). Consider the inversion centered at Z with radius DZ, meaning $ZA\cdot ZM=ZE\cdot ZF=ZD^2=ZB\cdot ZC\implies(ADC)\leftrightarrow(MDB)$. Let $U=(EDX)\cap(ADC),V=BM\cap AC$; observe that $\bullet\quad\operatorname{pow}_{(BDM)}V=BV\cdot VM\stackrel{(!)}{=}AV\cdot VC=\operatorname{pow}_{(ADC)}$, so V lies on the radax UD; furthermore, $UV\cdot VD=AV\cdot VC=BV\cdot VM$, so BDMU is cyclic. This means U is fixed under this inversion; in particular, letting $O_3$ be the center of (BDMX), this means $ZO_1O_3$ is a line (since the circles invert to each other), but by coaxiality (DEUX is coaxial) $O_2$ lies on this line as well! $\textbf{Remark.}$ There were two main steps here. The first was defining Y to prove BCEF cyclic, then using a good diagram to conjecture the tangent circles; a flurry of information was suddenly figured out. The second step was observing all the cyclic quads, and btw idt the last few concyclicities was required: for example, u can also prove BEMX cyclic!

this is an insane problem, congratulations to the author

Pure unfiltered heat. Awesome. Let $K$ be the intersection of $AD$ and $BC$. $\textbf{Claim.}$ $BCEF$ is cyclic. $\textit{Proof.}$ Let the circle tangent to $CD$ at $D$ passing through $E$ intersect $AD$ at $P$. Let $EP$ intersect $AB$ at $F'$. Note now that it is sufficient to prove that $(BDK)$ is tangent to $DF$. Note that $\measuredangle APE = \measuredangle DPE = \measuredangle CDE = \measuredangle CKD = \measuredangle CKA = \measuredangle BKA,$ whereby we conclude that $\measuredangle AEF' = \measuredangle CBA$. Therefore $BCEF'$ is cyclic. Also note that from a relatively easy angle chase $\triangle DPE \sim \triangle CKD$. Now, \[ \frac{PD}{PF'} = \frac{PD}{\frac{AF' \cdot PE}{AE}} = \frac{PD}{PE} \cdot \frac{AB}{AC} = \frac{CK}{DK} \cdot \frac{BK}{CK} = \frac{BK}{DK}, \]and since $\angle BKA = \angle F'PD$ we conclude that $\triangle F'PD \sim \triangle DKB$. This establishes that $F' = F$. $\square$ Define $T = \overline{EF} \cap \overline{BC}$ and $M = \overline{AT} \cap (ABC)$. Note that $M$ is the miquel point of complete quadrilateral $BCEFAT$. $\textbf{Claim.}$ $MBEX$ is cyclic. $\textit{Proof.}$ $\measuredangle BME = \measuredangle BMT + \measuredangle AME = \measuredangle BFT + \measuredangle AFE = 2\measuredangle BCA = \measuredangle BXE. \square$ Let $W = \overline{BM} \cap \overline{AC}$. By radical axes on $(ABC)$, $(BMEX)$, and $(EXD)$, and also $(ABC)$, $(EXD)$, $(ADC)$, we conclude that the radical axis of $(ADC)$ and $(EXD)$ passes through $W$. Let $K$ be the second intersection of $(EXD)$ and $(ADC)$, and hene $MKBD$ is cyclic. $\textbf{Claim.}$ $TD^2 = TB \cdot TC$. $\textit{Proof.}$ Suffices to prove that $\measuredangle FED + \measuredangle DCB = \measuredangle FDB$, as then the two circles $(DEF)$ and $(BCD)$ are tangent. This is equivalent to showing \[ \measuredangle ADC + \measuredangle DCB = \measuredangle AKB = \measuredangle FDB \]but this is very easy since we know that $\measuredangle DBC = \measuredangle FDA$. $\square$ Now, \[ \textbf{Invert about $T$ with radius $\sqrt{TB \cdot TC}$}. \](incredible!!!) $D$ stays fixed, $(B,C)$, $(E,F)$ $(M,A)$ swap. Then $(MBD)$ and $(ADC)$ swap, which implies that $K$ is fixed. Thus $TK = TD$, and so the perpendicular bisector of $KD$, which is $O_1O_2$, passes through $T$. Whew.

What an adventure! Let $D'$ be the isogonal conjugate of $D$ wrt $\triangle ABC$. Consider the following claim: Claim: Quadrilateral $BCEF$ is cyclic. Proof. $\angle D'DE = \angle ADE = \angle BCD = \angle ACD' = \angle ECD'$, thus $CDD'E$ is cyclic. Similarly $BDD'F$ is cyclic. Hence we get $AB \cdot AF = AD \cdot AD' = AE \cdot AC$, so we're done. $\blacksquare$ Claim: Circles $(BCD), (EFD)$ are tangent to each other. Proof. It suffices to prove that $\angle EFD + \angle CBD = \angle CDE$. Note that $\angle EDF + \angle BDC = 180^{\circ}$, so $\angle CDE = 180^{\circ} - \angle BDF = 180^{\circ} - \angle BD'F = \angle BFD' + \angle D'BF = \angle BFD' + \angle CBD$. Therefore we only have to prove $\angle EFD = \angle BFD'$, or equivalently $\angle EFD' = \angle BFD$. Now compute $\angle BFD' + \angle BFD = 180^{\circ} - \angle ADB + \angle BD'D = \angle BAD + \angle DBA + \angle BAD' + \angle ABD' = \angle BAC + \angle ABC = 180^{\circ} - \angle BCE = \angle BFE$. Therefore $\angle EFD' = \angle BFD$, as desired. $\blacksquare$ Let $P = EF \cap BC$. Then we have $PE \cdot PF = PC \cdot PB$, so $P$ lies on the radical axis of circles $(EFD), (BCD)$. Thus $PD$ tangents $(EFD), (BCD)$. Hence we get $PE \cdot PF = PC \cdot PB = PD^2$. Now perform an inversion centered $P$ with radius $PD$. Let $A', X'$ be the images of $A, X$ under the inversion, respectively. Then since $PA \cdot PA' = PD^2 = PB \cdot PC$, so $A'$ is Miquel point of $BCEF$. The inversion swaps $(ADC)$ to $(A'DB)$, so $(ADC) \cap (A'DB)$ is fixed under the inversion. Let $Y$ be the intersection of $(ADC)$ and $(A'DB)$ other than $D$. Then $PD = PY$. Claim: $Y$ lies on $(EXD)$. Proof. We only need to prove that $Y$ lies on $(FX'D)$. Note that $(A'EB)$ passes through the center of $(BCEF)$, so $X$ lies on $(A'EB)$. Therefore $X'$ lies on $(AFC)$. Since $X$ lies on $AC$, therefore $X'$ lies on $(PA'B)$. Hence $X' = (AFC) \cap (PA'B)$. Now note that the radical axes of circles $(ACD), (A'DB), (ABC)$ are concurrent, so $AC, A'B, DY$ are concurrent. Let $R = AC \cap A'B \cap DY$. The radical axes of circles $(AFCX'), (PX'A'FB), (ABC)$ are concurrent, so $FX', AC, AB$ are concurrent, which means $FX'$ passes through $X$. Therefore $YR \cdot RD = AR \cdot RC = FR \cdot X'R$. Hence $FDX'Y$ is cyclic, as needed. $\blacksquare$ Since $Y$ lies on $(EXD), (ADC)$ and $PD = PY$ means $P$ lies on $O_1O_2$. This completes proof. $\blacksquare$

Here's a sketch: This felt staightforward

CLaim 1: $BCEF$ is cyclic

Claim 2 $(EDF),(DBC)$ are tangent to $D$; $EF \cap BC \cap DD = T$

Let $(AEF) \cap (ABC) = I$, clearly $T-I-A$, let $H = (DIB) \cap (DEC)$, $U=BI \cap AC \cap DH$ Claim 3:$IEXB$ is cyclic

Claim 4: $ADHC$ is cyclic

By this we conclude $TD=TH$ and we're done.

lol Let $D'$ be the isogonal conjugate of $D$ in $\triangle ABC$. From our angle condition, clearly $BD'DF$ and $CD'DE$ are cyclic. Radical axis theorem, as $D'$ still lies on the angle bisector, implies $BFEC$ is cyclic. Note that the angle condition is equivalent to, if we let $P$ be the foot of the bisector from $A$, $ED$ tangent to $PDC$ and $FD$ tangent to $PDB$ at $D$. Hence, $\measuredangle CDE=\measuredangle CPD=\measuredangle BPD=\measuredangle BDF$ thus $D$ lies on the isoptic cubic of $BCEF$ thus it has an isogonal conjugate. Let $D'$ is already the intersection of two isogonal lines, hence $D'$ is precisely the isogonal conjugate. This also implies that $D'$ is the Clawson-Schmidt conjugate of $D$ in $BCEF$, and from its definition, we conclude that $(FDE)$ is tangent to $(BDC)$. Radical axis theorem on $(BDC), (FDE)$ and $(BFEC)$ gives $EF,BC,$ and the tangent to $(BDC)$ at $D$ concur, say at $Q$ Now invert at $Q$ fixing $(BCEF)$. $D$ is fixed by Power of a Point. Now the line $AO_1$ must contain $O_1'$ thus we want to show $O_1',O_2,Q$ collinear. Let $\omega$ be the circle we are inverting about. We want to show that $\omega, (ADC), (XED)$ are coaxial. as then their centers are collinear. First of all, all of these circles pass through $D$, hence we only need to characterize some other point that has equal power to these three circles. $XB$ is tangent to $(BFQ)$ as $\measuredangle XBC=\measuredangle BCE=\measuredangle BFE$, now use Desargues Involution Theorem on degenerate quadrilateral $BBFQ$ and line $AC$. $BF$ intersects $AC$ at $A$, $BQ$ intersects $AC$ at $C$. Hence $\{A,C\}$ is one of the pairs of involution. Now, consider the intersection of $(BFQ)$ with $\omega$, these intersections lie on $AC$ as $AC$ is the inverse of $(BFQ)$ upon inversion. Hence, letting these be $M,N$ in some order, $\{M,N\}$ is another pair of involution. Finally, $BB\cap AC=X$, and $FQ\cap AC=E$, hence $\{E,X\}$ are also a pair of involution. Such pairs are an involution on a line hence an inversion about some fixed point. That point has equal power with respect to the three circles, so we are done.

Let $\Omega$ be the circumcircle of $\triangle ABC$. Let $M$ be the arc midpoint of $BC$. Let $N$ be the arc midpoint of $BAC$. Let $K$ be the intersection of $MC$ with $(ADC)$. Let $Z$ be the intersection of $KA$ with $\Omega$. Let $\ell_C$ be the tangent to $\Omega$ at $C$. Let $U$ be the intersection of $ZN$ with $\ell_C$. Then, we have \[X = AC \cap NM\]\[U = \ell_C \cap ZN\]\[K = MC \cap AZ.\]By Pascal's theorem on $NMCCAZ$, $X$, $U$, and $K$ are colinear. Let $T$ be on $MN$ such that $\angle DCT = \angle MAC$. Then, \[\angle ZKC = \angle AKC = \angle ADC = \angle MDC\]\[\angle CZK = \angle CZA = \angle CMA = \angle CMD\]Therefore, $\triangle ZKC \sim \triangle MDC$. We have \[\angle TCM = \angle DCM + \angle TCD = \angle KCZ + \angle CAM =\angle KCZ + \angle UCK = \angle UCZ\]\[\angle CMT = \angle CMN = \angle CZN = \angle CZU.\]Therefore, $\triangle MTC \sim \triangle ZUC$. Therefore, $MTDC\sim ZUKC$. Let $R$ be the intersection of $TD$ with $XK$. Then, \[\angle RKC = \angle UKC = \angle TDC = \angle RDC.\]So, $R$ is on $(KCD) = (ADC)$. We have $\angle DCT = \angle MAC = \angle DAC$. So, $TC$ is tangent to $(ADC)$. Therefore, \[TB^2 = TC^2 = TD \cdot TR.\]So, inversion at $T$ with radius $TC$ swaps $D$ and $R$ while keeping $B$ and $C$ fixed. So, it swaps $(BRC)$ with $(BDC)$. Let $P$ be the point such that $PD$ is tangent to $(BDC)$ and $PR$ is tangent to $(BRC)$. Because inversion preserves angles, $\angle TDP = \angle PRT$. So, $\triangle RPD$ is isosceles. Let $\omega$ be the circle tangent to $PD$ at $D$ and $PR$ at $R$. By the radical center theorem on $(BDC)$, $(BRC)$, and $\omega$, $P$ lies on $BC$. We have \[\angle XRD = \angle KRD = \angle KCD = \angle MCD = \angle MAB + \angle BCD\]\[= \angle CAM + \angle ADE = \angle EAD+\angle ADE = \angle AED = \angle XED.\]Therefore, $R$ is on $(EXD)$. So, $RD$ is the radical axis of $(EXD)$ and $(ADC)$. Therefore, $O_1O_2$ is the perpendicular bisector of $RD$. Because $PR = PD$ and $P$ is on $BC$, $P=O_1O_2\cap BC$. Let $D'$ be the point on $AD$ such that $DBD'C$ is cyclic. Then, \[\angle ADE = \angle BCD =\angle BD'D = \angle BD'A.\]Because $\angle EAD = \angle D'AB$, $\triangle BAD'\sim \triangle EAD$. Similarly, $\triangle CAD'\sim \triangle FAD$, so $AEDF\sim ABD'C$. So, $\triangle AEF\sim \triangle ABC$. Therefore, $EFBC$ is cyclic. We have \[\angle PDE = \angle PDD' + \angle ADE = \angle DBD' + \angle BCD = \angle DBD' + \angle BD'D = \angle BDD' = \angle BCD' = \angle DFE.\]So, $PD$ is tangent to $(DEF)$. By the radical center theorem on $(FECB)$, $(BCD)$, and $(EFD)$, $P$ lies on $EF$. Therefore, $BC$, $EF$, and $O_1O_2$ concur at $P$.

Another IMO3 YAY! My first actual solution involving some non-trivial isogonal conjugate argument. We denote by $R$ the intersection of lines $\overline{EF}$ and $\overline{BC}$. We start off by making the following observation. Claim : Points $B$ , $C$ , $E$ and $F$ are concyclic. Proof : Note that since \[\measuredangle CDB + \measuredangle FDE = \measuredangle CBD + \measuredangle DCB + \measuredangle FDA + \measuredangle ADE = \pi\]it follows that $D$ has an isogonal conjugate $D'$ with respect to quadrilateral $BFEC$. Further note that since $CD$ and $CD'$ are isogonal with respect to $\angle C$ , and $BD$ and $BD'$ are isogonal with respect to $\angle B$, $D'$ is also the isogonal conjugate of $D$ with respect to $\triangle ABC$. Since $D$ lies on the internal $\angle A-$bisector this implies that $D'$ also lies on the internal $\angle A-$bisector. Now, note that if we let $D'' = \overline{AD} \cap (CDE)$ and $K = \overline{BC} \cap (CED)$, we must have $AD \parallel EK$ since $\measuredangle KED = \measuredangle KCD = \measuredangle BCD = \measuredangle ADE$ so \[\measuredangle BCD'' = \measuredangle KCD'' = \measuredangle KED'' = \measuredangle DD''E = \measuredangle DCE\]so lines $CD$ and $CD''$ are isogonal in $\angle BCA$. Combined with the fact that $D''$ lies on the internal $\angle A-$bisector this implies that $D''$ is the isogonal conjugate of $D$ with respect to $\triangle ABC$, so $D''\equiv D'$ and thus $D'$ lies on $(DEC)$. An entirely similar argument shows that $D'$ also lies on $(BDF)$ which implies that $(BDF) \cap (CED)$ lies on $\overline{AD}$. Thus, $\overline{AD}$ must be the radical axis of $(BDF)$ and $(CDE)$ implying, \[AF \cdot AB = AD \cdot AD' = AE \cdot AC\]which concludes that $BFEC$ is a cyclic quadrilateral as desired. Now, let $M_Q$ denote the Miquel Point of cyclic quadrilateral $BFEC$. We then know that $\overline{AM_Q}$ also passes through $R$. Now, we can notice the following further cyclic quadrilaterals. Claim : Quadrilaterals $BEM_QX$ and $BDM_QS$ are cyclic. Proof : The first one is immediate. Note that, \[\measuredangle EM_QB = \measuredangle EM_QA+ \measuredangle AM_QB = \measuredangle EFA + \measuredangle ACB = 2\measuredangle ACB = \measuredangle CXB = \measuredangle EXB\]which implies that $BEM_QX$ is indeed cyclic. Further, let $L = \overline{AC} \cap \overline{BM_Q}$. Note that, \[LX \cdot LE = LB \cdot LM_Q = LA \cdot LC\]which implies that $L$ lies on the radical axis of circles $(XED)$ and $(ADC)$. Letting $S = (XED) \cap (ADC) \ne S$, this then implies $LB \cdot LM_Q = LD \cdot LS$ so quadrilateral $BDM_QS$ must be cyclic as well, as desired. We now prove a claim which will come in very important later on. Claim : Line $\overline{DL}$ is a common tangent to circles $(DBC)$ , $(DEF)$ and $(DAM_Q)$. Proof : We start off by showing that circles $(DBC)$ and $(DEF)$ are tangent to each other. Note that it suffices to check that $\measuredangle EDC = \measuredangle DBC + \measuredangle FDE$ which we shall do as follows. Let $P = \overline{AD} \cap \overline{BC}$. \begin{align*} \measuredangle EFD + \measuredangle DBC & = \measuredangle D'FB + \measuredangle DBC\\ & = \measuredangle D'DB + \measuredangle DBP \\ &= \measuredangle DPB \\ & = \measuredangle EKB \\ &= \measuredangle EKC \\ &= \measuredangle EDC \end{align*}so indeed circles $(BDC)$ and $(EDF)$ are tangent to each other at $D$. Further, by Radical Center theorem on circles $(DEF)$ , $(BCD)$ and $(BCEF)$ their common tangent must pass through $R$. Again, \[AR^2 = RB \cdot RC = RA \cdot RM_Q\]which further implies that $\overline{RD}$ is tangent to $(ADM_Q)$ finishing the proof of the claim. To finish off we need to show the following key claim. Claim : Circles $(SBC)$ and $(SAM_Q)$ are tangent at $S$, with common tangent $\overline{SR}$. Proof : As before, we first show that circles $(SBC)$ and $(SAM_Q)$ are tangent at $S$, for which it suffices to check $\measuredangle SAM_Q + \measuredangle CBS = \measuredangle CSM_Q$. This is a really long angle chase. I will only include a sketch here but it is detailed enough for anyone interested to be able to fill in the details. Note that since \[\measuredangle CSM_Q = \measuredangle CSD + \measuredangle DSM_Q = \measuredangle CAD + \measuredangle DBM_Q = \measuredangle CAD + \measuredangle CBM_Q + \measuredangle DBC \]it suffices to show that, \[\measuredangle SAC + \measuredangle CBS = \measuredangle CAD + \measuredangle CBS\]which we shall do as follows. Note that since due to our previous claim that $\overline{DR}$ is tangent to $(AM_QR)$ and $(DBC)$, we have $\measuredangle M_QDR = \measuredangle M_QAD$ and $\measuredangle RDC = \measuredangle DBC$. Thus, \[\measuredangle DBC = \measuredangle M_QDR + \measuredangle RDC + \measuredangle DAM_Q = \measuredangle M_QDC + \measuredangle DAM_Q\]Further, \[\measuredangle M_QDC + \measuredangle CBM_Q = \measuredangle M_QAD + \measuredangle CAM_Q + \measuredangle DBC\]which implies $\measuredangle CAD + \measuredangle DBC = \measuredangle M_QDC + \measuredangle CBM_Q$. Now we are almost done as $(BDM_QS)$ cyclic implies, \[\measuredangle CAD + \measuredangle DBC = \measuredangle SDM_Q + \measuredangle M_QDC + \measuredangle M_QBS + \measuredangle CBM_Q = \measuredangle SAC + \measuredangle CBS\]which is a result which we have already confirmed. Thus, circles $(SBC)$ and $(SAM_Q)$ are indeed tangent at $S$. Further, by Radical Center Theorem on circles $(SBC)$ , $(SAM_Q)$ and $(ABCM_Q)$ it follows that the common tangent to $(SBC)$ and $(SAM_Q)$ at $S$ passes through $R$. Now we simply need to connect all the pieces. Note that due to the last two claims, \[RS^2 = RA \cdot RM_Q = RD^2\]so $R$ must lie on the perpendicular bisector of segment $SD$, which is well known to be the line joining the centers of the circles $(XED)$ and $(ADC)$ which is simply line $\overline{O_1O_2}$ implying that this line indeed passes through $R$ as we set out to show.

The above admitted orz while solving

We'll use Animation. But first let's make a proper setup. One can easily observe that if $D'$ is the isogonal conjugate of $D$ wrt $ABC$ , then $C,D',D,F$ and $B,D',D,E$ and so $B,C,E,F$ are concyclic and if $T=EF \cap BC$ then $TD$ is tangent to $(BDC)$ , all by radical axis theorem. Let $S$ be the midpoint of arc $(BAC)$ in $(ABC)$. Let $J = (ADC) \cap (ASX)$. Now let's start animating: Move $D$ on $BC$ with degree 1. Then by inverting through $A$ , $J$ has degree 2. And by inverting through $B,C$ , both $E,F$ have degree 2. We'll now prove that $D,E,X,J$ are concyclic. By inverting through $X$ , $E*,D*$ have degree 2 and $J*$ has degree 1 so we need 6 special cases to check the claimed concyclicity which are the items 1,2,3,4,7 below+ the case for $D$ where $E$ goes to infinity. Now , It's enough to show that $T$ lies on the perpendicular bisector of $DJ$. But before that , note that there's two cases where $E=F$ so the line $EF$ has degree at most $2 = deg(E)+deg(F)-k$ where $k = 2$ is the number of coincidences of $E,F$. Let $M$ be the midpoint of $DJ$ which has degree at most $deg(D)+deg(J)=3$ and note that since there exists one case(no.8 below) where $D=J$ , $DJ$ has degree at most $2 =deg(D)+deg(J)-k$.So perpendicular bisector of $DJ$ , which is the perpendicular through $M$ to $DJ$ has degree at most $3+2=5$ and $O_1O_2 \cap BC$ would have degree at most 5. So to prove that the recent intersection , is $T$ , a degree 2 point we need $8=5+2 +1$ special cases which are: 1,2,3,4 . The infinity point of the direction $l$ , $l \cap (ABC)$ , $l \cap BC$ , $A$ 5,6 . The two cases , where $TJ$ turns to be tangent to $(DEX)$ which exists by IVT 7. The one case where the perpendicular bisector of $DJ$ coincides with $BC$ which exists by IVT 8.The one case where $D=J$ which exists by IVT. Where $l$ is the angle-bisector of $A$...And we're done.

Let $M$ be the intersecting point of lines $EF$ and $BC$, and $K$ be the intersecting point of lines $AD$ and $BC$. Since $\angle CDE=\angle ADC-\angle ADE=\angle ADC-\angle DCB=\angle DKC$, we have $\frac {CE}{EA}=\frac {DCsinCDE}{DAsinADE}=\frac {DCsinDKC}{DAsinDCB}$. Similarly we have $\frac{BF}{FA}=\frac{DBsinDKB}{DAsinDBC}$. Using Menelaus theorem we get $$\frac{CM}{MB}=\frac{CE}{EA}\cdot\frac{FA}{BF}=\frac{DCsinDKC}{DAsinDCB}\cdot\frac{DAsinDBC}{DBsinDKB}=\frac{DCsinDBC}{DBsinDCB}=\frac{DC^2}{DB^2}$$From this, it is easy to show that line $MD$ is tangent to the circumcircle of $DBC$. We need to show that $M$ is on line $O_1O_2$. Define $\omega$ as the circle centered at $M$ with radius $MD$. It suffices to show that $\omega$ and the circumcircles of triangles $ADC$ and $EXD$ are coaxial. We can show this by finding a point $T(\neq D)$ such that the power of $T$ to the three circles are equal. Define $T$ as the point on segment $AE$ such that $XT\cdot TE=AT\cdot TC$. We will prove that $MT^2-MD^2=-AT\cdot TC$. Note that this solves the problem, because $T$ has same power on $\omega$ and the circumcircles of triangles $ADC$ and $EXD$. Clearly $T\neq D$, so the three circles would be coaxial. We have $\frac{TA}{TE}=\frac{TX}{TC}=\frac{XA}{CE}$, so $TA=\frac{XA\cdot AE}{XA+CE}$ and $TC=AC-TA=\frac{XA\cdot AC+CE\cdot AC-XA\cdot AE}{XA+CE}=\frac{CE\cdot XC}{XA+CE}$. We have $MT^2-MD^2=TC^2+MC^2+2TC\cdot MCcosC-MB\cdot MC=TC^2+2TC\cdot MCcosC-MC\cdot BC$, so we have $$MT^2-MD^2=-AT\cdot TC \iff TC^2+2TC\cdot MCcosC-MC\cdot BC=-AT\cdot TC \iff TC\cdot AC+2TC\cdot MCcosC=MC\cdot BC$$Using $TC=\frac{CE\cdot XC}{XA+CE}$ as we noted above, and $2XCcosC=BC$ which is obvious, we get $$TC\cdot AC+2TC\cdot MCcosC=MC\cdot BC \iff \frac{CE\cdot XC\cdot AC+BC\cdot CE\cdot MC}{XA+CE}=MC\cdot BC$$$$\iff CE\cdot XC\cdot AC=\cdot XA\cdot MC\cdot BC\iff \frac{CE}{MC}=\frac{XA\cdot BC}{XC\cdot AC}$$ So proving $\frac{CE}{MC}=\frac{XA\cdot BC}{XC\cdot AC}$ will solve the problem. RHS is easy to calculate. $XC=\frac{a}{2cosC}=\frac{a^2b}{a^2+b^2-c^2}$ and $XA=XC-b=\frac{(c^2-b^2)b}{a^2+b^2-c^2}$, so $\frac{XA\cdot BC}{XC\cdot AC}=\frac{c^2-b^2}{ab}$. For LHS, we will show that $\angle CEM=\angle B$. (The motivation for this is that LHS has to be fixed as $D$ moves on the bisector of $\angle A$, which means $\angle CEM$ has to be fixed. Considering the case $D=A$, we can see that this fixed angle has to be $\angle CEM=\angle B$.) Many people in this forum have proved that $BCEF$ is cyclic, e.g. using the isogonal conjugate of $D$, so it follows that $\angle CEM=\angle AEF=\angle B$. However I failed to see this, so below is what I did. It is not elegant, but it is quite straightforward.

We can easily finish off. We have $\frac{CE}{MC}=\frac{sinECM}{sinCEM}=\frac{sin(C-B)}{sinB}=cosB\cdot\frac{c}{b}-cosC=\frac{c^2-b^2}{ab}$, therefore $\frac{CE}{MC}=\frac{XA\cdot BC}{XC\cdot AC}$ as desired. So $T$ is indeed a point which the power to the three circles are equal. The problem is solved.