My problem. Not terribly original, though: you may notice that it is just a generalization of ELMO SL 2013 G3, USA TST 2008/7, and USA December TST 2012/1.

a1267ab wrote: Let $ABC$ be a triangle and let $P$ be a point not lying on any of the three lines $AB$, $BC$, or $CA$. Distinct points $D$, $E$, and $F$ lie on lines $BC$, $AC$, and $AB$, respectively, such that $\overline{DE}\parallel \overline{CP}$ and $\overline{DF}\parallel \overline{BP}$. Show that there exists a point $Q$ on the circumcircle of $\triangle AEF$ such that $\triangle BAQ$ is similar to $\triangle PAC$. Andrew Gu Fix $ABCP$ and animate $D$ on $BC$ with constant velocity. Since $D \mapsto E$ and $D \mapsto F$ are linear, $\odot(AEF)$ passes through a fixed point. So, we show for $D=B$ and $D=C$ that $Q$ lies on this circle. Both cases are analogous, so effectively, we just check $D=B$. Then $E=B$ and say $F=F'$ with $\overline{BF'} \parallel \overline{CP}$. Then $\measuredangle BQA=\measuredangle PCA=\measuredangle BF'A$ so $Q$ lies on $\odot(ABF')$. The result follows. $\blacksquare$

anantmudgal09 wrote: Fix $ABCP$ and animate $D$ on $BC$ with constant velocity. Since $D \mapsto E$ and $D \mapsto F$ are linear, $\odot(AEF)$ passes through a fixed point. You can also use barycentric coordinates to show this, which is what I did.

a1267ab wrote: My problem. Not terribly original, though: you may notice that it is just a generalization of ELMO SL 2013 G3, USA TST 2008/7, and USA December TST 2012/1.

In Solution 1. I think the correct is $\angle QFA=\angle QEC $ following the spiral similarity.

Perform $\sqrt{bc}$ inversion in $ABC$. We have that $Q$ is mapped to $P$. Hence if $E\rightarrow E', F\rightarrow F'$ then we need to show $E',P,F'$ collinear. Observe that $D\rightarrow E, D\rightarrow F$ are perspectivities while $E\rightarrow E', F\rightarrow F'$ are projectivities. Hence $E'\rightarrow F'$ is a projectivity. When $D$ is to infinity along $BC$, $E,F$ are also at infinity along respective lines and hence $E',F'$ are at $A$. Hence the projectivity fixes a point, hence it is a perspectivity. When $D=B$ line $E'F'$ is line $CP$ and when $D=C$, line $E'F'$ is line $BP$. Hence $P$ is the centre of perspectivity. We are done.

Here is an amusing solution using Generalized Reim's theorem. Lemma (Generalized Reim's): Let $\mathcal{C}$ be a conic and $A,B$ be points on it. Let $\omega_1$ and $\omega_2$ be circles through $A,B$ which intersect $\mathcal{C}$ again at $\{C,D\}$ and $\{E,F\}$ respectively. Then $CD\parallel EF$. Proof: Let $I,J$ be circle points. Consider a projective transformation which takes $A,B$ to $I,J$ respectively. Denote the image with primes. Then $\mathcal{C}', \omega_1', \omega_2'$ are all circles. Moreover, since $\omega_1, \omega_2$ intersect at $A,B,I,J$, we get the other two intersection of $\omega_1', \omega_2'$ are $I',J'$. Thus by radical axis theorem, $C'D', E'F', I'J'$ are concurrent. Taking the projective transformation back, we deduce the lemma. $\blacksquare$ Back to the main problem. Let $Q$ be the point such that $\triangle BAQ\stackrel{+}{\sim}\triangle PAC$. Clearly $X=CQ\cap DF$ lie on $\odot(AQF)$ and $Y=BQ\cap DE$ lie on $\odot(AQE)$. Moreover, by converse of Pascal's theorem on $AFXQYE$, these six points lie on a conic. Assume for the contradiction that this conic is not a circle. Then by Generalized Reim's theorem, $FX\parallel EY$ which is a contradiction as two lines intersect at $D$!

10 min solve using moving points (Although I didn't realize the much easier solution given by Anant above ). Anyway, here goes nothing: Fix $A,B,C,D,F$ and animate $P$ on the line $\ell$ through $B$ parallel to $DF$. Then $$P \mapsto \infty_{CP} \mapsto D\infty_{CP} \mapsto E$$is a projective map. Also, it's easy to see that $P \mapsto Q$ is also a projective map (Try thinking about spiral similarities). In fact, $\angle AQC$ is fixed (since $\angle ABP$ is also fixed), amd so $Q$ moves on a fixed circle through $A$ and $C$. Thus, it suffices to prove $A,E,F,Q$ concyclic for 3 positions of $P$ (Note that 3 is sufficient, say by inversion at $A$). Taking $P=B$ gives $E=Q=C$, taking $P=C\infty_{AD} \cap \ell$ gives $E=A$, and finally $P=\infty_{\ell}$ makes $Q=A$. Thus, all three cases degenerate to given result. Hence, done. $\blacksquare$

anantmudgal09 wrote: a1267ab wrote: Let $ABC$ be a triangle and let $P$ be a point not lying on any of the three lines $AB$, $BC$, or $CA$. Distinct points $D$, $E$, and $F$ lie on lines $BC$, $AC$, and $AB$, respectively, such that $\overline{DE}\parallel \overline{CP}$ and $\overline{DF}\parallel \overline{BP}$. Show that there exists a point $Q$ on the circumcircle of $\triangle AEF$ such that $\triangle BAQ$ is similar to $\triangle PAC$. Andrew Gu Fix $ABCP$ and animate $D$ on $BC$ with constant velocity. Since $D \mapsto E$ and $D \mapsto F$ are linear, $\odot(AEF)$ passes through a fixed point. So, we show for $D=B$ and $D=C$ that $Q$ lies on this circle. Both cases are analogous, so effectively, we just check $D=B$. Then $E=B$ and say $F=F'$ with $\overline{BF'} \parallel \overline{CP}$. Then $\measuredangle BQA=\measuredangle PCA=\measuredangle BF'A$ so $Q$ lies on $\odot(ABF')$. The result follows. $\blacksquare$ How do you localized the point $Q $ in this step: $$\measuredangle BQA=\measuredangle PCA=\measuredangle BF'A$$. Also you have a typo using $F $ like $E $.

MarkBcc168 wrote: Lemma (Generalized Reim's): Let $\mathcal{C}$ be a conic and $A,B$ be points on it. Let $\omega_1$ and $\omega_2$ be circles through $A,B$ which intersect $\mathcal{C}$ again at $\{C,D\}$ and $\{E,F\}$ respectively. Then $CD\parallel EF$. Does the converse hold? I mean the following: Quote: Let $\mathcal{C}$ be a conic and $A$ $\in$ $\mathcal{C}$. Let $C,D,E,F$ $\in$ $\mathcal{C}$, such that $CD||EF$. Then, $\odot (ACD)$ $\cap$ $\odot (AEF)$ $\in$ $\mathcal{C}$ If this holds, then we can use it here

A possible proof Quote: Let $\mathcal{C}$ be a conic and $A$ $\in$ $\mathcal{C}$. Let $C,D,E,F$ $\in$ $\mathcal{C}$, such that $CD||EF$. Then, $\odot (ACD)$ $\cap$ $\odot (AEF)$ $\in$ $\mathcal{C}$ Let $T=\odot{AEF}\cap \odot{ACD}$.It suffices to show \[(TC,TD,TE,TF)=(AC,AD,AE,AF)\]to show that $T\in \mathcal{C}$.Now \begin{align*}M=AD\cap\odot{AEF} \\ N=AC\cap \odot{AEF} \\ K=TC\cap \odot{AEF} \\ L=TD\cap \odot{AEF} \end{align*}.Now by Reim's Theorm we have $NL\parallel CD\parallel EF\parallel KM$.Thus we have $NMEF$ and $LKFE$ are just reflections of each other in the perpendicular bisector of $EF$.In particular we have $(N,M,E,F)=(K,L,E,F)$.Thus we have \[(AC,AD,AE,AF)\overset{A}{=}(N,M,E,F)=(K,L,E,F)\overset{T}{=}(TC,TD,TE,TF)\]so we conlcude.$ T\in \mathcal{C} \blacksquare$ [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(12cm); real labelscalefactor = 0.25; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -10.15829267367829, xmax = 5.7088690931314545, ymin = -10.187082943603759, ymax = 3.1611399809851473; /* image dimensions */ pen ttttff = rgb(0.2,0.2,1); pen qqzzqq = rgb(0,0.6,0); pen ffvvqq = rgb(1,0.3333333333333333,0); pen ccqqqq = rgb(0.8,0,0); draw(circle((-2.7649821719223664,-0.7014638853057545), 2.721795572670584), linewidth(0.7) + ttttff); draw(circle((-1.9932436628718546,-4.1746288993265255), 6.1092791987257336), linewidth(0.7) + qqzzqq); /* draw figures */ draw((-5.141944381050513,-2.0274434461534434)--(-0.10406669660888554,-1.2739147326685858), linewidth(0.8)); draw((-7.92556230736818,-5.634384984480556)--(-2.084770743483372,1.933964647595522), linewidth(0.7) + blue); draw((-2.084770743483372,1.933964647595522)--(-6.3517682426391415,-8.455581575489546), linewidth(0.8) + ffvvqq); draw((-2.084770743483372,1.933964647595522)--(3.4271343960244307,-6.992925197911663), linewidth(0.7) + blue); draw((-2.084770743483372,1.933964647595522)--(4.106571525019676,-3.8347068471575865), linewidth(0.8) + ffvvqq); draw((-4.497267715956468,1.3979069569562004)--(-6.3517682426391415,-8.455581575489546), linewidth(0.8) + ccqqqq); draw((-4.497267715956468,1.3979069569562004)--(4.106571525019676,-3.8347068471575865), linewidth(0.8) + ccqqqq); draw((-8.069941667734536,-4.804734990140489)--(4.001877157867984,-2.9991210632341323), linewidth(0.8)); draw((-8.069941667734536,-4.804734990140489)--(-7.92556230736818,-5.634384984480556), linewidth(0.7) + linetype("4 4")); draw((3.4271343960244307,-6.992925197911663)--(-7.92556230736818,-5.634384984480556), linewidth(0.7) + linetype("4 4")); draw((3.4271343960244307,-6.992925197911663)--(4.001877157867984,-2.9991210632341323), linewidth(0.7) + linetype("4 4")); draw((4.001877157867984,-2.9991210632341323)--(4.106571525019676,-3.8347068471575865), linewidth(0.7) + linetype("4 4")); draw((-6.3517682426391415,-8.455581575489546)--(4.106571525019676,-3.8347068471575865), linewidth(0.7) + linetype("4 4")); draw((-8.069941667734536,-4.804734990140489)--(-6.3517682426391415,-8.455581575489546), linewidth(0.7) + linetype("4 4")); draw((-7.92556230736818,-5.634384984480556)--(-6.3517682426391415,-8.455581575489546), linewidth(0.7) + linetype("4 4")); /* dots and labels */ dot((-2.084770743483372,1.933964647595522),dotstyle); label("$A$", (-1.9986531762279593,2.1492585657340526), NE * labelscalefactor); dot((-0.10406669660888554,-1.2739147326685858),dotstyle); label("$D$", (-0.017949129353473085,-1.0586208145300553), NE * labelscalefactor); dot((-5.141944381050513,-2.0274434461534434),dotstyle); label("$C$", (-5.055826813795101,-1.812149528014913), NE * labelscalefactor); dot((-8.069941667734536,-4.804734990140489),dotstyle); label("$E$", (-7.983824100479125,-4.58944107200196), NE * labelscalefactor); dot((-4.497267715956468,1.3979069569562004),dotstyle); label("$T$", (-4.409945059379508,1.6110237703877257), NE * labelscalefactor); dot((4.001877157867984,-2.9991210632341323),linewidth(4pt) + dotstyle); label("$F$", (4.094164707092471,-2.8240309432660076), NE * labelscalefactor); dot((-7.92556230736818,-5.634384984480556),linewidth(4pt) + dotstyle); label("$N$", (-7.833118357782153,-5.472146136369935), NE * labelscalefactor); dot((3.4271343960244307,-6.992925197911663),linewidth(4pt) + dotstyle); label("$M$", (3.5128711281184373,-6.828497820642679), NE * labelscalefactor); dot((4.106571525019676,-3.8347068471575865),linewidth(4pt) + dotstyle); label("$K$", (4.201811666161737,-3.6636772240062774), NE * labelscalefactor); dot((-6.3517682426391415,-8.455581575489546),linewidth(4pt) + dotstyle); label("$L$", (-6.2614727553708756,-8.292496463984689), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy][/asy]

This geo is giving away a big hint. $D$ could be any point on $BC$, hence we know that $\odot(AEF)$ has to pass through a fixed point. The problem is much easier when $D$ is either $B$ or $C$, so we wish to prove that $\odot(AEF)$ passes through a fixed point. Let $O$ be the circumcenter of $\triangle AEF$. $\triangle OEF$ is always similar to a fixed triangle. It's isosceles, and $\angle EOF=2\angle A$. Notice that, as $D$ moves on $BC$ with velocity $v$, $E$ and $F$ both move with constant velocity as well, and using a little complex numbers it's easy to see that $O$ also moves with constant velocity $\Longrightarrow$ $\odot(OEF)$s are coaxial. We're done.

Different sketch from all others, this solution mainly focusses on Isogonal Lines. USATST for EGMO 2020 P2 wrote: Let $ABC$ be a triangle and let $P$ be a point not lying on any of the three lines $AB$, $BC$, or $CA$. Distinct points $D$, $E$, and $F$ lie on lines $BC$, $AC$, and $AB$, respectively, such that $\overline{DE}\parallel \overline{CP}$ and $\overline{DF}\parallel \overline{BP}$. Show that there exists a point $Q$ on the circumcircle of $\triangle AEF$ such that $\triangle BAQ$ is similar to $\triangle PAC$. Andrew Gu Let $DE\cap \odot(ABC)=X$ and $DF\cap\odot(ABC)=Y$. We claim that the point $Q=XB\cap YC\in\odot(ABC)$. $XB\cap YC\in\odot(ABC)$ by Pascal's Theorem on $AFYQXE$. Now $\angle ACP=\angle AEX=\angle AQB$. So we just need to prove that $\{AP,AQ\}$ are isogonal WRT $\triangle AEF$. Let $XQ\cap PC=T$ and $PB\cap CQ=J$. Claim:- $APBT$ and $APCJ$ are cyclic quadrilaterals. Let $TC\cap YK=K$. So, $\angle ACT=\angle AEX=\angle AQT\implies ACQT$ is a cyclic quad. So, $\angle ATC=\angle AQY=\angle AFY$ and as $PB\cap FY$. So, by Reim's Theorem we get that $APBT$ is a cyclic quadrilateral. Similarly we get that $APCJ$ is a cyclic quadrilateral. So, $\angle TAB=\angle TPB=\angle JPC=\angle JAC\implies \{AT,AJ\}$ are isogonal WRT $\triangle AFE$ so as $\{AB,AC\}$. So, by Isogonal Line Lemma we get that $\{AP,AQ\}$ are also Isogonal WRT $\triangle AEF$. Hence, $\triangle BAQ\sim\triangle CAP$. So, $Q=CY\cap BC$ is the desired point. $\blacksquare$

alifenix- wrote: anantmudgal09 wrote: Fix $ABCP$ and animate $D$ on $BC$ with constant velocity. Since $D \mapsto E$ and $D \mapsto F$ are linear, $\odot(AEF)$ passes through a fixed point. You can also use barycentric coordinates to show this, which is what I did. Claim: Let the reflection of $AP$ over the angle bisector of $\angle BAC$ meet $(AEF)$ at a point $Q'.$ If we fix $P$ and vary $D$ along $BC,$ then $Q'$ is fixed too. Proof: We use barycentric coordinates with reference triangle $\triangle ABC.$ Let $P=(t_1, t_2, t_3)$ and $D=(0,d,1-d).$ Note that the point at infinity along $PC$ must have coordinates $(t_1:t_2:-t_2-t_1),$ so $E=(x_1,0,1-x_1)$ must satisfy \begin{align*} 0&=\begin{vmatrix} 0 & d & 1-d \\ x_1 & 0 & 1-x_1 \\ t_1 & t_2 & -t_2 - t_1 \\ \end{vmatrix} \\ 0&=d((1-x_1)t_1+x_1(t_2+t_1))+(1-d)(x_1t_2) \\ 0&=dt_1+x_1t_2 \\ x_1&=-\frac{dt_1}{t_2}, \\ \end{align*}so $E=(-dt_1:0:t_2+dt_1).$ Similarly, $F=(x_2,1-x_2,0)$ satisfies \begin{align*} 0&=\begin{vmatrix} 0 & d & 1-d \\ x_2 & 1-x_2 & 0 \\ t_1 & -t_1 - t_3 & t_3 \\ \end{vmatrix} \\ 0&=dx_2t_3+(1-d)(x_2(t_1+t_3)+t_1(1-x_2))\\ 0&=x_2t_3+(1-d)t_1 \\ x_2&=-\frac{(1-d)t_1}{t_3}, \\ \end{align*}so $F=(-(1-d)t_1:t_3+(1-d)t_1:0).$ $(AEF)$ is parametrized by $a^2yz+b^2zx+c^2xy=(x+y+z)(ux+vy+wz);$ plugging in $A$ gives us $u=0,$ and plugging in $E$ and $F,$ we get \begin{align*} -b^2dt_1(t_2+dt_1)&=t_2w(t_2+dt_1) \\ w&=-\frac{b^2dt_1}{t_2} \\ \end{align*}and \begin{align*} -c^2(1-d)t_1(t_3+(1-d)t_1)&=t_3v(t_3+(1-d)t_1) \\ v&=-\frac{c^2(1-d)t_1}{t_3}, \\ \end{align*}respectively. Finally, plugging in $Q'=(t:\tfrac{b^2}{t_2}:\tfrac{c^2}{t_3})$ gives us \[b^2c^2\left(\frac{a^2}{t_2t_3}+\frac{t}{t_3}+\frac{t}{t_2}\right)=-\frac{b^2c^2}{t_2t_3}\left(t+\frac{b^2}{t_2}+\frac{c^2}{t_3}\right)((1-d)t_1+dt_1),\]which in particular implies that $t$ is fixed because the terms with $d$ cancel out. $\Box$ Back to the main problem: Let $BP\cap CQ' = G$ and $BQ'\cap CP = H.$ We could probably do this next step with some easy angle chasing, but instead we'll use DDIT because it's a crime not to use it when you can. By Dual of Desargues' Involution about $A$ with respect to $BQ'CP,$ we find that $\{AB,AC\},$ $\{AP,AQ'\},$ and $\{AG,AH\}$ are swapped under some involution about the pencil of lines through $A,$ which clearly must be reflection across the angle bisector of $\angle BAC.$ This implies that $\measuredangle GAB=\measuredangle GQ'B=\measuredangle CQ'H=\measuredangle CAH,$ so $A$ is the Miquel Point of $BQ'CP.$ Thus, $\measuredangle Q'GA=\measuredangle HBA=\measuredangle HPA,$ whence $\triangle BAQ'\stackrel{+}{\sim}\triangle PAC$ by AA, so $Q'$ is the point we want. $\blacksquare$

If we construct the point $Q$ one can notice that $\angle AQB = \angle ACP$ by definition and $\angle AQC = \angle ABP$ by $\triangle ABP\sim \triangle AQC$ which follows from the spiral similarity with center $A$. Denote these two angle relations by $(1)$. Let $\Gamma = (AEF)$ , $R = \Gamma \cap ED$, $S = \Gamma \cap FD$, $Q' = RB \cap SC$. We have $B = RQ' \cap FA$, $D = RE\cap FS$, $C = SQ'\cap AE$ are collinear. From Pascal's theorem $\implies$ $A,R,F,Q',E,S$ lie on a conic but 5 of them lie on $\Gamma \implies Q'\in \Gamma$. Then $\angle AQ'B = \angle AQ'R = \angle AER = \angle AED = \angle ACP$, $\angle AQ'C = \angle AQ'S = \angle AFS = \angle AFD = \angle ABP$ which means that $Q'$ satisfies $(1)$. But there is only one such point (It is intersection of two circles through $AB$ and $AC$ and it is $\neq A$) $\implies Q\equiv Q' \implies Q\in \Gamma$

Basically the first solution in post #2. Assume all angles are directed. Let $G$ be the point on line $\overline{AC}$ such that $\overline{BG}\parallel\overline{CP},$ let $H$ be the point on line $\overline{AB}$ such that $\overline{CH}\parallel\overline{BP},$ and redefine $Q$ as the point such that $\triangle ABQ\sim\triangle APC.$ $\textbf{Claim: }$ $(ABG)$ and $(ACH)$ intersect at $Q.$ $\emph{Proof: }$ Let $X$ be the second intersection of $(ABG)$ and $(ACH).$ Note that $$\angle BXA=\angle BGA=\angle BGC=\angle PCA.$$Similarly, $$\angle CXA=\angle CHA=\angle CHB=\angle PBA.$$Therefore, $A$ is the center of the spiral similarity sending $\overline{XB}$ to $\overline{CP},$ so $X=Q,$ as desired. $\blacksquare$ Therefore, $Q$ is the center of the spiral similarity sending $\overline{GB}$ to $\overline{CH}.$ Hence, to show that $Q$ lies on $(AEF),$ it suffices to show that $\frac{BF}{FH}=\frac{GA}{AC}.$ Since both $\overline{DF}$ and $\overline{CH}$ are parallel to $\overline{BP},$ we have $\overline{DF}\parallel\overline{CH},$ so $\frac{BF}{FH}=\frac{BD}{DC}.$ Similarly, we can show that $\frac{GA}{AC}=\frac{BD}{DC},$ so we are done.

Amazing problem. Let $M$ denote the second intersection of $PB$ and $(PAC)$, define $N$ similarly. Let $BN\cap CM=Q$. Note that $A$ is the Miquel point of $PBQCNM$, so there is a spiral similarity about $A$ taking $BQ$ to $PC$. It remains to demonstrate that $Q$ lies on $(AEF)$. Take $D=C$, then $C=E$ and $\measuredangle DQA=\measuredangle MQA=\measuredangle MBA=\measuredangle DFA$, so $Q$ lies on $(AEF)$. Similarly, $Q$ lies on $(AEF)$ when $D=B$. Let the value of $F$ in the case $D=C$ be $F^*$ and the value of $E$ in the case $D=B$ be $E^*$. Now, consider some intermediate value of $D$, and note that by the Mean Geometry Theorem (as $E$ and $F$ move linearly with respect to $D$), $\triangle QEF\sim\triangle QCF^*\sim\triangle QE^*B$, hence $Q$ is the center of spiral similarity between $EF$ and $CF^*$, and so lies on $(AEF)$, as desired.

Let the lines parallel to $BP$ and $CP$ through $C$ and $B$ hit at $C'$ and $B'$. Let $Q$ be the intersections of the circumcircles of $BAB'$ and $CAC'$. Note that since $AQ$ is a radical axis of the two such circles, then $\triangle QB'C \sim \triangle QBC'$. Now, since $\frac{C'F}{FB} = \frac{CD}{DB} = \frac{CE}{EB}$, then $\triangle QB'E \sim \triangle QBF$ and $\triangle QCE \sim \triangle QC'F$ and $\angle EQF = \angle CQC' = 180^{\circ} - \angle A$ so $AEQF$ is cyclic. Letting $B''$ and $C''$ be where $CP$ hits $BQ$ and where $BP$ hits $CQ$ respectively, we see $\angle QB''C = \angle QBB' = \angle QAB' = \angle QAC$ so $QB''AC$ is cyclic and so is $QC''AB$. Now, $\angle B''AB = \angle B''AC' = \angle B''CC' = \angle B''PB$ so $B''BPA$ is cyclic. Finally, $$\angle BAP = \angle BB''P = \angle BB''C = \angle QB''C = \angle QAC$$which means since $\angle ABP = \angle AC'C = \angle AQC$, then $\triangle BAP \sim \triangle QAC$ and due to similarity, $\triangle BAQ = \triangle QAC$ and we are done.

[asy][asy] size(6cm); defaultpen(fontsize(10pt)+linewidth(0.4)); dotfactor *= 1.5; pair A = dir(110), B = dir(210), C = dir(330), D = B+dir(B--C)*abs(B-C)*0.4, P = B+dir(B--C)*abs(B-C)*0.55+dir(90)*0.4, P1 = A+dir(A--P)*abs(A-B)*abs(A-C)/abs(A-P), Q = 2*foot(P1,A,dir(270))-P1, E = extension(D,D+dir(C--P),A,C), F = extension(D,D+dir(B--P),A,B), X = extension(B,Q,D,E); draw(D--F--B--C--E--X--B--P, heavyblue); draw(A--B--Q--A--P--C--A, heavyred); draw(circumcircle(A,E,F), heavygreen+dashed); dot("$A$", A, dir(120)); dot("$B$", B, dir(180)); dot("$C$", C, dir(30)); dot("$D$", D, dir(270)); dot("$E$", E, dir(315)); dot("$F$", F, dir(200)); dot("$X$", X, dir(150)); dot("$P$", P, dir(270)); dot("$Q$", Q, dir(225)); [/asy][/asy] Let $Q$ be the point such that $\triangle ABQ \sim \triangle APC$ and $X = \overline{BQ} \cap \overline{DE}$. Since $$\measuredangle XQA = \measuredangle BQA = \measuredangle PCA = \measuredangle XEA,$$$AXQE$ is cyclic, so it suffices to show that $AXFQ$ is cyclic, which can be done by length chasing: \begin{align*} \frac{BF}{BX} &= \frac{\sin \angle BDF}{\sin \angle BDX} \cdot \frac{\sin \angle BXD}{\sin \angle BFD} \quad (\text{law of sines}) \\ &= \frac{\sin \angle PBC}{\sin \angle PCB} \cdot \frac{\sin \angle PAB}{\sin \angle PBA} \quad (\text{angle chase}) \\ &= \frac{PC}{PA} \quad (\text{trig ceva}) \\ &= \frac{BA}{BQ} \quad (\text{similarity})\end{align*}$\blacksquare$

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Let $Q$ be the point such that $\triangle BAQ\sim\triangle PAC$. This also means that $\triangle QAC\sim\triangle BAP$. Now, let $G=DE\cap BQ$, then we have $\measuredangle AEG=\measuredangle AED=\measuredangle ACP=\measuredangle AQB=\measuredangle AQG$, which means that $AEQG$ is cyclic. Redefine $F,H$ so that $F=AB\cap (AEQG)$ and $H=QC\cap (AEQG)$. By Pascal's theorem on hexagon $FAEGQH$, we get that $D$ lies on $FH$. Therefore, $\measuredangle PBA=\measuredangle CQA=\measuredangle HQA=\measuredangle HFA=\measuredangle DFA$, thus $DF\parallel PB$. We are done. [asy][asy]import olympiad; size(10cm);defaultpen(fontsize(10pt)); pair O,A,B,C,P,Q,D,E,F,G,H; O=(0,0);A=dir(115);B=dir(195);C=dir(345);P=dir(255)*0.74;Q=-(C-P)/(P-A)*(A-B)+B;D=0.6B+0.4C;F=extension(A,B,D,B+D-P);E=extension(A,C,D,C+D-P);G=extension(E,D,B,Q);H=extension(F,D,Q,C); draw(A--B--C--cycle,red);draw(circumcircle(A,E,F),royalblue);draw(B--Q--C,orange);draw(F--H,orange);draw(G--E,orange);draw(B--P--C,red); dot("$A$",A,dir(A)); dot("$B$",B,dir(B)); dot("$C$",C,dir(C)); dot("$P$",P,dir(P)); dot("$Q$",Q,dir(Q)); dot("$F$",F,dir(F)); dot("$E$",E,dir(E)); dot("$D$",D,dir(D)); dot("$G$",G,dir(G)); dot("$H$",H,dir(H)); [/asy][/asy]

The key (and difficulty) of this seems to be messing around with the symmetry and getting it to work. Denote $X \neq D = (DEF) \cap \overline{BC}$. We claim that the Miquel Point of $EFX$ with respect to $ABC$ is our desired point $Q$. Observe that \begin{align*} \measuredangle BQA &= \measuredangle BQF + \measuredangle FQA \\ &= \measuredangle BXF + \measuredangle FEA \\ &= \measuredangle DEF+ \measuredangle FEA \\ &= \measuredangle DEA \\ &= \measuredangle PCA, \end{align*}and for similar reasons $\measuredangle CQA = \measuredangle PBA$. However, the point $Q'$ such that $\triangle BAQ' \sim \triangle PAC$ also satisfies $\triangle CAQ \sim \triangle PAB$ by spiral similarity lemma, and thus fulfills the above angle conditions. In fact, $$Q \neq A = (AQ'C) \cap (AQ'B)$$by our angle conditions, so $Q = Q'$ must hold. The conclusion follows naturally because $Q$ lies on $(AEF)$ by definition.

bary/mmp solution by someone who does not actually know bary nor mmp Fix $P$ and note that the choice of $Q$ is uniquely determined. Now vary $D:=(0,t,1-t)$ linearly along $\overline{BC}$, so the (normalized) components of the coordinates of $E$ and $F$ vary linearly as well because $\overline{DE},\overline{DF}$ are parallel to fixed lines. Now consider the "normalized" formula for the circle $(AEF)$, which is of the form $-a^2yz-b^2zx-c^2xy+(ux+vy+wz)=0$. Since $A$ always lies on this circle, we have $u=0$, and then by plugging in $E$ and $F$ we find that both $v$ and $w$ vary linearly in terms of $t$. We wish to prove that $Q$ always lies on this circle. Since it is fixed, this is equivalent to some linear function in $t$ being identically zero, so we only have to check it at two points. At $D=B$, we have $E=B$, and because $\measuredangle AQB=\measuredangle ACP=\measuredangle AFB$, $ABQF=AEFQ$ is cyclic. Similarly, when $D=C$, $ACQE=AEFQ$ is cyclic, so we are done. $\blacksquare$

Ok @v_Enhance forgive me for what i'm about to show u. Let $Q$ the inverse of $P$ down $\sqrt{bc}$ inversion, now let $CQ \cap FD=K$ and $BQ \cap DE=L$. (Note that $\triangle APB \sim \triangle ACQ$ and $\triangle APC \sim \triangle ABQ$) Claim 1: $ALQE, AFQK$ are cyclic. Proof: By simple angle u just need: $$\angle AFD=\angle ABP=\angle AQK \; \text{and} \; \angle AED=\angle ACP=\angle AQL$$Claim 2: $ALFQEK$ lies in a conic $\mathcal C$ Proof: Follows from the converse of Pascal's theorem. Finishing: Assume FTSOC that $\mathcal C$ isn't a circle, then by Degenerate 3 conics theorem over $\mathcal C, (ALQE), (AFQK)$ we get that $LE \parallel FK$, contradiction!. Hence $\mathcal C$ is a circle, so $Q$ lies in $(AEF)$ as desired, thus we are done .

Construct $X$ on $AB$ with $BX \parallel DE$, and $Y$ analogously. Claim 1: $(ABC)$, $(AEF)$, and $(AXY)$ are coaxial. Simply note there exists a spiral similarity sending $BX \mapsto FE \mapsto YC$, as \[\frac{BF}{FY} = \frac{BD}{DC} = \frac{XE}{EC}. \quad {\color{blue} \Box}\] Claim 2: The second concurrency point is the desired point $Q$. Let $Q'$ be the desired point such that $\triangle BAQ' \sim \triangle PAC$. Then \[\measuredangle BQA = \measuredangle PCA = \measuredangle BXA, \quad \measuredangle AQC = \measuredangle ABP = \measuredangle AYC,\] so $Q' \ne A$ lies on $(ABX)$ and $(AYC)$, and thus $Q' = Q$. $\blacksquare$

Solved with Janhavi Kulkarni Let $F' \in \overline{AB}$ be such that $CF' \parallel \overline{BP}$ Define $E'$ analogously. Thus, $\frac{BF}{FF'} = \frac{BD}{DC} = \frac{E'E}{EC}$ Consider $Q$ such that $A$ is spiral center taking $\overline{BQ}$ to $\overline{PC}$ $\implies \triangle AQB \sim \triangle APC$ $\implies \angle AQB = \angle ACP = \angle AE'B \implies (AQE'B)$ and similarly $(AQF'C)$ Now, consider spiral similarity centered at $Q$ taking $\overline{BF'}$ to $\overline{E'C}$ Note that the same spiral similarity takes $F$ to $E$ $\implies \measuredangle QFA = \measuredangle QFF' = \measuredangle QEC = \measuredangle QEA \implies (AQEF)$

So basically, we want $Q$, the image of $P$ under $\sqrt{bc}$ inversion, to be on $(AEF)$. First, check this for $D=B$. Then call this $E$ $E_1$. Then $DE_1\parallel PC$, so the image of $(ADE_1)$ will be the line through $C$ and $CP\cap AB$, that is, line $CP$. Similarly, this also works for $D=C$. Now note that $E$ and $F$ are moving in the way of the ratio thing of the Miquel point config. So basically, indeed $Q$ will be the Miquel point of self-intersecting quadrilateral $BF_1CE_1$, and the ratio thing gives that it lies on all such circles $(AEF)$. $\blacksquare$

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