This problem was proposed by Anant Mudgal, India

Nice problem First note $\angle BQC=\angle BQP+\angle CQP = \angle BFP+\angle CEP = \frac{1}{2}\overarc{EDF} = \angle BIC$ so $BQIC$ is cyclic. Now let $PQ\cap (BIC) = Y$. The main claim is that $FREP, BICY$ are similar. Both quadrilaterals are cyclic, so it suffices to show the measures of the corresponding arcs on the two circles are equal. Clearly $\overarc{FR}, \overarc{BI}$ are both equal to $\angle C$, and similarly $\overarc{RE},\overarc{CI}$ are also equal. Also, $\overarc{FP} = 2\angle BFP=2\angle BQP = \overarc{BY}$ and similarly $\overarc{EP}=\overarc{CY}$, so $FREP, BICY$ are similar. Now we ignore $Q$ and let $X\in DI$ with $AX\perp AI$; it suffices to show $P,X,Y$ are collinear. Let $D'$ be the antipode to $D$ on $\omega$ and $A_1,B_1,C_1$ be the midpoints of $EF,DF,DE$. Note that $(R,P;E,F)$ is obviously harmonic, so similarity implies $(B,C;I,Y)$ is too. It's well-known that $D',M,P$ are collinear (say, by projecting harmonic bundle $(R,P;E,F)$ through $D'$ onto $EF$). Meanwhile, inversion with respect to $\omega$ sends $B,C$ to $B_1,C_1$ and sends $I$ to infinity, so to preserve the harmonic bundle $(B,C;I,Y)$ the inversion must send $Y$ to the midpoint $Y'$ of $B_1C_1$. Finally, let $X'$ be the foot of the projection from $A_1$ to $DI$, so clearly $X',X$ correspond in the inversion. Now after inversion we wish to show $P,Y',X',I$ are concyclic. Let $D_1 = DA_1\cap \omega$; clearly $D_1D'X'A_1$ is cyclic with diameter $A_1D'$, so $DA_1\cdot DD_1 = DX'\cdot DD'$. Since $I,Y'$ are midpoints of $DD', DA_1$ this implies $DI\cdot DX' = DY'\cdot DD_1$ so $D_1\in (IX'Y')$. Now note that $IY'||D'P$ so $\angle PIY' = \angle IPD' = \angle ID'P = \angle DD'P =\angle DD_1P=\angle Y'D_1P$, so $P\in (D_1IX'Y')$ and we're done.

So congrats Anant for this great prob

We use complex numbers with $D=x$, $E=y$, $F=z$. Then $A = \frac{2yz}{y+z}$, $R = \frac{-yz}{x}$ and so \[ P = \frac{A-R}{1-R\overline{A}} = \frac{\frac{2yz}{y+z} + \frac{yz}{x}} {1 + \frac{yz}{x} \cdot \frac{2}{y+z}} = \frac{yz(2x+y+z)}{2yz+x(y+z)}. \]We now compute \begin{align*} O_B &= \det \begin{bmatrix} P & P \overline P & 1 \\ F & F \overline F & 1 \\ B & B \overline B & 1 \end{bmatrix} \div \det \begin{bmatrix} P & \overline P & 1 \\ F & \overline F & 1 \\ B & \overline B & 1 \end{bmatrix} = \det \begin{bmatrix} P & 1 & 1 \\ z & 1 & 1 \\ \frac{2xz}{x+z} & \frac{4xz}{(x+z)^2} & 1 \end{bmatrix} \div \det \begin{bmatrix} P & 1/P & 1 \\ z & 1/z & 1 \\ \frac{2xz}{x+z} & \frac{2}{x+z} & 1 \end{bmatrix} \\ &= \frac{1}{x+z} \det \begin{bmatrix} P & 0 & 1 \\ z & 0 & 1 \\ 2xz(x+z) & -(x-z)^2 & (x+z)^2 \end{bmatrix} \div \det \begin{bmatrix} P & 1/P & 1 \\ z & 1/z & 1 \\ 2xz & 2 & x+z \end{bmatrix} \\ &= \frac{(x-z)^2}{x+z} \cdot \frac{P-z}{(x+z)(P/z-z/P)+2z-2x + \frac{2xz}{P}-2P} \\ &= \frac{(x-z)^2}{x+z} \cdot \frac{P-z}{ (\frac xz-1) P - 2(x-z) + (xz-z^2) \frac 1P } \\ &= \frac{x-z}{x+z} \cdot \frac{P-z}{P/z + z/P - 2} = \frac{x-z}{x+z} \cdot \frac{P-z}{\frac{(P-z)^2}{Pz}} = \frac{x-z}{x+z} \cdot \frac{1}{\frac 1z - \frac 1P} \\ &= \frac{x-z}{x+z} \cdot \frac{y(2x+y+z)}{y(2x+y+z) - (2yz+xy+xz)} = \frac{x-z}{x+z} \cdot \frac{yz(2x+y+z)}{xy+y^2-yz-xz} \\ &= \frac{x-z}{x+z} \cdot \frac{yz(2x+y+z)}{(y-z)(x+y)}. \end{align*}Similarly \[ O_C = \frac{x-y}{x+y} \cdot \frac{yz(2x+y+z)}{(z-y)(x+z)}. \]Therefore, subtraction gives \[ O_B-O_C = \frac{yz(2x+y+z)}{(x+y)(x+z)(y-z)} \left[ (x-z)+(x-y) \right] = \frac{yz(2x+y+z)(2x-y-z)}{(x+y)(x+z)(z-y)}. \]It remains to compute $T$. Since $T \in \overline{ID}$ we have $t/x \in {\mathbb R}$ so $\overline t = t/x^2$. Also, \begin{align*} \frac{t - \frac{2yz}{y+z}}{y+z} \in i {\mathbb R} \implies 0 &= \frac{t-\frac{2yz}{y+z}}{y+z} + \frac{\frac{t}{x^2}-\frac{2}{y+z}}{\frac1y+\frac1z} \\ &= \frac{1+\frac{yz}{x^2}}{y+z} t - \frac{2yz}{(y+z)^2} - \frac{2yz}{(y+z)^2} \\ \implies t &= \frac{x^2}{x^2+yz} \cdot \frac{4yz}{y+z} \end{align*}Thus \begin{align*} P-T &= \frac{yz(2x+y+z)}{2yz+x(y+z)} - \frac{4x^2yz}{(x^2+yz)(y+z)} \\ &= yz \cdot \frac{(2x+y+z)(x^2+yz)(y+z) - 4x^2(2yz+xy+xz)} {(y+z)(x^2+yz)(2yz+xy+xz)} \\ &= -yz \cdot \frac{(2x-y-z)(x^2y+x^2z+4xyz+y^2z+yz^2)} {(y+z)(x^2+yz)(2yz+xy+xz)}. \end{align*}This gives $\overline{PT} \perp \overline{O_B O_C}$ as needed.

Is there any solution using famous lemmas related to incircle and circumcircle?

Last year, there was also a Geometry problem on P6 of the day 2. Similarly, This year, there is also another geometry problem on P6.

JND wrote: Last year, there was also a Geometry problem on P6 of the day 2. Similarly, This year, there is also another geometry problem on P6. There were 2 options for P6- Algebra or Geometry. I really wished for an Algebra, but P6 is rarely Algebra. So, it was pretty obvious that it would be a Geo

Dear MLs, For a variable P on incircle the locus of Q is Mension circle (BCI), if it is of any help. Friendly, M.T.

Oh my god feeling soooo good to see an Indian proposing a problem, more specifically a CMI undergrad

Let $DI$ meet line through $A$ and perpendicular to $AI$ at at $L$. From angle chase $ALDP$ is cyclic. It allows us to make next Generalization. $ABC$ given $E, F$ are any points on sides $AC, AB$. $D$ any point. $I$ is the second intersection point of the circumcircles of $DFB$, $DCE$. $L$ is the intersection point of $DI$ and a line through $A$ and parallel to $EF$. $P$ is the second intersection point of the circumcircles of $ALD$, $DEF$. $Q$ is second intersection point of the circumcircles of $PBF, PCE$. Prove that $P$, $Q$, $L$ are collinear.

Comment. It is nice problem, and the more general theory of quadric surfaces ( https://en.wikipedia.org/wiki/Hyperboloid ) gives a nice proof of it and better understanding. Actually this theory is a way how I found my generalization at previous post.

Any solutions using mixtilinear incircles?

@above Here is my solution found during the contest. It uses Mixtilinear Incircle extensively. Lemma 1: $AP$ pass through mixtilinear touchpoint $T$ of $\triangle ABC$. Proof: Let $R'$ be the reflection of $R$ across $AI$. Clearly $R'$ is the antipode of $D$ w.r.t. $\omega$ so $AR'$ is $A$-Nagel cevian of $\triangle ABC$. Thus $AR$ pass through isogonal conjugate of Nagel point so it passes through $T$. Lemma 2: $B,I,C,Q$ are concyclic. Proof: Angle chasing \begin{align*} \measuredangle BQC &= \measuredangle BQP + \measuredangle PQC \\ &= \measuredangle BFP + \measuredangle CEP \\ &= \measuredangle FDP + \measuredangle PDE \\ &= \measuredangle FDE \\ &= \measuredangle BIC \end{align*} Extend $PQ$ to meet $\odot(BIC)$ again at $K$. Then note that $$\angle KBC = \angle KQC = \angle PQC = \angle PEC = \angle PFE$$so $\triangle KBC\sim\triangle PFE$. Further angle chasing gives $\angle RFE = 0.5\angle B$ so $\triangle BKC\cup I\sim\triangle PFE\cup R$. The latter quadrilateral is harmonic so the former is harmonic too. Thus if $N$ be the midpoint of arc $BAC$, then $K\in IN$. Let $M$ be the other midpoint of arc $BC$ and $T$ be the mixtilinear touch-point. Thus $\angle MTN = 90^{\circ}$ which means $T$ is the midpoint of $IK$. Now, let $S$ be the center of $A$-mixtilinear incircle and $X=DI\cap AN$. So by Lemma 1 and homothety, $IP\parallel ST$. By Menelaus theorem on $\triangle ATN$ , it suffices to show that $$\frac{AP}{PT}\cdot\frac{TK}{KN}\cdot\frac{NX}{XA}=1$$We compute the first term first, note that $$\frac{AP}{PT} = \frac{AI}{IS} = \cot^2\frac{\angle A}{2}$$For the second term, we use Power of Point twice. $$\frac{TK}{KN} = \frac{IT}{KN} = \frac{IT\cdot IN}{NK\cdot NI} = \frac{AI\cdot IM}{NB^2}$$And the last term note that $MN\parallel DI$ so, $$\frac{NX}{XA} = \frac{MI}{IA}$$Thus multiplying all together, the resulting terms $$\cot^2\frac{\angle A}{2}\cdot \frac{MI^2}{NB^2} = \cot^2\frac{\angle A}{2}\cdot \frac{MB^2}{NB^2}$$as $\tfrac{MB}{NB} = \tan\tfrac{\angle A}{2}$, we are done.

Assume J the intersection of DI and ex-anglebisector of A , then it's sufficient to prove the power of J to (BPF),(CPE) are the same, using Casey Theorem, sufficient to prove O_1L= O_2K, where O1, O2 the center of (BPF),(CPE), K,L the midpoint of BI,CI. The rest is easy, consider triangle IO_1L and IO_2K, with sine law, sin(IO1L)/sin(IO2L) = PF/PE=RF/RE = BI/CI done

Let D' symmetric D wrt AI, AI cuts BC at V, DM perp EF. (PID') cuts (BIC) at S, D'I cuts (I) at G then D,R,G collinear so PSC=PSI+ISC=PD'I+IBC=PDR+RDE =180-PRE=180-PEC so S lies on (PEC) similarly P lies on (PFB) so S=Q. Notice AGD+APD=180-ID'D+AVD=180 so G lies on (APD). Call S be orthocentre of DEF, P'=(DS) cuts (I) hence P'E/P'F=SE/SF=RE/RF so A,R,P' collinear so P'=P. From PV,PD perp PS,PM and SM//IM' with M' istomic M hence PMS~PVD thus PMS=PGD so P,M,G collinear. Let (ID'F) cuts EF at K. Notice KED'~CED so KE/DF=CE/DE so KE/FB=EC/ED.FD/FB=RF/RE=PF/PE so PFB~PEK so KBFP cyclic similarly HCPE. Now M lies on radical center of (FBP) and (EPC) hence M,P,Q,G collinear so q.e.d

Here is a solution only using spiral similarity and its properties. By angle chasing, $\triangle REF \sim \triangle ICB$. Consider the spiral similarity sending $\triangle REF$ to $\triangle ICB$. Let $K$ be the center of this spiral similarity. So $K,B,C,A$ are concyclic because $A$ is the intersection of $BF$ and $CE$. Let $G$ be the image of $D$ under the spiral similarity. Then $RD \perp FE \Longrightarrow IG \perp BC$, so $G \in DI$. Let $S$ be the image of $A$ under the spiral similarity. Then $\angle BSC = \angle FAE = \angle BAC$ and $FA=AE \Longrightarrow BS=SC$, so $S$ is the midpoint of arc $BC$ containing $A$ on the circumcircle of $\triangle ABC$. Therefore $AS \perp AI$. Denote the intersection of $DG$ and $AS$ by $Z$. Then since $K$ is the center of spiral similarity sending $D$ to $G$ and $A$ to $S$, we have $K,Z,D,A$ concyclic. On the other hand, by angle chasing we have $\angle APD + \angle AZD = 180^{\circ}$, so $A,D,P,Z$ are concyclic. Therefore, $A,K,P,D,Z$ are all on the same circle. Let $T$ be the image of $P$ under the same spiral similarity. Since $K$ is the center of spiral similarity sending $P$ to $T$ and $A$ to $S$, the points $K,P,A,PT \cap AS$ are concyclic. False position gives $Z=PT \cap AS$, i.e. $Z \in PT$. Now by angle chasing, $Q$ lies on the circumcircle of $\triangle BIC$, and notice $T$ must lie on the same circle (because $P$ lies on the circumcircle of $\triangle FRE$. By angle chasing we have $\angle BQT=\angle BQP$ and $\angle CQT = \angle CQP$, so $P$, $Q$, $T$ are collinear. Combining, $Z$ lies on line $DIG$, line $PQT$ and line $AS$ which is perpendicular to $AI$, as desired.

@above, what software did you use for that diagram? It is really good.

lminsl wrote: Let $I$ be the incentre of acute triangle $ABC$ with $AB\neq AC$. The incircle $\omega$ of $ABC$ is tangent to sides $BC, CA$, and $AB$ at $D, E,$ and $F$, respectively. The line through $D$ perpendicular to $EF$ meets $\omega$ at $R$. Line $AR$ meets $\omega$ again at $P$. The circumcircles of triangle $PCE$ and $PBF$ meet again at $Q$. Prove that lines $DI$ and $PQ$ meet on the line through $A$ perpendicular to $AI$. Proposed by Anant Mudgal, India Nice configuration. Let $T = DI \cap AN$ where $N$ is the midpoint of $\overarc{BAC}$. It is enough to prove that $T, P, Q$ are collinear. Claim 1: $ B, Q, I, C$ are concyclic. Note that $$\angle{BQC} = \angle{BFP} + \angle{CEP} = 180^{\circ} - \angle{FPE} = 90^{\circ} + \angle{A} = \angle{BIC} \square$$ Invert w.r.t $(DEF)$. Denote the inverse of a point $U$ by $U^*$. Now it is enough to prove that $T^*, Q^*, P^*, I$ are concyclic. Let $S$ be the midpoint of $C^*B^*$ and $X$ be the $D$-antipode of $\triangle DEF$. Claim 2: $S^*, P, Q$ are collinear. Proof: Note that $-1 = (S,B^*;P_{\infty},C*) = (B,S^*;C,I)$. Let $PQ \cap (BIC) = S’$. Also note that $$\angle{S’BC} = \angle{S’QC} = \angle{PEC} = \angle{PFE}$$Similarly, we have $\angle{S’CB} = \angle{PEF}$. Moreover, $$\angle{ICB} = \angle{RDF} = \angle{REF}$$. Similarly, we have $\angle{IBC} = \angle{RFE}$. So, the spiral similarity mapping $EF$ to $BC$ maps $R$ to $I$ and $D$ to $S’$. But $(I,B;S’;C) = (R,E;D;F) = -1$. So, $S^* \equiv S’$. $\square$ Claim 3: $L, I, S, T^*$ are concyclic. Proof: Note that $\angle{A^*TX} = \angle{A^*L X} = 90^{\circ}$ where $L = DA^* \cap (DEF)$. So, $$DT^* \cdot DI = \frac{DT^* \cdot DX}{2} = \frac{DL \cdot DA^*}{2} = DS \cdot DL$$Thus, $L, I, S, T^*$ are concyclic. $\square$ Claim 4: $L, I, S, P$ are concyclic. Proof: Recognize $P$ as $DH \cap (DEF)$ where $H$ is the orthocentre of $\triangle DEF$. Note that $P, A^*, X$ are collinear. So, $XP \parallel IS$. Thus, $$\frac{\angle{PID}}{2} = \angle {PLD} = \angle{PXD} = \angle{SID}$$. So, $L, I, S, P$ are concyclic.$\square$ By combining Claims 3 and 4, we can see that $T,Q^*,P^*,I$ are concyclic and so $T, P, Q$ are collinear.

MathGenius_ wrote: @above, what software did you use for that diagram? It is really good. Geogebra XD

Elementary solution:

WLOG let $AB<AC$. We claim that $BQIC$ is cyclic. This is true by simple angle chasing: note that \begin{align*}\angle BQC &= \angle BQP+\angle PQC\\ &=\angle BFP + \angle CEP \\ &= \angle FEP + \angle EFP \\ &= 180^\circ - \angle FPE \\ &= 180^\circ - \angle FDE \\ &= \angle BIC \end{align*}as desired. $~$ Let $S$ be the Miquel point of $EFBC$. Consider $\mathcal{S}$, the spiral similarity centered at $S$ that sends $EF$ to $BC$. Evidently, since $\angle FRE=\angle BIC$ the incircle is mapped to circle $(BQIC)$ after $\mathcal{S}$. Let $G$ be the point such that $\mathcal{S}(G)=D$. Note that \[\frac{FG}{GE}=\frac{BD}{DC}=\frac{BF}{CE}\]and since $\angle BFG = \angle GEC$, we have $\triangle BFG \sim \triangle CEG$ so $\angle FGB=\angle EGC$. Additionally, \[\frac{GB}{GC}=\frac{FB}{EC}=\frac{BD}{DC}\]so $DG$ bisects $\angle BGC$. Adding this up with our previous conclusion, we get that $DG\perp EF$ so $D$, $G$, $R$ are collinear. Thus, $\mathcal{S}(R)=I$. $~$ Let $PQ$ intersect $(BIC)$ at $Y\neq Q$ then note that since \[\angle BIY=\angle BQY=\angle BQP=\angle BFP=\angle FRP\]so $\mathcal{S}(P)=Y$. Now, it is time to finish the problem. $~$ Let $X$ be on $ID$ such that $AX\perp AI$. We claim that $APDX$ is cyclic. Indeed, let $AI$ intersect $BC$ at $Z$ then \begin{align*} \angle AXD &= 90^\circ - \angle AIX \\ &= \angle IZD \\ &= \angle C + \frac{1}{2}\angle A \\ &= 90^\circ + \frac{1}{2}(\angle C - \angle B) \\ &= 90^\circ + \angle BCI - \angle CBI \\ &= 90^\circ + \angle FER-\angle EDR \\ &= \angle FER + \angle FED \\ &= \angle RED \\ &=180^\circ - \angle RPD \end{align*}as desired. Next, we claim that $ASDX$ is cyclic. Recall that there is a spiral similarity centered at $S$ taking $BF$ to $CE$. Thus, \[\frac{BD}{DC}=\frac{BF}{CE}=\frac{SB}{SC}\]so $SD$ bisects $\angle BSC$. Thus, we have \begin{align*} \angle SDX &= 90^\circ - \angle SDB \\ &= \frac{1}{2}(\angle SBC -\angle SCB) \\ &= \frac{1}{2}(\angle SFE - \angle SEF) \\ &= \frac{1}{2}(\angle AFE + \angle SFA -\angle FEA + \angle SEA) \\ &= \angle SFA \\ &= 180^\circ - \angle SAX \end{align*}which proves the claim. We have that $SPXD$ is cyclic. Let $ID$ intersect $(BIC)$ again at $J$, then $\mathcal{S}D=J$, so $\mathcal{S}(PD)=YJ$. Thus, $S$ is the Miquel point of $PDJY$ and so $S$, $P$, $D$, and $PQ\cap DI$ are concyclic, so $X=PQ\cap DI$, and we are done.

Define the linear functions \[ f(\bullet) := \text{Pow}(\bullet, (PBF)) - \text{Pow}(\bullet, (PCE)) \]and \[ g(\bullet) := \text{Pow}(\bullet, (BDIF)) - \text{Pow}(\bullet, (CDIE)). \]Let $L=\overline{DI} \cap \overline{PQ}$. Note that $f(A)=g(A)$ and $f(L)=g(L)$. We want to show that $\overline{AL}$ is the external angle bisector $\ell$ of $\angle BAC$, so it suffices to find some point on $\ell$ that has an equal $f$- and $g$-value. Select $K=\ell \cap \overline{BC}$. Observe that $KB/KC=AB/AC$. Let $(PBF)$ intersect $BC$ at $B_1$ and $(PCE)$ intersect $BC$ at $C_1$. Since we need to show that $f(K)=g(K)$, we can reduce this (by PoP) to \[ \frac{DB_1}{DC_1} = \frac{KC}{KB} = \frac{AC}{AB}. \]Fortunately, this is something very much in our hands. Claim: In fact, we have \[ \frac{DB_1}{DC_1} = \frac{AC}{AB}. \]Proof. First, we need to understand what $P$ is with respect to standard configs. Do everything with respect to $\triangle DEF$, so that $R$ has a simple definition. Redefine $P$ as the intersection of the $R$-symmedian in $\triangle REF$ with $(REF)$. Thus, we have \[ \frac{PE}{PF} = \frac{RF}{RE} = \frac{\cos(\angle F)}{\cos(\angle E)}. \]Second, redefine $B_1$ and $C_1$ with respect to $P$ and $\triangle DEF$. Note that the center of the spiral similarity that maps $BF \mapsto FE$ is $P$. Thus, we readily have $\triangle B_1PC_1 \sim \triangle EPF$, so taking similarity ratios we have \[ \frac{PB_1}{PC_1} = \frac{PE}{PF} = \frac{\cos(\angle F)}{\cos(\angle E)} = \frac{\sin(\angle C/2)}{\sin(\angle B/2)}. \]Finally, we apply LoS on $\triangle DPB_1$ and $\triangle DPC_1$ to involve $D$, from which we find that it suffices to prove that \[ \angle DPB_1 = 90^{\circ} + \angle B/2. \]This is easy angle chasing, in that we write \[ \angle DPB_1 = 360^{\circ} - (\angle B_1PF + \angle DPF) = 360^{\circ} - ((180^{\circ}-\angle B)+(90^{\circ}+\angle B/2)) = 90^{\circ}+\angle B/2, \]as desired. Remark: I think the hardest part is actually the ratio computation at the very end. If you want to force linearity of PoP onto the problem, then that is not hard, since you just need to bring another pair of circles whose radical axis is $DI$. Then the rest of the solution flows until the point at which you have to do the ratio computation, which, rather than requiring bash, needs some intuition with American configs.

let $DI$ intersect $\omega$ at $Y$, let $PR$ intersect $EF$ at $H$, let $DR$ intersect $EF$ at $G$, and let $DI$ and $PQ$ intersect at $X$ because $D$ is a point of tangency, $YDB=YRD$, $YR||EF$, $YFE=REF$, and $YF=RE$ because $E$ and $F$ are points of tangency, $AEF$ is isoceles, and we can prove $AY=AR$, $AI||DR$, $AR||YI$, and that $AYIR$ is a rhombus then, we prove that $AYIP$ is and isoceles trapezoid, and that $AYX$ and $PID$ are congruent, and from that we can prove that $AX \perp AI$

chat is this real We will complex bash with $(DEF)$ the unit circle, where $x, y, z$ represent $D, E, F$. Note that $a=\frac{2yz}{y+z}$ and cyclic permutations. Let $X$ correspond to $kx$ for some real constant $k$. As $\frac{x-a}a$ must be pure imaginary, \begin{align*} \frac{kx-\frac{2yz}{y+z}}{\frac{2yz}{y+z}} + \frac{\frac kx-\frac 2{y+z}}{\frac 2{y+z}} &= 0 \\ \iff \frac{kx(y+z)-2yz}{2yz}+\frac{k(y+z)-2x}{2x} &= 0 \\ \iff kx^2(y+z)+kyz(y+z)-4xyz &= 0 \\ \iff \frac{4xyz}{(y+z)(x^2+yz)} &= k. \end{align*}So $X$ is located at $\frac{4x^2yz}{(y+z)(x^2+yz)}$. Now, compute $$p = \frac{-\frac{yz}x-\frac{2yz}{y+z}}{-\frac{yz}x \cdot \frac 2{y+z}-1} = \frac{yz(y+z)+2xyz}{2yz+x(y+z)}.$$It follows that the circumcenter $o_B$ of $(BFP)$ is located at \begin{align*} o_B &= z -\frac{(b-z)(p-z)(\overline b - \overline p)}{(b-z)(\overline{p-z}) - (\overline{b-z})(p-z)} \\ &= z + \frac{-\left(\frac{2xz}{x+z}-z\right)\left(\frac{yz(y+z)+2xyz}{2yz+x(y+z)} - z\right)\left(\frac 2{x+z} - \frac{xy+xz+2yz}{yz(2x+y+z)}\right)}{\left(\frac{2xz}{x+z}-z\right)\left(\frac{xy+xz+2yz}{yz(2x+y+z)} - \frac 1z\right)-\left(\frac 2{x+z}-\frac 1z\right)\left(\frac{yz(y+z)+2xyz}{2yz+x(y+z)}+z\right)} \\ &= z + \frac{\frac{z(x+y)(z-x)(y-z)(x^2y+x^2z-xyz+xz^2-2y^2z)}{y(x+z)^2(2x+y+z)(xy+xz+2yz)}}{\frac{(x+y)^2(x-z)(y-z)^2}{y(x+z)(2x+y+z)(xy+xz+2yz)}} \\ &= z + \frac{z(x^2y+x^2z-xyz+xz^2-2y^2z)}{(x+y)(x+z)(y-z)} \\ &= \frac{yz(z-x)(2x+y+z)}{(x+y)(x+z)(z-y)}. \end{align*}Intermediate steps are omitted so the reader does not need to suffer the factorization process like I did. Then we have $$o_B-o_C = \frac{yz(z-x)(2x+y+z)}{(x+y)(x+z)(z-y)} -\frac{yz(y-x)(2x+y+z)}{(x+y)(x+z)(y-z)} = -\frac{yz(2x+y+z)(y+z-2x)}{(x+y)(x+z)(y-z)}.$$It will suffice to show that $\frac{o_B-o_C}{p-x}$ is pure imaginary. In particular, $$p-x = \frac{yz(y+z)-2xyz}{2yz-x(y+z)} - \frac{4x^2yz}{(y+z)(x^2+yz)} = \frac{-yz(2x+y+z)(x^2y+x^2z-4xyz+y^2z + yz^2)}{(y+z)(xy+xz-2yz)(x^2+yz)}.$$So it suffices that the ratio $$\frac{(2x+y+z)(y+z)(x^2+yz)(xy+xz+2yz)}{(x+y)(x+z)(y-z)(x^2y+x^2z+4xyz+y^2z+yz^2)} \in i\mathbb R$$which is clear as the expression equals the negative of its own conjugate.

Observe that $$\measuredangle BQC = \measuredangle BQP + \measuredangle PQC = \measuredangle BFP + \measuredangle PEC = \measuredangle BIC$$ so $Q$ lies on $(BIC)$. Moreover, if $PQ$ intersects $(BIC)$ again at $T$, then $$\measuredangle BIT = \measuredangle BQT = \measuredangle BQP = \measuredangle BFP = \measuredangle FRP$$ so quadrilaterals $RFPE$ and $IBTC$ are similar. Since $P$ lies on the $R-$ symmedian of $\triangle IFE$, it follows that $T$ lies on the $I-$ symmedian of $\triangle IBC$, which passes through the midpoint $M$ of $\widehat{BAC}$. It suffices to show that $X$ lies on $PT$. Now, the spiral similarity centered at the Sharkydevil point $S$ mapping $EF$ to $BC$ maps $P$ to $T$ and $A$ to $M$. Hence, it suffices to show that $X$ lies on $(APS)$. Inverting about the incircle, $S$ gets sent to the foot form $D$ to $EF$, $A$ gets sent to the midpoint of $EF$, $X$ gets sent to the foot from $A^*$ to $DI$, and $P$ remains as the $D-$ Queue point of $\triangle DEF$, so all four inverted points lie on the circle with diameter $DA^*$, as desired.

[asy][asy] import geometry; size(10cm); defaultpen(fontsize(9pt)); pen pri; pri=RGB(31, 191, 184); pen sec; sec=RGB(5, 113, 108); pen tri; tri=RGB(25, 120, 165); pen qua; qua=RGB(3, 17, 99); pen fil=invisible; pen sfil=invisible; pen tfil=invisible; pen qfil=invisible; pair O, A, B, C, I, D, E, F, R, P, Q, K, D_prime; O=(0, 0); A=dir(125); B=dir(220); C=dir(320); I=incenter(A, B, C); D=foot(I,B,C); E=foot(I,A,C); F=foot(I,A,B); R=2*foot(I,D,foot(D,E,F))-D; P=2*foot(I,A,R)-R; Q=intersectionpoints(circle(B,F,P), circle(C,E,P))[1]; K=extension(D,I,P,Q); D_prime=2I-D; fill(D--E--F--cycle,sfil); filldraw((path)(A--B--C--cycle), white+0.1*pri, pri); filldraw(circumcircle(B, P, F), tfil, tri); filldraw(circumcircle(C, E, P), tfil, tri); filldraw(circumcircle(B, I, C), sfil, sec); filldraw(circumcircle(A, B, C), fil,pri); filldraw(incircle(A, B, C), fil, pri); draw(R--D,tri+dashed); draw(P--K,tri+dashed); draw(D--K,tri+dashed); draw(A--K,tri+dashed); dot("$A$",A,dir(110)); dot("$B$",B,dir(200)); dot("$C$",C,dir(-15)); dot("$K$",K,dir(-5)); dot("$I$",I,dir(-30)); dot("$D$",D,S); dot("$E$",E,dir(75)); dot("$F$",F,dir(130)); dot("$R$",R,NW); dot("$P$",P,dir(120)); dot("$Q$",Q,dir(-20)); dot("$D'$",D_prime,dir(45)); [/asy][/asy] Claim: $Q$ lies on $(BIC)$. Proof. Angle chasing we have, \begin{align*} \angle BQC = \angle BFP + \angle PEC &= 360 - \angle PFA - \angle PEA\\ & = 360 - (\angle PFE + \angle PEF) - (180 - \angle A)\\ &= 90 + \angle A/2 \end{align*}which proves the claim. $\square$ Let $D'$ be the antipode of $D$ in the incircle, and let $T$ be the mixintillinear touch point. Claim: $\overline{AP}$ and $\overline{AD'}$ are isogonal. Proof. Invert about the incircle, so that $A$ maps to the midpoint of $\overline{EF}$ and $T$ maps to the midpoint of $\overline{DH}$ where $H$ is the orthocenter of $\triangle DEF$, namely $T^*$ lies on $\overline{DR}$. Then it suffices to show that $A^*$, $T^*$, $I$ and $R$ are concyclic. This follows as upon reflection about $\overline{BC}$ we observe $R$ maps to $H$, and it is easy to show say by complex numbers that $A^*H \parallel T^*O$, which proves the claim. $\square$ From now on much of our work will deal with the inverted figure. [asy][asy] import geometry; size(7cm); defaultpen(fontsize(9pt)); pen pri; pri=RGB(31, 191, 184); pen sec; sec=RGB(5, 113, 108); pen tri; tri=RGB(25, 120, 165); pen qua; qua=RGB(3, 17, 99); pen fil=invisible; pen sfil=invisible; pen tfil=invisible; pen qfil=invisible; pair I, D, E, F, A, B, C, X, Y, P, Q, D_prime, K; I=(0, 0); D=dir(110); E=dir(220); F=dir(320); A=((E+F)/2); B=((D+F)/2); C=((D+E)/2); X=((D+A)/2); Y=intersectionpoints(circle(D,E,F),line(D, A))[0]; P=intersectionpoints(circle(X,abs(X-D)), circle(D, E, F))[1]; Q=intersectionpoints(circle(P, E, C), circle(B, P, F))[0]; D_prime=2I-D; K=foot(A,D,D_prime); filldraw((path)(D--E--F--cycle), white+0.1*pri, pri); filldraw(circumcircle(D, E, F), sfil, pri); filldraw(circumcircle(P,A,D), sfil, tri); filldraw(circumcircle(P,Q,Y), sfil, tri); draw(circumcircle(P,Q,C), dashed+sec); clipdraw(circumcircle(P,Q,B), dashed+sec); draw(D--Y,tri); draw(Q--B,tri); draw(D--D_prime,tri); draw(A--K,tri); draw(D_prime--P,tri+dashed); dot("$A^*$",A,dir(-120)); dot("$B^*$",B,dir(40)); dot("$C^*$",C,dir(-110)); dot("$I$",I,dir(20)); dot("$D$",D,N); dot("$E$",E,dir(-100)); dot("$F$",F,dir(-40)); dot("$P$",P,dir(130)); dot("$Q^*$",Q,dir(W)); dot("$X$",X,dir(240)); dot("$Y$",Y,dir(-40)); dot("$D'$",D_prime,dir(-50)); dot("$K$",K,dir(-15)); [/asy][/asy] First however, we provide a characterization of $P$. Claim: $\angle DPA^* = 90$. Proof. Upon inverting back to our original figure this is equivalent to showing $D'PIA$ is cyclic. This follows as, \begin{align*} \angle AD'I = \angle ARD' + \angle RD'D &= 180 - \angle D'RP + \angle RD'D \\ &= 180 - (180 - \angle D'DP) + (90 - \angle D'DR)\\ &= \angle D'DP - \angle D'DR + 90 \\ &= \frac{1}{2}\angle RIP + 90\\ &= 180 - \angle IPA \end{align*}which proves the claim. $\square$ Now define $K$ as the intersection of the external angle bisector of $\angle BAC$ and $\overline{DI}$, let $X$ be the center of $(PDA^*)$ and $Y = \overline{DA^*} \cap (DEF)$. Note immediately that $X$ is the midpoint of $\overline{B^*C^*}$ by homothety. Claim: $X$ and $Y$ both lie on $(PQ^*I)$. Proof. First note that $Y \in (PQ^*I)$ as, \begin{align*} \angle IYP = 90 - \frac{1}{2}\angle PIY &= 90 - \frac{1}{2}(180 - \angle PID - \angle YID')\\ &= \angle PD'D + \angle YDD'\\ &= \angle PFB + \angle XDD'\\ &= \angle PQB + \angle XPI \\ &= \angle PQB + \angle XQI = \angle PQI \end{align*}where we have $\angle XDD' = \angle XPI$ as both $X$ and $I$ lie on the perpendicular bisector of $\overline{PD}$ because $\overline{PD}$ is a chord in circles centered at $X$ and $I$. Now $X \in (PQ^*I)$ because, \begin{align*} \angle PQX = \angle PFB = \angle PYD = \angle PYX \end{align*}so we have proven the claim. $\square$ Now we may delete both $P$ and $Q^*$ from the picture. Claim: $K^ *$ lies on $(XIY)$. Proof. Note that, \begin{align*} \angle IKX = \angle DKX = \angle KDA^* = \angle IYD \end{align*}and thus the claim is proven. $\square$ Inverting back the claim follows as $K \in (PQ^*I)$ implies $K$, $P$ and $Q$ are collinear.

We will split the solution into 2 parts. Part 1) We will prove that lines $PQ,IR,EF$ are concurrent. (Sketch of) proof: Let $K = IR \cap EF$, $B_1=(BPF) \cap EF$, $C_1 = (CPE) \cap EF$. Also let $\angle BAC=\alpha, AB=c$ and so on. Define $f(X)=P(X,(BPF))-P(X,(CPE))$, there $P(X, \omega)$ is power of point $X$ onto circle $\omega$. Note that since $f$ is linear function, we have $f(K)=\frac{1}{EF} \left( KE \cdot f(F) + KF \cdot f(E) \right)=KF \cdot EB_1-KE \cdot FC_1$. So we want to prove that $\frac{EB_1}{FC_1}=\frac{KE}{KF}$. Note that $\frac{KE}{KF}=\frac{AC}{AB}$. Let $Q'=BC_1 \cap CB_1$. By easy angle chasing $Q'=Q$. So $\Delta PBB_1 \sim \Delta PC_1C$ and $\Delta PBC_1 \sim \Delta PB_1C$. Now we can find $AB \cdot EB_1 = \frac{AB \cdot FB \cdot CB_1}{BC_1}$ and similar equality with segment $CB_1$. After this good approach will be using trig-Ceva to triangle $ABC$ and point $P$ and not hard bashing. Part 2): If $X=PK \cap DI$, then $AX \perp AI$. Sketch of proof: If $U=AP \cap EF, V=DI \cap EF$, then $FREP$ is harmonic quadrilateral, so $-1=^(A,U;R,P)=^K(AI \cap DI, V; I, X)$ and it ie enough to prove that $AI$ is bisector of $\angle KAV$ or $FK=VE$, which is too not hard (now we can delete all points but $D,E,F,R,I,K,V$).

Our main idea is incircle inversion and the angle chasing. Define $O$ as the intersection of the external bisector of $\angle CAB$ and $DI$. Define $M = (\omega \cap DI) - D$ On inverting about $\omega$ we have that $O \iff O'$ $Q \iff Q'$ $A \iff A'$ $B \iff B'$ $C \iff C'$ $\therefore PQ \iff (Q'PI)$ $(BPQF) \iff (B'PQ'F)$ $(CPQE) \iff (C'PQ'E)$ It is easy to see that $\angle CQB = 90^\circ + \frac{\alpha}{2}$ Which implies $Q\in (CIB)$ $(CIQB) \iff C'B'Q'$ Define $S = ((BIC)\cap ID) - I$ $S\iff S'$ $S' = B'C'\cap ID$ Now we claim that $Q\in PO$ Proof: We have $\angle DMP = \angle DFP \equiv \angle B'FP = \angle B'Q'P$ Define $W = ((Q'PQ)\cap B'C') - Q'$ Let $W' \iff W$ $\because$ $(Q'PWQ)$ is orthogonal $W' = ((Q'PWQ)\cap (CIQB)) - Q$ $\angle IAP = \angle DRP = \angle B'FP = \angle B'Q'P \equiv \angle WQ'P = \angle WW'P$ $\implies W', P, I, A$ are concyclic $\because P, I, M, A$ are concyclic, $W', P, I, M, A$ are concyclic Also $(W'PIA) \iff PWM$ $\angle S'MP\equiv\angle DMP = \angle B'Q'P\equiv\angle S'Q'P$ $\therefore S', M ,Q', P$ are concyclic $\therefore S, M, Q, P$ are concyclic Claim inside claim: $WS \parallel A'D$ Proof: $\angle ERD = \angle EFD = 90^\circ - \frac{\gamma}{2} = \angle CID\equiv\angle CIS$ $\angle RDE = 90^\circ - \angle DEF = \frac{\beta}{2} = \angle IBC = \angle ISC$ $\Delta ERD \sim \Delta CIS$ Define $T = DR\cap EF$ and $T' = DR\cap B'C'$ $\therefore \frac{MD}{DS} = \frac{2\cdot ID}{DS} = \frac{2\cdot RT}{TD} = \frac{RT}{TT'}$ Now $\frac{RT}{TT'} = \frac{MK}{A'W}$ $\therefore A'D\parallel WS$ We know that $A, O, D, P$ are concyclic (this is obvious) $\therefore A', O', D, P$ are concyclic $\angle OPM = \angle MPO' = \angle A'DO' = \angle WSM\equiv\angle QSM = \angle QPM$ $\therefore Q\in PO$

this is honestly not that bad Let $J$ be $IR\cap EF$, $M$ be the midpoint of $EF$, and $J'$ as $DI\cap EF$. Let $D'$ be on $\omega$ such that $DD' \parallel EF$ (so the antipode of $R$), $G$ as $AD'\cap \omega$, $K=D'P\cap EF$, $N$ the midpoint of $AK$. Define $T$ to be $DI$ intersect the $A$ external bisector. We first note a few properties of the configuration. Note that $-1=(PR;EF)\overset{D'}{=}(KJ;EF)$ and $-1=(GD';EF)\overset{R}{=}(RG\cap EF,J;E,F)$. Thus $R,G,K$ collinear. As $RG\perp AD'$ we get $R$ is the orthocenter of $AKD'$. Note that $IJ$ and $IJ'$ are symmetric about line $AI$. It is well known $AJ'$ is a median so $AJ$ is a symmedian. This implies that $AK$ is tangent to $(ABC)$ by projecting $(KJ;EF)$ onto $(ABC)$. As $AN$ is tangent to $(ABC)$, $AM$ is the bisector, and $NA=NM$ from $\angle AMK=90 ^{\circ}$, we conclude that $NM \parallel BC$. We are now ready to tackle the problem. Let $Q'$ and $T'$ denote the inverses of $Q$ and $T$ about $\omega$. Claim: Line $PQ$ inverts to the nine point circle of $AKD'$. Proof. It would suffice to show $Q$ goes to the midpoint of $KD'$, since $PQ$ passes through the midpoint of $RD'$ and the foot from $A$ to $KD'$. Let $B',C'$ be the midpoints of $DF$, $DE$, so $Q'PB'F$ and $Q'PC'E$ are cyclic. Then, \[\measuredangle PQB'=\measuredangle PFB'=\measuredangle PFD=\measuredangle PED=\measuredangle PEC'=\measuredangle PQC'\]so $Q',B',B'$ collinear. Also, $\measuredangle D'PE=\measuredangle FED=\measuredangle B'C'D=\measuredangle Q'C'E=\measuredangle Q'P'E$ so $D',P,Q'$ collinear. A homothety at $D'$ with scale factor $2$ now sends line $B'C'$ to $EF$, so $Q'$ goes to $K$. $\blacksquare$ Thus it suffices to show $T'$ lies on the nine point circle of $AKD'$. But as $MT'\parallel BC$ as both are perpendicular to $DI$, we have $\angle NT'I=90^{\circ}$ as desired.

It is well known that $P$ is the $D$ queue point in $\Delta DEF$ . We now invert with respect to the incircle and restate in terms of $\Delta DEF$. Quote: In $\Delta ABC$ with center $O$ and $A$ queue point $P$, midpoint of $AC$ is $M$ and midpoint of $AB$ is $N$. $(PBN) \cap (PCM)=Q \neq P$. If $K$ is midpoint of $BC$ and $D \in AO$ such that $KD \perp AO$, then prove that $(PQOD)$ is cyclic. First notice by angle chasing that $Q$ lies on the $A$ midline of $\Delta ABC$ and that $APKD$ is a cyclic quadrilateral with diameter $AK$. Notice that the center of this quadrilateral lies on the $A$ midline of $BC$. Let the center be called $E$. Angle chase some more to show that $(PEOD)$ are cyclic and finally even more angle chasing shows that $(PQODE)$ is cyclic.

Let $DI$ and the perpendicular from $A$ to $AI$ intersect at $S$. $DI\cap (ABC)=K$, let $M$ be the midpoint of $EF$. Claim: $P,M,K$ are collinear. Proof: Since $PM$ is median on $\triangle PEF$ and $PA$ is $P-$symmedian, we have \[\measuredangle KPE=\measuredangle KDE=\frac{\measuredangle C}{2}=\measuredangle FPA=\measuredangle MPE\]Hence $P,M,K$ are collinear.$\square$ Invert the diagram from $I$ with radius $ID$. $A$ swaps with $M$ and $B,C$ swap with the midpoints of $DF,DE$ respectively. $S^*$ is the foot of the altitude from $M$ to $ID$. New Problem Statement: $ABC$ is a triangle whose circumcenter is $O$. $D,E,F$ are the midpoints of $BC,CA,AB$ respectively and $AO$ intersects $(ABC)$ at $S$ for second time. $SD$ intersects $(ABC)$ at $P$ and $Q$ is the intersection of $(PBF)$ and $(PEC)$. If $H$ is the altitude from $D$ to $AS,$ then $P,Q,O,H$ are concyclic. Let $T$ be the point on $(ABC)$ where $AT\parallel BC$. Claim: $Q$ lies on $EF$. Proof: \[\measuredangle PQF=\measuredangle PBF=\measuredangle PBA=\measuredangle PCA=\measuredangle PCE=\measuredangle PQE\]Thus, $Q,E,F$ are collinear.$\square$ Claim: $Q,P,T$ are collinear. Proof: Let $T'=PQ\cap (ABC)$. \[\measuredangle T'QF=\measuredangle PQF=\measuredangle PBA=\measuredangle PT'A\]This yields $AT'\parallel BC\iff T'=T$.$\square$ Claim: $QF$ is the angle bisector of $\measuredangle PQH$. Proof: Let the perpendicular from $H$ to $BC$ intersect $TP$ and $EF$ at $X,Y$. \[-1\overset{?}{=}(Q(XY)_{\infty},\overline{QFE};\overline{QPT},QH)\overset{XY}{=}(XY_{\infty},Y;X,H)\]$-1\overset{?}{=}(XY_{\infty},Y;X,H)\iff YX\overset{?}{=}YH$. Let the reflection of $H,D,T$ with respect to $EF$ be $H',D',T'$. Note that $D'$ lie on $AT$ and $T'$ lie on $BC$. Also $S,T,T'$ are collinear. Since $\measuredangle SHD=90=\measuredangle ST'D,$ we get that $D,H,T',S$ are concyclic. Also $DD'H'H$ and $TT'HH'$ are isosceles trapezoids. By $DHT'\sim D'H'T$ we have \[\measuredangle PTA=\measuredangle PSA=\measuredangle DSH=\measuredangle DT'H=\measuredangle H'TD'=\measuredangle H'TA\]This yields $T,H',P$ are collinear. Hence $YX=YH$.$\square$ Claim: $P,Q,O,H$ are concyclic. Proof: \[\measuredangle PQH=2\measuredangle PQF=2\measuredangle PBA=\measuredangle POA=180-\measuredangle HOP\]As desired.$\blacksquare$

Invert at $I$. Let $M$ be the midpoint of $\overline{BC}$. It suffices to show that $QPMIX$ is cyclic. Points $P$, $M$, $I$ and $X$ are all feet/midpoints of $\triangle DAD'$, where $D'$ is the antipode of $D$ WRT $(DEF)$, meaning that they lie on the nine-point circle of $\triangle DAD'$. Now we show that $Q$ lies on this circle too. Firstly, point $Q$ lies on line $BC$ since \[\angle PQC = \angle PEC = \angle PFB = \angle PQB.\]Now, $\angle PIM = \angle IPD = \angle DD'P = \angle PEC = \angle PQM$ finishes.

Solved with emotional support from cosdealfa. I had a horrible time solving this problem because I though it was going to be harder than it actually was. Let $T$ be the $A$-mixtilinear touch point, let $S$ be the foot from the $A$-excenter to $BC$, let $N$ be the midpoint of major arc $BC$, let $IM$ meet $(BIC)$ again at $M$, let $DI$ meet $AN$ at $K$ and let $H$ be the orthocenter of $\triangle DEF$. We need to prove that $K$ lies on $PQ$. In fact, we will show that $M$ also lies on $PQ$. It is well known that $\overline{N-I-T}$ and we will also prove $\overline{A-R-T}$. Now note that $AT$ and $AS$ are isogonal so it suffices for \[\frac{\sin(\angle BAR)}{\sin(\angle CAR)}=\frac{\sin(\angle BAT)}{\sin(\angle CAT)}=\frac{\sin(\angle CAS)}{\sin(\angle BAS)}\]By ratio lemma \[\frac{\sin(\angle CAS)}{\sin(\angle BAS)}=\frac{CS}{BS}\cdot\frac{AB}{AC}=\frac{p-b}{p-c}\cdot\frac{AB}{AC}\]By trig ceva \[\frac{\sin(\angle BAR)}{\sin(\angle CAR)}=\frac{\sin(\angle AFR)}{\sin(\angle RFE)}\cdot\frac{\sin(\angle FER)}{\sin(\angle REA)}=\left(\frac{\sin(\angle FDR)}{\sin(\angle EDR)}\right)^2=\left(\frac{\sin C/2}{\sin B/2}\right)^2\]One can easily check that these are equal so we get $\overline{A-R-T}$. Now note that \[\measuredangle BQC=\measuredangle BQP+\measuredangle PQC=\measuredangle BFP+\measuredangle PEC=\measuredangle EDF= \measuredangle BIC\]So $Q$ lies on $(BIC)$. Invert around the incircle and denote $X$'s inverse by $X'$. Clearly, $A'$, $B'$ and $C'$ are the midpoints of $EF$, $DF$ and $DE$. $T'$ is the midpoint of $DH$, $M'$ is the midpoint of $B'C'$, $K'$ and $N'$ are the intersections of $DI$ and $DN$ with the circle with diameter $IA'$ and $Q$ lies on $(PB'F)$, $(PC'E)$ and $B'C'$. Also note that $IT'PR$ is cyclic. Now look at the problem from the perspective of $\triangle DEF$. Clearly, $T'C'A'B'$ is the nine-point circle. The problem translates to $Q'IPK'$ being cyclic. Claim: $M'$ lies on $(IPK')$. Proof: We firstly prove that $PA'\parallel IM'$. To do this, redefine $P$ to be the intersection of the circle with diameter $DA'$ (which is centered at $M'$) with $(DEF)$. By orthocenter config, we know that $H$ lies on $PA'$ and since $T'IA'G$ is a parallelogram, we get that $PA'\parallel IM'$. Since $\measuredangle DPH=90^\circ$ we have $PT'=T'H=IA'$ so $PT'IA'$ is an isosceles trapezoid, which suffices. From the above, we also know that $PT'=TD$ so $M'T'$ is actually the perpendicular bisector of $DP$. This implies \[\measuredangle DIM'=\measuredangle M'IP= \measuredangle K'IN'\]And since $M'P=M'K'$ we get that $PM'IK'$ is cyclic (well known lemma). $\square$ We are finally ready to finish the problem. Claim: $Q'$ lies on $(IPK'M')$. Proof: Note that $\measuredangle C'Q'P=\measuredangle DEP$ and \[\measuredangle PDK'=\measuredangle PDE+\measuredangle EDK'=\measuredangle PDE+90^\circ - \measuredangle DFE=90^\circ-\measuredangle PED\]Therefore, using that $M'$ is the circumcenter of $\triangle DPK'$ we get \[\measuredangle M'K'P=90^\circ-\measuredangle PDK'=\measuredangle PED=\measuredangle C'Q'P. \ \ \ \blacksquare\]

Let $M$ be the midpoint of $EF$, let $Z$ be the reflection of $D$ over $AI$, let $S$ be the reflection of $D$ over $I$, let $P_B,P_C$ be the feet from $B,C$ onto $AI$, let $Q_B,Q_C$ be the intersections of $BI,CI$ with $EF$, let $R_B,R_C$ be the second intersections of $(BPF),(CPE)$ with $EF$. We claim $MQ_B=MR_B$ and $MQ_C=MR_C$. First, since $PR$ is a symmedian of $\triangle PEF$ we have $P,M,S$ are collinear. Next, $P_B\in DE$ by Iran lemma, and by symmetry $P_B\in ZF$. Thus arc chasing implies $\measuredangle FPM=\measuredangle FP_BM$,so $FPP_BM$ is cyclic and $\angle FPP_B=90^\circ$. Let $X_C=FP\cap CP_C$. Then $P,P_C$ lie on the circle with diameter $P_BX_C$. We show $Z$ lies on this circle as well. First $P_C\in DF$ by Iran lemma, so by symmetry $P_C\in ZE$ and thus \[\measuredangle PZP_C=\measuredangle PZE=\measuredangle PFE=\measuredangle PX_CP_C\]as desired. Now observe \[FR_C\cdot FE=FP\cdot FX_C=FP_B\cdot FZ=EP_B\cdot ED=EQ_C\cdot EF,\]since $Q_C\in(BDP_BIF)$ by Iran lemma. This implies $MQ_C=MR_C$, and similarly we get $MQ_B=MR_B$. Next notice $\measuredangle PQC=\measuredangle PEC=\measuredangle PFE=\measuredangle PFR_B=\measuredangle PQR_B$, so $Q\in CR_B$. Similarly $Q\in BR_C$. Now apply DIT on points $B,Q,I,C$ and line $EF$. Since $BQ\cap EF=R_C,CI\cap EF=Q_C$ and $BI\cap EF=Q_B,CQ\cap EF=R_B$ we get the involution is reflection over $M$. Thus the conic through $B,Q,I,C,E$ must also pass through $F$. Let $\mathcal I,\mathcal J$ be the circular points at infinity. Construct the cubic $\mathcal C$ through $A,B,C,D,E,F,\mathcal I,\mathcal J,P$. By Cayley-Bacharach with \[((BDIF\mathcal I\mathcal J)\cup\overline{AEC})\cap((CDIE\mathcal I\mathcal J)\cup\overline{AFB})\]we see $I\in\mathcal C$. Next since \[B+F+P+\mathcal I+\mathcal J=P+\mathcal I+\mathcal J-A=C+E+P+\mathcal I+\mathcal J\]we get $Q\in\mathcal C$. Additionally, since \[B+F+\mathcal I+\mathcal J+P+Q=0=B+F+\mathcal I+\mathcal J+D+I,\]we get $X=PQ\cap DI\in\mathcal C$. We also get \[B+D+I+F+\mathcal I+\mathcal J=0=B+F+A+D+I+X,\]so $A+X=\mathcal I+\mathcal J$. Finally, note \[\measuredangle BQC=\measuredangle BQP+\measuredangle PQC=\measuredangle BFP+\measuredangle PEC=\measuredangle FEP+\measuredangle PFE=\measuredangle FPE=\measuredangle FDE=\measuredangle BIC,\]so $B+Q+I+C+\mathcal I+\mathcal J=0$. Since $B,Q,I,C,E,F$ are coconic, $B+Q+I+C+E+F=0$, so \[E+F=\mathcal I+\mathcal J=A+X.\]Thus $EF$ and $AX$ intersect on the line at infinity, as desired.

This is the best problem of all time, perfect for my 1700th post. Solved in 2.5 hours, no hints. We identify that $P$ is the $D$-queue point of $\triangle DEF.$ Let $D'$ be the point diametrically opposite $D$ on $\omega.$ Let $M$ be the midpoint of segment $EF,$ and let $DM$ meet $\omega$ again at point $K.$ We start with a preliminary claim: Claim 1: Lines $PK$ and $DI$ meet on the line through $A$ perpendicular to $AI.$ Proof: By reciprocating poles/polars, this amounts to showing that if the tangents at $P$ and $K$ to $\omega$ meet at point $L,$ then $LM \parallel BC.$ Let $\ell$ be the line through $M$ parallel to $BC,$ and let $L$ intersect $\omega$ at two points $V_1, V_2.$ It is clear that $(D,D';V_1, V_2) = -1.$ Projecting this harmonic bundle through $M$ gives $(K,P;V_1,V_2) = -1,$ so $V_1 V_2 = \ell$ passes through $L,$ as claimed. Thus it suffices to show that $P,Q,K$ are collinear. Now, we will reflect a number of objects across the line $AI.$ The point $D$ is sent to the point $X$ such that $DX \parallel EF.$ The point $P$ gets sent to the second intersection of $AD'$ and $\omega,$ say $P'.$ Some simple angle-chasing yields that the point $B$ gets sent to the second intersection of $(BIC)$ with $AC,$ say $B'.$ Similarly define $C'.$ Obviously, points $E$ and $F$ are sent to each other. By symmedian properties, if $AD$ intersects $\omega$ again at point $K',$ then $K$ is sent to $K'.$ We must show that if $(B'EP')$ intersects $(C'FP')$ at $Q',$ then $P', Q', K'$ are collinear. We will explicitly construct a $Q'$ which satisfies that $B'EP'Q'$ and $C'FP'Q'$ are cyclic and $P', Q', K'$ are collinear. Let $I_A$ be the $A$-excenter, so that $B,I,C,B',C',I_A$ are concyclic, and let $N$ be the midpoint of $BC.$ Redefine $Q'$ as the intersection of $I_A N$ with $(BIC).$ We will show that this $Q'$ satisfies the desired properties. Claim 2: $DHP'Q'I$ is cyclic. Proof: We show that they lie on the circle with diameter $IH.$ Clearly $D$ lies on this circle, as does $Q'$ since $\angle IQ'H = \angle IQ' I_A = 90^\circ.$ Finally, say by the incircle diameter lemma, we see that $NP'$ is tangent to $\omega,$ so $\angle IP'N = 90^\circ.$ This proves our claim. Claim 3: $P'Q'EB'$ is cyclic. Proof: Invert about $\omega.$ By symmetry, $\overline{B'XC'}$ is tangent to $\omega,$ so $B'$ is sent to the midpoint of $EX.$ The circle $(BIC)$ is sent to the $D$-midline of $\triangle DEF$, and by Claim 2, the inverse of $Q'$ lies on $DP'.$ This if $DP' \cap EF = Y,$ then $Q'$ is sent to the midpoint of $DY.$ Therefore, our inverted claim is as follows: Let $\triangle ABC$ be a triangle, and let $A'$ be on $(ABC) = \omega$ such that $AA' \parallel BC.$ Let $Q$ be the $A$-queue point of $BC,$ and let $Y = A'Q \cap BC.$ Let $D, E$ be the midpoints of $A'Y, AB,$ respectively, Prove that $DEBQ$ is cyclic. Since $DE \parallel AA',$ the claim thus follows by Reim's Theorem. Similarly, $C'FQ'P'$ is cyclic. This equates our two definitions of $Q'$. Finally: Claim 4: $P',Q', K'$ are collinear. Proof: Note that $\triangle DEF \sim \triangle I_A BC,$ so symmedian properties give $\angle EDA = \angle CI_A Q'.$ But $$\angle CI_A Q' = 180^\circ - \angle CB'Q' = \angle EB'Q' = \angle EP'Q'$$and $\angle EDA = \angle EDK' = \angle EP'K'.$ Therefore, $\angle EP'Q' = \angle EP'K',$ and we are done!