Let $\ell_A$ denote the perpendicular bisector of $AD$. Define $\ell_B$ and $\ell_C$ similarly. Define $X= \ell_B \cap \ell_C$. Define $Y,Z$ similarly. Claim : $\omega_A,\omega_B, \omega_C$ concur at a point $Q\neq P$ Proof : We show that the orthocenter $H$ of $\triangle ABC$ has the same power wrt all the three circles. Let $H_A,H_B,H_C$ denote the feet of altitude from $A,B,C$ to their opposite sidelines. Note that since $\omega_A$ is symmetric in the $A$-midline, $H_A$ must lie on $\omega_A$. Hence the power of $H$ wrt $\omega_A$ is $HH_A\cdot HA$ and similarly for others. Since $HH_A\cdot HA= HH_B \cdot HB = HH_C \cdot HC$, we are done. $\square$ Let $\ell$ denote the line through the centers $O_A$, $O_B$ and $O_C$ of $\omega_A$, $\omega_B$, $\omega_C$ respectively. Note that $Q$ is the reflection of $P$ in $\ell$. Claim : $Q$ is the miquel point of the quadrilateral formed by the lines $\ell, \ell_A, \ell_B, \ell_C$. Proof : We prove $Q\in (O_BO_CX)$. Two similar other cyclic quads imply the result. Angle Chase : $$ \measuredangle O_CQO_B = \measuredangle (O_CQ, \ell) + \measuredangle (QO_B, \ell) = \measuredangle PBQ + \measuredangle QCP = \measuredangle CQB + \measuredangle BPC = \measuredangle PQB + \measuredangle CQP + \measuredangle BAC = \measuredangle H_BEB + \measuredangle FH_CC+\measuredangle (BH_B,CH_C) = \measuredangle (BE,CF) = \measuredangle (\ell_B, \ell_C) = \measuredangle O_CXO_B $$ The above claim implies $Q \in \odot (XYZ)$. $\square$

Let $H$ be the orthocenter and $O$ be the circumcenter of $\triangle ABC$. Let $H_A,H_B,H_C$ be the altitudes from $A,B,C$ and $M_A,M_B,M_C$ be the midpoints of $BC,CA,AB$. Notice that $AH_APD$, $BH_BPE$ and $CH_CPF$ are all cyclic being isosceles trapezoids. Let $Q$ be the point on $PH$ such that $QH\times HP=AH\times HH_A$, then $$QH\times HP=AH\times HH_A=BH\times HH_B=CH\times HH_C$$Therefore, $Q$ lies on $\omega_A,\omega_B,\omega_C$ by power of a point. 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Claim. $Q,Z,O_A,O_C$ are concyclic. Proof. This is just angle chasing. \begin{align*} \angle QO_AZ&=\angle AO_AZ-\angle AO_AQ\\ &=180^{\circ}-\angle APD-2\angle APH\\ &=\angle AHP-\angle APH \end{align*}\begin{align*} \angle QO_CZ&=\angle FO_CZ-\angle FO_CQ\\&=\angle FPC-2\angle FPQ\\&=\angle QPC-\angle FPQ \end{align*}[asy][asy] size(10cm); 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 = -11.087065689421976, xmax = 0.6710017679701562, ymin = -9.681903556798279, ymax = 2.6405511385486715; /* image dimensions */ pen zzttqq = rgb(0.6,0.2,0); pen zzttff = rgb(0.6,0.2,1); pen ttzzqq = rgb(0.2,0.6,0); /* draw figures */ draw(circle((-4.037913311076221,-2.2566115038288466), 2.8107234008805477), linewidth(0.8) + zzttqq); draw((-1.6898775280504954,-0.7116538999159499)--(-5.893742741384873,-3.0100930250914564), linewidth(0.8) + linetype("4 4") + blue); draw((-5.277489523809524,0.266009523809525)--(-6.4,-3.78), linewidth(0.8) + zzttff); draw((-6.4,-3.78)--(-1.58,-3.62), linewidth(0.8) + zzttff); draw((-1.58,-3.62)--(-5.277489523809524,0.266009523809525), linewidth(0.8) + zzttff); draw((-3.884080522551288,-1.6921101181166989)--(-1.160240268822567,-6.674037707108505), linewidth(0.8) + red); draw(circle((-6.304981146793863,-3.9320460032951523), 1.009511921723559), linewidth(0.8) + linetype("4 4") + blue); draw(circle((-4.770208019710931,-2.532101009100448), 1.220985928730586), linewidth(0.8) + linetype("4 4") + red); draw(circle((-3.8569832421468813,-5.268386404975224), 3.041098197586366), linewidth(0.8) + linetype("4 4") + red); draw(circle((-3.3799031645392175,-4.652039837831397), 3.002562031662692), linewidth(0.8) + linetype("4 4") + red); draw((-3.884080522551288,-1.6921101181166989)--(-6.368668526955157,-4.939546989751335), linewidth(0.8) + ttzzqq); draw((-6.858049709277972,-4.77657515335082)--(-1.160240268822567,-6.674037707108505), linewidth(0.8) + ttzzqq); draw((-6.858049709277972,-4.77657515335082)--(-3.5848021187049826,-2.2394929298914716), linewidth(0.8) + ttzzqq); /* dots and labels */ dot((-5.277489523809524,0.266009523809525),dotstyle); label("$A$", (-5.184515825811127,0.5005828613033043), NE * labelscalefactor); dot((-6.4,-3.78),dotstyle); label("$B$", (-6.313290301720771,-3.5441923440395877), NE * labelscalefactor); dot((-1.58,-3.62),dotstyle); label("$C$", (-1.4924826441899974,-3.379579399636098), NE * labelscalefactor); dot((-1.6898775280504954,-0.7116538999159499),dotstyle); label("$P$", (-1.5865471838491345,-0.4870948051176345), NE * labelscalefactor); dot((-1.629686949426607,-2.5248950809605866),dotstyle); label("$D$", (-1.539514914019566,-2.2978371935560222), NE * labelscalefactor); dot((-7.088942846384126,-5.8487971134333065),dotstyle); label("$E$", (-6.995258214249515,-5.613612216540602), NE * labelscalefactor); dot((-4.421699276839515,0.04625299922881543),dotstyle); label("$F$", (-4.337934968878893,0.288937647070246), NE * labelscalefactor); dot((-5.181662901657082,-2.6207674685327818),linewidth(4pt) + dotstyle); label("$H$", (-5.09045128615199,-2.4389340030447277), NE * labelscalefactor); dot((-5.893742741384873,-3.0100930250914564),linewidth(4pt) + dotstyle); label("$Q$", (-5.7959353335955175,-2.815192161681276), NE * labelscalefactor); dot((-1.160240268822567,-6.674037707108505),linewidth(4pt) + dotstyle); label("$O_B$", (-1.0691922157238807,-6.48370920838762), NE * labelscalefactor); dot((-3.884080522551288,-1.6921101181166989),linewidth(4pt) + dotstyle); label("$O_A$", (-3.7970638658388554,-1.4982886064533576), NE * labelscalefactor); dot((-3.5848021187049826,-2.2394929298914716),linewidth(4pt) + dotstyle); label("$O_C$", (-3.49135411194666,-2.06267584440818), NE * labelscalefactor); dot((-5.349204909943023,-3.6070750735595207),linewidth(4pt) + dotstyle); label("$Z$", (-5.2550642305554796,-3.4266116694656663), NE * labelscalefactor); dot((-6.368668526955157,-4.939546989751335),linewidth(4pt) + dotstyle); label("$X$", (-6.266258031891203,-4.743515224693585), NE * labelscalefactor); dot((-6.858049709277972,-4.77657515335082),linewidth(4pt) + dotstyle); label("$Y$", (-6.760096865101673,-4.578902280290095), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy][/asy] Meanwhile \begin{align*} \angle AHP+\angle FPQ&=\angle AHP+\angle PHC\\ &=\angle AHC\\ &=180^{\circ}-\angle ABC\\ &=\angle APC\\ &=\angle APH+\angle QPC \end{align*}Therefore $\angle QO_AZ=\angle QO_CZ$ and we are done. $\blacksquare$ Now by symmetry $Q,Y,O_A,O_C$ are concyclic and $Q,X,O_B,O_A$ are concyclic. Therefore, $Q$ is the miquel point of $ZXQ_BQ_C$, and so $Q,Z,Y,X$ are concyclic as desired. Therefore $\omega$ passes through $Q$ as well and the proof is thus completed.

First, we will show that $\omega_A,\omega_B,\omega_C$ (circumcircles of $\triangle ADP,\triangle BEP,\triangle CFP$ respectively) are coaxial. Let $A_1,B_1,C_1$ be foot of altitude from $A,B,C$ respectively. Since $\square AA_1DP$ is a trapezoid, $A_1$ must lie on $\omega_A$. Let $Q$ be an image of $P$ by the negative inversion which center at $H$ and sent $A,B,C$ to $A_1,B_1,C_1$. Hence, $AH\cdot HA_1=BH\cdot HB_1=CH\cdot HC_1=PH\cdot HQ$. Therefore, $Q$ lies on $\omega_A,\omega_B,\omega_C$. The rest is to show that $Q$ lies on $\omega$. Let $\ell_A,\ell_B,\ell_C$ be perpendicular bisectors of $AD,BE,CF$ respectively, $Q_A,Q_B,Q_C$ be a reflection of $Q$ across $\ell_A,\ell_B,\ell_C$. Claim: $P$,$Q_A$,$Q_B$,$Q_C$ are collinear. It is sufficient to prove that $P,Q_A,Q_B$ are collinear, or equivalently, $\measuredangle QQ_AP-\measuredangle QQ_BP=\measuredangle Q_AQQ_B$. ($Q_C$ can be done similarly.) In fact, This can be proved by angle chasing. Let $R$ be an isogonal conjugate of $P$ wrt $\triangle A_1B_1C_1$. Notice that $QQ_AAD$ is an trapezoid, so $Q_A$ lies on $\omega_A$ and $QQ_A//AD$. By angle chasing, \begin{align*} \measuredangle Q_AQQ_B &=\measuredangle (AD,BE)\\ &=\measuredangle (AD,AA_1)+\measuredangle (AA_1,BB_1)+\measuredangle (BB_1,BE)\\ &=\measuredangle (AA_1,A_1P)+\measuredangle (BC,CA)+\measuredangle (B_1P,BB_1)\\ &=\measuredangle AA_1P+\measuredangle PB_1B+\measuredangle BCA \end{align*}, and \begin{align*} \measuredangle QQ_AP-\measuredangle QQ_BP &=\measuredangle QAP-\measuredangle QBP \\ &=\measuredangle AQP+\measuredangle QPA-\measuredangle BQP -\measuredangle QPB\\ &=\measuredangle AA_1P- \measuredangle BB_1P+\measuredangle BPA\\ &=\measuredangle AA_1P+\measuredangle PB_1B+\measuredangle BCA. \end{align*}Therefore, $\measuredangle Q_AQQ_B=\measuredangle QQ_AP-\measuredangle QQ_BP$ which means that $P,Q_A,Q_B$ are collinear. Similarly, $P,Q_A,Q_B,Q_C$ lie on a line. By Simson's line, $Q$ must lie on circumcircle of a triangle formed by $\ell_A,\ell_B,\ell_C$.

Solution by L567 We will first characterize $D,E,F$ in a nice way that does not involve those weird midpoints. Let $XYZ$ be the orthic triangle. Claim: $ADPX$ is cyclic Proof: Note that $X$ and $A$ are reflections across the midline and so are $P,D$. So, $AXDP$ is an isosceles trapezoid and hence cyclic. So henceforth, we will only look at $D$ as the point on $(AXP)$ such that $PD \perp BC$ Now, to characterize the common point. Let $H$ be the orthocenter of $\triangle ABC$ Claim: Let $PH$ meet the nine point circle of $ABC$ for the second time at $Q$. Then, $Q$ lies on $\omega_A, \omega_B, \omega_C$ Proof: Its enough to show it lies on $\omega_A$ since the others follow similarly. Let $HX, HQ$ meet $(ABC)$ at $R,S$. Then, since $QX || SR$, ($Q,X$ are midpoints of $HR, HS$ by definition of nine point circle), we have $\angle XQP = \angle RSP = \angle RAP = \angle XAP$ and so $Q \in (APX) = \omega_A$, as desired. Now, we must show that $Q \in \omega$ as well, which is the hard part of the problem. Let $O_1, O_2$ be circumcenters of $\omega_A, \omega_C$. Let the perpendicular bisectors of $AD, CF$ meet at $T$. Claim: $O_1O_2TQ$ is cyclic Proof: This is going to be a very long chase of angels to show $\angle O_1TO_2 = \angle O_1QO_2$ Let $K$ be an arbitrary point on $O_2O_1$ beyond $O_1$ which I'll use for the angle chase. Its probably not necessary but whatever. We have $\angle O_1TO_2 = \angle TO_1K - \angle TO_2K$ Now, see that $\angle TO_1K = \angle AO_1T - \angle AO_1K$ and $\angle TO_2K = \angle CO_2K - \angle CO_2T$ So, $\angle TO_1K - \angle TO_2K = \angle AO_1T - \angle CO_2K - \angle AO_1K + \angle CO_2T$ But $\angle AO_1T = \frac{\angle AO_1D}{2} = \angle APD$ and similarly, $\angle CO_2T = 180 - \angle CPF$ So this means $\angle O_1TO_2 = 180 - \angle CPF + \angle APD - \angle AO_1K - \angle CO_2K$ Observe that $\angle APD = 180 - \angle XAP$ and $\angle CPF = \angle PFZ = \angle PCZ$. So, $\angle APD - \angle CPF = 180 - (\angle XAP + \angle PCZ) = 180 - ((\angle BAP - \angle BAX) + (\angle PCB - \angle BCZ)) = 180 - ((\angle BAP + \angle PCB) - \angle BCZ - \angle BAX) = \angle BCZ + \angle BAX = 180 - 2 \angle ABC$ So, $\angle O_1TO_2 = 360 - 2 \angle ABC - \angle AO_1K - \angle CO_2K$ Now, we need a way to deal with this annoying $AO_1K$ and $CO_2K$ Let $AO_1$ meet $BC$ at $U$, $CO_2$ meet $AB$ at $V$ and $AO_1$ meet $CO_2$ at $G$ So, $\angle AO_1K + \angle CO_2K = (\angle O_1GO_2 + \angle O_1O_2G) + (\angle GO_1O_2 + \angle O_1GO_2) = 180 + \angle O_1GO_2$ So, we have $\angle O_1TO_2 = 360 - 2 \angle ABC - (180 + \angle O_1GO_2) = 180 - 2 \angle ABC - \angle O_1GO_2$ Leave this for now and look at the other angle. $\angle O_1PO_2 = \angle APC - \angle APO_1 - \angle O_2PC = 180 - \angle ABC - \angle PAU - (90 - \angle PVC) = 90 - \angle ABC - \angle PAU + \angle PVC$ So, showing $\angle O_1PO_2 = \angle O_1QO_2$ is equivalent to $90 - \angle ABC = \angle O_1GO_2 + \angle PVC - \angle PAU$ See that $\angle O_1GO_2 = \angle AUB - \angle VCB$ Let $W = PV \cap BC$ and $M = PA \cap BC$ We have $\angle PAU - \angle AUB = \angle PMC$ and $\angle PVC - \angle VCB = \angle PWC$ So, we just want to show $90 - \angle ABC = \angle PWC - \angle PMC = \angle MPW = \angle APV$ But $\angle APV = \angle APC - \angle VPC = 180 - \angle ABC - 90 = 90 - \angle ABC$, as desired. So after all of that insane angle chase, $O_1O_2TQ$ is indeed cyclic, as claimed. Now, we're basically done. Let $O_3$ be circumcenter of $(BEP)$. Let perpendicular bisectors of $BE, CF$ meet at $I$ and those of $AD, BE$ meet at $J$. By symmetry, we also have $(O_2O_3IQ)$ is also cyclic. So, we have that $Q$ must be the miquel point of the quadrilateral $O_3O_1JI$. So, $\omega = (IJT)$ also passes through $Q$. So, $\omega_A, \omega_B, \omega_C, \omega$ all have the common point $Q$ and so we are done.

It takes time for me to understand that it's actually A-median line

Let $\triangle XYZ$ be the orthic triangle of $\triangle ABC$. Let $A'$ be the point on $\overline{BC}$ such that $PA \perp PA'$. Define $B'$ and $C'$ respectively. Finally, let $O_A$, $O_B$ and $O_C$ be the centers of $\omega_A$, $\omega_B$ and $\omega_C$ respectively. It is well-known that $A'$, $B'$, $C'$ lie on a single line $\ell$ (This line is called the orthotransveral of $P$.) We will first show that $\omega_A$, $\omega_B$ and $\omega_C$ are concurrent at a point. Since $X$ is the reflection of $A$ over the $A$-midline in $\triangle ABC$, $APDX$ is an isosceles trapezoid and hence $X$ lies on $\omega_A$. Now as $\angle AXA' = \angle APA' = 90^\circ$, it follows that $A'$ also lies on $\omega_A$. Thus $O_A$ is the midpoint of $AA'$. Similarly, $O_B$ and $O_C$ are midpoints of $BB'$ and $CC'$. This means that $O_A$, $O_B$ and $O_C$ all lie on the Gauss line $k$ of quadrilateral $\mathcal{P}(\overline{AB}, \overline{BC}, \overline{CA}, \ell)$, and hence $\omega_A$, $\omega_B$ and $\omega_C$ are concurrent at the point $P'$ which is the reflection of $P$ over $k$. Let the perpendicular bisectors of $AD$, $BE$ and $CF$ be $\ell_A$, $\ell_B$ and $\ell_C$ respectively. We now claim that $P'$ is the Miquel point of quadrilateral $\mathcal{P}(\ell_A, \ell_B, \ell_C, k)$. To see this, let $U = \ell_A \cap \ell_C$. We just need to show that $O_A$, $O_C$, $U$, $P'$ are concyclic. This follows from angle chasing like so: \begin{align*} \measuredangle O_AP'O_C &= \measuredangle O_CPO_A\\ &= \measuredangle O_CPC + \measuredangle CPA + \measuredangle APO_A\\ &= 90^\circ - \measuredangle CZP + \measuredangle AHC + \measuredangle AXP - 90^\circ \\ &= \measuredangle AHC + \measuredangle DAX - \measuredangle FCZ\\ &= \measuredangle AHC + \measuredangle CAH + \measuredangle DAC - \measuredangle ACH - \measuredangle FCA\\ &= \measuredangle (\overline{AD}, \overline{CF}) = \measuredangle O_AUO_C. \end{align*}Therefore $P'$ lies on the circumcircle of the triangle formed by $\ell_A$, $\ell_B$ and $\ell_C$.

Similar to above solutions, but whatever. [asy][asy] size(12cm); defaultpen(fontsize(10pt)); pair A = dir(100), B = dir(220), C = dir(320), P = dir(60), M = (B+C)/2, N = (C+A)/2, K = (A+B)/2, X = foot(A,B,C), Y = foot(B,C,A), Z = foot(C,A,B), H = orthocenter(A,B,C), Oa = circumcenter(A,P,X), Ob = circumcenter(B,P,Y), Oc = circumcenter(C,P,Z), D = 2*foot(P,N,K)-P, E = 2*foot(P,K,M)-P, F = 2*foot(P,N,M)-P, P1 = intersectionpoint(H--H+dir(P--H)*100, circumcircle(X,Y,Z)), R = extension((B+E)/2,Ob,(C+F)/2,Oc), S = extension((C+F)/2,Oc,(A+D)/2,Oa), T = extension((A+D)/2,Oa,(B+E)/2,Ob); draw(A--B--C--A, heavyblue); draw(unitcircle, heavyblue); draw(circumcircle(X,Y,Z), heavyblue); draw(circumcircle(A,X,P)^^circumcircle(B,Y,P)^^circumcircle(C,Z,P), purple); draw(P--P1, red); draw(A--X^^B--Y^^C--Z, heavyblue+dotted); draw(R--S--T--R, magenta); draw(circumcircle(R,S,T), magenta+dashed); draw(Oc--S--Oa, magenta); draw(R--Ob, magenta); draw(Ob--Oc, magenta); draw(circumcircle(P1,Ob,Oc), heavycyan); dot("$A$", A, dir(100)); dot("$B$", B, dir(240)); dot("$C$", C, dir(315)); dot("$M$", M, dir(270)); dot("$N$", N, dir(45)); dot("$K$", K, dir(135)); dot("$X$", X, dir(270)); dot("$Y$", Y, dir(30)); dot("$Z$", Z, dir(180)); dot("$P$", P, dir(45)); dot("$P'$", P1, dir(270)); dot("$H$", H, dir(120)); dot("$D$", D, dir(315)); dot("$E$", E, dir(225)); dot("$F$", F, dir(45)); dot("$R$", R, dir(180)); dot("$S$", S, dir(90)); dot("$T$", T, dir(180)); dot("$O_A$", Oa, dir(60)); dot("$O_B$", Ob, dir(135)); dot("$O_C$", Oc, dir(270)); [/asy][/asy] Let $H$, $MNK$, and $XYZ$ be the orthocenter, medial triangle, and orthic triangle of $ABC$, respectively. Let $O_A$ denote the center of $\omega_A$; similar for $O_B$ and $O_C$. Let the perpendicular bisectors of $\overline{BE}$ and $\overline{CF}$ meet at $R$; similar for $S$ and $T$. To show that $\omega_A$, $\omega_B$, and $\omega_C$ are coaxial, note that by symmetry $APDX$ is an isosceles trapezoid, so $X$ lies on $\omega_A$. Similarly, $Y$ lies on $\omega_B$ and $Z$ lies on $\omega_C$. Since $AH \cdot HX = BH \cdot HY = CH \cdot HZ$, we know that $H$ has equal power wrt $\omega_A$, $\omega_B$, and $\omega_C$, so the circles must be coaxial. The second intersection point of these circles, $P'$, is the intersection point of line $\overline{PH}$ and $(XYZ)$ not on segment $\overline{PH}$. Now we prove that $P'$, $R$, $S$, $T$ are concyclic. I claim that $P'RO_BO_C$ is cyclic, so that by symmetry $P'SO_CO_A$ and $P'TO_AO_B$ are both cyclic, and we are done by Miquel points. Indeed \begin{align*} \measuredangle O_BRO_C &= \measuredangle(\overline{BE}, \overline{CF}) \\ &= (180^\circ - 2\measuredangle KMN) - \measuredangle YPZ \\ &= (180^\circ - 2\measuredangle CAB) - \measuredangle YPZ \\ &= (90^\circ - \measuredangle YBP) - (90^\circ - \measuredangle ZCP) - \measuredangle YPZ \\ &= (\measuredangle YPO_B - \measuredangle ZPO_C) - \measuredangle YPZ \\ &= \measuredangle O_CPO_B \\ &= \measuredangle O_BP'O_C. \end{align*}$\blacksquare$

Call the formed triangle $XYZ$. Also denote by $O_a$, $O_b$, $O_c$ the centers of $PAD$, $PBE$, $PCF$. Let $L_a$, $L_b$, $L_c$ be the altitude feet. Also, let ray $PH$ meet the nine-point circle again at $Q$. We contend that $Q$ is the desired point. [asy][asy] import graph; size(12cm); pen zzttqq = rgb(0.6,0.2,0); pen qqwuqq = rgb(0,0.39215,0); pen zzttff = rgb(0.6,0.2,1); pen ccqqqq = rgb(0.8,0,0); pen fuqqzz = rgb(0.95686,0,0.6); pen cqcqcq = rgb(0.75294,0.75294,0.75294); draw((-1.23265,1.71708)--(-2.19748,-0.20728)--(0.20251,-0.20728)--cycle, linewidth(1) + zzttqq); draw((-1.23265,1.71708)--(-2.19748,-0.20728), linewidth(1) + zzttqq); draw((-2.19748,-0.20728)--(0.20251,-0.20728), linewidth(1) + zzttqq); draw((0.20251,-0.20728)--(-1.23265,1.71708), linewidth(1) + zzttqq); draw(circle((-0.99748,0.39512), 1.34271), linewidth(1)); draw((-0.43383,-0.82356)--(-0.43383,2.33336), linewidth(1) + qqwuqq); draw((-0.43383,-0.82356)--(-0.56753,-0.92327), linewidth(1) + qqwuqq); draw((-0.43383,-0.82356)--(-1.82851,-0.12430), linewidth(1) + qqwuqq); draw((-1.23265,1.71708)--(-0.43383,2.33336), linewidth(1) + zzttqq); draw((-2.19748,-0.20728)--(-0.56753,-0.92327), linewidth(1) + zzttqq); draw(circle((0.14678,0.75490), 1.68186), linewidth(1) + linetype("2 2") + zzttff); draw(circle((-1.14082,-0.01508), 1.07399), linewidth(1) + linetype("2 2") + zzttff); draw(circle((-1.11506,0.45369), 0.67135), linewidth(1) + ccqqqq); draw((0.14678,0.75490)--(-1.14082,-0.01508), linewidth(1) + fuqqzz); draw((-0.43383,-0.82356)--(-1.51851,0.99031), linewidth(1)); draw((-0.45814,1.53901)--(-1.14082,-0.01508), linewidth(1) + fuqqzz); draw((-0.72900,1.89010)--(-0.79847,0.18963), linewidth(1) + fuqqzz); draw((-0.72900,1.89010)--(0.14678,0.75490), linewidth(1) + fuqqzz); dot("$A$", (-1.23265,1.71708), dir((1.108, 2.769))); dot("$B$", (-2.19748,-0.20728), dir((1.224, 2.749))); dot("$C$", (0.20251,-0.20728), dir((1.033, 2.749))); dot("$P$", (-0.43383,-0.82356), dir((0.978, 2.902))); dot("$D$", (-0.43383,2.33336), dir((0.978, 2.894))); dot("$E$", (-0.56753,-0.92327), dir((1.056, 2.904))); dot("$F$", (-1.82851,-0.12430), dir((1.157, 2.759))); dot("$Z$", (-0.45814,1.53901), dir((1.193, 2.300))); dot("$Y$", (-0.72900,1.89010), dir((1.141, 2.082))); dot("$X$", (-0.77259,0.82317), dir((1.070, 2.162))); dot("$H$", (-1.23265,0.51227), dir((1.108, 2.238))); dot("$Q$", (-1.51851,0.99031), dir((1.171, 2.894))); dot("$L_a$", (-1.23265,-0.20728), dir((1.108, 2.195))); dot("$P_a$", (-0.43383,0.40899), dir((0.978, 2.320))); dot("$O_a$", (0.14678,0.75490), dir((1.068, 2.343))); dot("$O_b$", (-1.14082,-0.01508), dir((1.063, 2.082))); dot("$O_c$", (-0.79847,0.18963), dir((1.166, 2.102))); dot("$L_b$", (-0.65526,0.94288), dir((1.244, 2.099))); [/asy][/asy] Claim: $AQL_aPD$ is cyclic, and so on. Proof. $ADPL_a$ is cyclic because it is an isosceles trapezoid, while $AQL_aP$ is cyclic by power of a point from $H$. $\blacksquare$ Claim: $Q$ is the Miquel point of complete quadrilateral $XYZO_aO_bO_c$. Proof. It is sufficient to show that $Q$ lies on $(O_a O_b Z)$. This is an angle chase: \begin{align*} \measuredangle O_b Q O_a &= \measuredangle O_b Q P + \measuredangle P Q O_a = (90^{\circ} - \measuredangle P L_b Q) + (90 ^{\circ} - \measuredangle Q L_a P) \\ &= \measuredangle Q L_b P + \measuredangle P L_a Q = \measuredangle L_a P L_b + \measuredangle L_b Q L_a \\ &= \measuredangle L_a P L_b + 2 \measuredangle ACB \\ \measuredangle O_b Z O_a &= \measuredangle(BE, AD) = \arg (AD) - \arg (BE) \\ &= (2\arg(BC) - \arg(PL_a)) - (2\arg(AC) - \arg(PL_b)) \\ &= 2\measuredangle ACB - \measuredangle L_b P L_a \end{align*}as needed. $\blacksquare$

Relabel the problem introducing new points. Quote: Let $P$ be a point on the circumcircle of acute triangle $ABC$. Let $P_A,P_B,P_C$ be the reflections of $P$ in the $A$-midline, $B$-midline, and $C$-midline. Let $M_A$ be the midpoint of $BC$, $M_B$ the midpoint of $AC$ and $M_C$ the midpoint of $AB$. Let $H_A$ be the foot from $A$ to $BC$, $H_B$ be the foot from $B$ to $AC$, $H_C$ be the foot from $C$ to $AB$. Let $O_A$ be the center of $(PP_AA)$, $O_B$ be the center of $(PP_BB)$, $O_C$ be the center of $(PP_CC)$. Let $X$ be the intersection of perpendicular bisector of $P_AA$ and $P_BB$, $Y$ the intersection of perpendicular bisector of $P_BB$ and $P_CC$ and $Z$ the intersection of perpendicular bisector of $P_CC$ and $P_AA$. Show that $(PP_AA), (PP_BB), (PP_CC),$ and $(XYZ)$ share a common point. [asy][asy]import olympiad;import geometry; size(13cm);defaultpen(fontsize(10pt));pen org=magenta;pen med=mediummagenta;pen light=pink;pen deep=deepmagenta; pair O,A,B,C,ma,mb,mc,ha,hb,hc,H,P,D,pa,pb,pc,X,Y,Z,oa,ob,oc; O=(0,0);A=dir(105);B=dir(200);C=dir(340);path w=circumcircle(A,B,C);ma=midpoint(B--C);mb=midpoint(A--C);mc=midpoint(A--B); ha=foot(A,B,C);hb=foot(B,A,C);hc=foot(C,A,B);H=orthocenter(A,B,C);P=dir(90);path x=circumcircle(ma,mb,mc);D=intersectionpoints(x,100H-99P--P)[1]; path c=circumcircle(C,P,hc); path a=circumcircle(A,P,ha);path b=circumcircle(B,P,hb); pc=2foot(P,ma,mb)-P;pa=2foot(P,mb,mc)-P;pb=2foot(P,ma,mc)-P;line ap=perpendicular(midpoint(A--pa),line(pa,A));line bp=perpendicular(midpoint(B--pb),line(pb,B));line cp=perpendicular(midpoint(C--pc),line(pc,C));X=intersectionpoint(ap,bp);Y=intersectionpoint(bp,cp);Z=intersectionpoint(ap,cp); oa=circumcenter(A,P,ha);ob=circumcenter(B,P,hb);oc=circumcenter(C,P,hc); draw(A--B--C--cycle,deep);draw(w,deep);draw(b,red);draw(a,red);draw(c,red);draw(x,deep);draw(circumcircle(X,Y,Z),lightblue+dashed); draw(P--D,med);draw(C--hc,med);draw(B--hb,med);draw(A--ha,med);draw(ma--mb--mc--cycle,deep); draw(Y--midpoint(C--pc),deep);draw(Y--midpoint(B--pb),deep);draw(X--midpoint(A--pa),deep); draw(C--pc,deep);draw(B--pb,deep);draw(A--pa,deep); draw(ob--Y--Z--oa--cycle,deepblue);draw(Y--X--Z--oc--oa,blue); dot("$A$",A,dir(A)); dot("$B$",B,dir(B)); dot("$C$",C,dir(C)); dot("$M_A$",ma,dir(90)); dot("$M_B$",mb,dir(mb)); dot("$M_C$",mc,dir(mc)); dot("$H_A$",ha,dir(ha)); dot("$H_B$",hb,dir(hb)); dot("$H_C$",hc,dir(hc)); dot("$H$",H,dir(H)); dot("$P$",P,dir(P)); dot("$D$",D,dir(D)); dot("$P_A$",pa,dir(pa)); dot("$P_B$",pb,dir(pb)); dot("$P_C$",pc,dir(pc)); dot("$X$",X,dir(X)); dot("$Y$",Y,dir(Y)); dot("$Z$",Z,dir(Z)); dot("$O_A$",oa,dir(oa)); dot("$O_B$",ob,dir(ob)); dot("$O_C$",oc,dir(oc)); [/asy][/asy] Let $D$ be intersection of $PH$ and $(M_AM_BM_C)$. By reflections, $H_A$ lies on $(PP_AA)$, $H_B$ lies on $(PP_BB)$, $H_C$ lies on $(PP_CC)$. Claim. $D$ lies on $(PP_AAH_A),(PP_BBH_B),(PP_CCH_C)$. Proof. By inversion at $H$ sending $(M_AM_BM_C)$ to $(ABC)$, this takes $D$ to $P$. By PoP, we get desired, the claim follows. $\square$ Claim. $O_AO_CDZ$ is cyclic. Proof. We do angle chasing. We want $\measuredangle O_CDO_A=\measuredangle(YZ,XZ)$. Note that \begin{align*} \measuredangle(YZ,XZ)&=\measuredangle(CP_C,AP_A)\\&=\measuredangle P_CCB-\measuredangle(AP_A,BC)\\&=\measuredangle P_CCB-\measuredangle CH_AP\\&=\measuredangle P_CCB+\measuredangle PH_AA-90^\circ\\&=\measuredangle PH_AA+\measuredangle P_CCP-\measuredangle BAP-90^\circ \end{align*}and \begin{align*} \measuredangle O_CDO_A&=\measuredangle CDP-\measuredangle CDO_C-\measuredangle O_ADP\\&=\measuredangle CDP+\measuredangle DP_CC+\measuredangle PP_AD-180^\circ\\&=\measuredangle CDP+\measuredangle DP_CC-\measuredangle DAP. \end{align*} Let $\ell$ be the parallel to $AB$ through $P$. \begin{align*} \measuredangle CDP+\measuredangle DP_CC+\measuredangle PP_AD-180^\circ&=\measuredangle P_CCB-\measuredangle CH_AP\Longleftrightarrow \\ \measuredangle DP_CC+\measuredangle PP_AD+\measuredangle CH_AP&=\measuredangle P_CCB+\measuredangle PP_CC\Longleftrightarrow \\ \measuredangle DP_CC+\measuredangle PP_AD+90^\circ-\measuredangle PH_AA&=\measuredangle P_CCB+\measuredangle PP_CC\Longleftrightarrow \\ \measuredangle DP_CC+90^\circ+\measuredangle AP_AD&=360^\circ-\measuredangle CBP-\measuredangle BPP_C\Longleftrightarrow \\ \measuredangle DP_CC+\measuredangle A_PAD&=270^\circ-\measuredangle CBP-\measuredangle BPC-\measuredangle CPP_C\Longleftrightarrow \\ \measuredangle DP_CC+\measuredangle CPP_C+\measuredangle A_PAD&=90^\circ+\measuredangle PCB\Longleftrightarrow \\ \measuredangle DPP_C+\measuredangle AP_AD&=90^\circ+\measuredangle PCB\Longleftrightarrow \\ \measuredangle AP_AC&=90^\circ+\measuredangle PCB\Longleftrightarrow \\ \measuredangle PCB&=\measuredangle (AP,\ell)=\measuredangle APP_C-90^\circ.\\ \end{align*}The claim follows. $\square$ In the same manner, we get that $XO_BO_CD$ and $YO_BO_CD$ are cyclic. Note that $O_A,O_B,O_C$ are collinear as they all lie on the perpendicular bisector of $PD$. Thus, $D$ is the Miquel point of $O_BYZO_A$, hence $D$ lies on $(XYZ)$. $\blacksquare$

Denote by $\ell_A,\ell_B,\ell_C$ perpendicular bisectors of $AD,BE,CF.$ Note for example that projection of $A$ onto $BC$ lies on $\odot (ADP),$ so orthocenter of $\triangle ABC$ has equal powers wrt all $\odot (ADP),\odot (BEP),\odot (CFP),$ hence these circles concur at $P,Q.$ Denote by $O_A,O_B,O_C$ centers of respective circles. By $$(\ell_A;\ell_B)=(AD;BE)=\measuredangle DAB+\measuredangle ABE=90^{\circ}+\measuredangle CBA+\measuredangle ADP+90^{\circ}+\measuredangle BAC+\measuredangle PEB=$$$$=\measuredangle BPA+\frac{\measuredangle AO_AP}{2}+\frac{\measuredangle PO_BB}{2}=\measuredangle BPA+\measuredangle APO_A+\measuredangle O_BPB=\measuredangle O_BPO_A=\measuredangle O_AQO_B$$and analogous equalities $Q$ is the Miquel point of complete quadrilateral on $\overline{O_AO_BO_C},$ $\ell_A,$ $\ell_B,$ $\ell_C.$ The conclusion follows.

Solved with: MrOreoJuice, Geometry06, Fakesolver19, KILLER1234, Pranav1056, SatisfiedMagma, Arifa Alam, trigocalc, youngmind_007, Executioner230607, Bahnhofstrasse($\text{he rickrolled us when we fell asleep}$) and L567($\text{who helped us in almost every single moment}$).

We start by noticing that $ADY_aT$ is an isosceles trapezium as $AD$ and $Y_a$ share the same perpendicular bisector. So $\measuredangle (PY_a,PA)=\measuredangle Y_aPA =\measuredangle DPT= 90^\circ+\measuredangle (PT,BC)$. Now, in a similar fashion, we have that $\measuredangle (PY_b,PB)=90^\circ+\measuredangle (PT,CA)$. As, $ABCP$ is cyclic, $\measuredangle (PA,PB)= \measuredangle (CA,BC)$. Now, we will move to the critical part. Notice the following for finish: \begin{align*} \measuredangle (PY_a,PY_b)&= \measuredangle (PA,PB)+ \measuredangle (PB,Y_bB)+\measuredangle (PY_a,PA) \\ &=\measuredangle (CA,CB)+ 90^\circ +\measuredangle (CA,PT)+ 90^\circ + \measuredangle (PT,BC) \qquad \text{(}\bigstar\text{)} \\ &= 0^\circ \end{align*}This all comes down to zero if you properly focus on the triangle formed by the angles in ($\bigstar$). As these points are now collinear, we are done by symmetry. $\blacksquare$

@above tbh I haven't still done the chase :joy: due to IGO :blobsweat: . I'll do it in this week uff too much hardwork

WLOG,Assume $AB<AC$.Let $H$ ,$Z$ and $T$ be the feet from $A$,$B$ nd $C$ onto the opp sides.Let $U$ be the orthocentre. Let $X$ and $Y$ be the midpoints of $AB$ and $AC$ respectively. Also let the perpendicular bisectors of $AD$ and $CF$ meet at $L$ , of $FC$ and $BE$ at $K$,and of $AD$ and $BE$ at $J$.Then let the centres of circles $(APD),(CFP) ,(BEP)$ be $M,N,O$ respectively. Now,let $AP$ meet $HD$ at $G$.Triangles $GPD$ and $GAH$ are homothetic.The midpoints of $DP$,$AH$ and $G$ are thus collinear,which implies $G$ lies on $XY$ and thus $GPD$ is isosceles .$PA=DH$ and thus $D$ lies on $(APD)$.Similar conditions follow for $T$ and $Z$. Now,$UA.UH=CU.UT=BU.UZ$,it follows that $U$ is the radical center of the circles $(APD),(CFP) ,(BEP)$,but $P$ is also the radical center ,implying that these circles are coaxial. Let the second intersection of these circles be $I$.Also,$O--M--N$ Now,$ITH+IHT=TCP+HAP=90-A+ACP+90-C+PAC=2B$ Now,$MIN=NIP-MIP=PIT+PHI-180=PTC+AHP-B$ $=PTC+AHP+AUC-180$. Also,$PADH$ and $PFTC$ are isc. traps. So,$PTC+AHP+AUC-180=AUC+FCT+DAH-180$ $=FCU-(CT,AD)=(FC,AD).$ And $MLN=ALN-ALM=ALF+90-LFC-90+LAD=(AD,CF)$ Thus,$MNLI$ is cyclic. Similalry, $MOIJ$ is cyclic. Now, $ILK=IMN=IJK$,thus $IJKL$ is cyclic,which finishes. Remark:The hours that I spent with the problem were the most obnoxious and unsavoury I have ever spent.The words that uttered out of my mouth after the completion of the angle chase were behench...

Eyed wrote: Let $P$ be a point on the circumcircle of acute triangle $ABC$. Let $D,E,F$ be the reflections of $P$ in the $A$-midline, $B$-midline, and $C$-midline. Let $\omega$ be the circumcircle of the triangle formed by the perpendicular bisectors of $AD, BE, CF$. Show that the circumcircles of $\triangle ADP, \triangle BEP, \triangle CFP,$ and $\omega$ share a common point. Extraordinary problem!! Solved with MathsCrazy No idea how many hours this took lol, but I think less than 4.5 hours. First of all let \(H_a, H_b, H_c\) be the foot of the \(A, B, C\) altitudes of triangle \(ABC\). First of all, \(PH_aAD\) is an isosceles trapezoid because the \(A\) midline bisects \(AH_a\) and \(PD\), and hence \(PH_aAD\) is cyclic. Similarly, \(PH_bBE\) and \(PH_cCF\) are cyclic. Note that the power of \(H\) wrt these three circles is equal because \[AH\cdot HH_a=BH\cdot HH_b=CH\cdot HH_c\]and so the three circles are concurrent at a point, say \(L\). We now do something more interesting. Let \(L_1,L_2,L_3\) be the reflections of \(L\) over the perpendicular bisectors of \(AD, BE, CF\) respectively. If we prove \(L_1,L_2,L_3\) collinear, we will be done because of Steiner. Infact we will prove something stronger, which is \(P\) belongs in this line. Now if we prove \(P,L_1,L_3\) collinear, the other results are analogous. We shall prove that \(\angle APL_1+\angle CPL_3=\angle APC\), if we do so, we are done. Now, \(\angle APL_1=\angle ADL_1=\angle LPD\). Analogously, \(\angle CPL_3=\angle LPF\), so we want to prove \(\angle APC=\angle FPD\) or \(\angle APD=\angle CPF\), which is true because of symmetry, so we are done! PS: Why is the title so strange? Where is fish here?

Let $M_aM_bM_c$ and $H_aH_bH_c$ be the medial and orthic triangle of $ABC$, and $H$ the orthocentre of $ABC$. Furthermore, let $Q$ be the point by a negative inversion in $H$ with power $-AH\cdot HH_a$. In other words, $Q$ belongs to the nine point circle $(H_aH_bH_cM_aM_bM_c)$ and is on the opposite side of $P$ wrt to $H$ on line $HP$. By symmetry wrt to the $A$-midline (from now on call the three midlines $m_a,m_b,m_c$), $H_a$ belongs to $(ADP)$, and so $QH\cdot HP=AH\cdot HH_a$, which implies that $Q$ also belongs to $(ADP)$, and similarly to $(BEP)$ and $(CFP)$. Therefore their circumcenters $O_a$ $O_b$ and $O_c$ lies on the same line $\ell$, which is the axis of $PQ$ (since all three circles pass through $P$ and $Q$). Furthermore call the three axes of $AD,BE,CF$ $\ell_a,\ell_b,\ell_c$ respectively and $\ell_b\cap \ell_c=X$ and cyclically for $Y,Z$. Finally, let the three axes of $H_aP,H_bP,H_cP$ be called $\ell_a',\ell_b',\ell_c'$. Let us now establish two lemmas: 1) $$\measuredangle (\ell_a,\ell)=\measuredangle (\ell_a,m_a)+\measuredangle (m_a,\ell)=\measuredangle (m_a,\ell_a')+\measuredangle (m_a,\ell)= (HH_a,H_aP)+\measuredangle (HH_a,PQ)=\measuredangle HH_aP+\measuredangle H_aHP$$where in the second to last equality we use the fact that $\measuredangle (\ell,m)=\measuredangle (\ell',m')$ if $\ell'\perp \ell$ and $m\perp m'$. 2) $\measuredangle (QO_a,\ell)=\measuredangle QAP=\measuredangle QH_aP$ where in the first equality we used the fact that $\measuredangle (QO_a,\ell)$ is half the central angle corresponding to arc $\overarc QP$ of $(AQPH_a)$ By the first lemma, we have $$\measuredangle O_bXO_c=\measuredangle (\ell_b,\ell_c)=\measuredangle (\ell_b,\ell)-\measuredangle (\ell_c,\ell)=(\measuredangle HH_bP+\measuredangle H_bHP)-(\measuredangle HH_cP+\measuredangle H_cHP)=(\measuredangle (H_bH,HP)-\measuredangle (H_cH,HP))+(\measuredangle HH_bP+\measuredangle PH_cH)=\measuredangle H_bHH_c+2\pi-(\measuredangle H_bPH_c+H_cHH_b)=2H_bHH_c-\measuredangle H_bPH_c=2(\pi-\measuredangle A)-H_bPH_c=-2\measuredangle A-\measuredangle H_bPH_c$$Instead, by the second lemma, $$\measuredangle O_bQO_c=\measuredangle QO_b,\ell)-\measuredangle (QO_c,\ell)=\measuredangle QH_bP-\measuredangle QH_cP=\measuredangle QH_bP+\measuredangle PH_cQ=2\pi-(\measuredangle H_bPH_c+\measuredangle H_cQH_b)=-\measuredangle H_bPH_c+\measuredangle H_bH_aH_c=\pi-2\measuredangle A-\measuredangle H_bPH_c=-2\measuredangle A-\measuredangle H_bPH_c$$Since these two expressions are the same, it follows that $O_bO_cXQ$ is cyclic, and similarly for the other two quadrilaterals. Therefore, since $YZ\cap O_bO_c=O_a$ and $Q$ belongs to both $(O_aO_bZ)$ and $(O_aO_cY)$, it follows that $Q$ is the miquel point of $O_bO_cYZ$, which means that $Q\in (XYZ)$, or in other words $\omega$ also passes through $Q$, as wanted.

Solved with apotosaurus and sixoneeight. This took over 4 hours, but it was totally worth it! Claim: The circumcircles of $\triangle ADP$, $\triangle BEP$, and $\triangle CFP$ are coaxial. Proof. We will show that the orthocenter $H$ of $\triangle ABC$ has equal power with respect to the three circles, which suffices since $H \neq P$. Let $h_A$, $h_B$, $h_C$ be the foot of the altitudes from $A$, $B$, $C$ in $\triangle ABC$. Note that $h_A \in (ADP)$, and analogous concyclicities hold for $B$, $C$, as $Ph_AAD$ is an isosceles trapezoid. Thus the powers of $H$ with respect to $(ADP)$, $(BEP)$, and $(CFP)$ are \[ AH \cdot Hh_A; BH \cdot Hh_B; CH \cdot Hh_C \]respectively, and elementary calculations show that these quantities are equal, finishing. By the above claim, the three circumcircles are concurrent at some point $Q \neq P$. Let $\Gamma$ denote the nine point circle of $\triangle ABC$. Claim: $Q$ lies on $\Gamma$. Proof. Let $Q'$ be the intersection of $\overline{PH}$ and $\Gamma$ different from $Q$. Noting the homothety of scale factor 2 centered at $H$ that maps $\Gamma$ to $(ABC)$, it's easy to see that $Q'$ is the midpoint of $PH$. Thus the PoP equation $AH \cdot Hh_A = QH \cdot HP$ rewrites as $(AH/2)\cdot Hh_A = QH \cdot (HP/2)$, so $Q$ is concyclic with $Q'$, the midpoint of $AH$, and $h_A$, which means that $Q$ lies on $\Gamma$, as required. Let $O_A$, $O_B$, $O_C$ be the circumcenters of $\triangle ADP$, $\triangle BEP$, $\triangle CFP$ respectively. As the circumcircles are coaxial, $O_A$, $O_B$, and $O_C$ are collinear. Let $X$ be the intersection of the perpendicular bisectors of $\overline{BE}$ and $\overline{CF}$, and similarly define $Y$, $Z$. Claim: $Q$, $X$, $O_B$, and $O_C$ are concylic. Proof. We omit this proof since I am lazy. However, this is nothing other than an angle chase which has steps that are rather straightforward. Using cyclic variants of the above claim, $Q$ is the Miquel point of the complete quadrilateral induced by the three perpendicular bisectors given in the problem statement and $\overline{O_AO_BO_C}$, and this implies $Q \in \omega$, as desired.

[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(21.5cm); real labelscalefactor = 1; /* changes label-to-point distance */ pen dps = linewidth(0.4) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = 1.5, xmax = 23., ymin = -17.16082195484213, ymax = 6.403086819243258; /* image dimensions */ pen zzttqq = rgb(0.6,0.2,0.); pen ffwwzz = rgb(1.,0.4,0.6); pen ffqqtt = rgb(1.,0.,0.2); pen zzwwff = rgb(0.6,0.4,1.); pen zzccqq = rgb(0.6,0.8,0.); draw((8.,5.)--(4.,-7.)--(16.,-7.)--cycle, linewidth(1.) + zzttqq); /* draw figures */ draw((8.,5.)--(4.,-7.), linewidth(1.) + zzttqq); draw((4.,-7.)--(16.,-7.), linewidth(1.) + zzttqq); draw((16.,-7.)--(8.,5.), linewidth(1.) + zzttqq); draw(circle((10.,-2.3333333333333335), 7.6011695006609195), linewidth(1.)); draw(circle((11.25859729693003,-1.), 6.827770964491974), linewidth(1.) + ffwwzz); draw(circle((11.640433313973183,-9.46064997095977), 8.026893515230881), linewidth(1) + ffqqtt); draw(circle((11.34700753997859,-2.958977380064229), 6.162808016468519), linewidth(1.) + zzwwff); draw((xmin, 3.051341078506704*xmin-35.353820418487146)--(xmax, 3.051341078506704*xmax-35.353820418487146), linewidth(1.) + zzccqq); /* line */ draw((xmin, 0.050255999895993386*xmin-10.045651586376124)--(xmax, 0.050255999895993386*xmax-10.045651586376124), linewidth(1.) + zzccqq); /* line */ draw((xmin, 1.455274712652225*xmin-19.4719905172692)--(xmax, 1.455274712652225*xmax-19.4719905172692), linewidth(1.) + zzccqq); /* line */ draw(circle((7.423363104467918,-6.726919991206829), 3.0659345971159127), linewidth(1.) + zzccqq); draw(circle((9.,-3.3333333333333335), 3.8005847503304597), linewidth(1.)); draw((xmin, -22.157810141843235*xmin + 248.46586136884505)--(xmax, -22.157810141843235*xmax + 248.46586136884505), linewidth(1.)); /* line */ draw((8.,5.)--(8.,-7.), linewidth(1.)); draw((5.367817773118896,-4.452125858859596)--(17.436390166623063,-3.907461367366656), linewidth(1.)); draw((4.,-7.)--(12.307692307692294,-1.4615384615384392), linewidth(1.)); draw((5.2,-3.4)--(16.,-7.), linewidth(1.)); /* dots and labels */ dot((8.,5.),dotstyle); label("$A$", (8.065570049789427,5.174886064078719), NE * labelscalefactor); dot((4.,-7.),dotstyle); label("$B$", (4.065144732967769,-6.826389886386216), NE * labelscalefactor); dot((16.,-7.),dotstyle); label("$C$", (16.066420683432742,-6.826389886386216), NE * labelscalefactor); dot((17.436390166623063,-3.907461367366656),dotstyle); label("$P$", (17.505170139482637,-3.7383422734010865), NE * labelscalefactor); dot((17.436390166623063,1.9074613673666603),dotstyle); label("$D$", (17.505170139482637,2.0868384510935893), NE * labelscalefactor); dot((4.285198890406509,-12.674922218177695),dotstyle); label("$E$", (4.363422059222016,-12.49365908521688), NE * labelscalefactor); dot((5.906411046281557,-0.06413499391948552),dotstyle); label("$F$", (5.977628766009702,0.10417151775654603), NE * labelscalefactor); dot((6.709048673381286,-9.70848163694446),linewidth(4.pt) + dotstyle); label("$X$", (6.784732119403545,-9.563522997895763), NE * labelscalefactor); dot((8.433006119182366,-9.621842431727583),linewidth(4.pt) + dotstyle); label("$Z$", (8.504213176633908,-9.475794372526867), NE * labelscalefactor); dot((9.950607469079369,-4.991123091989649),linewidth(4.pt) + dotstyle); label("$Y$", (10.013145532978918,-4.8437229530491726), NE * labelscalefactor); dot((5.367817773118896,-4.452125858859596),linewidth(4.pt) + dotstyle); label("$J$", (5.433711288722547,-4.317351200835798), NE * labelscalefactor); dot((8.,-7.),linewidth(4.pt) + dotstyle); label("$A_0$", (8.065570049789427,-6.861481336533775), NE * labelscalefactor); dot((5.2,-3.4),linewidth(4.pt) + dotstyle); label("$C_0$", (5.275799763058534,-3.26460769640905), NE * labelscalefactor); dot((12.307692307692294,-1.4615384615384392),linewidth(4.pt) + dotstyle); label("$B_0$", (12.381818417939108,-1.3170322132195647), NE * labelscalefactor); dot((8.,-4.333333333333334),linewidth(4.pt) + dotstyle); label("$H$", (8.065570049789427,-4.194531125319345), NE * labelscalefactor); dot((10.487540294650579,-6.830713540893745),linewidth(4.pt) + dotstyle); label("$G$", (10.557063010266072,-6.6860240857959825), NE * labelscalefactor); dot((11.25859729693003,-1.),linewidth(4.pt) + dotstyle); label("$O_A$", (11.329074913512358,-0.8608433613013069), NE * labelscalefactor); dot((11.640433313973183,-9.460649970959771),linewidth(4.pt) + dotstyle); label("$O_B$", (11.7150808651355,-9.317882846862855), NE * labelscalefactor); dot((11.347007539978591,-2.9589773800642294),linewidth(4.pt) + dotstyle); label("$O_C$", (11.416803538881252,-2.825964569564571), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Let $J$ be the intersection of $PH$ past $H$ and the nine point circle of $(ABC)$. Also, let $XYZ$ be the triangle formed and $A_0,B_0,C_0$ be the feet from the altitudes. Claim: $J$ lies on $(ADP)$, $(BEP)$, $CFP$. Proof. By symmetry $A_0$ lies on $(ADP)$. Then, power of a point and scaling by $2$ implies the result. $\blacksquare$ Let $O_A,O_B,O_C$ be the circumcenters of the three triangles. Claim: $J$ is the miquel point of the quadrilateral formed by lines $AX$, $BY$, $CZ$, $O_AO_BO_C$. Proof. It suffices to show that $O_AO_CYJ$ is cyclic as symmetry will finish. This can be done by noting \[\measuredangle O_CJO_A=\measuredangle JO_AO_C+\measuredangle O_AO_CJ =\measuredangle JAP+\measuredangle PCJ\]from the circumcenters. Now, from the quadrilateral, $\measuredangle JAP+\measuredangle PCJ=\measuredangle APC+\measuredangle CJA$. Furthermore, $\measuredangle CJA=\measuredangle CJP+\measuredangle PJA=\measuredangle FCC_0+\measuredangle A_0AD$ and adding this back to $\measuredangle APC=\measuredangle CHA$ we get $\measuredangle (CF,AD)$. Then $\measuredangle(O_CY,CF)=90^{\circ}=\measuredangle(O_AY,AD)$ so $\measuredangle(CF,AD)=\measuredangle(O_CY,O_A,Y)=\measuredangle O_CYO_A$ as desired. $\blacksquare$ Then this miquel point lies on $(XYZ)$ as desired.

Let $\omega_A$, $\omega_B$, $\omega_C$ by $(ADP)$, $(BEP)$, $(CFP)$. Note that since the $A$-midline is a diameter of $\omega_A$, the reflection of $A$ over the $A$-midline is on $\omega_A$. Indeed, the foot of altitude from $A$ is on $\omega_A$. Similarly, the feet of altitudes from $B$ and $C$ are on $\omega_B$ and $\omega_C$, respectively. Now, let $H$ be orthocenter, and $H_A$, $H_B$, $H_C$ be the feet of altitudes from $A$, $B$, and $C$ respectively. Then $HA\cdot HH_A=HB\cdot HH_B=HC\cdot HH_C$, so \[\text{Pow}_{\omega_A}(H)=\text{Pow}_{\omega_B}(H)=\text{Pow}_{\omega_C}(H)\]which implies $\omega_A$, $\omega_B$, and $\omega_C$ are coaxial. Let $Q\neq P$ be their concurrency point. $~$ Let the perpendicular bisector of $AD$ intersect the perpendicular bisectors of $BE$ and $CF$ at $Z$ and $Y$ respectively. Let the perpendicular bisectors of $BE$ and $CF$ intersect at $X$. Note that \begin{align*} \measuredangle QO_AY &= \measuredangle QO_AA - \measuredangle YO_AA \\ &= 2 \measuredangle QPA - 90^\circ - \measuredangle DAO_A \\ &= 2 \measuredangle HPA - \measuredangle DPA \\ &= 2 \measuredangle HPA - \measuredangle HAP \\ \end{align*}Similarly, $\angle QO_CY = 2\measuredangle HPC - \measuredangle HCP$. The rest is trivial triangle and circumcircle angle chasing. $~$ Hence, $Q$ is on $(O_CO_AY)$. Similarly, it is on $(O_BO_CX)$ and $(O_AO_BZ)$ so by Miquel point, $Q$ is on $(XYZ)$ as desired.

Let $l_A$ be the perpendicular bisector of $AD$ and define $l_B, l_C$ analogously. Let $X = l_B \cap l_C, Y = l_A \cap l_C, Z = l_A \cap l_B$. Claim: $\omega_A, \omega_B, \omega_C$ concur at point $Q$ other than $P$. Proof. Let $AA_1, BB_1, CC_1$ be the altitudes of $\triangle ABC$ and let $H$ be the orthocenter of $\triangle ABC$. Note that $A_1 \in \omega_A$, so the power of $H$ wrt $\omega_A$ is $AH \cdot HA_1$. Similarly the power of $H$ wrt $\omega_B, \omega_C$ are $BH \cdot HB_1, CH \cdot HC_1$. Hence $HP$ is the radical axis of $\omega_A, \omega_B, \omega_C$, so we're done. $\blacksquare$ By Simson's theorem, it suffices to show that to prove that the reflections $Q_A, Q_B, Q_C$ of Q in lines $l_A, l_B, l_C$, respectively, are collinear. We'll prove that $P, Q_A, Q_B, Q_C$ are collinear, completing the proof. Thus we only need to consider the following claim: Claim: $P, Q_A, Q_B$ are collinear. Proof. We'll prove that $\measuredangle(PQ_A, PQ) = \measuredangle(PQ_B, PQ)$. Note that $\measuredangle(PQ_A, PQ) = \measuredangle(PQ_A, PA) + \measuredangle(PA, PQ) = \measuredangle(DQ_A, DA) + \measuredangle(PA, PQ) = \measuredangle(Q_AD, Q_AQ) + \measuredangle(PA, PQ) = \measuredangle(PD, PQ) + \measuredangle(PA, PQ) = \measuredangle(BC, PQ) + 90^{\circ} + \measuredangle(PA, PQ)$. Similarly $\measuredangle(PQ_B, PQ) = \measuredangle(AC, PQ) + 90^{\circ} + \measuredangle(PB, PQ)$. So we only need to prove that $\measuredangle(BC, PQ) + \measuredangle(PA, PQ) = \measuredangle(AC, PQ) + \measuredangle(PB, PQ)$ and $\measuredangle(BC, PQ) + \measuredangle(PA, PQ) = \measuredangle(BC, AC) + \measuredangle(AC, PQ) + \measuredangle(PA, PQ) = \measuredangle(BC, AC) + \measuredangle(AC, PQ) + \measuredangle(PA, PB) + \measuredangle(PB, PQ) = \measuredangle(AC, PQ) + \measuredangle(PB, PQ)$. $\blacksquare$ Therefore $P, Q_A, A_B$ are collinear and we're done. $\blacksquare$ Remark: This problem is very hard. Firstly I let $O_A, O_B, O_C$ be the center of $\omega_A, \omega_B, \omega_C$, respectively and trying to show $Q$ is the miquel point of the quadrilateral $O_BYZO_A$. But I couldn't prove that . Then I came with a nasty idea: Showing the projections of $Q$ onto the sidelines $l_A, l_B, l_C$ are collinear. That's how I came with the above solution.

Cool Problem

Let $\gamma$ be the nine point circle Let $L,O,K$ be the foot of altitudes Let $M,N,R$ be the midpoints Let $S,T,U$ be the centers. Let $Q$ be the orthocenter Let $G,H,I$ be the intersections of perpendicular bisectors Let $J = \gamma \cap QP$ Claim 1: $\gamma,(ADP),(BEP),(CFP)$ concurr

Then notice $S-T-U$ Claim 2: $J$ is the miquel point of $TGISUH$

Nice problem! We divide the problem into two parts. Part(i):We’ll show that $(ADP),(BEP),(CFP)$ have a common point $T\neq P$ Define $A_1,B_1,C_1$ to be foot of altitudes from $A,B,C$ to its opposite sides and denote $H$ by the orthocenter of $ABC$ Note that $(ADP)=(APA_1),(BEP)=(BPB_1),(CFP)=(CPC_1)$ Observe that $Pow(H,(APA_1))=AH \cdot HA_1=BH\cdot HB_1=Pow(H,(BPB_1))=CH\cdot HC_1=Pow(H,(CPC_1))$ which means H has equal power wrt. these three circles. We have that $P$is the common point of these three circles, they must be coaxial circles which has $HP$ as the common radical axis. Thus, $(ADP),(BEP),(CFP)$ have a common point $T\neq P$ s.t. $T\in HP$. $\square$ _______________________________________________________________________________________________________________ Let $l_1,l_2,l_3$ the perpendicular bisectors of $AD,BE,CF$, respectively. Define $A’=l_2\cap l_3, B’=l_3\cap l_1, C’=l_1\cap l_2$ Define $O_a,O_b,O_c$ be the centers of $(ADP),(BEP),(CFP)$, respectively. By part(i) it is obvious that $O_a,O_b,O_c$ are collinear. Observe that $O_a\in B’C’$, $O_b\in C’A’$ and $O_c\in A’B’$. _______________________________________________________________________________________________________________ Part(ii): We’ll show that $(A’B’C’)$ pass through T First, we prove the following claim. Claim: $T\in (O_aO_cB’)$(Similary, $T\in(O_cO_bA’),(O_bO_aC’)$) Proof: We proceed by angle chasing $\measuredangle{O_aB’O_c}=\measuredangle{(l_1,AA_1)}+\measuredangle{AHC}+\measuredangle{(CC_1,l_3)}=90^\circ-\measuredangle{A_1AD}+\measuredangle{CBA}+90^\circ-\measuredangle{FCC_1}=\measuredangle{DAA_1}+\measuredangle{CBA}+\measuredangle{C_1CF}=\measuredangle{AA_1P}+\measuredangle{CBA}+\measuredangle{PC_1C}=\measuredangle{ATP}+\measuredangle{TPA}+\measuredangle{CPT}+\measuredangle{PTC}=\measuredangle{TAP}+\measuredangle{PCT}=90^\circ+\measuredangle{TAP}+90^\circ+\measuredangle{PCT}=\measuredangle{O_aTP}+\measuredangle{PTO_c}=\measuredangle{O_aTO_c} \square$ By the claim we can conclude that $T$ is the Miquel point formed by {$l_1,l_2,l_3,\overline{O_aO_bO_c}$} So, $T\in(A’B’C’)$, as desired $\blacksquare$ _______________________________________________________________________________________________________________ Remark : First, it is natural to consider only $(ADP),(BEP),(CFP)$ which we can view those three circle as $(APA_1),(BPB_1),(CPC_1)$, respectively. Then we just consider power of $H$ wrt. those three circles. Finally, The key step is to claim that the intersection point is the Miquel point which has the same idea as IMO 2011/6 (In 2011/6 the tangency point is the Miquel point of complete quadrilateral formed by lines $l_a,l_b,l_c$ and the Steiner line of the tangency point) Final Remark : When you see a problem which introduce a triangle forms by three lines it is a good idea to consider Miquel point of some complete quadrilateral.

Longest angle chase I've ever done Let $\Gamma$ be the circumcircle of $\triangle ABC$, and let $\gamma$ be its nine-point circle. Further let $H_A$, $H_B$, $H_C$ be the feet of the perpendiculars from each vertex to the opposite side, and let $H$ be the orthocenter of $\triangle ABC$. Let $A'$, $B'$, $C'$, and $P'$ be the midpoints of $\overline{AH}$, $\overline{BH}$, $\overline{CH}$, and $\overline{PH}$, which all lie on $\gamma$ by a homothety at $H$. Let $Q = HP' \cap \gamma \neq P'$, and let $\ell_A$, $\ell_B$, and $\ell_C$ be the three midlines of $\triangle ABC$ parallel to $BC$, $CA$, and $AB$, respectively. Note that $\ell_A$, $\ell_B$, and $\ell_C$ are the perpendicular bisectors of segments $\overline{AH_A}$, $\overline{BH_B}$, and $\overline{CH_C}$, respectively. Therefore $ADPH_A$, $BEPH_B$, and $CFPH_C$ are (possibly self-intersecting) isosceles trapeziums, which implies that they are cyclic. Claim: $(ADP)$, $(BEP)$, and $(CFP)$ pass through $Q$. Proof. Note \[HA \cdot HH_A = 2 HA' \cdot HH_A = 2 HP' \cdot HQ = HP \cdot HQ,\]and cyclic variations thereof also hold. $\blacksquare$ Now let $m$, $m_A$, $m_B$, and $m_C$ be the perpendicular bisectors of segments $\overline{PQ}$, $\overline{AD}$, $\overline{BE}$, and $\overline{CF}$. Let $X'$, $Y'$, and $Z'$ be the circumcenters of cyclic pentagons $ADPH_AQ$, $BEPH_BQ$, and $CFPH_CQ$. Clearly $X'$ is the intersection of the perpendicular bisectors of $\overline{AD}$, $\overline{AH_A}$, and $\overline{PQ}$, and therefore $\ell_A$, $m_A$, and $m$ concur at $X'$. Similarly, $\ell_B$, $m_B$, and $m$ concur at $Y'$, and $\ell_C$, $m_C$, and $m$ concur at $Z'$. Now we contend the following, which solves the problem: Claim: $Q$ is the Miquel point of the complete quadrilateral formed by lines $m$, $m_A$, $m_B$, and $m_C$. Proof. By symmetry it suffices to prove that $Q$ lies on the circumcircle of the triangle determined by lines $m$, $m_B$, and $m_C$, which is $(XY'Z')$. Now \begin{align*} \measuredangle Y'XZ' &= \measuredangle(m_B, m_C) \\ &= - \measuredangle(CF, BE) \\ &= \measuredangle(BE, CF) \\ &= \measuredangle(BE, \ell_B) + \measuredangle(\ell_B, \ell_C) + \measuredangle(\ell_C, CF) \\ &= \measuredangle(\ell_B, H_B P) + \measuredangle CAB + \measuredangle (H_C P, \ell_C) \text{ (reflection over }\ell_B\text{ and }\ell_C\text{)}\\ &= \measuredangle H_C PH_B - \measuredangle(\ell_C, \ell_B) + \measuredangle CAB \\ &= \measuredangle H_C PH_B + 2 \measuredangle CAB. \end{align*}However, \begin{align*} \measuredangle Y'QZ' &= \measuredangle Y'QP + \measuredangle PQZ' \\ &= 90^\circ - \measuredangle PH_B Q + 90^\circ - \measuredangle QH_C P \\ &= \measuredangle QH_B P + \measuredangle PH_C Q \\ &= - \measuredangle H_B PH_C - \measuredangle H_C QH_B \text{ (angle sum in a quadrilateral)} \\ &= \measuredangle H_C PH_B + \measuredangle H_B QH_C \\ &= \measuredangle H_C PH_B + \measuredangle H_B A'H_C \\ &= \measuredangle H_C PH_B + 2 \measuredangle H_B AH_C \\ &= \measuredangle H_C PH_B + 2 \measuredangle CAB, \end{align*}as $A'$ is the circumcenter of $\triangle AH_B H_C$. $\blacksquare$

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Let $\omega_a, \omega_b$ and $\omega_c$ denote circles $\triangle ADP, \triangle BEP, \triangle CFP$, let $H_1,H_2.H_3$ denote the $A,B,C$ feet of altitude, and let $H$ denote the orthocenter of $\triangle ABC$. Firstly, we want to show that $\omega_a, \omega_b$ and $\omega_c$ concur at the same point. We claim that the $N_9$ circle pass through that point. It's easy to show that $H_1 \in \omega_a$ similarly for $H_2,H_3$. Now by power of a point we get that $HH_1\cdot HA= HH_2 \cdot HB = HH_3 \cdot HC$ hence $\omega_a, \omega_b$ and $\omega_c$ concur at the same point, call it $Q$. Claim: We claim that $Q$ is the miquel point of $JIO_2O_3$. Proof: Huge angle chase $\blacksquare$ This implies that $Q \in \omega $ hence we are done.