First, by Menelaus theorem, we see that $X = \overline{HI} \cap \overline{AB}$ and $Y = \overline{H'I'} \cap \overline{AB}$ are symmetric around the point $D$. Let $T = \overline{CC} \cap \overline{DD}$. [asy][asy] size(10cm); pair C = dir(50); pair D = dir(270); pair H = dir(205); pair A = extension(C, H, D, D+dir(0)); pair Y = 2*C*D/(C+D); pair X = 2*D-Y; pair B = 2*D-A; pair I = extension(X, H, C, B); pair F = midpoint(A--C); pair G = midpoint(B--C); pair Hp = A+C-H; pair Ip = B+C-I; pair M = extension(F, G, Hp, Ip); pair N = extension(C, D, F, G); filldraw(unitcircle, invisible, blue); filldraw(A--B--C--cycle, invisible, blue); draw(X--A, blue); draw(Y--B, blue); draw(F--G, dotted+blue); draw(C--D, dotted+blue); draw(X--I, red); draw(Hp--Y--C, red); dot("$C$", C, dir(C)); dot("$D$", D, dir(D)); dot("$H$", H, dir(160)); dot("$A$", A, dir(A)); dot("$Y$", Y, dir(Y)); dot("$X$", X, dir(X)); dot("$B$", B, dir(B)); dot("$I$", I, dir(I)); dot("$F$", F, dir(F)); dot("$G$", G, dir(G)); dot("$H'$", Hp, dir(Hp)); dot("$I'$", Ip, dir(20)); dot("$M$", M, dir(300)); dot("$N$", N, dir(300)); /* Source generated by TSQ: !size(10cm); C = dir 50 D = dir 270 H = dir 205 R160 A = extension C H D D+dir(0) Y = 2*C*D/(C+D) X = 2*D-Y B = 2*D-A I = extension X H C B F = midpoint A--C G = midpoint B--C H' = A+C-H I' = B+C-I R20 M = extension F G Hp Ip R300 N = extension C D F G R300 unitcircle 0.1 lightblue / blue A--B--C--cycle 0.1 lightcyan / blue X--A blue Y--B blue F--G dotted blue C--D dotted blue X--I red Hp--Y--C red */ [/asy][/asy] We now present the key claim that lets us eliminate most of the points, and give three proofs. Claim: The points $T$ and $Y$ coincide. Proof. [First proof using cross ratios, Colin Tang] Let $\overline{XE}$ be the second tangent from $E$ to $\Gamma$. Then $(ED;HI) = -1$. Projecting from $C$ onto line $AB$ gives $-1 = (A,B; D, \overline{CE} \cap \overline{AB})$ which implies $\overline{CE} \parallel \overline{AB}$. Then reflecting the whole picture about the perpendicular bisector of $\overline{AB}$ shows that $\overline{CY}$ is tangent as well. $\blacksquare$ Proof. [Second proof using involutions, Michael Kural] Apply Desargues involution theorem on $CCHI$ with line $AB$. We get an involutive pairing which sends $D$ to itself, swaps $A$ and $B$, and swaps $X$ and $T$. The first two are enough to imply it is reflection across $D$, so $T = Y$. $\blacksquare$ Proof. [Third proof using homography, Evan Chen] Let $\infty$ be the harmonic conjugate of $D$ to $\overline{AB}$. It suffices then to prove $(XT;D\infty) = -1$. (This lets us forget about the midpoints; same trick as with Butterfly.) Take a homography sending $CHDI$ to a rectangle (and hence $T$ to infinity) while preserving $\Gamma$. [asy][asy] size(7cm); pair C = dir(90); pair D = -C; pair H = dir(200); pair I = -H; draw(C--H--D--I--cycle, blue); filldraw(unitcircle, invisible, blue); pair A = extension(C, H, D, D+dir(0)); pair B = extension(C, I, A, D); draw(H--A--B--I, blue); pair X = extension(H, I, A, B); pair infty = 2*X-D; draw(infty--C, heavygreen); draw(X--I, heavygreen); draw(infty--A, blue); draw((C+2*dir(0))--(C-3*dir(0)), blue); draw(C--D, blue); dot("$C$", C, dir(C)); dot("$D$", D, dir(D)); dot("$H$", H, dir(H)); dot("$I$", I, dir(I)); dot("$A$", A, dir(-90)); dot("$B$", B, dir(-90)); dot("$X$", X, dir(-90)); dot("$\infty$", infty, dir(-90)); /* Source generated by TSQ: !size(7cm); C = dir 90 D = -C H = dir 200 I = -H C--H--D--I--cycle blue unitcircle 0.1 lightblue / blue A = extension C H D D+dir(0) R-90 B = extension C I A D R-90 H--A--B--I blue X = extension H I A B R-90 \infty = 2*X-D R-90 infty--C heavygreen X--I heavygreen infty--A blue (C+2*dir(0))--(C-3*dir(0)) blue C--D blue */ [/asy][/asy] Observe that now $\overline{HI} \parallel \overline{C\infty}$, thus $X$ is the midpoint of $\overline{D\infty}$. So $(XT;D\infty)=-1$ as needed. $\blacksquare$ With this new information, we may now delete points $A$, $B$, $F$, $G$, $H$, $I$, $H'$, $I'$, $X$ from the picture, retaining only the fact that $\overline{NM} \parallel \overline{DY}$, where $N = \overline{FG} \cap \overline{CD}$ is the midpoint of $\overline{CD}$. [asy][asy] size(9cm); pair C = dir(70); pair D = conj(C); pair Y = 2*C*D/(C+D); pair N = midpoint(C--D); pair O = origin; filldraw(unitcircle, invisible, blue); pair K = dir(130); pair L = -K+2*foot(O, K, Y); pair M = extension(N, N+D-Y, K, L); pair T = 2*K*L/(K+L); pair Q = extension(C, D, M, Y); pair P = -C+2*foot(O, C, M); draw(C--M, red); draw(T--Y, red); draw(O--Q, red); draw(C--K--D--L--cycle, heavygreen); draw(C--D, blue); draw(N--M, heavycyan); draw(D--Y, heavycyan); draw(C--Y, blue); draw(O--Y, blue); draw(K--Y, blue); draw(L--T--K, orange+dotted); draw(T--C, blue+dotted); dot("$C$", C, dir(30)); dot("$D$", D, dir(D)); dot("$Y$", Y, dir(Y)); dot("$N$", N, dir(325)); dot("$O$", O, dir(315)); dot("$K$", K, dir(K)); dot("$L$", L, dir(L)); dot("$M$", M, dir(M)); dot(T); dot("$Q$", Q, dir(Q)); dot("$P$", P, dir(P)); /* Source generated by TSQ: !size(10cm); C = dir 70 R30 D = conj(C) Y = 2*C*D/(C+D) N = midpoint C--D R325 O = origin R315 unitcircle 0.1 lightcyan / blue K = dir 130 L = -K+2*foot O K Y M = extension N N+D-Y K L T .= 2*K*L/(K+L) Q = extension C D M Y P = -C+2*foot O C M C--M red T--Y red O--Q red C--K--D--L--cycle heavygreen C--D blue N--M heavycyan D--Y heavycyan C--Y blue O--Y blue K--Y blue L--T--K orange dotted T--C blue dotted */ [/asy][/asy] Let line $\overline{QMY}$ meet $\Gamma$ at $K$, $L$. Then we need only show the polar of $Q$ (to $\Gamma$) is parallel to $\overline{CM}$. At this point there are four non-Euclidean conditions: The conic $\Gamma$ is a circle. $N$ is the midpoint of $\overline{CD}$; that is, the harmonic conjugate $\infty_1$ of $N$ with respect to $CD$ is at infinity. We want $\overline{NM} \parallel \overline{YD}$; that is, $\infty_2 = \overline{NM} \cap \overline{YD}$ should be at infinity. We want $\overline{CM}$ parallel to the $Q$-polar; that is, the intersection of $\overline{CM}$ with the $Q$-polar should be at infinity. To eliminate these we collate them together projectively: we just want $\overline{\infty_1\infty_2}$ and $\overline{CM}$ to be concurrent with the polar of $Q$. Now take a homography sending $CKDL$ to a square centered at $Q$, while preserving the circle $\Gamma$. The points $N$ and $M$ become arbitrary on $\overline{CD}$ and $\overline{KL}$, while $\infty_2 = \overline{NM} \cap \overline{DD}$, and the polar of $Q$ is mapped to the line at infinity. So we wish to show $\overline{\infty_1\infty_2} \parallel \overline{CM}$ in the new picture. [asy][asy] size(6cm); pair C = dir(90); pair K = dir(180); pair D = dir(-90); pair L = dir(0); pair Q = origin; pair M = 1.4*L; pair N = 0.6*D; pair _infty_2 = extension(M, N, D, D+K-L); pair _infty_1 = 1/conj(N); filldraw(unitcircle, invisible, blue); draw(C--_infty_1, blue); draw(K--M, blue); draw(C--M, red); draw(C--K--D--L--cycle, heavygreen); draw(_infty_2--_infty_1, red); draw(M--_infty_2--D, blue); dot("$C$", C, dir(C)); dot("$K$", K, dir(K)); dot("$D$", D, dir(315)); dot("$L$", L, dir(45)); dot("$Q$", Q, dir(45)); dot("$M$", M, dir(45)); dot("$N$", N, dir(315)); dot("$\infty_2$", _infty_2, dir(_infty_2)); dot("$\infty_1$", _infty_1, dir(_infty_1)); /* Source generated by TSQ: !size(6cm); C = dir 90 K = dir 180 D = dir -90 R315 L = dir 0 R45 Q = origin R45 M = 1.4*L R45 N = 0.6*D R315 \infty_2 = extension M N D D+K-L \infty_1 = 1/conj(N) unitcircle 0.1 lightcyan / blue C--infty_1 blue K--M blue C--M red C--K--D--L--cycle heavygreen infty_2--infty_1 red M--infty_2--D blue */ [/asy][/asy] Notice that \[ \frac{N\infty_2}{NM} = \frac{ND}{NQ} = \frac{N\infty_1}{NC} \]and so we're done.

It was Romani TST Day 4 Problem 2. See https://www.facebook.com/ssmr.ro/ for solution.

Since $ AH $ $ = $ $ CH', $ $ BI $ $ = $ $ CI', $ so $ CH' $ $ \cdot $ $ CF $ $ = $ $ CI' $ $ \cdot $ $ CG $ $ \Longrightarrow $ $ F, $ $ G, $ $ H', $ $ I' $ are concyclic. Let $ DF, $ $ DG $ cuts $ \Gamma $ again at $ Y, $ $ Z, $ respectively From $ DF $ $ \cdot $ $ YF $ $ = $ $ CF $ $ \cdot $ $ HF $ $ = $ $ AF $ $ \cdot $ $ H'F $ $ \Longrightarrow $ $ Y $ $ \in $ $ \odot (ADH'), $ so $ AB, $ $ H'Y $ are antiparallel WRT $ \angle AFD $ $ \Longrightarrow $ $ YH' $ $ \parallel $ $ H'I' $ $ \Longrightarrow $ $ Y $ $ \in $ $ H'I'. $ Analogously, we can prove $ Z $ lies on $ \odot (BDI') $ and $ H'I'. $ Obviously, $ FG, $ $ YZ $ are antiparallel WRT $ \angle GDF, $ so $ F, $ $ G, $ $ Y, $ $ Z $ are concyclic $ \Longrightarrow $ $ MC $ $ \cdot $ $ MP $ $ = $ $ MF $ $ \cdot $ $ MG $ $ = $ $ MH' $ $ \cdot $ $ MI' $ $ \Longrightarrow $ $ C, $ $ H', $ $ I', $ $ P $ are concyclic. Since $ \measuredangle H'CQ $ $ = $ $ \measuredangle ZDC $ $ = $ $ \measuredangle QYC, $ so $ {QC}^2 $ $ = $ $ QH' $ $ \cdot $ $ QY. $ Similarly, we can prove $ {QC}^2 $ $ = $ $ QI' $ $ \cdot $ $ QZ, $ so perform the inversion with center $ Q $ and radius $ QC $ we conclude that $ QC$ $ = $ $ QP. $

Let $O$ be the center of $\Gamma$ $X=CD\cap FG,Y=AB\cap H'I'$ Since $CH'\cdot CA=AH\cdot AC=AD^2=BD^2=BI\cdot BC=CI'\cdot CB$ So $A,B,I',H'$ are concyclic $\Rightarrow YC^2=YB\cdot YA+CI'\cdot CB=YD^2-BD^2+BD^2=YD^2$ $\Rightarrow YC$ is tangent to $\Gamma$ Since $M(D,Y;CP\cap AB,\infty_{AB})=(D,Q;C,X)$ So $(CD,CY;CP,C\parallel AB)=(OD,OQ;OC,OX)=(OX,OC;OQ,OD)$ $\Rightarrow OQ\perp CP\Rightarrow CQ=QP$

First we claim that $AH'I'B, FH'I'G, DCI'H', PCFG$ are all cyclic quadrilaterals. First observe that $AD^2=AH\cdot AC=CH'\cdot CA$ and similarly $BD^2=CI'\cdot CB$. But $AD=BD$, this $CH'\cdot CA=CI'\cdot CB$, so $AH'I'B$ is cyclic. Since $AB\parallel FG$, then by Reim's Theorem, $GFH'I'$ is also cyclic. Now consider the radical axis of $(CHI)$ and $(ABI'H')$. The power of $F$ wrt both circles is $FH\cdot FC$ and $FH'\cdot FA$ respectively, and clearly they are equal since $H, H'$ are just reflections about the midpoint of $AC$. So $F$ lies on radical axis of both circles, likewise we deduce the same result for $G$, implying $FG$ is the radical axis of $(CHI)$ and $(ABI'H')$. Now if $FG\cap (CHI)=U, V$, then since $M\in FG$, $MP\cdot MC=MU\cdot MV=MH'\cdot MI'=MF\cdot MG$, so $PCFG, DCH'I', FH'I'G$ are all cyclic quadrilaterals. Now we direct attention to the points $C, F, G, H', I', M, P$ only, with reference triangle $\triangle CFG$. Now suppose $N$ is midpoint of $FG$, $O$ circumcenter of $\triangle CFG$, and $CN\cap (CFG)=X\neq C$. Consider the inversion centered at $C$ with power $CH'\cdot CF$, and we call $\phi(k)$ to be the image of object $k$ under this inversion. Clearly $Q$ maps to a point in $CN$ and lies on $\phi(H'I')=(CFG)$, so $\phi(Q)=CN\cap (CFG)=X$. Likewise $\phi(P)$ lies on $\phi((CH'I'))=FG$ and $\phi((CFG))=H'I'$, so $\phi(P)=H'I'\cap FG=M$. Consequently $CP\cdot CM=CH'\cdot CF=CQ\cdot CX$, so $MPQX, PMFH'$ are both cyclic. Now $\triangle CPQ \sim \triangle CMX$, so $QP=QC\iff MX=MC$, but $OC=OX$, so it suffices to prove that $OM$ is the perpendicular bisector of $CX$, which is enough to prove $OM\perp CN$. Now, $$OM\perp CN\iff CM^2-MN^2=OC^2-ON^2=OF^2-ON^2=FN^2$$But recall that $$CH'\cdot CF=\frac{CH'\cdot CA}{2}=\frac{AD^2}{2}=\frac{FG^2}{2}=2FN^2$$So $$CM^2-(MN^2-FN^2)=CM^2-MF\cdot MG=CM^2-MP\cdot MC=MC\cdot PC=CP\cdot CM=CH'\cdot CF=2FN^2 $$So this proves that $CM^2-MN^2=FN^2$, which solves the problem.

WLOG $AC>BC$. Since $AH\cdot AC=BI\cdot BC=AD^2$, $\frac{CH'}{CI'}=\frac{AH}{BI}=\frac{BC}{AC}=\frac{CG}{CF}$, so $FGI'H'$ is cyclic. Now, note that the Miquel point of $FGI'H'$ lies on $CM$ and the circumcircle of $CHI$ (since the spiral similarity centered at the Miquel point mapping $H'F$ to $I'G$ also maps $H$ to $I$ by ratios), so $P$ is the Miquel point of $FGI'H'$. Let $N$ be the midpoint of $FG$, $CQ$ intersect the circumcircle of $CFG$ (centered at $O$) at $Q'$, and $P'$ be the midpoint of $CP$. Note that by homothety, the circumcircle of $CNP'$ is tangent to $FG$. Also, $ON\perp BC$ and $OP'\perp CP$ because $O$ lies on the perpendicular bisectors of $FG$ and $CP$. Hence, $MNOP'$ is cyclic, so $\angle NOM=\angle NP'M=180^\circ-\angle CP'N=180^\circ-\angle CNF=\angle CNM$. This means that $OM\perp CQ'$ since $ON\perp FG$, so $MC=MQ'$. But an inversion centered at $C$ fixing $(FGI'H')$ sends $Q'$ to $Q$ and $M$ to $P$, so $CQ=PQ$, as desired.

My solution : Because circle $\Gamma$ tange to $BC$ at $D$ so we have : $AD^2 = AH.AC = CH'.CA = BD^2 = BI.BC = CI'.CB$ then $H'I'BA$ is cyclic so $H'I'FG$ is also cyclic , too We have : $FH'.FA = FH . CF$ so $F$ lies on the radical-axis of $\Gamma$ and $HI'BC$. Prove similary with $G$ we have $FG$ is the radical-axis of $\Gamma$ and $HI,BA$ $\Gamma$ intersects $HI'BA$ at $Z,T$ so : $MP.MC = MZ.MT = MH'.MI' = MF.MR$ $\implies PCGF$ is cyclic Note that $\angle MPF = \angle CGF = \angle MH'F$ so $MPH'F$ is also cyclic Denote the center of $FGI'H'$ by $O$ Taking inversion centre $C$ wiht power $CH'.CF = CI'.CG = CP.CM$, point $Q$ becomes the intersection of $CD$ with $H'I'GF$ so $MPGR$ is cyclic. We have : $\angle MCR = \angle MCO - \angle OCR = \angle CPO - \angle ORC = \angle ORM - \angle ORC = \angle CRM$ . Then $\angle MCR$ is an isoceles triangle $\implies \triangle CPQ $ is isoceles then $PQ = PC$

CantonMathGuy wrote: Let $ABC$ be a triangle with $CA \neq CB$. Let $D$, $F$, and $G$ be the midpoints of the sides $AB$, $AC$, and $BC$ respectively. A circle $\Gamma$ passing through $C$ and tangent to $AB$ at $D$ meets the segments $AF$ and $BG$ at $H$ and $I$, respectively. The points $H'$ and $I'$ are symmetric to $H$ and $I$ about $F$ and $G$, respectively. The line $H'I'$ meets $CD$ and $FG$ at $Q$ and $M$, respectively. The line $CM$ meets $\Gamma$ again at $P$. Prove that $CQ = QP$. Another problem: Let $(CAB) \cap (CH'I')=X$, prove that $XC$ is tangent to $\Gamma$.

Solution 1 We use barycentric coordinates. All is very straightforward up to the point where we compute $Q(c^2:c^2:4a^2+4b^2-2c^2)$ and $P((2a^2-c^2)(b^2-a^2):(c^2-2b^2)(b^2-a^2):-2(a^2+b^2-c^2)^2)$. We can even compute $|CQ|^2=\frac{c^4}{16(a^2+b^2)^2}(2a^2+2b^2-c^2)$ from the distance formula quite easily. Unfortunately, computing $|QP|^2$ from the distance formula is way too messy to do by hand, but we can circumvent this by computing the equation of the circle centred at $Q$ radius $CQ$ (let this circle be $\Omega$) and showing $P$ lies on that. We have \[ \Omega : -a^2yz-b^2zx-c^2xy + (x+y+z)(ux+vy+wz)=0, \]where $u,v,w$ are the powers of $A,B,C$ w.r.t. $\Omega$. From the distance formula we get \[ |AQ|^2 = \frac{1}{16(a^2+b^2)^2}(-a^2c^2(4a^2+4b^2-2c^2)-b^2(c^2-4(a^2+b^2))(4(a^2+b^2)-2c^2)-c^4(c^2-4(a^2+b^2))),\]so \[ u=|AQ|^2-|CQ|^2=\frac{1}{4(a^2+b^2)} (c^4-(a^2+3b^2)c^2+4b^2(a^2+b^2)). \] We get $v$ by swapping around $b$ and $c$ in $u$, and $w=0$ as $C\in \Omega$. Hence \[ \Omega : -a^2yz-b^2zx-c^2xy+\frac{1}{4(a^2+b^2)}(x+y+z)((c^4-(a^2+3b^2)c^2+4b^2(a^2+b^2))x+(c^4-(b^2+3a^2)c^2+4a^2(a^2+b^2))y)=0. \] Now showing $P$ lies on $\Omega$ is fairly straightforward. Solution 2 This solution was communicated to me via a friend of mine. Firstly we show that $H'I'GF$ is cyclic by Power of a Point: $CH'\cdot CF / CI'\cdot CG = CH'\cdot CA / CI'\cdot CB = AH\cdot AC / BI\cdot BC = AD^2 / BD^2 = 1$. Now if we let $P'\neq C$ be the intersection of $(CH'I')$ and $(CHI)$, we get by spiral symmetries that it must also lie on $(CFG)$ ($F,G$ are midpoints of $H'H, I'I$). By Radical Axis on $(CH'I'), (CFG), (H'I'GF)$, it follows that $H'I',FG,CP'$ are concurrent, so $P' \in CM$, and thus $P'=P$. Now we're done if we can show that $H',I'$ lie on a circle coaxal with $(CFG)$ and $\Omega$, where $\Omega$ is defined in Solution 1. By the Coaxal Circles Lemma, it suffices to show that $(H'F\cdot CH')/(H'Q^2-CQ^2) = (I'G\cdot CI')/(I'Q^2-CQ^2)$, which can be done by some computations, with the important step being that $CQ$ is a symmedian of $\triangle CH'I'$, so $CQ$ can be computed from Stewart's theorem.

Let $P'$ be the foot of the altitude from $Q$ on $P$, $T$ the foot of the altitude from $C$ on $H'I'$, and $D'$ the midpoint of $CD$. We will show that $P'$ is on the circle that passes $C$ and is tangent to $FG$ at $D'$. $$M'P' \cdot MC = MD^2 \iff MQ \cdot MT = MD^2 \iff \angle QDG = \angle GTD$$Clearly $CT$ passes $O$, the circumcenter of $CFG$, and $OTMD$ is cyclic. $$\angle QTD = \frac{\pi}{2} - \angle OQD = \frac{\pi}{2} - \angle OMD$$It suffices to show that $OM$ and $CD$ are perpendicular. Let $x = \angle CDG$. We will show that $\tan x = \frac{OM}{OD}$. From $\frac{MG}{FM} \cdot \frac{I'C}{GI'} \cdot \frac{H'F}{CH'} = 1$ and $CI' \cdot CG = CH' \cdot CF = \frac{FG^2}{2}$, we get $\frac{OM}{OD} = \frac{4S}{CF^2 - CG^2}$ where S is the area of $CFG$, which equals to $\tan x$.

CantonMathGuy wrote: Let $ABC$ be a triangle with $CA \neq CB$. Let $D$, $F$, and $G$ be the midpoints of the sides $AB$, $AC$, and $BC$ respectively. A circle $\Gamma$ passing through $C$ and tangent to $AB$ at $D$ meets the segments $AF$ and $BG$ at $H$ and $I$, respectively. The points $H'$ and $I'$ are symmetric to $H$ and $I$ about $F$ and $G$, respectively. The line $H'I'$ meets $CD$ and $FG$ at $Q$ and $M$, respectively. The line $CM$ meets $\Gamma$ again at $P$. Prove that $CQ = QP$. Proposed by El Salvador Note that $$CH' \cdot CF=\frac{1}{2}AH \cdot AC=\frac{1}{2}AD^2$$and $$CI' \cdot CG=\frac{1}{2}BI \cdot BC=\frac{1}{2}BD^2.$$Put that together with $AD=BD=FG$ to obtain $H'I'GF$ is cyclic. Define $P' \ne C$ to be the intersection of $(CH'I')$ and $(CFG)$. Then $P'$ is the center of the spiral similarity $\overline{H'I'} \mapsto \overline{FG}$. But $H \mapsto I$, so, $P'$ lies on $(CHI)$ and hence $P'$ coincides with $P$. Apply radical axis theorem on $(CH'I'), (H'I'GF), (CFG)$ to get $\overline{CP}, \overline{H'I'},$ and $\overline{FG},$ meet at $M$. Let $\overline{CD}$ meet $(CHG)$ again at $R$. Consider inversion at $C$ with power $\sqrt{\tfrac{1}{2}} \cdot FG$, then $P \mapsto M$ and $Q \mapsto R$, so it suffices to show $MC=MR$. Let $CF=y, CG=x, FG=z$. By Menelaus Theorem on $\overline{H'I'}$ and $\triangle CFG$ we have $$\frac{FM}{MG}=\frac{FH'}{H'C} \cdot \frac{CI'}{I'G}=\frac{2y^2-z^2}{2x^2-z^2}.$$Let $N$ be the mid-point of $\overline{FG}$, $O$ be the center of $(CFG)$ and $L$ be the projection of $C$ on $\overline{FG}$. Observe that $$\tan \angle CNL=\frac{CL}{LN}=\frac{\frac{xy \sin C}{z}}{\frac{(x^2-y^2)}{2z}}=\frac{2xy\sin C}{(x^2-y^2)}$$and $$\tan \angle MON= \frac{MN}{ON}=\frac{\frac{z(x^2+y^2-z^2)}{2(x^2-y^2)}}{\frac{z \cos C}{2\sin C}}=\frac{2xy\sin C}{(x^2-y^2)}.$$Evidently, $\angle MON=\angle CNL$ so $\overline{OM} \perp \overline{CN} \implies MC=MR$ so we are done! $\blacksquare$

ThE-dArK-lOrD wrote wrote: Another problem ; Let $(CAB) \cap (CH'I')=X$, prove that $XC$ is tangent to $\Gamma$ As in post 5 we show that $YC$ is tangent to $\Gamma$ where $Y:=AB\cap H'I'$. By definition $Y$ is the radical center of $(AH'I'B), (ABC), (CH'I')$ so $(CAB)\cap (CH'I') \in YC$, as desired.

[asy][asy] unitsize(0.3inches); /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(0cm); 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 = -16.06212214675547, xmax = 17.479403440194773, ymin = -10.855564451469904, ymax = 11.469152254129908; /* image dimensions */ pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); /* draw figures */ draw((-8.680507284770137,-5.779151072769558)--(6.4599460149698364,-5.695961768924833), linewidth(2) + wrwrwr); draw((6.4599460149698364,-5.695961768924833)--(-5.302697705227275,8.2021205411153), linewidth(2) + wrwrwr); draw((-5.302697705227275,8.2021205411153)--(-8.680507284770137,-5.779151072769558), linewidth(2) + wrwrwr); draw(circle((-1.1519715326432554,1.8501869683979415), 7.587857923814066), linewidth(2) + wrwrwr); draw((-4.4101762578353325,5.2345101031356425)--(-7.341768375553144,6.2390368081043155), linewidth(2) + wrwrwr); draw((-5.302697705227275,8.2021205411153)--(-1.1102806349001502,-5.737556420847195), linewidth(2) + wrwrwr); draw((-3.269260820353354,5.799525183260173)--(-4.4101762578353325,5.2345101031356425), linewidth(2) + wrwrwr); draw((-4.4101762578353325,5.2345101031356425)--(-6.238396075452949,4.3291215591640935), linewidth(2) + wrwrwr); draw((-6.238396075452949,4.3291215591640935)--(-12.595911495758553,1.1806918275752896), linewidth(2) + wrwrwr); draw((-12.595911495758553,1.1806918275752896)--(-7.341768375553144,6.2390368081043155), linewidth(2) + wrwrwr); draw((-7.341768375553144,6.2390368081043155)--(-5.302697705227275,8.2021205411153), linewidth(2) + wrwrwr); draw((-12.595911495758553,1.1806918275752896)--(-6.9916024949987055,1.2114847341728714), linewidth(2) + wrwrwr); draw((-6.9916024949987055,1.2114847341728714)--(0.5786241548712807,1.253079386095234), linewidth(2) + wrwrwr); draw(circle((-5.291421584813609,6.149866625828135), 2.052284893406148), linewidth(2) + wrwrwr); draw(circle((-3.221696558728432,4.00002679711304), 4.68915318635598), linewidth(2) + wrwrwr); draw((-12.595911495758553,1.1806918275752896)--(-3.9711130962013317,3.77463658950205), linewidth(2) + wrwrwr); /* dots and labels */ dot((-8.680507284770137,-5.779151072769558),dotstyle); label("$A$", (-8.896432589543373,-6.434181533190137), NE * labelscalefactor); dot((6.4599460149698364,-5.695961768924833),dotstyle); label("$B$", (6.5457373073057985,-5.475852230705852), NE * labelscalefactor); dot((-5.302697705227275,8.2021205411153),dotstyle); label("$C$", (-5.215576859546885,8.419922655316274), NE * labelscalefactor); dot((-1.1102806349001502,-5.737556420847195),linewidth(4pt) + dotstyle); label("$D$", (-1.0337762668881534,-5.562973076386242), NE * labelscalefactor); dot((0.5786241548712807,1.253079386095234),linewidth(4pt) + dotstyle); label("$G$", (0.6650802238794564,1.4284747894650165), NE * labelscalefactor); dot((-6.9916024949987055,1.2114847341728714),linewidth(4pt) + dotstyle); label("$F$", (-7.328257367296349,1.4720352123052112), NE * labelscalefactor); dot((-7.744808914544462,-1.9061520908183507),linewidth(4pt) + dotstyle); label("$H$", (-7.654960538597812,-1.7296558664491033), NE * labelscalefactor); dot((4.426509130095916,-3.293366411069705),linewidth(4pt) + dotstyle); label("$I$", (4.520177645236725,-3.1235893973353357), NE * labelscalefactor); dot((-6.238396075452949,4.3291215591640935),linewidth(4pt) + dotstyle); label("$H'$", (-6.849092716054202,4.368803331178162), NE * labelscalefactor); dot((-3.269260820353354,5.799525183260173),linewidth(4pt) + dotstyle); label("$I'$", (-3.190017197477812,5.980538976265368), NE * labelscalefactor); dot((-4.4101762578353325,5.2345101031356425),linewidth(4pt) + dotstyle); label("$Q$", (-4.322588191322885,5.414253479342837), NE * labelscalefactor); dot((-12.595911495758553,1.1806918275752896),linewidth(4pt) + dotstyle); label("$M$", (-13.012892547941812,1.4066945780449192), NE * labelscalefactor); dot((-7.341768375553144,6.2390368081043155),linewidth(4pt) + dotstyle); label("$P$", (-7.698520961438007,6.459703627507511), NE * labelscalefactor); dot((-3.2064891700637124,1.2322820601340527),linewidth(4pt) + dotstyle); label("$J$", (-3.1246765632175193,1.4066945780449192), NE * labelscalefactor); dot((-4.168776339450625,4.431861696624795),linewidth(4pt) + dotstyle); label("$N$", (-3.7998631172405437,4.325242908337968), NE * labelscalefactor); dot((-2.639528487175389,-0.6528473621111971),linewidth(4pt) + dotstyle); label("$O$", (-2.5583910662949827,-0.48818381550355283), NE * labelscalefactor); dot((-3.9711130962013317,3.77463658950205),linewidth(4pt) + dotstyle); label("$R$", (-3.734522482980251,3.3451333944335855), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Let $J$ be the midpoint of $FG$, let $O=CJ\cap(CFG)$. Redefine $P$ as $(CH'I')\cap(CFG)$. We'll show that $P=CM\cap\Gamma$ first, so the definitions match up. Note that there is a spiral similarity at $P$ sending $H'F\mapsto I'G$, so by the gliding principle, it also maps $H'H\mapsto I'I$. Thus, $P=(CH'I)\cap(CHI)$, so all we have to show is that $P\in CM$. Note that $CH'\cdot CF=\frac{1}{2}CH'\cdot CA=\frac{1}{2}AH\cdot AC=\frac{1}{2}AD^2=\frac{1}{2}FG^2$, and similarly $CI'\cdot CG=\frac{1}{2}FG^2$. Thus, an inversion at $C$ with power $\frac{1}{2}FG^2$ maps $(CH'I')$ to $FG$ and $(CFG)$ to $H'I'$. Thus, $\phi(P)=FG\cap H'I'=M$, so $C,P,M$ collinear. Now, we have that $\phi(Q)=O$ since $\phi(H'I')=(CFG)$, so $PQOM$ cyclic. Therefore, $\angle CPQ=\angle COM$, so it suffices to show that $\angle COM=\angle OCM$, or that $MO=MC$. Unfortunately, I could not finish synthetically, but a computation does do it. Let $c=FG$, $f=CG$, $g=CF$, $\theta=\angle CJF$, $d=CJ$, and $\ell=CO$. We'll show that the foot $R$ from $M$ to $CO$ is the midpoint of $CO$. We have that $CH'=c^2/2g$, $H'F=(2g^2-c^2)/2g$, $CI'=c^2/2f$, and $GI'=(2f^2-c^2)/2f$, since $\phi(H')=F$ and $\phi(I')=G$. Therefore, by Menalaus on $I',H',M$, we have that \[\frac{c^2}{2g^2-c^2}\cdot\frac{MF}{MG}\cdot\frac{2f^2-c^2}{c^2},\]so $MF/MG=\frac{2g^2-c^2}{2f^2-c^2}$. Let $x=MJ$. Then, \[\frac{x-c/2}{x+c/2}=\frac{2g^2-c^2}{2f^2-c^2},\]so solving this gives $x=\frac{c}{2}\frac{f^2+g^2-c^2}{f^2-g^2}$. Now, by law of cosines on $CJF$ and $CJG$, we have that \[\cos\theta=\frac{(c/2)^2+d^2-g^2}{cd}=-\frac{(c/2)^2+d^2-f^2}{cd}.\]Adding the two equations for $\cos\theta$, we get \[2\cos\theta=\frac{f^2-g^2}{2d}.\]Now, \[RJ=MJ\cos\theta=x\cos\theta=\frac{1}{4d}(f^2+g^2-c^2).\]To compute $\ell$, we will use Ptolemey on $CFOG$. Note that $FO/\sin\theta=(c/2)/\sin G$, and $OG/\sin\theta=(c/2)/\sin F$, by law of sines on $OMF$ and $OMG$. Therefore, Ptolemey gives us \[\left(\frac{c/2}{\sin G}\sin\theta\right)f+\left(\frac{c/2}{\sin F}\sin\theta\right)g=c\ell.\]Thus, \[\ell=\frac{1}{2}\sin\theta\left(\frac{f}{\sin G}+\frac{g}{\sin F}\right)=R\sin\theta\left(\frac{f^2+g^2}{fg}\right)\]where $R$ is the circumradius of $FCG$. Note that $[FGH]=\frac{1}{2}cd\sin\theta=\frac{fgc}{4R}$, so $R\sin\theta=fg/2d$, so \[\ell=\frac{f^2+g^2}{2d}.\]Now, $OJ=\ell-d$, so $OR=\ell-d+RJ$. We want $OR=\ell/2$, so note that \[OR=\ell/2\iff\ell/2=d-RJ\iff\ell/2=d-\frac{1}{4d}(f^2+g^2-c^2).\]Also, \[\ell/2=d-\frac{1}{4d}(f^2+g^2-c^2)\iff f^2+g^2=4d^2-f^2-g^2+c^2\iff 4d^2=2f^2+2g^2-c^2,\]which is true by stewart's theorem. Therefore, $R$ is the midpoint of $CO$, so $MO=MC$, as desired. $\blacksquare$

Here's a different approach: Invert about $C$ with radius $\sqrt{CA \cdot CB}$ followed by reflection in the angle bisector of $\angle ACB$. Then, using the inversive distance formula, and re-labeling with respect to the vertex $A$ we get the following equivalent problem:- Inverted problem wrote: Let $AF$ be the $A$-symmedian of $\triangle ABC$, where $F$ lies on $\odot ABC$. Suppose the tangent to $\odot (ABC)$ at $F$ meets $AB$ at $G$ and $AC$ at $H$. Let $X$ and $Y$ be the reflections of $A$ in $B$ and $C$ respectively, and $K \in AB$ and $L \in AC$ be points satisfying $(A,X;G,K)=-1$ and $(A,Y;H,L)=-1$. Let $M=\odot (AXY) \cap \odot (AKL)$ and $P=AM \cap GH$. Also suppose that $AF$ meets $\odot (AKL)$ again at $Q$. Then show that $PA=PQ$. Redefine $P$ as the point where $GH$ meets $KL$. Then $$-1=(A,X;G,K) \overset{P}{=} (A,PX \cap AC;H,L) \Rightarrow P,X,Y \text{ are collinear.}$$Also, by some trivial length chase (using the given harmonic bundles), one gets that $AX \cdot AK=AY \cdot AL$, i.e. $KXYL$ is cyclic. Applying Radical Axes Theorem to $\odot (AXY),\odot (AKL), \odot (KXYL)$, we get that $P \in AM$, which means that our definition of $P$ is correct (Yay ). Now, Let $N$ be the midpoint of segment $KL$. As $KL$ is antiparallel wrt $\angle BAC$, so $AF$ passes through $N$. Also, $\angle AFB=\angle ACB=\angle AKN$, which gives that $BFNK$ is cyclic. So we have that $$\angle PNF=\angle KBF=\angle ACF=\angle GFA=\angle PFN \Rightarrow PF=PN$$Thus, it suffices to show that $AF=QN$. Let $R=AM \cap \odot (ABC)$. Then $R$ is the midpoint of $AM$. Using $PF=PN$, we get that $$PR \cdot PA=PF^2=PN^2 \Rightarrow (PM+MR)PA=(PK-KN)(PL+LN)$$$$\Rightarrow PM \cdot PA+AR \cdot AP=PK \cdot PL+NK(PK-PL)-NK^2 \Rightarrow AR \cdot AP=NK^2$$ Now, $\angle ARF=\angle KBF=\angle PNF$, which gives that $FNPR$ is cyclic, i.e. $AP \cdot AR=AF \cdot AN$. Thus, $$NA \cdot NQ=NK \cdot NL=NK^2=AP \cdot AR=AF \cdot AN \Rightarrow AF=QN \quad \blacksquare$$

Let $D'$ be the midpoint of $FG$, and $CD' \cap (CFG) = Q'$. Since $AH \cdot AC = BI \cdot BC = AD^2$ from power of a point, we have $CH' \cdot AC = CI' \cdot BC$, so $ABI'H'$ is cyclic which implies $FGI'H'$ is cyclic from $FG \parallel AB$. Now from radical axis theorem applied to $H'I'$, $FB$, and $CP$, we have $CPH'I'$ is cyclic, so $P$ is the miquel point of quadrilateral $FGI'H'$. Now consider an inversion at $C$ with power $\sqrt{CH \cdot CF}$. Since $P$ is the miquel point of $FGI'H'$, $P \mapsto M$. Similarly, $I' \mapsto G$ and $H' \mapsto F$. Finally, $Q \mapsto Q'$. Claim: $MC = MQ'$. Proof: Let $X$ be the midpoint of $CP$. We have that $D'OXM$ is cyclic, so $\measuredangle MOD' = \measuredangle MXD' = \measuredangle CD'F$, whence $MO \perp CD'$, which implies $CM = CQ'$ as desired. $\square$ Thus after inversion the lengths are equal, so we conclude that $CP = CQ$. $\blacksquare$

PSJL wrote: Let $O$ be the center of $\Gamma$ $X=CD\cap FG,Y=AB\cap H'I'$ Since $CH'\cdot CA=AH\cdot AC=AD^2=BD^2=BI\cdot BC=CI'\cdot CB$ So $A,B,I',H'$ are concyclic $\Rightarrow YC^2=YB\cdot YA+CI'\cdot CB=YD^2-BD^2+BD^2=YD^2$ $\Rightarrow YC$ is tangent to $\Gamma$ Since $M(D,Y;CP\cap AB,\infty_{AB})=(D,Q;C,X)$ So $(CD,CY;CP,C\parallel AB)=(OD,OQ;OC,OX)=(OX,OC;OQ,OD)$ $\Rightarrow OQ\perp CP\Rightarrow CQ=QP$ Pl explain last 2 lines

Hybrid solution Note by Power of Point that $CH'\cdot CA = AH\cdot AC = AD^2$ and similarly $CI'\cdot CB = DB^2$. Thus $CH'\cdot CA = CI'\cdot CB$ which means $H',I',A,B$ are concyclic. Thus by Reim's, $H',I',F,G$ are concyclic. Let $P'$ be the Miquel point of $FH'I'G$. Then clearly $P'\in CM$. Moreover, by Coaxial Circles lemma, circles $\odot(CH'I')$, $\odot(CFG)$, $\Gamma$ are coaxial. Thus $P'=P$ or $P\in\Gamma$. Now, we redefine $P$ as the point on $\Gamma$ such that $CQ = QP$. We aim to show that $C,P,M$ are colinear. To do that, we turn in to computational method. Set $CF = b$, $CG = a$ and $FG=c$. Then $AH = \tfrac{AD^2}{AC} = \tfrac{c^2}{2b}$ which means $HF = \tfrac{2b^2-c^2}{2b}$. Similarly, $GI = \tfrac{2a^2-c^2}{2a}$. By Menelaus theorem, $$\frac{H'M}{MI'} = \frac{H'F}{FC}\cdot\frac{CG}{GI'} = \frac{a^2(2b^2-c^2)}{b^2(2a^2-c^2)}$$Finally, we use barycentric coordinate with respect to $\triangle CH'I'$. To do that, we need one more synthetic observation. Let $O$ be the circumcenter of $\triangle CH'I'$ and let $H'H'\cap I'I' = U$, $CC\cap H'I' = V$. By Pole-Polar, $CP\perp OQ\perp UV$. Therefore we need to show that $CM\parallel UV$. We easily evaluate $M = (0 : b^2(2a^2-c^2) : -a^2(2b^2-c^2))$, $V = (0 : CI'^2 : -CH'^2) = (0 : b^2 : -a^2)$ and $U = (-c^2 : b^2 : a^2)$. The displacement vector $VU$ is \begin{align*} \overrightarrow{VU} &= (-- : b^2(a^2+b^2-c^2) - b^2(b^2-a^2) : -a^2(a^2+b^2-c^2) - a^2(b^2-a^2)) \\ &= (-- : b^2(2a^2-c^2) : -a^2(2b^2-c^2)) \end{align*}which is clearly parallel to $CM$.

Here's an incomplete solution We begin with a few easy observations. Claim: $FH'IG$ is cyclic. Proof: First, observe that $$H'C \cdot AC = AH \cdot AC = AD^2 = BD^2 = BI \cdot BC = CI' \cdot BC \implies \frac{H'C}{I'C} = \frac{BC}{AC},$$hence $\triangle H'I'C \sim \triangle BAC$, so $\angle CH'I' = \angle FGI' = \angle ABC$. Let $L$ and $K$ be the intersections of line $FG$ with $(ABI'H')$. Note that $M$ lies on the radical axis of $(ABI'H')$ and $(FGI'H')$, so $ML \cdot MK = MF \cdot MG$. Again by Power of a Point, $$IG \cdot GC = I'G \cdot GB = LG \cdot GK \implies I \in (KLC)$$$$HF \cdot FC = AF \cdot FH' = LF \cdot FK \implies H \in (KLC)$$which implies that $L$ and $K$ lie on $\Gamma$. Moreover, $MI' \cdot MH' = ML \cdot MK = MP \cdot MC$, so $I'H'CP$ is cyclic. By radical axis $FGPC$ is also cyclic, so $P$ is the Miquel Point of $(FGI'H')$. Invert around $C$ with radius $\sqrt{CI' \cdot CG}$. Notice that this inversion swaps $P$ and $M$, $I'$ and $G$, and $H'$ and $F$. Finally, it sends $Q$ to a point $Q' = CD \cap (FGPC)$. Note that it suffices to show $\angle MQ'C = \angle MCQ'$. Can someone give me a hint on how to proceed from here? I haven't looked at the solutions on this thread yet, and I feel I'm close, but idk

Since, $CH' \cdot CA = AH \cdot AC = AD^2 = DB^2 = BI \cdot BC = CI' \cdot BC$. So, $A,B,H',I'$ are concyclic. By Reim's theoram, $F,G,H',I,$ are also concyclic. Because $FH \cdot FC = FH' \cdot FA$ , $F$ lies on radical axis of $\Gamma,\odot(ABH'I')$, Similarly to $G$. Therefore, $M$ lies on the radical axis. $MP \cdot MC = MH' \cdot MI'$. Hence, $P,C,H',I'$ are concyclic. Which follow that $C,P,F,G$ are concyclic. Let $\alpha$ be a transformation that sends any point $X$ to $X'$ by invert with radius $\sqrt{CH'\cdot CF}$. It suffices to show that $MC = MQ'$ (Since $P'=M$) and $Q'=CD \cap \odot(CFG)$. Let $N=CM \cap AB$ , $C_1$ be the antipode of $C$ (wrt. $\odot(CFG)$) , $C_2$ lie on $\odot(CFG)$ such that $CC_2 \parallel AB , C_3 = NC_1 \cap \odot(CFG)$. Obviously, $D,C_1,C_2$ are collinear on perpendicular bisector of $AB$ . Since $N$ lies on radical axis of $\Gamma,\odot(CPFG)$, $ND^2=NC_3 \cdot NC_1$. Hence, $\angle NC_3D = 90^{\circ}= \angle C_1C_2C$. So, $A,C_3,D$ are collinear and $C_3=Q'$. But $M$ is midpoint of $CN$. Hence, $MC=MQ'$. [asy][asy] size(300); import graph; pair A,B,C,F,G,H0,I0,M,N,Q,Q0,D; C=dir(110); A=dir(205); B=dir(335); D=midpoint(A--B); F=midpoint(A--C); G=midpoint(C--B); pair D_1 = rotate(90, D)*B; pair D_2 = (C+D)/2; pair D_3 = rotate(90, D_2)*C; pair O=extension(D,D_1,D_2,D_3); path oo = circle(O,length(O-C)); pair[] H=intersectionpoints(A--C,oo); pair[] I=intersectionpoints(B--C,oo); H0=2*F-H[0]; I0=2*G-I[0]; M = extension(H0,I0,F,G); pair[] P =intersectionpoints(C--M,oo); N = extension(M,C,A,B); Q = IP(C--D,H0--I0); path ooo = circumcircle(C,F,G); pair D_4 = 3*D_1-2*D; pair[] E = intersectionpoints(D--D_4,ooo); pair[] R = intersectionpoints(E[0]--N,ooo); draw(oo,deepblue); draw(ooo,deepmagenta); draw(A--B--C--cycle,blue); draw(N--A,blue); draw(M--G,heavymagenta); draw(C--N,heavycyan); draw(C--D,lightblue); draw(D--E[1],purple+dashed); draw(N--E[0],fuchsia); draw(M--I0,magenta); draw(C--E[1],purple+dashed); draw(M--R[1],heavycyan); dot("$A$",A,dir(220)); dot("$B$",B,dir(B)); dot("$C$",C,dir(C)); dot("$F$",F,dir(145)); dot("$G$",G,dir(0)); dot("$D$",D,dir(270)); dot("$H$",H[0],dir(190)); dot("$I$",I[0],dir(10)); dot("$H'$",H0,dir(110)); dot("$I'$",I0,dir(10)); dot("$P$",P[0],dir(120)); dot("$M$",M,dir(120)); dot("$N$",N,dir(180)); dot("$Q$",Q,dir(245)); dot("$C_1$",E[0],dir(-45)); dot("$C_2$",E[1],dir(10)); dot("$Q'$",R[1],dir(240)); [/asy][/asy] *the first and second paragraph is the same as @above comment. The last one can complete your solution.

Denote by $O$ center of $\Gamma,$ let $AB$ meet $HI$ and tangent to $\Gamma$ at $C$ by $X,X'.$ Also let $Y=CX'\cap FG,$ $Z=CD\cap MY.$ By DIT on $CCHI$ there exist involution on $AB$ which swaps $(D;D),(A;B),(X;X')$ and clearly it's a symmetry wrt $D,$ so by isotomic lines $X'\in H'I'.$ Now since $$(CQZD)\stackrel{X'}{=}(YMZ\infty_{MZ})$$and $OC\perp CY,OZ\perp CZ,OD\perp C\infty_{MZ}$ we obtain $OQ\perp CM,$ so we are done.

Let $\omega = (CFG)$. Note that $AH = CH'$ and $BI = CI'$ by reflection. Claim 1: $H'ABI'$, $H'FGI'$, $H'I'PC$ cyclic. Proof: $CH' \cdot CA = AH \cdot AC = Pow_{\Gamma}(A) = AD^2 = BD^2 = Pow_{\Gamma}(B) = BI \cdot BC = CI' \cdot CB \Rightarrow$ by converse of Pop $H'ABI'$ cyclic. Then also $CH' \cdot CF = 1/2 \cdot CH' \cdot CA = CI' \cdot CG$ and $H'FGI'$ cyclic. $GI \cdot GC = GI' \cdot GB$, and $FH \cdot FB = FH' \cdot FA$. Thus $FG$ is the radical axis of $\Gamma$ and $(H'ABI')$. Since $M$ lies on it, we have $MI' \cdot MH' = Pow_{(H'ABI')}(M) = Pow_{\Gamma}(M) = MC \cdot MP$. Thus, $H'I'PC$ cyclic. $\square$ Claim 2: $P$ is the Miquel point of $H'FGI'$. Proof: Since $H'FGI'$ is cyclic, its Miquel point can be characterised by $(H'I'C) \cap MC, \neq C$. This is exactly $P$ by Claim 1. $\square$ We now have $CFGP$, $GMPI'$, $FMPH'$ also cyclic. Invert about $C$ radius $\sqrt{CH' \cdot CF}$. $H' \leftrightarrow F$, $P \leftrightarrow M$, $Q \leftrightarrow Q'$. Thus $H'I'M \leftrightarrow (CFGP) = \omega$, $FGM \leftrightarrow (CH'I'P)$, and $Q' = CD \cap \omega$. It suffices to prove the following statement: Inverted problem 1 wrote: Let $ABC$ be a triangle with $CA \neq CB$. Let $D, F, G$ be the midpoints of $AB, AC, BC$, respectively. Let $\Gamma$ be the circle through $C$ and tangent to $AB$ at $D$. Let $\omega = (CFG)$. Let $P = \Gamma \cap \omega$, $P \neq C$, and let $M = CP \cap FG$. Let $Q' = CD \cap \omega$. Then $MQ' = MC$. Now invert about $C$ radius $\sqrt{CF \cdot CB}$, and reflect about the $C$ internal angle bisector. $F \leftrightarrow B$, $G \leftrightarrow A$. $FG \leftrightarrow (ABC)$, $AB \leftrightarrow \omega$ so $D \leftrightarrow X$, the intersection of the $C$-symmedian with $\omega$. Thus $\Gamma \leftrightarrow$ the tangent to $\omega$ at $X$, $\lambda_{X}$. $P' = AB \cap \lambda_{X}$, and $M' = CP' \cap (ABC)$. $Q' \leftrightarrow Y$, the intersection of the $C$-symmedian with $AB$. We want to prove $YC = YM'$. If we define phantom point $R$ as the point on $(ABC)$ such that $YR = YC$, note that it suffices to show $CR, \lambda_{X}, AB$ concur. $YO$ is the perpendicular bisector of $CR$ and hence meets it at $T$ such that $\measuredangle CTO = 90$. Now note that $\omega$ has diameter $OC$ since $\measuredangle OGC = \measuredangle CFO = 90$, and so $T$ lies on $\omega$. Thus, it suffices to prove: Inverted problem 2 wrote: Let $ABC$ be a triangle with $CA \neq CB$. Let $\omega$ be the circle diameter $OC$. Let the $C$-symmedian of $\triangle ABC$ meet $AB$ at $Y$ and $\omega$ again at $X$. Let $YO$ meet $\omega$ again at $T$. Then $AB, CT$ and the tangent at $X$ to $\omega$, $\lambda_X$, concur. Let $E$ be the foot of the altitude from $C$ onto $AB$, and $N$ the midpoint of $FG$, where $D, F, G$ are as defined before. Let $O'$ be the midpoint of $\omega$. Claim 3: $E, X, N$ collinear. Proof: Consider $\sqrt{BC}$-inversion at $C$ wrt. $\triangle CFG$. $X \leftrightarrow N$, and since $E$ is the reflection of $C$ in $FG$ by homothety scale factor 2, $E \leftrightarrow O'$. It suffices to prove $NXCO'$ cyclic. $O'$ is the centre of $\omega$ hence lies on the perpendicular bisector of $XC$. It's well-known that $FG$ is the internal bisector of $\angle CNX$, and hence $NO'$ is the external one. Thus $O'$ is the intersection of the perpendicular bisector of $XC$ with the external bisector of $\angle CNX$, thus $NXCO'$ is cyclic. $\square$ Claim 4: $NE = ND$. Proof: By homothety scale factor 2 from $C$, $N$ is the midpoint of $CD$. But $\angle DEC = 90$ means $N$ is the centre of $(DEC)$ and hence $NE = ND$. $\square$ Claim 5: $(YEX)$ is tangent to $\lambda_X$ at $X$. Proof: $\measuredangle(YX, \lambda_X) = \measuredangle(CX, \lambda_X) = \measuredangle(CF, FX)$. Since $N, X$ are inverses under $\sqrt{BC}$-inversion at $C$ in $\triangle CFG$, we have $\measuredangle(CF, FX) = \measuredangle(CN, NG)$. Thus $\measuredangle(DX, \lambda_X) = \measuredangle(CD, DB) = \measuredangle(ND, DE) = \measuredangle(DE, EN) = \measuredangle(YE, EX)$. Thus by converse of alternate segment, $(YEX)$ tangent to $\lambda_X$ at $X$. $\square$ Claim 6: $AB, CT, \lambda_X$ concur. Proof: $\measuredangle YEC = 90 = \measuredangle CTY$ hence $YECT$ cyclic. Now consider the radical centre of $(YEX), (YECY), (XCT)$. This is where $AB, CT, \lambda_X$ concur. $\blacksquare$

This problem was quite fun , it should be tried by more people. Point label: So define $(ADH') \cap (BDI')=J$, let $DF,DG$ meet $H'I'$ at $K,L$ respectivily, let $O$ the circumcenter of $\triangle ABC$, let the tangents from $C,D$ to $(CHDI)$ meet at $T$ and let $CM \cap AB=R$. Claim 1: $K,L$ lie in $(CHDI)$ Proof: Note that by PoP $CH' \cdot CA=AH \cdot AC=AD^2=BD^2=BI \cdot BC=CI' \cdot CB$ hence $AH'I'B$ is cyclic now by converse of Reim's theorem twice on $(AH'I'B)$ at $AH'$ and $I'B$ we get that $AKH'D$ and $BLI'D$ are cyclic, now by PoP again we have that $KF \cdot FD=AF \cdot FH'=CF \cdot FH$ and $LG \cdot GD=BG \cdot GI'=CG \cdot GI$ hence $K,L$ lie in $(CHDI)$ as desired. Claim 2: $T$ lies in $H'I'$ and $P$ is the miquel point of $FH'I'G$ Proof: Projecting we have $-1=(F, G; CD \cap FG, \infty_{BC}) \overset{D}{=} (K, L; C, D)$ so $KL$ passes through $T$ meaning that $H'I'$ passes through $T$. Now since $AH'I'B$ is cyclic we get $FH'I'G$ cyclic and note that $\angle LKF=\angle CAB=\angle FGD$ so we have that $KLGF$ is cyclic so by PoP $MP \cdot MC=MK \cdot ML=MF \cdot MG=MH' \cdot MI'$ so $CPH'I', CPFG$ are cyclic meaning that indeed $P$ is the miquel point of $FH'I'G$. Claim 3: $\triangle TQD$ and $\triangle CRD$ are orthologic. Proof: First note that $O$ lies in $(CFG)$ and $CO$ is its diameter hence $\angle CPO=90=\angle ODR$ so $ODRP$ is cyclic and now by and inversion centered at $C$ with radius $\sqrt{CP \cdot CM}$ + homothety centered at $C$ with scale factor 2 we can get that $RPJOD$ is cyclic becuase before that we can use miquel theorem on $\triangle ACB$ at $J$ to get that $CPH'JI'$ is cyclic. Now $\angle RJD=\angle RPD=\angle JDR$ meaning that $JR=DR$ but since $RO$ is diameter in $(RPJOD)$ we get that $OJ=OD$ and $RO \perp QD$, now since $H'I'$ are anti-paralels we get that $CO \perp TQ$ and its clear that $OD \perp TD$ so our claim is true. Finish: By Claim 3 we can get that the perpendiculars from $T$ to $CD$, $D$ to $RD$ and $Q$ to $CR$ must meet at a single point, but the first 2 perpendiculars meet at the center of $(CH'DI')$ hence $Q$ lies in the perpendicular bisector of $CP$ meaning that $CQ=QP$ thus we are done

From \[CH' \cdot CA = AH \cdot AC = AD^2 = BD^2 = BI \cdot BC = CI' \cdot CB,\]$(AH'I'B)$ is cyclic. Let $E$ be the second intersection of $(ADH')$ and $(BDI')$. Since $CH' \cdot CA = CI' \cdot CB$, $C$ lies on the radical axis of these two circles, i.e. $DE$, so $E$ lies on line $CD$. Let $J$ be the second intersection of $(ABC)$ with $CD$. Then \[JD = \frac{AD^2}{CD} = \frac{AH \cdot AC}{CD} = \frac{CH' \cdot CA}{CD} = \frac{CD \cdot CE}{CD} = CE,\]so the midpoint of $DE$, which we will call $K$, is the same as the midpoint of $JC$. A homothety centered at $C$ with factor $\frac12$ sends $(JABC)$ to $(KFCG)$ giving $(KFCG)$ cyclic. F lies on the radical axis of $(AH'I'B)$ and $(CIDH)$ since $FA \cdot FH' = FH \cdot FC$. Similarly $G$ lies on the radical axis, so the two intersections of $(AH'I'B)$ and $(CIDH)$ both lie on $FG$; let them be $X$ and $Y$ in some order. Also, \[\angle GFH' = \angle BAH' = 180^\circ - \angle BI'H' = 180^\circ - \angle GI'H'\]gives $(FH'I'G)$ cyclic. This implies \[MF \cdot MG = MH' \cdot MI' = MX \cdot MY = MC \cdot MP,\]giving $(KFCPG)$ cyclic and $(H'CPI')$ cyclic. By Miquel's theorem on $\triangle ABC$, $(H'CEPI')$ is cyclic. \[\angle FH'Q = \angle FH'I' = 180^\circ - \angle FGI' = 180^\circ - \angle FGC = 180^\circ - \angle FKC = 180^\circ - \angle FKQ\]gives $(FH'QK)$ cyclic. By Miquel's theorem on $\triangle FMC$, $(MGI'P)$ is cyclic. So \[CK \cdot CQ = CF \cdot CH' = CG \cdot CI' = CM \cdot CP,\]giving $(KQPM)$ cyclic. Let $L = AB \cap CM$. Then $M$ is the midpoint of $LC$ from $\triangle CGM \sim \triangle CBL$, and \[\angle EPL = 180^\circ - \angle EPC = 180^\circ - \angle EI'C = \angle EI'B = 180^\circ - \angle EDB = 180^\circ - \angle EDL,\]giving $(DEPL)$ cyclic. Then \[CK^2 - DK^2 = (CK + DK)(CK - DK) = (CK + DK)(CK - EK) = CD \cdot CE = CL \cdot CP = 2CM \cdot CP = 2CM \cdot (MC - MP) = 2MC^2 - 2MP \cdot MC.\]Rearranging, $2MP \cdot MC - DK^2 = 2MC^2 - CK^2$, $(MP + MC) \cdot 2MC - DK^2 = 4MC^2 - CK^2$, $LP \cdot LC - DK^2 = LC^2 - CK^2$, and $LD^2 - DK^2 = LC^2 - CK^2$. By the Pythagorean theorem, the condition is satisfied if $K$ is the foot of $L$ to $DE$. Moreover, this is the unique choice of $K$ on segment $DE$ that satisfies the condition, by monotonicity, so $K$ is indeed the foot of $L$ to $DE$. Then $MK = ML = MC$, so \[\angle QPC = 180^\circ - \angle QPM = \angle QKM = \angle CKM = \angle KCM = \angle QCP,\]implying $CQ = QP$, as desired.

Note that \[CH'\cdot CA = AH\cdot AC = AD^2 = BD^2 = BI\cdot BC = CI'\cdot CB\]so $ABI'H'$ is cyclic. Since $FH\cdot FA=FH'\cdot FC$ and similarly for $G$, $FG$ is radical axis of $(ABI'H')$ and $(CHDI)$. Hence, $M$ has equal power to those circles. We have $MP\cdot MC=MH'\cdot MI'$ implying $PCI'H'$ cyclic. Thus, $P$ is Miquel point of $HII'H'$ and thus is center of spiral similarity taking $HH'$ to $II'$, and the same spiral similarity which takes $F$ to $G$. Thus, $P$ is also Miquel point $H'FGI'$. Since $P$ is on $MC$, we also have $H'FGI'$ cyclic. Let $DF$, $DG$ meet $(CHI)$ at $Y$ and $X$ then \[GD\cdot GX=GI\cdot GC=GI'\cdot GB\]so $XI'DB$ is cyclic. Then \[\angle XI'B=\angle XDB=\angle CAB=\angle CI'H'\]which implies that $X$ lies on $H'I'$. Similarly, $Y$ lies on $H'I'$ and $ADH'$. Now, \[\angle I'XC=\angle YDC=\angle DCG=\angle QCI'\]so $QC^2=QI'\cdot QX=QH'\cdot QY$. The inversion with center $Q$ and radius $QC$ transposes $H'$ and $Y$, $I'$ and $X$. Thus, $(CH'I')$ and $(CYX)$ are also transposed, so their second intersection, $P$, stays put under the inversion. We are done.

CantonMathGuy wrote: Let $ABC$ be a triangle with $CA \neq CB$. Let $D$, $F$, and $G$ be the midpoints of the sides $AB$, $AC$, and $BC$ respectively. A circle $\Gamma$ passing through $C$ and tangent to $AB$ at $D$ meets the segments $AF$ and $BG$ at $H$ and $I$, respectively. The points $H'$ and $I'$ are symmetric to $H$ and $I$ about $F$ and $G$, respectively. The line $H'I'$ meets $CD$ and $FG$ at $Q$ and $M$, respectively. The line $CM$ meets $\Gamma$ again at $P$. Prove that $CQ = QP$. Proposed by El Salvador Claim (1): $A,B,I',H'$ are cyclic and also $F,G,I',H'$are cyclic Proof: By power point we have: $CI'*CB=BI*BC=BD^2=AD^2=AH*AC=CH'*AC\Rightarrow A,B,I',H'$ are cyclic and $<AH'I'=180-<ABI'=180-<FGI'\Rightarrow F,G,I',H'$ are cyclic Let now $P'$ be the Miquel point of $HII'H'$ then: Claim (2): $P'\equiv P$ Proof: First $F,C,P',G$ are cyclic by spiral similarity Then by radical center at $(FGI'H'),(FCPG),(I'H'CP')$ gives $P'\equiv P$ Let now $Z=CD\cap (CI'H')$ Claim (3): $Z,D,B,I'$are cyclic and also $Z,D,A,H'$ are cyclic Proof: Consider the invertion with center $C$ and power $CI'*CB=CH'*CA$ then $I'\leftrightarrow B,H'\leftrightarrow A\Rightarrow (I'H'C)\leftrightarrow AB$ and since $Z \varepsilon (I'H'C)\Rightarrow Z\leftrightarrow D$ we get the result Let $GF$ intersect $(DIC)$ at $K,L$ Claim (4): $K,I',H',L,A,B$ are cyclic Proof: By power point we have :$GK*GL=GI*GC=GB*GI'\Rightarrow K,L,B,I'$ are cyclic similary $K,L,H',A$ are cyclic an by claim (1) $A,B,I',H'$ are cyclic If they are three diferen circle then there is no radical center so they must be the same. Let $X=DG\cap (DIC),Y=DF\cap (DIC)$ Claim (5): $X\varepsilon (DZI'B),Y\varepsilon (ADZH')$ Proof: $GX*GD=GC*GI=GI'GB\Rightarrow $ the result Claim (6): $X,Y\varepsilon I'H'$ Proof: $<I'XC=<YDC=<DCG=<QCI'\Rightarrow X\varepsilon I'H'$ similar for $Y$ Claim (7): $QC$ is tangent to $(CI'X),(CH'Y)$ Proof: $<I'XC=<YXC=<YDC=<DCB=<QCI'$ which proves the claim Claim (8): $QC=QP$ Proof: By claim(7) we have $QI'*QX=QC^2=QH'*QY$ so if we invert with center $Q$ and this power we have : $I'\leftrightarrow X,H'\leftrightarrow Y,C\leftrightarrow C\Rightarrow (I'H'C)\leftrightarrow (XYC)\Rightarrow (I'H'C)\cap (XYC)$ stay the same so $QC^2=QP^2\Rightarrow QC=QP$

I think I found a really silly solution. Step 1: Proving the collinearity. Let $CC\cap DD=T$. I claim that $H',I',T$ are collinear. I'm sure there's a much smarter way to do this, but here's a Menelaus "bash" motivated by length conditions. We have: \[(C,D;H,I)\stackrel{C}{=}(T,D;A,B)\implies \frac{CH}{CI}\div \frac{DH}{DI}=\frac{TA}{TB}\div \frac{DA}{DB}\]and therefore \begin{align*} \frac{TA}{TB}\cdot \frac{H'C}{H'A}\cdot \frac{I'B}{I'C} &= \frac{TA}{TB}\cdot \frac{HA}{HC}\cdot \frac{IC}{IB} \\ &= \frac{HA}{HC}\cdot \frac{IC}{IB}\cdot \frac{CH}{CI}\cdot \frac{DI}{DH} \\ &= \frac{DI}{IB}\cdot \frac{HA}{DH} \\ &= \frac{DC}{DB}\cdot \frac{DA}{DC} = 1 \end{align*}proving the claim. Step 2: Reducing the problem completely. I think the motivation for this step is noticing how points are now defined. We can draw $N=FG\cap CD\cap OT$ where $O$ is the center of $\Gamma$. Also define $K$ as the midpoint of $TC$. We can change the reference triangle to $\triangle TCD$. Then $O$ is the $T$-antipode, $Q$ varies on $CD$, and $M=KN\cap TQ$. We need to show that $CM\perp OQ$. Define $L\in OT$ where $QL\parallel TC$, and notice by Ceva's that $TQ,NK,CL$ are concurrent at $M$. As a result, $Q$ is the orthocenter of $\triangle LCO$ and $CM\perp OQ$. Done! As a final note to myself, the construction of $L$ was sitting in my head for around 5 minutes before I managed to extract it. I think I should keep up this mindset of constantly constructing new points and lines and trying to transform the problem, and not hold back.

Here goes G5: Let $Y=H'I' \cap AB$ with $X = HI \cap AB$. Notice that, by Menelaus theorem, $DX=DY$. Let $Y' = CC \cap AB$. Claim: $Y'=Y$. Let $W$ be the point of second tangency from point $X$. Then: $(ED; HI)=-1$. Taking perspective from $C$ gives $CW \parallel AB$. Note that $C, W$ are symmetric with-respect to $D$ and so are $X, Y$. Therefore $CY$ is also tangent and therefore $Y=Y'$. Let $N = HG \cap CD$. Note that: $MN \parallel AB$ with $N$ being the midpoint of $CD$. Note that, it suffice to show that the polar of $Q$ is parallel to $CM$. Let $\infty_1, \infty_2, \infty_3$ denote the point of infinity of lines $MN, CD, CM$ respectively. We have the following conditions converted from Euclidean to Projective: - $N$ is the midpoint of $CD$: It can be written as $(CD; N \infty_2) = -1$. - $MN \parallel YD$: It can be written as $\infty_1 = MN \cap YD$> - we have to prove that $Q$ polar is parallel to $CM$: It can be written as $\infty_3 = CM \cap (\infty_1 \infty_2)$. Let $U, V$ denote the intersections of $YM$ with $\Gamma$. Then, there exists a homography which maps $CUDV$ to a rectangle (basically $Q$ is the center of $\Gamma$) while preserving $\Gamma$ as a circle. Note that $M, N$ are two independent points of $UV, CD$ respectively. We construct points $\infty_1, \infty_2$ by: $(CD; N \infty_2)=-1$ and $\infty_1 = DD \cap MN$. Note that: $Y = CC \cap DD$ is point at infinity along the lines $CC, DD$. Note that the $Q$-polar (with respect to $\Gamma$) is point at infinity. It suffice to show $CM \parallel \infty_1 \infty_2$ which is evidently true.