Let $ \Psi $ be the inversion with center $ H $ that fixed $ \odot (ABC). $ Since $ Q, H, M $ and the antipode of $ A $ in $ \odot (ABC) $ are collinear, so $ \Psi(Q) $ is the antipode of $ A $ in $ \odot (ABC) $ and note that $ MH=M\Psi(Q) $ we get $ \Psi(M) $ is the reflection of $ H $ in $ Q. $ Clearly, $ \Psi(K) $ is the antipode of $ Q $ in $ \odot (ABC) $ and $ \Psi(F) $ is the reflection of $ H $ in $ A, $ so $ AQ\Psi(Q) \Psi(K) $ is a rectangle and $ A\Psi(K) $ is the perpendicular bisector of $ \Psi(F)\Psi(M), $ hence $ \Psi(Q)\Psi(K) $ is the tangent of $ \odot (\Psi(F)\Psi(K)\Psi(M)) $ $ \equiv $ $ \Psi(\odot (FKM)). $ I.e., $ \odot (HQK) $ is tangent to $ \odot (FKM). $ $ \qquad \blacksquare $

Let $AE$ be the diameter of $\Gamma$. Extend $AF$ to meet $\Gamma$ again at $D$. Suppose $\Gamma$ has center $O$. Step 1: $Q,H,M,E$ are collinear. Proof: $BECH$ is a parallelogram, so $E,M,H$ are collinear. $AE$ is the diameter, and $\angle AQE=90$, so $Q,H,M,E$ are collinear. Step 2: circumcircle of $\triangle DHK$ is tangent to $QE$. Proof: Let $QP$ be the diameter of $\Gamma$, then $P,H,K$ are collinear. Now $\angle HKD = \angle PQD = \angle OQD = 90-\angle QBD$. Note that $\angle QMC=\angle DMC$, so $QBDC$ is harmonic, so $\angle QBD = \angle QMC$. Thus $\angle HKD = 90-\angle QMC=\angle MHD$, proving step 2. Now let $X$ be the circumcenter of $\triangle KHD$. Then $X$ lies on $BC$ because it is the perpendicular bisector of $HD$, and $XH\perp QE$. So $XK^2 = XH^2 = XF\cdot XM$, which means $XK$ is tangent to circumcircle of $\triangle MFK$. Also since $XK=XH$ and $XH$ is tangent to circumcircle of $\triangle QHK$, $XK$ is also tangent to circumcircle of $\triangle QHK$, and the result follows.

This can be done with complex numbers with $\Gamma$ as the unit circle. We construct the tangent to $(KQH)$ at $K$ and let it intersect $\Gamma$ at $X\neq K$. Now we can prove that if $A', Q'$ are the points diametrically opposite $A,Q$, respectively, then $A'MHQ, Q'HK$ are collinear. $\angle XKH = \angle KQH \implies \frac{(x-k)(h-q)}{(h-k)(k-q)} \in \mathbb{R} \implies \frac{(x-k)(q+a)}{(q+k)(q-k)} \in \mathbb{R} \implies \frac{(x-k)(q+a)}{(q+k)(q-k)} = \frac{\frac{k-x}{xk}\frac{q+a}{qa}}{\frac{q+k}{qk}\frac{k-q}{qk}} \implies x=\frac{qk}{a}$. EDIT: Just realised that since $\angle AQK = \angle QKX$, $QK || AX$ so $x=\frac{qk}{a}$. So the above paragraph is unnecessary. We want to prove that $\angle FKX=\angle FMK \iff \frac{(f-k)(k-m)}{(x-k)(b-c)} \in \mathbb{R} \iff \frac{(f-k)(k-m)}{(x-k)(b-c)} = \frac{\overline{(f-k)}\overline{(k-m)}}{\frac{k-x}{xk}\frac{c-b}{bc}} \iff (f-k)(k-m) = xkbc\overline{(f-k)}\overline{(k-m)} \iff (f-k)(k-m) = \frac{qk^2bc}{a} \overline{(f-k)}\overline{(k-m)}$. Now we can calculate $q=\frac{bc(2a+b+c)}{2bc+ab+ac}$ from $aq=-\frac{q-a}{\overline{(q-a)}} = \frac{h+a}{\overline{(h+a)}} = \frac{2a+b+c}{\frac{2}{a}+\frac{1}{b}+\frac{1}{c}}$. We calculate $k=a\frac{(a+b+c)(3bc+ab+ac)+abc}{(ab+bc+ca)(3a+b+c)+abc}$ from $kq=-\frac{k-q}{\overline{(k-q)}} = \frac{-q-h}{\overline{(-q-h)}} =q\frac{q+h}{1+q\overline{h}}$. The hard part is calculating $f-k$ and $k-m$. $k-m = \frac{(2a+b+c)[(b+c)(a^2-bc)-a(b-c)^2]}{2[(ab+bc+ca)(3a+b+c)+abc]}$ and $f-k = \frac{[(b+c)(a^2-bc)+a(b-c)^2](a+b)(a+c)}{2a[(ab+bc+ca)(3a+b+c)+abc]}$. The main step was really finding the factorisation for the numerator of $k-m$, which I did by somewhat guessing the factor $2a+b+c$ by inspection and luck (mostly luck). Then you can guess the form of $f-k$ so that the RTP statement holds. Here's how I sort of guessed that $2a+b+c$ was a factor: The numerator of $k-m$ is \begin{align*} & 2a[(a+b+c)(3bc+ab+ac)+abc] - (b+c)[(ab+bc+ca)(3a+b+c)+abc] \\ &= 2a(a+b+c)(ab+bc+ca) + abc(6a+3b+3c) - (ab+bc+ca)(b+c)(3a+b+c) \\ &= 3abc(2a+b+c) - (ab+bc+ca)(b+c)(2a+b+c) - a(ab+bc+ca)(b+c) + 2a(a+b+c)(ab+bc+ca) \\ &= 3abc(2a+b+c) - (ab+bc+ca)(b+c)(2a+b+c) +a(ab+bc+ca)(2a+b+c). \end{align*} The first manipulation was done with the intention of possibly factoring out $ab+bc+ca$, but then the $3abc(2a+b+c)$ appeared and it seemed as if the $3a+b+c$ was "pretty close" to $2a+b+c$ and that possibly $2a+b+c$ would factor out...and it did. The remainder of the solution is basically conjugating these expressions and substituting everything in.

Great idea to draw circumcenter of triangle $KHD$ dear One Plus One. So we must prove that $(KHD)$ is tangent to $QE$. I do a little different from you. We easily seen $\triangle KQH$ and $\triangle KAE$ is right and $\angle KQH=\angle KQE=\angle KAE$, we deduce $\angle KHQ=\angle KEA=\angle KHD$. So we are done. I propose a general problem and solution Let $ABC$ be an actue triangle inscribed in circle $(O)$ and $P$ is a point inside it such that $\angle BPC=180^\circ-\angle A$. $BP,CP$ cut $CA,AB$ at $E,F$. Circle $(AEF)$ cuts $(O)$ again at $G$. Circle diameter $PG$ cuts $(O)$ again at $K$. $D$ is projection of $P$ on $BC$ and $M$ is midpoint of $BC$. Prove that the circle $(KPG)$ and circle $(KDM)$ are tangent. Solution. Let $Q$ are symmetric of $P$ through $D$, easily seen $Q$ lie on $(O)$. Let $GP$ cuts $(O)$ again at $N$. We see $\angle NPC=\angle FPG=\angle FAG=\angle BNP$ so $BN\parallel PC$ similarly, $CN\parallel BP$. Let $AS,NR$ are diameter of $(O)$. We eaily seen $\angle PQN=90^\circ$ so $P,Q,R$ are collinear. So $M$ is midpoint of $PN$. We have $\angle KPG=90^\circ-\angle KGP=90^\circ-\angle KAN=\angle AKN+\angle ANK-90^\circ=\angle NKS+\angle KQA=\angle NAS+\angle KQA=\angle ANR+\angle KQA=\angle AQR+\angle KQA=\angle KQP.$ From this, $GN$ is tangent to $(KPQ)$. Let tangent at $K,P$ of $(KPG)$ intersect at $T$. We have $\angle KTP=180^\circ-2\angle KGP=2\angle KPG=2\angle KQP$ and $TK=TP$, so we have $T$ is circumcenter of triangle $KPQ$, but $BC$ is perpendicular bisector of $PQ$, therefore $T$ lies on $BC$. Now, $TK^2=TP^2=TD.TM$, this means $TK$ is common tangents of $(KGP)$ and circle $(KHM)$ or they are tangent. We are done.

Another generalization: Let $ABC$ be a triangle with circumcircle $\odot (O)$. A arbitrary circle $\odot (D)$ passing through $B, C$ intersects $CA, AB$ at $E, F$, respectively. $K$ is the orthogonal projection of $D$ on $AP$. Construct the diameter $AP$ of $\odot (AEF)$. $\odot (AEF) \cap \odot (O)=\{A, G\}$. $\odot (GP) \cap \odot (O)=\{G, J\}$. Prove that $\odot (JGP)$ and $\odot (JKD)$ are tangent.

Another generalization : Let $ABC$ be a triangle with circumcircle $\Gamma$ . A arbitrary circle $(O)$ passing through $B, C$ intersects $CA, AB$ at $D, E$, respectively. Let $H=BD\cap CE$. Let $Q $ be the point on $ \Gamma $ such that $\angle HQA =90^{\circ}$ and let $K $ be the point on $\Gamma $ such that $\angle HKQ=90^{\circ}$. Let $F$ is the orthogonal projection of $H$ on $BC$, and let $M=KO\cap BC$. Prove that the circumcircles of triangles $KQH $ and $FKM $ are tangent to each other.

THVSH wrote: Another generalization: Let $ABC$ be a triangle with circumcircle $\odot (O)$. A arbitrary circle $\odot (D)$ passing through $B, C$ intersects $CA, AB$ at $E, F$, respectively. $K$ is the orthogonal projection of $D$ on $AP$. Construct the diameter $AP$ of $\odot (AEF)$. $\odot (AEF) \cap \odot (O)=\{A, G\}$. $\odot (GP) \cap \odot (O)=\{G, J\}$. Prove that $\odot (JGP)$ and $\odot (JKD)$ are tangent. The solution by oneplusone also works for this generalization I just want to note that step 2 in the solution simplified follows from Reim theorem ($ \because AQEP $ is rectangle $ \Longrightarrow QE \parallel AP $) .

Probably equivalent to the other inversion solutions, but anyways. . . [asy][asy]size(6cm); defaultpen(fontsize(9pt)); size(8cm); pair A = dir(80); pair B = dir(220); pair C = dir(-40); draw(unitcircle, blue); pair O = circumcenter(A, B, C); pair H = orthocenter(A, B, C); pair Q = IP(CP(midpoint(A--H), H), unitcircle); pair N = midpoint(H--Q); pair T = midpoint(A--H); pair N_9 = midpoint(O--H); pair M = midpoint(B--C); pair F = foot(A, B, C); draw(circumcircle(M, F, N), darkcyan); pair L = M+T-N; pair K = OP(CP(N, H), unitcircle); draw(A--B--C--cycle, red); draw(A--F); draw(T--M--L--N, heavygreen); draw(M--Q--A, magenta); draw(Q--K--L, heavymagenta); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$O$", O, dir(O)); dot("$H$", H, dir(H)); dot("$Q$", Q, dir(Q)); dot("$N$", N, dir(N)); dot("$T$", T, dir(T)); dot("$N_9$", N_9, dir(N_9)); dot("$M$", M, dir(M)); dot("$F$", F, dir(F)); dot("$L$", L, dir(L)); dot("$K$", K, dir(K)); /* Source generated by TSQ */ [/asy][/asy] Let $N$, $T$ be midpoints of $HQ$, $AH$. Let $L$ be on the nine-point circle with $\angle HML = 90^{\circ}$. The negative inversion at $H$ swapping $\Gamma$ and nine-point circle maps $A$ to $F$, $K$ to $L$, $Q$ to $M$. AS $LM \parallel AQ$ we just need to prove $LA = LQ$. But $MT$ is a diameter, hence $LTNM$ is a rectangle, so $LT$ passes through the center of $\Gamma$.

Let $R$ be the reflection of $H$ onto $BC$ and $KR$ intersects $BC$ at $T$. We easily prove that $KQMT$ is cyclic. If $KF$ intersects circle $KQH$ at $P$ , then $Q,P,T$ are collinear. => let $l$ be line which tangents circle $KQH$ at $K$. Then it is not hard to see line $l$ also tangents to the circle $KFM$ at $K$. Done!

If $O$ is cicumcenter of $ABC$ and $V,S,T$ are midpoints of segments $OH,HK,HQ$ respectively, then $VS$ and $VT$ are midlines of triangles $HOK$ and $HOQ$ resprectively. So $S,T$ lie on Euler's circle of $ABC$. If $t$ is a line through $H$ which is perpendicular to $HM$, then $t,MF,ST$ are concurrent at radical center $X$ of Euler's circle of $ABC$ and circumcircles of triangles $STH$ and $MFH$. Triangles $XKS$ and $XHS$ are congruent, so $XK^2=XH^2=XS\cdot XT=XM\cdot XF$ and $\measuredangle XKS=90$, which means that the line $XK$ is common tangent of circumcircles of triangles $KFM$ and $HKQ$.

I found this problem quite pretty. First of all, it is well known that $Q$, $H$ and $M$ are collinear. Assuming this, we call $H_{A}$ the second intersection of $AH$ with $\Gamma$, and we perform an inversion of center $H$ and power $HA \cdot HH_{A}$. After very basic angle chasing + application of the fact stated above, we get the problem transformed into the following: let $ABC$ be a triangle and let $H'$ be its incenter. Let $F'$ be its A-excenter (we know that $BF'CH'$ is cyclic for obvious reasons, we call this circle $\omega$), let $Q'$ be the midpoint of the circumcircle arc containing $A$, let $M'$ be the second intersection of $Q'H$ with $\omega$ and let $K'$ be the intersection between $\Gamma^{'}$ and the perpendicular line wrt $HQ'$ passing through $Q'$. We want to show that the circumcircle of $F'M'K'$ is tangent to $Q'K'$ at $K'$. To do this, we'll show that $K'$ lies on the AXIS of $F'M'$. To show this is enough , once we defined $D$ as the center of $\omega$, to show that $K'D$ is orthogonal to $M'F'$. But $M'F'$ is orthogonal to $M'Q'$ ($HF'$ being a diameter) and $M'Q'$ is orthogonal to $Q'K'$ by definition. But $Q'D'$ is a diameter of $\Gamma^{'}$, hence done!!

I think it's easy for Problem 3: We have that two lemmas: \textbf{Lemma 1.} The points $Q$, $H$ and $M$ are collinear. \textbf{Lemma 2.} Let $N$ is the midpoint of $QH$ and a point $R$ lies on the line $BC$ such that $RH\parallel AQ$. Then $\angle RNO= 90^{\circ} $, where $O$ is the circumcenter of the triangle $ABC$. We can use by two lemmas and we easily can see that the circumcircles of the triangles $KHQ$ and $KFM$ are tangent.

Here's a solution with inversion WRT $Q$. First, let $A_0$ denote the antipode of $A$ in $\Gamma$; we know that $Q, H, M, A_0$ are collinear and $M$ is the midpoint of $HA_0$, so that $M'$ becomes the harmonic conjugate of $Q$ WRT $A'_0, H'$ (Note that $A', M', F'$ are collinear since $(QAMF)$ is cyclic): [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(15cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -4.800874036810980, xmax = 34.89250840395043, ymin = -7.788624316942659, ymax = 11.96084728234451; /* image dimensions */ pen qqttqq = rgb(0.000000000000000,0.2000000000000002,0.000000000000000); pen aqaqaq = rgb(0.6274509803921575,0.6274509803921575,0.6274509803921575); pen qqwuqq = rgb(0.000000000000000,0.3921568627450985,0.000000000000000); draw((558.4074274619950,-167.6931808709421)--(558.4848138116372,-168.2773169816810)--(559.0689499223762,-168.1999306320388)--(558.9915635727340,-167.6157945212999)--cycle, qqwuqq); draw((11.94814897609412,-2.701786992093843)--(11.83986216803928,-3.280991282635144)--(12.41906645858058,-3.389278090689988)--(12.52735326663543,-2.810073800148687)--cycle, qqwuqq); draw((12.24038502123578,-5.722950743110142)--(11.76393233897595,-5.376257096575377)--(11.41723869244118,-5.852709778835205)--(11.89369137470101,-6.199403425369972)--cycle, qqwuqq); draw((18.65735096727185,3.095730151442312)--(19.13380364953167,2.749036504907546)--(19.48049729606644,3.225489187167374)--(19.00404461380661,3.572182833702140)--cycle, qqwuqq); /* draw figures */ draw(circle((11.92149259612108,3.614427735914360), 7.082678004468662), linewidth(1.200000000000002) + qqttqq); draw((11.89369137470101,-6.199403425369972)--(16.80965888055248,-1.511014497759957), aqaqaq); draw((7.004365667404278,-1.483237463161803)--(11.89369137470101,-6.199403425369972), aqaqaq); draw((19.00404461380661,3.572182833702140)--(11.89369137470101,-6.199403425369972)); draw((11.53043691294743,-3.457446355699668)--(11.89369137470101,-6.199403425369972)); draw((5.406156668775245,-1.478709958916395)--(11.89369137470101,-6.199403425369972)); draw((5.406156668775245,-1.478709958916395)--(26.02563863289654,-1.537122089126370)); draw((19.00404461380661,3.572182833702140)--(26.02563863289654,-1.537122089126370)); draw((26.02563863289654,-1.537122089126370)--(8.649924021738130,-3.839056692143185)); draw((5.406156668775245,-1.478709958916395)--(14.14050284768400,-3.111664830980842)); draw((11.89369137470101,-6.199403425369972)--(12.52735326663543,-2.810073800148687)); /* dots and labels */ dot((19.00404461380661,3.572182833702140),dotstyle); label("$Q$", (19.11515275219922,3.738845041291063), NE * labelscalefactor); dot((7.004365667404278,-1.483237463161803),dotstyle); label("$C'$", (7.143250840395042,-1.205467117180268), NE * labelscalefactor); dot((16.80965888055248,-1.511014497759957),dotstyle); label("$B'$", (16.42078039617833,-1.122136013385808), NE * labelscalefactor); dot((11.89369137470101,-6.199403425369972),dotstyle); label("$H'$", (12.25422520645530,-6.177556310249753), NE * labelscalefactor); dot((26.02563863289654,-1.537122089126370),dotstyle); label("$A'$", (26.14274250553207,-1.372129324769190), NE * labelscalefactor); dot((15.30832042658575,-1.506761414320957),dotstyle); label("$A'_0$", (14.55971907810204,-1.177690082582115), NE * labelscalefactor); dot((14.14050284768400,-3.111664830980842),dotstyle); label("$M'$", (14.19861762832604,-3.455406919630706), NE * labelscalefactor); dot((11.53043691294743,-3.457446355699668),dotstyle); label("$F'$", (10.92092754574392,-3.260967677443630), NE * labelscalefactor); dot((5.406156668775245,-1.478709958916395),dotstyle); label("$K'$", (5.504405799103985,-1.316575255572882), NE * labelscalefactor); dot((8.649924021738130,-3.839056692143185),dotstyle); label("$M_0$", (8.143224085928571,-4.233163888379004), SW * 0.1*labelscalefactor); dot((12.52735326663543,-2.810073800148687),dotstyle); label("$X$", (12.64310369082944,-2.649872916284253), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy][/asy] Note that $A'(Q, M'; A'_0, H')$ is a harmonic pencil and $A'Q \parallel H'K' \Longrightarrow A'M'$ cuts $H'K'$ at its midpoint $M_0$. Define $X$ as the foot of the perpendicular from $H'$ to $M'K'$, so that $(H'M'XF')$ all lie on the circle with diameter $H'M'$. Now $\angle K' XF' = \angle F'H'M' = 180 - \angle K'M_0 F'$, i.e. $(K' X F' M_0)$ is cyclic. But note that $M_0 X^2 = M_0 H'^2 = M_0 F' \cdot M_0 M'$, i.e. $M_0 X$ is tangent to $(M' X F')$ and $\angle H' K' F' = \angle M_0 X F' = \angle K' M' F'$, i.e. $H'K'$ is tangent to $(F'K'M')$, i.e. $(KQH)$ is tangent to $(FKM)$ at $K$, as desired.

I can be found quickly that $Q$ is the intersection of $\Gamma$ and the ray $MH$. What happens if, by accident, we take the other intersection point? Instead of $Q$ and $K$, we will get the reflections of $H$ about $M$ and $F$, respectively and find that the two circles in the problem statement are trivially tangent to each other. Now let us draw both sides of the problem in the same picture. We can realize what object can connect the two parts...

I found out that $K$ is the reflection of $H$ in $AC$ and that the midpoint of $HQ$ is the reflection of the midpoint of $AC$ in the foot of the altitude from $B$. Is that correct?

My solution is quite similar to onetoone. I just posted as another way to prove that $XK$ is the common tangent of $(FKM)$ and $(QKH)$. Still call $D$ as the intersection of $AF$ and $(ABC)$. Let $E$ be the other endpoint of diameter through $A$ of $(ABC)$. We know that $DE$ parallel to $BC$. Let $L$ be the intersection of $QK$ and $DE$. Then we have $LK.LQ = LD.LE$, so L must be in axe radical of $(HDE)$ and $(HQK)$. But $(HDE)$ and $(HQK)$ tangent with each other at $H$, so $LH$ must be perpendicular to $QE$. So $LH$ tangent with $(QKH)$ Let $X$ as intersection of $BC$ and $LH$, then $X$ is midpoint of $HL$. Then we can go on like the proof of onetoone

My solution Of course $M, H, Q$ are collinear. Let $D, E$ be foot of altitudes from $B, C$. Notice that $EMCQ$ is cyclic quadrilateral and $MC=ME$, so we get $MC^2=MH\cdot MQ$.Let $J=AQ \cap BC$.We get that $E, D, J$ are collinear by power of point. Hence $(B, C; F, J)=1$, so $MF \cdot MJ= MC^2$. Let $P$ be second intersection of $MK$ with the circumcircle of $HKQ$. Consider inversion with center in $M$ and radius $MC$. This inversion takes the circumcircle of $MFK$ to the line $JP$ so it sufficies to show that $JP$ is tangent to the circumcircle of $HKQ$, which is mapped on its own. Notice that from Pascal theorem for $PPKQQH$ we need to show that the intersection of $HP$ and $QK$ lies on $BC$ since $AQ$ is tangent to circumcircle of $HKQ$. It is equivalent to the concyclity of $BCPH$. Let $R=BC \cap QP$. We get that $\angle PHF= \angle PRC$ since $ QP \perp HP$. Now we need to prove that $\angle BHF= \angle FPC$. Hovewer $\angle MCK= \angle MPC$ and $\angle ACK= \angle KQJ= \angle KPQ = \angle MPF$, so we are done.

This problem can obviously be crushed when one knows the configuration... But even if one doesn't here is a solution that doesn't really use anything: Let $KM$ meet $\odot HBC$ at $K'$, $QH$ meet $\odot HBC$ at $Q'$. $\odot HBC$ is simply $\odot ABC$ reflected about the midpoint of $BC \implies Q'K' \parallel QK \implies \odot HBC \cap \odot KQH = T \in KM$. Hence, by Radical-Axis theorem on $\odot ABC, \odot HBC, \odot KQH \implies KQ, HT, BC$ concur. Hence, the tangents at $K, H$ to $\odot KQH$ concur at $R$ on $BC$ (polarity). By similarity, $RK^2 = RH^2 = RF \cdot RM \implies RK$ is tangent to $\odot FKM$. I am not sure what the idea behind having this problem was, but certainly it was not there to challenge the proficient geometer. PS: By a bit of angle chasing, you can extend the first paragraph easily to solve the generalisation posted by LeVietAn.

another easy solution: its well known that it suffices to prove ^KFB+^MKH=^KQH and this is equivalent to prove ^QHK+^MKH=^KFH. we know that Q,H,M are collinear and QH intersects the circumcircle in antidiameter of A, name it A'.also we know that the reflection of H in to F is on the circumcircle,name it T.AQ=A'D because ADA'Q is a parallelogrom (D is diametrically opposite of Q) so ^KHQ=^KTA.so it suffices to prove that ^MKH=^TKF. it mean that we should prove that KM is antimedian in triangle KHT.and this is trivial because MH=MT and MH is tangent to KHT.done!!! I think this problem was just a few popular lemmas .

Notice that $Q$ is the $A$-Queue point. Take a negative inversion at $H$ swapping $A,F.$ This also swaps the other vertices with the feet of the altitudes from those vertices by PoP, so $(ABC)$ is sent to the nine-point circle. By a well-known lemma, $Q$ and $M$ are swapped, and $(KQH)$ is sent to the line through $M$ perpendicular to $MH.$ Thus, $K$ is sent to $K',$ the intersection of this line with the nine-point circle. Let $E$ be the midpoint of $AH,$ and let $D$ be the midpoint of $HQ.$ Then $M,E$ are antipodes on the nine-point circle. By homothety at $H,$ $D$ is on the nine-point circle, and $\angle K'MD=90$ so $K',D$ are antipodes as well. Therefore $K'E \parallel MD \perp AQ,$ and since $AE=EQ$ we get $AK'=K'Q,$ and $K'M \parallel AQ,$ so $K'M$ is tangent to $(AK'Q).$ This is the inverse of $(FKM)$ being tangent to $(KQH),$ so we are done.

Awesome! Let $A'$ be the $A$ antipode. Now the two properties of $K$ (on the circle through $Q$ and $H$ tangent to $AQ$ and on $ABC$) plus the well-known fact that $QHA'$ are collinear imply that it is the Miquel point of $QAA'H$. Now consider what happens to the points $M$ and $F$ under this $90$ degree spiral. $M$ gets mapped to the midpoint $F'$ of $AQ$, and $F$ goes to the foot from $Q$ to the line through $R$ parallel to $AH$ $M'$. Now it suffices to check $(KM'F')$ and $(KQH)$ are orthogonal. Let $G$ be the foot from $Q$ to $AH$. Then $M'$ is the midpoint of $QG$ and $G \in KQH$. Now since $QG \perp F'M'$, it suffices to check that $KM'$ and $KF'$ are isogonal. $KM'$ is a median, so it suffices to show $KF'$ is a symmedian. However, this is clear since $F'Q$ is tangent to $(QGHK)$ and $F'G=F'Q$ since $F'$ is circumcenter of $(AQG)$.

We consider an inversion around $H$ followed by a reflection around $H$, and we have the radius of the inversion such that the circumcircle of $ABC$ maps to the nine-point circle of $ABC$. Notice that we have $A$ mapping to $F$, $Q$ mapping to $M$, and we let $K$ map to $X$ on the nine-point circle. We let $Y$ and $Z$ be the midpoint of $AH$ and $QH$ respectively. We can see that $M$ is the point closest to $H$ thus $HM \perp MX$. We simply need to show that the circumcircle fo $QAX$ is tangent to $MX$. Thus since $QA \parallel MX$, we simply need that $YX \perp MX$. We claim that $YZMX$ is a concyclic and thus a rectangle. This is because $Y$ and $Z$ get mapped to the circumcircle of $ABC$. Thus We have shown that $YX \perp MX$. $\blacksquare$

Invert at $H$ with radius $\sqrt{HA \cdot HF}$ composed with a reflection about $H$, which takes the circumcircle to corresponding points on the nine-point circle. Then $A$, $F$ as well as $Q$, $M$ are swapped. $K$ is mapped to $KH \cap (\text{9 pt})$, which we denote $L$, and not on ray $HK$. $\angle HML = \angle HKQ = 90$. $(KQH) \mapsto ML$, $(FKM) \mapsto (AQL)$. We claim $\triangle ALQ$ is isosceles, which is equivalent to proving the perpendicular bisector and altitude to $AQ$ are the same. Define $O$, $U$, and $V$ as the centers of $(ABC)$, $(HQA)$, and $(KQH)$. We find $U$, $V$ lie on the nine-point circle by homothety at $H$. $MLUV$ is a rectangle, which then implies that $O$ lies on $LU$. $LO \perp UV \parallel AQ$, which finishes. Thus $\triangle ALQ$ is isosceles, from which we deduce that $ML$ is a tangent. $\blacksquare$

Let $X$ be the antipode of $A$. It can easily be shown that $X,M,H,Q$ are collinear. Let $P$ be the intersection of $AF$ with $\Gamma$. Claim $1$ : $(XHP)$ and $(MHF)$ are internally tangent, and also externally tangent to $(HKQ)$. Follows from the fact that $XH$, $MH$ and $HQ$ are the diameters of the respective circles and all four points are collinear. Claim $2$ : $(HKP)$ is tangent to line $XQ.$ Let $I$ be the antipode of $Q$ w.r.t $\Gamma.$ Then $Q,O,I$ and $K,H,I$ are collinear. Note that $AQXI$ is a rectangle. Thus, $$ \angle XHP = \angle MHP = \angle IAP = \angle IKP = \angle HKP,$$ as needed. Let $N$ be the center of $(PHK).$ Note that $N$ lies on $BC$ as $MH = MP$. Furthermore, by claim $2$ it follows $NH$ and $NK$ is tangent to $(HKQ)$. Thus $NH^2 = NK^2$. Furthermore, by claim $1$, we get that $NH^2 = NF \cdot NM$. Hence, $NK^2 = NF \cdot NM.$ Which gives our desired tangency.

Let $QT,AD$ be diameters of $\Gamma$ and let $S$ be the midpoint of $HT$. Since $BHCD$ is a parallelogram $D,M,H$ are collinear and $\angle DQA=90$ so $D,M,Q,H$ are collinear. Since $\angle TKQ=90$ we have $T,H,K$ collinear. We have $HQ\cdot HM=HQ\cdot HD/2=HK\cdot HT/2=HK\cdot HS$ so $QKMS$ is cyclic. Also $\angle MQA=\angle MFA=90$ so $MFQA$ is cyclic. Since $OS||QH$ and $\angle HQA=90$, $OS$ is the perpendicular bisector of $AQ$. Thus $\angle SAQ=\angle SQA=\angle MSQ$ (since $MS||TD||AQ$). Let $AH$ meet $\Gamma$ again at $L$. We have $HF\cdot HA=HL\cdot HA/2=HK\cdot HT/2=HK\cdot HS$ and so $KFSA$ is cyclic. Now we have that $$\angle KMF+\angle FKH=\angle HMF-\angle HMK+\angle FAS=\angle HAQ-\angle QSK+\angle FAS=\angle SAQ-\angle QSK=\angle QSM-\angle QSK=\angle MSK=\angle KQH$$. Now if $l$ is tangent to the circumcircle of $KFM$ at $K$ with a point $P$ on $l$ on the same side of $KF$ as $L$ we get $\angle PKH=\angle FKH+\angle FKP=\angle FKH+\angle KMF=\angle KQH$ and hence $l$ is tangent to the circumcircle of $KQH$ so the $2$ circles are tangent.

Define $AH \cap BC = D$, $BH \cap AC = E$ and $CH \cap AB = F$, $EF \cap BC = T$, $A´$ the $A$-antipode, $U = BB \cap CC$. Also, let $S$ be the non-$A$ point on $(ABC)$ such that $TA=TS$, $TS \cap (ABC) = R$ and $TK \cap (ABC) = L$. By simple angle chasing, we obtain: $(QKH)$ tangent to $(KDML)$ $\leftrightarrow$ $\angle KDL = \angle QHK + \angle QKL $. So, it´s enough to prove that D is $LR \cap KS$, because it´s implies $\angle KDL = \widehat{RS} + \widehat{KL} \implies \angle KDL = \angle QHK + \angle QKL$. Now, let´s again reduce the problem: $D$ is $LR \cap KS$ $ \leftrightarrow$ $KR \cap LS$ = $U$ (Consequence of Brocard´s theorem) And $KR \cap LS = U \leftrightarrow K-R-U$ (Can easily be proved projecting through T or by ratio lemma) Then, our main objective is to prove $K-R-U$ i.e. prove that line $KR$ is symmedian of $\triangle BKC$ $\leftrightarrow \angle BKR = \angle MKC \leftrightarrow \widehat{BR} = \angle MKH - \angle CKH \leftrightarrow \widehat{BR} = \angle MKH - (\widehat{QA} - \widehat{CA`}$ $(@)$ Notice that we can obtain $\widehat{BR}$ as follows: $\angle TAS = 90 - \angle ATO \implies \widehat{SQ} = 90 - \angle ATO \implies \widehat{BR} = 90 - \angle ATO - \widehat{QB} - \widehat{RS} \implies \widehat{BR} = 90 - \angle ATO - \widehat{QB} - \widehat{QA} \implies \widehat{BR} = 90 - \angle ATO - \angle C$ Replacing in $(@)$, we need to prove: $ 90 - \angle ATO - \angle C = \angle MKH - (\widehat{QA} - 90 + \angle B) \leftrightarrow 90 - (\angle ATC - \angle OTM) - \angle C = \angle MKH - \widehat{QA} + 90 - \angle B \leftrightarrow 90 - (\angle B - \widehat{QB} -\angle OTM) - \angle C = \angle MKH -\widehat{QA} + 90 - \angle B \leftrightarrow 90 + \widehat{QB} - \angle B + \angle OTM = 90 + (\angle C - \widehat{QA}) - \angle B + \angle MKH \leftrightarrow \angle OTM = \angle MKH $ But it is well known that D is the ortocenter of $\triangle OTU \implies \angle OTM = \angle DUM $, so it´s suffices prove $\angle MUD = \angle MKH$!!!!! Final part: Trigonometry $\angle QCA = x$, $\angle MKH = y$, $\angle DLM = z$ and WLOG the radius of $(ABC) = 1/2$ LS=Law of sines 1) Calculating $cotgy$ in function of $cotg x$ Classic ratio lemma in $\triangle HKA'$ with $KM$; $1 = \frac{KA´}{KH} \cdot \frac{sen(x-y)}{seny}$ ($\frac{KA´}{KH} = \frac{AA´}{QH}$ because $\triangle AKA´ \sim \triangle KQH$) $\implies 1 = \frac{AA´}{QH} \cdot \frac{sen(x-y)}{seny} \implies QH = \frac{sen(x-y)}{seny}$ But $ QH \cdot HM = AH \cdot HD \implies QH = \frac{AH}{HA´} \cdot 2HD$, using LS in $\triangle AHA´$, $\implies \frac{senx}{sen(\angle B - \angle C)} \cdot 2HD = \frac{sen(x-y)}{seny}$ (EZ to get $HD = cos\angle B \cdot cos \angle C$) $\implies \frac{2senx}{sen(\angle B - \angle C)} \cdot cos\angle B \cdot cos\angle C = senx \cdot cotgy - cosx$ Finally: $cotgy = \frac{2cos\angle B \cdot cos\angle C}{sen(\angle B - \angle C)} + cotgx$. 2) Calculating $cotgx$ LS in $\triangle CEQ$; $\frac{QE}{senx} = \frac{QC}{sen(\angle B - \angle C + x)} \implies \frac{sen(\angle B - \angle C + x)}{senx} = \frac{QC}{QE}$ But $\triangle QEF \sim \triangle QCB \implies \frac{QC}{QE} = \frac{BC}{EF} = \frac{1}{cos \angle A}$, doing the same generic procedure: $cotgx = \frac{\frac{1}{cos \angle A} - cos(\angle B -\angle C)}{sen(\angle B - \angle C)}$ Consequently $cotgy = \frac{2cos \angle B \cdot cos \angle C - cos(\angle B - \angle C) + \frac{1}{cos \angle A}}{sen(\angle B - \angle C)}$ $\implies cotgy = \frac{sen \angle A}{cos \angle A \cdot sen(\angle B - \angle C)}$ 3) Calculating $cotgz$ Ratio lemma in $\triangle LMB$ with $LD$; $\frac{BD}{DM} = \frac{BL}{BM} \cdot \frac{sen(\angle (90 - A) - z)}{senz}$ $...$ $\implies cotgz = sen \angle A \cdot \frac{\frac{2sen\angle C cos\angle B}{sen \angle A - 2sen \angle C sen \angle B} + 1}{cos \angle A}$ 4) Concluding $y = z$ $\leftrightarrow cotgy = cotgz$ $\leftrightarrow sen \angle A \cdot \frac{\frac{2sen\angle C cos\angle B}{sen \angle A - 2sen \angle C sen \angle B} + 1}{cos \angle A} = \frac{sen \angle A}{cos \angle A \cdot sen(\angle B - \angle C)}$ $\leftrightarrow 2sen\angle C cos\angle B sen(\angle B - \angle C) = (sen\angle A - 2sen\angle C cos\angle B)(sen\angle A - sen\angle (\angle B - \angle C))$ Using $sen\angle A = sen\angle (\angle B + \angle C)$: $\leftrightarrow 2sen\angle C cos\angle B sen(\angle B - \angle C) = (sen\angle (\angle B + \angle C) - 2sen\angle C cos\angle B) \cdot 2sen\angle C cos \angle B$ $\leftrightarrow sen\angle (\angle B - \angle C) = sen\angle B cos\angle C - sen\angle C cos\angle B$, which is true $\implies$ in fact $\angle MUD = \angle MKH$

Let $X$ be the second intersection of $(QKH)$ and $(BHC)$. An inversion $\phi$ at $M$ with radius $MB$ swaps $(BHC)$ with $(BAC)$. It is well-known that $Q$, $H$, and $M$ are collinear. Thus the inversion sends $(QKH)$ to $(QKH)$. Therefore, $K$ and $X$ are swapped, so $K$, $X$, and $M$ are collinear. Thus, $M=QH\cap KX$. Let $P=QX\cap KH$. Let $N=QK\cap HX$. By the radical center theorem, $N$ is on $BC$. By Brocard's theorem, the polar of $P$ wrt $(QKH)$ is $MN$, which is $BC$. Let $F'$ be the pole of $QX$ with respect to $(QKH)$. By La Hire's theorem, since $P$ is on $QX$, $F'$ is on $BC$. Since $F'Q$ is tangent to $(QKH)$, we have $\angle MQF'=\angle HQF'=90=\angle MFH$. Since $\phi$ swaps $H$ and $Q$, it also swaps $F$ and $F'$. Also, $\phi$ sends $K$ to $X$. Therefore, $\phi$ sends $(KFM)$ to $F'X$. It suffices to show $F'X$ is tangent to $(QKH)$. This is true because $F'$ is the pole of $QX$.

2015 IMO p3 Consider negative inversion at $H$, Let $R$ be the midpoint of $HQ$, let $H’$ be the midpoint of $AF$, Let $M$ be the midpoint of $BC$, $N$ be a point on nine point circle such that $QH \perp NM$. Our inversion will send $A$ to $F$, $K$ to $N$ $Q$ to $M$. We want to show that $MN$ is tangent to $(QAN)$. We claim that $NMRH’$ is a rectangle, proving that we get that $NA=NQ$, as $MN \| AQ$, this implies the problem.

Basically the same as Evan's solution in $EGMO$. I originally had a different finish after the inversion but then I never got around to writing the solution until $OTIS$ causing me to forget what I'd done so this is just for storage basically.

20 minute solve, took time to find the right inversion. Note that $Q$, $H$ and $M$ are collinear from reflecting $H$ about $M$ to $H'$ and noting that $AH'$ is a diameter of $(ABC)$. Now consider the inversion with power $AH \cdot HF$, which swaps the following pairs of points: The pairs $(A, F)$, $(B, E)$ and $(C, D)$ where $D$ and $E$ are the feet from $B$ and $C$ to $AC$ and $AB$ respectively. The pair $(Q, M)$, as $\angle HQA = \angle HFM = 90$. The pair $(K, N)$, where $N$ denotes the point on $(DEF)$ such that $\angle HMN = 90$. Hence we wish to show $MN$ is tangent to $(NAQ)$ - however noting that $MN \parallel AQ$ as they are both perpendicular to $QM$, it follows that it suffices to show $NA = NQ$. This would follow immediately if we could show $2MN = AQ$, yet this is clear from the homothety at $H$ with scale factor $2$ sending $(DEF)$ to $(ABC)$, finishing. $\square$

I inverted at H

So this took about 10 minutes. Let $O$ be the circumcenter, $A'$ be the $A$ antipode, $N$ be the midpoint of $AH$ and $T$ lies on the nine-point circle such that $TM\perp HM$. It is known that $\overline{A'-M-H-Q}$. Now invert at $H$ with radius $\sqrt{-HA\cdot HF}$. It's easy to see that $Q\mapsto M$, $K\mapsto T$ and $A\mapsto F$. After inversion, it suffices to prove that $MT$ is tangent to $(AQT)$. From middle lines, $ON$ is parallel to $A'Q$, hence $ON\perp AQ$ so $ON$ is the perpendicular bisector of $AQ$. Now notice that $NM$ is a diameter in the nine-point circle so $\angle NTM=\angle TMH=90^\circ$ , or $NT\parallel MH$. Therefore, we get $\overline{T-O-N}$ and since $ON$ is the perpendicular bisector of $AQ$, we get the desired conclusion. $\blacksquare$

Invert at $H$. Let $N^*$ be the minor arc midpoint of arc $B^{*}C^*$ in $(A^*B^*C^*)$. Since $N^*$ and $Q^*$ are antipodal arc midpoints in $(A^*B^*C^*)$, it follows that $\angle N^* K^* Q^* = 90^{\circ}$. We also have $\angle K^* Q^* H = 90^{\circ}$ by definition. It's well known that the uninverted $M$, $H$ and $Q$ are collinear, so $M^*$, $H$ and $Q^*$ are collinear – since $M^*$ lies on $(HB^*C^*)$, it follows that $M^*$ is the foot from $F^*$ to line $HQ^*$. Now, what we want to show is clearly true: $\overline{Q^*K^*} \parallel \overline{M^* F^*}$ and $M^* K^* = F^* K^*$ since line $N^* K^*$ is the perpendicular bisector of $\overline{M^* F^*}$. Therefore, $(F^* K^* M^* )$ is tangent to line $KQ$.

reim's spam lol Note that it suffices to prove that $\angle{MFK}-\angle{QHK}=\angle{MKQ}=90^\circ+\angle{HKM}$. Let $O$ be the circumcenter of $\triangle{ABC}$. Define $A',Q'$ to be the antipodes of $A,Q$, respectively .Notice that $Q$ is the $A$-queue point of $\triangle{ABC}$, so it's well known that $\overline{Q,H,M,A'}$. Moreover, $\overline{K,H,Q'}$. Let $J$ be the midpoint of $Q'H$. Note that $AF\cap(ABC)$ is the reflection of $H$ over $BC$ and $A'$ is the reflection of $H$ over $M$. Using Reim's Theorem 2 times, we conclude that $AJFK$ and $MJQK$ are cyclic. Let $\angle{HQK}=\alpha$ and $\angle{JQM}=\beta$. Note that $J$ must be lying on the mid parallel line of $AQ$ and $A'Q'$ (which passes through $O$), so $\angle{AJQ}=2\angle{JQM}=2\beta$. We have $JO\parallel HQ$, so $\angle{AFK}=\angle{AJK}=\angle{AJO}+\angle{OJK}=\frac{\angle{AJQ}}{2}+\angle{QHK}=90^\circ-\alpha-\beta$, then $\angle{MFK}=90^\circ+\angle{AFK}=180^\circ-\alpha+\beta$. Lastly, using information that $\angle{QHK}=90^\circ-\alpha$, we can have $\angle{MFK}-\angle{QHK}=90+\beta=90^\circ+\angle{JQM}=90^\circ+\angle{JKM}=90^\circ+\angle{HKM}$, as desired. $\blacksquare$ P2 of GOWACO 2021 was definitely much harder than this one