Absolutely trivial for G4

Here's a kind of long, but "novel" solution I found Let $J$ be the excenter opposite $A$ and $K$ the second intersection of $AD$ and $\omega$. We first show that $P, M, Q, K$ are concyclic. To this effect we define $X$ to be the midpoint of $DK$. Then $JX$ is perpendicular to $DK$ and $X$ lies on the circumference with diameter $AJ$, which is the circumcircle of $AEF$. Thus by Power of a Point from $D$ to $\odot AFE$ and $\omega$ we have $$DK \cdot DM = 2DX \cdot DM = DX \cdot DA = DP \cdot DQ$$ Thus $K$ lies on circle $MPQ$, as desired. We now wish to show that circle $KPQ$ is tangent to $\omega$. Let $M$ be the midpoint of $EF$ and let $KM$ cut $\omega$ again at $D'$. Because $DEKF$ is harmonic, we have $DD' \parallel EF$. Now invert about $\omega$. Let $P_0$ and $Q_0$ be the inverses of $P$ and $Q$, we now wish to prove that circle $KP_0Q_0$ is tangent to $\omega$. These points lie on the circumcircle of $JEF$, which maps to line $EF$, and on the tangent to $\omega$ at $D$, which maps to the circle with diameter $JD$. Let $DP_0$ and $DQ_0$ cut $\omega$ again at $P'$ and $Q'$. Then because $JP_0$ and $JQ_0$ are perpendicular to $DP'$ and $DQ'$ we find that $P_0$ and $Q_0$ are the midpoints of $DP'$ and $DQ'$. Now let line $KM$ cut $P'Q'$ at $H$. We know that $M$ is the midpoint of $EF$ and that $P_0Q_0$ is the $D$-midline in triangle $DP'Q'$. From this is it easy to show that $H$ is the foot of the $D$-altitude in triangle $DP'Q'$. However, it is known from the configuration of 2011 G4 applied to triangle $DP'Q'$ that the circumcircle of $KP_0Q_0$ is tangent to $\omega$, so we are now done.

Notice that $MP=MQ$. This follows from the fact that the center of $\odot (AEF)$ is the mid point of $AI_A$ where $I_A$ is the $A-$ excenter, and from mid point theorem. Let $X$ be the foot of perpendicular from $I_A$ to $AD$. Notice that $X$ lies on $\odot (AEF)$. So, if $AD$ intersects $\omega$ at another point $D'$, we see that $X$ is the mid point of $DD'$. Now obviously, from power of point, $DP\cdot DQ=DX\cdot DA = DM\cdot DD'$. So, $M,P,Q,D'$ are cyclic. Also, from the well known fact that the mid point of arc lies on the chord of contact of a circle inscribed in a segment, we see that $\odot (MPQ) $ and $\omega$ touch at $D'$.

Let $\overline{AD}$ meet $\omega$ again at $T$. Then $DFET$ is a harmonic quadrilateral due to the tangents at $A$, and we may let $R = \overline{EF} \cap \overline{DD} \cap \overline{TT}$. Moreover, let $N$ be the midpoint of $\overline{DT}$. [asy][asy]size(320); pair A = dir(130); pair B = dir(210); pair C = dir(330); pair I = incenter(A, B, C); pair Z = dir(270); pair J = 2*Z-I; pair D = foot(J, B, C); pair E = foot(J, C, A); pair F = foot(J, A, B); path w = circumcircle(A,E,F); path l = (100*B-99*C)--(100*C-99*B); pair P = IP(l, w); pair Q = OP(l, w); pair N = foot(J, A, D); pair M = midpoint(A--D); pair T = 2*N-D; draw(A--B--C--cycle, red); draw(w, heavycyan); draw(C--E, red); draw(F--B, red); draw(circumcircle(D, E, F), heavygreen); pair R = extension(P, Q, E, F); draw(P--R, red); draw(F--R, heavygreen); draw(T--R, heavygreen); draw(J--R, dashed+heavygreen); draw(A--T, red); draw(circumcircle(M, P, Q), dashed+blue); dot("$A$", A, dir(A)); dot("$B$", B, dir(120)); dot("$C$", C, dir(50)); dot("$J$", J, dir(J)); dot("$D$", D, dir(45)); dot("$E$", E, dir(350)); dot("$F$", F, dir(220)); dot("$P$", P, dir(P)); dot("$Q$", Q, dir(30)); dot("$N$", N, dir(350)); dot("$M$", M, dir(230)); dot("$T$", T, dir(T)); dot("$R$", R, dir(R)); /* TSQ Source: A = dir 130 B = dir 210 R120 C = dir 330 R50 I := incenter A B C Z := dir 270 J = 2*Z-I D = foot J B C R45 E = foot J C A R350 F = foot J A B R220 ! path w = circumcircle(A,E,F); ! path l = (100*B-99*C)--(100*C-99*B); P = IP l w Q = OP l w R30 N = foot J A D R350 M = midpoint A--D R230 T = 2*N-D A--B--C--cycle 0.1 lightred / red w 0.1 lightcyan / heavycyan C--E red F--B red circumcircle D E F 0.1 lightgreen / heavygreen R = extension P Q E F P--R red F--R heavygreen T--R heavygreen J--R dashed heavygreen A--T red circumcircle M P Q dashed blue */ [/asy][/asy] Note that $MPQT$ is cyclic by power of a point since $DM \cdot DT = \frac{1}{2} AD \cdot DT = DN \cdot DA$. Now $RP \cdot RQ = RE \cdot RF = RT^2$ by power of a point again, hence $\overline{RT}$ is tangent to the circumcircle of $\triangle MPQ$.

Here's a way to motivate the tangency point if you misread the question like I did; this is probably much longer than most other solutions but tbh I was just happy to solve a modern G4 . WLOG let the points be $P$, $B$, $C$, $Q$ in that order. Suppose that instead we look at the circle $\Gamma$ with center $A$ and radius $AE=AF$; let it intersect $BC$ at $P'$ and $Q'$. An inversion $\Phi$ about $\Gamma$ fixes $\omega$ (due to orthogonality) and line $BC\equiv P'Q'$ is swapped with $\gamma := \odot(AP'Q')$. Since $BC$ is tangent to $\omega$ at $D$, $\gamma$ is tangent to $\omega$ at $\Phi(D)=: T$, which is the second intersection point of $AD$ with $\omega$. Futhermore, $AP'=AQ'$, so $A$ is the top point of $\gamma$ when $BC$ is horizontal. At this point we have our circles $\omega$ and $\gamma$ internally tangent at $T$ and with top points $D$ and $A$ respectively. We wish to prove that a circle passing through $M$, the midpoint of $\overline{AD}$, is also tangent to $\omega$. Combined with a (maybe not even) careful diagram this highly suggests that perhaps $\odot(MPQ)$ is the "halfway" point in the homothety centered at $T$ sending $\omega$ to $\gamma$ (vis a vis gliding principle). It turns out this is the case. Now for more details. In order to prove this, it suffices to show the following two assertions: $M$, $P$, $Q$, and $T$ are concyclic; $MP=MQ$. (The first one asserts that indeed $\odot(MPQ)$ passes through $T$, while the second implies that $M$ is the top point of said circle.) To prove (1), let $N := \odot(APQ)\cap DT$, and note that $N$ is the midpoint of $\overline{DT}$ since $\angle ANI_A = 90^\circ$ (for $I_A$ the $A$-excenter). Thus \[DM\cdot DT = (\tfrac12 AD)\cdot (2DN) = AD\cdot DN = BD\cdot DC,\]so the converse of Power of a Point establishes the concyclicity. Showing (2) is a bit harder, and I honestly couldn't find a good way to do it, but whatever. Let $K$ be the foot of the altitude from $A$ to $BC$. Because $MP$ and $MQ$ are medians of $\triangle APD$ and $\triangle AQD$ respectively, proving $MP=MQ$ is equivalent to showing that \begin{align*}\tfrac14(AP^2+PD^2 - 2AD^2) = \tfrac14(AQ^2+QD^2 - 2AD^2)\quad&\Leftrightarrow\quad AP^2 + PD^2 = AQ^2 + QD^2\\ &\Leftrightarrow \quad PD^2 - QD^2 = AQ^2 - AP^2 = KQ^2 - KP^2\\ &\Leftrightarrow\quad PD - QD = KQ - KP. \end{align*}To compute this, use Power of a Point on $B$ and $C$ with respect to $\odot(APQ)$ to get $b(s-b) = CQ(a+BP)$ and $c(s-c) = BP(a+CQ)$. As a result, \begin{align*} PD - QD &= (PB - QC) + (BD - DC) = \frac{c(s-c) - b(s-b)}a + (s-c) - (s-b)\\ &= \frac{b(s-b) - c(s-c)}a + \frac{b^2-c^2}a = (QC-BP) + (KC - KB) = KQ - KP. \end{align*}Huzzah. $\blacksquare$

Define $U$ to be the midpoint of $AI_A$ and $R$ to be the projection of $M$ on $BC$. Also, let $I_AD \cap \odot(AFE)=K$, where $I_A$ is the $A-\text{excenter}$. Finally, let $AM$ meet $\omega$ again at $X$ We will show that $X$ is the desired tangency point. Key Lemma: $X$ lies on $\odot(MPQ)$. Proof: Clearly, it suffices to show $DM \cdot DX=DP \cdot DQ$. But $\odot(AFE)$ gives $DP \cdot DQ=DI_A \cdot DK$. Hence, it suffices to show $\triangle MKD \sim \triangle I_AXD$. But since $M$ is the midpoint of the hypotenuse of $\triangle KAD$, hence $MKD$ is isosceles, and the desired similarity follows since $I_AXD$ is also isosceles. $\square$ Now there are three ways to finish: Method 1: Clearly, $U$ is the center of $\odot(AEF)$. Also, note that $MR \parallel I_AD$, and since $MU$ is the $A-\text{midline}$ of $\triangle AI_AD$, hence $U$ lies on $MR$. Thus $MR \equiv UR$ is the perpendicular bisector of $PQ$.

$\blacksquare$ Method 2: Let $V=AX \cap FE$ and $Z=FE \cap BC$. Let $\omega$ and $\odot(MPQ)$ intersect for the second time at $X'$. Then $$(F,E;D,X)=-1=(B,C; D,Z) \overset{A}{=} (F, E; V, Z) \overset{X}{=} (F, E; D, X')$$Hence $X=X'$, as desired. $\blacksquare$ Method 3: Since $FEDX$ is harmonic, hence $ZX$ is tangent to $\omega$. Also, the radical axis of $\omega, \odot(AFE), \odot(MPQ)$ concur at $PQ \cap EF=Z$, implying that $Z$ lies on the radical axis of $\omega, \odot(MPQ)$. But by the above result, we get that $ZX$ is tangent to $\omega$, and hence is also tangent to $\odot(MPQ)$ $\blacksquare$

Communicated to me by tarzanjunior: Invert it once to get mixtilinear incircle. Then, invert it again(with centre as touchpoint). End it by PoP(the tangency pt. becomes mid pt.) It was India TST too(marks were not given to partial solution of inversion though)

Let the $A$-excenter be $I_A.$ We claim that the $A$-excircle and $\odot(MPQ)$ are tangent on $\overleftrightarrow{AD}.$ Indeed, let $\overleftrightarrow{AD}$ intersect the $A$-excircle again at $X$ and $\odot(AEF)$ again at $Y.$ Then from $\angle AI_AY=90^{\circ},$ we get $DY=YX.$ Thus, by considering power at point $D,$ we see that $$DP\cdot DQ=DA\cdot DY=2DM\cdot \frac{DX}{2}=DM\cdot DX,$$implying that $PMQX$ is cyclic. We now show that $MP=MQ,$ which would imply the tangency by Archimedes' Lemma. Let the perpendiculars from $M,A$ to $\overline{BC}$ be $U,V$ respectively. By considering power of points $B,C$ WRT $\odot (AEF),$ we get $$\begin{cases}BP\cdot (a+CQ)=BP\cdot BQ=BA\cdot BE=c\cdot (s-c) \\ CQ\cdot (a+BP)=CP\cdot CQ=CA\cdot CF=b\cdot(s-b)\end{cases}\implies \boxed{BP-CQ=\frac{(c-b)(s-c-b)}{a}}$$upon subtraction. Also, $$BU=\frac{BV+BD}{2}=\frac{c\cos B+s-c}{2}=\boxed{\frac{2a^2+(a-b-c)(b-c)}{4a}},$$and similarly $CU=\frac{2a^2+(a-b-c)(c-b)}{4a}.$ Finally, \begin{align*}PU-QU&=(PB-QC)+(BU-CU) \\&=\frac{(c-b)(s-c-b)}{a}+\frac{(a-b-c)(b-c)}{2a} \\&=\dfrac{(c-b)(a-b-c)+(b-c)(a-b-c)}{2a}=0,\end{align*}implying $MP=MQ,$ as desired.

This is the same as the outline in post 9 but I will post this anyway. Invert about a circle centered at $A$ with radius $\sqrt{AB \cdot AC}$ followed by a reflection across the $A$-angle bisector. Denote $X'$ as the image of $X$ after this transformation. Then we have the following: $\omega$ maps to the $A$-mixtilinear incircle. $D$ maps to the $A$-mixtilinear touch point. $E$ and $F$ map to points on $AC$ and $AB$, respectively, tangent to $\omega'$. $M$ maps the the reflection of $A$ over $D'$. $P$ and $Q$ map to the intersections between $(ABC)$ and line $E'F'$. [asy][asy] size(7cm); pair A = dir(130); pair B = dir(210); pair C = dir(330); draw(A--B--C--cycle, red); pair I = incenter(A, B, C); pair O = circumcenter(A, B, C); pair M_A = -A+2*foot(O, A, I); pair M_B = -B+2*foot(O, B, I); pair M_C = -C+2*foot(O, C, I); draw(unitcircle, blue); pair L = dir(90); pair T = -L+2*foot(O, I, L); pair F= extension(T, M_B, A, C); pair E = extension(T, M_C, A, B); draw(circumcircle(T,E,F), heavygreen); pair D = B+C - foot(I,B,C); pair Q=intersectionpoints(((E-F)*3+F)--((F-E)*3+E), circumcircle(A,B,C))[0]; pair P=intersectionpoints(((E-F)*3+F)--((F-E)*3+E), circumcircle(A,B,C))[1]; pair M = (A+D)/2; pair M_1=2*T-A; draw(circumcircle(M_1,P,Q), heavygreen); draw(A--D^^A--M_1, orange); draw(Q--P, lightred); dot("$A$", A, NW); dot("$B$", B, SW); dot("$C$", C, SE); dot("$D$", D, SE); dot("$F'$", E, dir(120)); dot("$E'$", F, NE); dot("$P'$", P, NE); dot("$Q'$", Q, dir(210)); dot("$M$", M, W); dot("$M'$", M_1, SW); dot("$D'$", T, SW); [/asy][/asy] Then it suffices to show that $\omega'$ is tangent to $(P'Q'M')$. Now invert about $D'$ with arbitrary radius and denote $X_1$ as the image of $X$ after this transformation. Then we have the following: $(ABC)$ and $\omega'$ become parallel lines $F'_1E'_1$ and $B_1A_1C_1$, respectively. $AB$ and $AC$ map to the circumcircles of $A_1D'B_1F'_1$ and $A_1D'C_1E'_1$ (call them $\omega_b$ and $\omega_c$), respectively which are both tangent to line $F'_1E'_1$. $M'$ maps to the reflection of $A_1$ across $D'$. $P'$ and $Q'$ map to the intersections of $(D'E_1'F_1')$ with line $B_1C_1$. Now it suffices to show that $(P'_1M'_1Q'_1)$ is tangent to line $F'_1E'_1$. [asy][asy] size(9cm); pair F_1 = (-1,1); pair E_1 = (1,1); pair M = (0,1); draw(circle(E_1+(0,-0.85), 0.85), red); draw(circle(F_1+(0,-1.5), 1.5), red); draw((-3.5,1)--(3,1), heavygreen); pair A_1=intersectionpoints(circle(F_1+(0,-1.5), 1.5),circle(E_1+(0,-0.85), 0.85))[0]; pair D_1=intersectionpoints(circle(F_1+(0,-1.5), 1.5),circle(E_1+(0,-0.85), 0.85))[1]; draw(A_1-(3.7,0)--A_1+(2.8,0), blue); draw(circumcircle(E_1,F_1,D_1), orange); pair Q_1 = intersectionpoints(circumcircle(E_1,F_1,D_1),A_1-(3.7,0)--A_1+(2.8,0) )[0]; pair P_1 = intersectionpoints(circumcircle(E_1,F_1,D_1),A_1-(3.7,0)--A_1+(2.8,0) )[1]; pair M_1 = 2*D_1-A_1; draw(circumcircle(M_1,Q_1,P_1), heavygreen); pair X = 2*M-A_1; draw(M_1--X, orange); pair B_1 = intersectionpoints(circle(F_1+(0,-1.5), 1.5),A_1-(3.7,0)--A_1+(2.8,0) )[1]; pair C_1 = intersectionpoints(circle(E_1+(0,-0.85), 0.85), A_1-(3.7,0)--A_1+(2.8,0))[0]; dot("$A_1$", A_1, dir(35)); dot("$B_1$", B_1, NW); dot("$C_1$", C_1, NE); dot("$E_1'$", E_1, NE); dot("$F_1'$", F_1, NW); dot("$D'$", D_1, dir(-60)); dot("$P_1'$", P_1, SE); dot("$Q_1'$", Q_1, SW); dot("$X$", X, N); dot("$M$", M, NW); dot("$M'_1$", M_1, SE); draw(Q_1--M_1--P_1, red); label("$\omega_b$", circumcenter(B_1,F_1,A_1)+1.8*dir(200)); label("$\omega_c$", circumcenter(C_1,E_1,A_1)+1.2*dir(-20)); [/asy][/asy] Let line $A_1D'$ intersect $F'_1E'_1$ at $M$. Since $A'D'$ is the radical axis of $\omega_b$ and $\omega_c$, we see that $M$ is the midpoint of $F'_1$ and $E'_1$. Now let $X$ be the reflection of $A'$ over $M$. By PoP, \[MA_1 \cdot AD' = ME_1'^2 \iff MX \cdot MD' = MF_1' \cdot MD_1' \]so $XF'_1E'_1D'$ is cyclic. By PoP again, \[ XA_1 \cdot A_1D' = A_1Q_1' \cdot A_1P_1' \iff MA_1 \cdot A_1M_1' = A_1Q'_1 \cdot A_1P_1' \]so $M_1'P'_1MQ'_1$ is cyclic. Since $F'_1E'_1 \parallel Q_1P'_1$, quadrilateral $F'_1Q'_1P'_1E'_1$ is an isosceles trapezoid so $MQ'_1=MP'_1$ which implies that $MQ'_1=MP'_1$. Finally, this implies that \[\angle F'_1MQ'_1 = \angle MQ'_1P'_1 = \angle MP'_1Q'_1=\angle MM'_1Q'_1 \]so $F'_1E'_1$ is tangent to $(Q'_1P'_1M'_1)$ as desired. $\blacksquare$

Let $R=\overline{AD}\cap R,\ R\neq D$. We will prove $R$ is the desired tangency point. First, we prove that $PMQR$ is cyclic. Let $M=\overline{AD}\cap (PAQ)$. We need to show that, $$MD\cdot DR=PD\cdot DQ=AD\cdot DM$$i.e. it is enough to prove that $M$ is the midpoint of $\overline{DR}$. Since $\angle AFI_A=\angle AEI_A=90^\circ,\ I_A$ lies on $(EAF)$. Moreover, $\angle AMI_A=\angle AEI_A=90^\circ$, then $MD=MR$, as required. We proceed to show that $(MQRP)$ and $\omega$ are tangent at $R$. Let $S=\overline{BC}\cap \overline{I_AM}$. Line $MI_A$ is the perpendicular bisector of $\overline{DR}$ and $SD$ is tangent to $\omega$, hence $SR$ is tangent to $\omega$ too. Because $A$ lies on the $S$-polar $\omega$, $S$ must lie on the $A$-polar wrt $\omega$, thus $EF$ passes through $S$. Therefore, $$SR^2=SE\cdot SF=SQ\cdot SP$$then $SR$ is tangent to $(MQP)$ at $R$. The conclusion follows.

an other way consider $\cal{C}$ the image of $BC$ with the inversion wrt $\odot (A,AE)$ then $\cal{C}$ is tangent to $\omega$ at the second intersection of $AD$ and $\omega$ say $T$ let $X$ the pole of $AD$ we have $XT^2=XE.XF=XP.XQ$ then $\odot(TPQ) ,\cal{C}$ and $\omega$ are coaxal ,since $MA.MT=-MD.MT $ then it sufices to prove that the center of $\odot(TPQ)$ is the midpoint of the centers of $\cal{C}$ and $\omega$ . $P,Q$ are on $\odot(AI_a)$ implies the bisector of $PQ$ pass through the midpoint of $AI_a$ ;$\cal{C}$ cut $BC$ at two point equidistant from $A$ the their bisector pass through $A$ .Since their three parallel diameters pass through the ends and the midpoint of the segment $AI_a$ the result follows.

This problem was proposed by Mads Christensen, Denmark.

Let the radical center of $\omega, (MPQ),$ and $(AEF)$ be $N$ and let $R$ be the second tangency point of $N$ to $\omega$. By La Hire with respect to $\omega$, $N$ lies on the polar of $A$ so $AD$ is the polar of $N$. Thus, $R$ lies on $AD$. It suffices to show that $R$ also lies on $(MPQ)$. Call the A-excenter $I_A$ and suppose ray $I_AD$ hits $(AEF)$ at $S$; then $ASD$ is a right triangle and by angle chasing, $\angle MSD=\angle MDS=\angle I_ADR=\angle I_ARD$. Therefore, triangles $SMD$ and $DI_AR$ are similar, from which it follows that $SD\cdot I_AD=MD\cdot DR$. By Power of a Point, we have $PD\cdot DQ=SD\cdot I_AD=MD\cdot DR$, implying that $MPRQ$ is cyclic as desired.

Let $I_A$ be the center of $\omega$ and let $N$ be the midpoint of $\overline{AI_A}.$ Note that the circle $\gamma$ of diameter $\overline{AI_A}$ centered at $N$ passes through $E, F, P, Q.$ Because $(MN \parallel DI_A) \perp PQ$, we infer that $MP = MQ.$ Let $\Omega$ be the circle centered at $M$ passing through $P, Q.$ [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(20cm); 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 = 976.4444805508508, xmax = 1046.3798703597602, ymin = 995.5543000359224, ymax = 1031.3188886255336; /* image dimensions */ pen fsfsfs = rgb(0.9490196078431372,0.9490196078431372,0.9490196078431372); pen evefev = rgb(0.8980392156862745,0.9372549019607843,0.8980392156862745); pen qqwuqq = rgb(0.,0.39215686274509803,0.); filldraw((1002.3867582950736,1028.7129107607666)--(997.5041703427562,1015.4933869170767)--(1014.9751993422051,1015.6594613372235)--cycle, fsfsfs, linewidth(0.8) + gray); filldraw((1019.4609897924207,1016.108862051526)--(1019.0594009392324,1015.5648447119974)--(1019.6034182787611,1015.1632558588091)--(1020.0050071319494,1015.7072731983377)--cycle, evefev, linewidth(0.8) + qqwuqq); filldraw((1015.2651837526455,1010.424961164298)--(1014.8635948994573,1009.8809438247692)--(1015.4076122389861,1009.4793549715811)--(1015.8092010921742,1010.0233723111098)--cycle, evefev, linewidth(0.8) + qqwuqq); Label laxis; laxis.p = fontsize(10); xaxis(xmin, xmax, Ticks(laxis, Step = 2., Size = 2, NoZero),EndArrow(6), above = true); yaxis(ymin, ymax, Ticks(laxis, Step = 2., Size = 2, NoZero),EndArrow(6), above = true); /* draws axes; NoZero hides '0' label */ /* draw figures */ draw((1002.3867582950736,1028.7129107607666)--(997.5041703427562,1015.4933869170767), linewidth(0.8) + gray); draw((997.5041703427562,1015.4933869170767)--(1014.9751993422051,1015.6594613372235), linewidth(0.8) + gray); draw((1014.9751993422051,1015.6594613372235)--(1002.3867582950736,1028.7129107607666), linewidth(0.8) + gray); draw(circle((1005.3978430681511,1014.3557053916027), 14.669559554499537), linewidth(0.8) + red); draw((997.5041703427562,1015.4933869170767)--(993.777200913478,1005.4026805711303), linewidth(0.8) + gray); draw((1014.9751993422051,1015.6594613372235)--(1019.6364338696387,1010.8260440040347), linewidth(0.8) + gray); draw(circle((1008.4089278412287,999.9985000224386), 15.59783960332521), linewidth(0.8)); draw((1008.4089278412287,999.9985000224386)--(1020.0050071319494,1015.7072731983377), linewidth(0.8)); draw((1002.3867582950736,1028.7129107607666)--(1008.4089278412287,999.9985000224386), linewidth(0.8)); draw((1008.4089278412287,999.9985000224386)--(1008.26066610213,1015.5956349756209), linewidth(0.8)); draw((1002.3867582950736,1028.7129107607666)--(1008.26066610213,1015.5956349756209), linewidth(0.8)); draw((1008.26066610213,1015.5956349756209)--(1015.8092010921742,1010.0233723111098), linewidth(0.8) + dotted); draw((1002.3867582950736,1028.7129107607666)--(1020.0050071319494,1015.7072731983377), linewidth(0.8) + dotted); draw((997.5041703427562,1015.4933869170767)--(990.7676256418414,1015.4293513210553), linewidth(0.8) + gray); draw((1014.9751993422051,1015.6594613372235)--(1020.0050071319494,1015.7072731983377), linewidth(0.8) + gray); draw(circle((1005.3237121986018,1022.1542728681937), 16.03446992148692), linewidth(0.8) + linetype("2 2") + blue); draw((1005.3237121986018,1022.1542728681937)--(1005.3978430681511,1014.3557053916027), linewidth(0.8)); /* dots and labels */ dot((1002.3867582950736,1028.7129107607666),linewidth(3.pt) + dotstyle); label("$A$", (1002.1363329601401,1029.3425922863573), N * labelscalefactor); dot((1008.4089278412287,999.9985000224386),linewidth(3.pt) + dotstyle); label("$I_A$", (1008.352103704323,998.9331292609661), E * labelscalefactor); dot((1008.26066610213,1015.5956349756209),linewidth(3.pt) + dotstyle); label("$D$", (1008.2883522095109,1016.0504056180259), NE * labelscalefactor); dot((1019.6364338696387,1010.8260440040347),linewidth(3.pt) + dotstyle); label("$E$", (1020.0823787497554,1010.6634043064), NE * labelscalefactor); dot((993.777200913478,1005.4026805711303),linewidth(3.pt) + dotstyle); label("$F$", (992.8286147175687,1004.9895212681195), N * labelscalefactor); dot((1005.3237121986018,1022.1542728681937),linewidth(3.pt) + dotstyle); label("$M$", (1005.451410690371,1022.3299278570219), NE * labelscalefactor); dot((1005.3978430681511,1014.3557053916027),linewidth(3.pt) + dotstyle); label("$N$", (1005.7064166696196,1014.3609910055042), E * labelscalefactor); label("$\gamma$", (998.0562372921636,1026.2187690405624), E * labelscalefactor,red); dot((990.7676256418414,1015.4293513210553),linewidth(3.pt) + dotstyle); label("$Q$", (989.6410399769621,1015.7635238913713), N * labelscalefactor); dot((1020.0050071319494,1015.7072731983377),linewidth(3.pt) + dotstyle); label("$P$", (1020.4330119712222,1015.8272753861835), NE * labelscalefactor); label("$\omega$", (1000.6062970846489,1012.6397006455765), E * labelscalefactor); dot((1015.8092010921742,1010.0233723111098),linewidth(3.pt) + dotstyle); label("$P'$", (1016.2254133136214,1009.0058654412843), E * labelscalefactor); label("$\Omega$", (990.291219354418,1025.1439783142623), NE * labelscalefactor,blue); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Let $P'$ be the projection of $D$ onto $I_AP.$ Since $M$ is the midpoint of $\overline{AD}$, projecting onto $I_AP$ shows that $MP = MP'.$ Thus, $P' \in \Omega.$ Because $P, P'$ are inverses WRT $\omega$, it follows that $\omega$ and $\Omega$ are orthogonal.Therefore, inversion about $\Omega$ fixes $\omega$, and sends $BC \mapsto \odot(MPQ).$ As $BC$ is tangent to $\omega$, so is $\odot(MPQ).$

Let $\overline{AD}$ meet $(AEF)$ again at $Y$ and $\omega$ again at $X$. Let $J$ be the center of $\omega$ then $\overline{AJ}$ is the antipode of $A$ in $(AEF)$ so $\angle AYJ=90^{\circ}$. Thus $Y$ is the midpoint of $\overline{DX}$. Now $$DP \cdot DQ=DA \cdot DY=DM \cdot DX$$hence $X \in (MPQ)$. Now radical axes of $(MPQ), (AEF), \omega$ concur at a point $T$; then $T=\overline{EF} \cap \overline{PQ}$. Since $DEXF$ is harmonic, we conclude that $\overline{TX}$ is tangent to $\omega$ hence $\omega, (MPQ)$ touch at $X$.

Hello. We will use the converse of Casey's theorem.Let $I_A$ be the $A-$excenter of the triangle,$R_A$ the radius of the $A-$excircle,and let $O$ be a point on $AI_A$ such that $MO\perp BC$.It is obvious that $I_A$ lies on the circle $(AEF)$ and that $AI_A$ is a diameter of this circle.Since $MO\parallel DI_A$ and $M$ is the midpoint of $AD$,it follows that $O$ is the midpoint of $AI_A$,i.e. the center of the circle $(AEF)$.Hence,$MO$ is the perpendicular bisector of $PQ$,whence $MP=MQ$.The power of $M$ with respect to the $A-$excircle is equal to $MI_A^2-R_A^2=MI_A^2-DI_A^2$,thus,the length of the tangent from $M$ to this circle is equal to $\sqrt{MI_A^2-DI_A^2}$.According to the converse of Casey's theorem,it suffices to show that $PQ\cdot \sqrt{MI_A^2-DI_A^2}=MP\cdot DQ+MQ\cdot DP=MP\cdot PQ\Leftrightarrow MI_A^2-DI_A^2=MP^2$.The medians' theorem gives $MP^2=\frac{2AP^2+2PD^2-AD^2}{4}$ and $MI_A^2=\frac{2DI_A^2+2AI_A^2-AD^2}{4}$,thus,the latter relation becomes $2DI_A^2+2AI_A^2-AD^2-4DI_A^2=2AP^2+2PD^2-AD^2\Leftrightarrow DI_A^2+PD^2=AI_A^2-AP^2$ which is true,since both parts are equal to $ PI_A^2$ (Pythagorean Theorem). [asy][asy]import graph; size(13.cm); 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., xmax = 9., ymin = -7., ymax = 5.; /* image dimensions */ pen aqaqaq = rgb(0.6274509803921569,0.6274509803921569,0.6274509803921569); pen sqsqsq = rgb(0.12549019607843137,0.12549019607843137,0.12549019607843137); pen qqwuqq = rgb(0.,0.39215686274509803,0.); draw((1.2794562069853759,4.13627037589686)--(0.,0.)--(6.366644007995975,0.)--cycle, linewidth(0.8) + aqaqaq); draw((-1.8433865579556863,-0.4422839879365273)--(-1.4011025700191588,0.06207364141944567)--(-1.905460199375132,0.504357629355973)--(-2.347744187311659,0.)--cycle, linewidth(0.8) + qqwuqq); draw((3.6259600045187286,0.)--(3.6259600045187286,-0.6708142397673258)--(4.296774244286055,-0.6708142397673258)--(4.296774244286055,0.)--cycle, linewidth(0.8) + qqwuqq); /* draw figures */ draw((1.2794562069853759,4.13627037589686)--(0.,0.), linewidth(0.8) + sqsqsq); draw((0.,0.)--(6.366644007995975,0.), linewidth(0.8) + sqsqsq); draw((6.366644007995975,0.)--(1.2794562069853759,4.13627037589686), linewidth(0.8) + sqsqsq); draw(circle((4.296774244286055,-5.826746615486671), 5.826746615486671), linewidth(0.8)); draw((0.,0.)--(-1.2697454509612973,-4.104877107217038), linewidth(0.8)); draw((6.366644007995975,0.)--(7.9726460691033045,-1.3058017531155235), linewidth(0.8)); draw(circle((2.788115199155925,-0.8452381523623275), 5.204947566669243), linewidth(0.8)); draw((-2.347744187311659,0.)--(0.,0.), linewidth(0.8)); draw((6.366644007995975,0.)--(7.923974585623509,0.), linewidth(0.8)); draw((1.2794562069853759,4.13627037589686)--(4.296774244286055,0.), linewidth(0.8)); draw((-2.347744187311659,0.)--(2.7881152256357153,2.06813518794843), linewidth(0.8)); draw((2.7881152256357153,2.06813518794843)--(7.923974585623509,0.), linewidth(0.8)); draw((-1.2697454509612973,-4.104877107217038)--(-2.347744187311659,0.), linewidth(0.8)); draw((-2.347744187311659,0.)--(4.296774244286055,-5.826746615486671), linewidth(0.8)); draw((-2.347744187311659,0.)--(1.2794562069853759,4.13627037589686), linewidth(0.8)); draw((7.923974585623509,0.)--(4.296774244286055,-5.826746615486671), linewidth(0.8)); draw((4.296774244286055,-5.826746615486671)--(4.296774244286055,0.), linewidth(0.8)); draw((7.923974585623509,0.)--(1.2794562069853759,4.13627037589686), linewidth(0.8)); draw(circle((2.788115199155925,-5.342945811936877), 7.411080999885307), linewidth(0.4) + linetype("2 2")); draw((1.2794562069853759,4.13627037589686)--(4.296774244286055,-5.826746615486671), linewidth(0.8)); draw((2.7881152256357153,ymin)--(2.7881152256357153,ymax), linewidth(0.4) + linetype("2 2")); /* line */ /* dots and labels */ dot((1.2794562069853759,4.13627037589686),linewidth(3.pt) + dotstyle); label("$A$", (1.3991603267153,4.333873137795685), NE * labelscalefactor); dot((0.,0.),linewidth(3.pt) + dotstyle); label("$B$", (-0.4349438916611354,0.2545723762343191), NE * labelscalefactor); dot((6.366644007995975,0.),linewidth(3.pt) + dotstyle); label("$C$", (6.42,0.15), NE * labelscalefactor); dot((4.296774244286055,-5.826746615486671),linewidth(3.pt) + dotstyle); label("$I_A$", (4.4665415195172695,-6.291282334178105), NE * labelscalefactor); dot((4.296774244286055,0.),linewidth(3.pt) + dotstyle); label("$D$", (4.434919032993538,0.19132740318685607), NE * labelscalefactor); dot((7.9726460691033045,-1.3058017531155235),linewidth(3.pt) + dotstyle); label("$E$", (8.10312746974641,-1.105194544286136), NE * labelscalefactor); dot((-1.2697454509612973,-4.104877107217038),linewidth(3.pt) + dotstyle); label("$F$", (-1.9212007582765227,-4.488800602325408), NE * labelscalefactor); dot((-2.347744187311659,0.),linewidth(3.pt) + dotstyle); label("$P$", (-2.996365300083399,-0.030030002479264524), NE * labelscalefactor); dot((7.923974585623509,0.),linewidth(3.pt) + dotstyle); label("$Q$", (8.039882496698946,0.19132740318685607), NE * labelscalefactor); dot((2.7881152256357153,2.06813518794843),linewidth(3.pt) + dotstyle); label("$M$", (2.9170396798544194,2.2467890272294047), NE * labelscalefactor); dot((2.7881152256357153,-0.8452381197949074),linewidth(3.pt) + dotstyle); label("$O$", (2.9170396798544194,-0.6624797329538948), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy][/asy]

Let $O$ be the center of $\omega$, and $Y=AD\cap \omega$ and set $Z$ the midpoint of $DY$. Then $AO$ is the diameter of circle $(APFOZEQ)$. Set $X=FE\cap PQ$, then $X$ is on the polar of $A$ wrt $\omega$, hence $A$ is on the polar of $X$, thus $YX$ is tangent to $\omega$. Notice \[ DP \cdot DQ = DA \cdot DZ = \frac{DA \cdot DY}{2} = DM \cdot DY\]so $MPQY$ is cyclic. Finally, look at circles $\omega, (APFEQ), (MPQY)$ and the Radical Axis Theorem yields that the latter is tangent to $\omega$, as desired.

Harmonic Figure

Let $I_A$ the center of $ \omega$, $K= AD \cap (AEF)$ and $T=AD \cap \omega$. We have that $DI_A=TI_A$ because $I_A$ is the center of $\omega$ and $AI_A$ is a diameter in $(AEF$), hence $ \angle I_AKA=90$. So, we conclude that $K$ is the midpoint of $DT$. By power of point and because of the midpoints we get that: $DM \times DT= \frac {DA}{2} \times 2DK = DA \times DK = DP \times DQ $, and so $T$ lies on $(MPQ)$. We will show that $(MPQ)$ and $\omega$ are tangent in $T$. Let $O$ be the center of $(MPQ)$. It suffices to show that $T,I_A,O$ are collinear. Let $L=I_AD \cap (AEF)$. We have that $AI_A$ is a diameter and so $ALI_A=90 => AL \perp I_AD => AL \parallel BC => AL \parallel PQ => ALQP$ is an isosceles trapezoid and so $AL,PQ$ have the same perpendicular bisector. $M$ is the midpoint of the hypotenuse $AD$ in the right angle triangle $ALD$ and hence $MA=ML=>MP=MQ$, as $AL,PQ$ have the same perpendicular bisector. So we have that $OM$ is the perpendicual bisector of $PQ => OM \perp PQ => OM \perp BC => OM \parallel I_AD$. Using the parallel lines and that $OM=OT$, $I_AD=I_AT$ we get that: $\angle DTI_A = \angle TDI_A = \angle TMO = \angle MTO= \angle DTO$, hence $T,I_A,O$ are collinear, as needed.

We would like $MP^2 = MQ^2 = \text{Pow}_{\text{A-excircle}}(M)$ in order to apply shooting lemma. By Apollonius's theorem in $\Delta{APD}$: $AP^2 + PD^2 = 2 PM^2 + 2 AM^2$ $\implies PM^2 = \frac{AP^2 + PD^2 - 2 AM^2}{2} = \frac{(AI_A^2 - PI_A^2) + PD^2 - 2 AM^2}{2} = \frac{AI_A^2 - (PI_A^2 - PD^2) - 2 AM^2}{2} = \frac{AI_A^2 - r_a^2 - 2 AM^2}{2}$ which does not depend on $P$, so $PM = QM$ by symmetry. Further, if we let $T'$ be the intersection of $\overrightarrow{\rm AD}$ with the $A$-excircle, then $MP^2 = MQ^2 = \frac{AI_A^2 - r_a^2 - 2 AM^2}{2} = \frac{\text{Pow}_{\text{A-excircle}}(A) - 2 AM^2}{2} = \frac{AD \cdot AT' - 2 AM^2}{2} = \frac{(2MD) \cdot (MD + MT') - 2 AM^2}{2} = \frac{2 MD \cdot MT'}{2} = MD \cdot MT'$ and we are done by shooting lemma.

Note that $I_{A}\in (APFEQ)$ since $\angle I_{A}FA=90=\angle I_{A}EA$, and thus $AI_{A}$ is a diameter in $(AEF)$. Define $O$ as the center of $(AEF)$. Since $\frac{MA}{OA}=\frac{\frac{DA}{2}}{\frac{I_{A}A}{2}}=\frac{DA}{MA}$, then $MO\parallel DI_{A} \perp PQ$. And since $O$ is the center, $M$ must be the midpoint of arc $PQ$. Now define $K=\overline{AD} \cap \omega $ and $L=\overline{AD} \cap (AEF)$. Since $AI_{A}$ is a diameter, then $\angle ALI_{A}=90$, and since $I_A$ is the center of $\omega$, then $L$ must be the midpoint of $DK$. By power of a point from $D$ we have: $PD\cdot DQ=AD\cdot DL =2MD\cdot \frac{DK}{2} = MD\cdot DK$. Thus $MPQK$ is cyclic. Since $\omega$ is tangent to $PQ$ at $D$, and $M,D,K$ are collinear. Then there must exist a homothety from $K$ sending $\omega \mapsto (MPQK)$. Thus, $\omega$ is an circle inscribed in the segment $PQ$ and tangent to $MPQK$ at $K$, as required. $\blacksquare$

Let $R = EF \cap BC$, potentially the point at infinity, and define $S$ as the second intersection of $AD$ and $\omega$. Since $BC$ and $EF$ are the polars of $D$ and $A$, respectively, with respect to $\omega$, $R$ is the pole of the line $AD$ so $RS$ is also a tangent line to $\omega$. Now let the $A$-excenter be $I_a$. We draw $RI_a$ and let its intersection with $AD$ be $T$. Since $RI_a \perp AD$ and $AI_a$ is a diameter of the circumcircle of $\triangle AEF$, $T$ also lies on this circle. $T$ is also the midpoint of $DS$, so we have $MD\cdot DS = AD\cdot DT = PD\cdot PQ$ by PoP, implying that $M$, $P$, $Q$, $S$ are concyclic. We know that the radical center of $(AEF)$, $\omega$, and $(MPQ)$ is $PQ\cap EF = R$, so the radical axis of $(MPQ)$ and $\omega$ must be the line $RS$. Since this is a tangent line to $\omega$, we know that the circles must be tangent at $S$.

Let $T=AD \cap w.$ Since $DETF$ is harmonic, $DD, EF, TT$ are concurrent, say at $X.$ Since $\angle I_a EA = \angle I_aFA = 90,$ we know $AEI_aF$ is cyclic. Letting $Y=(AEI_aF) \cap AD$, we notice $\angle DYI_a=90$ so $DY=YT.$ Thus, $$DM \cdot DT = DA \cdot DY = DP \cdot DQ,$$so by PoaP $MQTP$ is cyclic. By Radical Axis Theorem, $PQ, EF, T?$ are concurrent, specifically at $X.$ Since $XT$ is tangent to $w,$ we know it is also tangent to $(MQP).$

Begin by noticing that $(AEF)$ has diameter $AI_A$, where $I_A$ is the center of $\omega$. Let $T \neq D = \overline{AD} \cap \omega$ and $N \neq A = \overline{AD} \cap (AEF)$. Note that $\overline{AD} \perp \overline{I_AN}$, which implies $N$ is the midpoint of $\overline{DT}$. Hence, \[DP \cdot DQ = DA \cdot DN = 2DM \cdot \frac{1}{2} DT = DM \cdot DT.\] Thus, $T$ lies on $(MPQ)$, so we just need to prove tangency. Note that $DETF$ is harmonic, so we can let $R = \overline{DD} \cap \overline{EF} \cap \overline{TT}$. Finally, we have \[RQ \cdot RP = RE \cdot RF = RT^2,\] whence we are done. $\square$

Let $I_A$ be the center of $\omega$, and let $AD$ intersect $(AEF)$ again at $K$ and $\omega$ again at $Y$. Let $BC \cap EF = Z$. Since $\angle DXI_A = \angle AXI_A = \angle AFI_A = 90^\circ$ and $I_AD = I_AY$, $X$ is the midpoint of $DY$. A dilation centered at $D$ with factor $1/2$ applied to $\omega$ results in a circle $\Omega$ through $X$ and $I_A$ tangent to $D$. Since $ZD^2 = ZE \cdot ZF$, $Z$ has equal power with respect to $\Omega$ and $(AEF)$, so $Z$ lies on their radical axis, $I_AX$, which is the perpendicular bisector of $DY$ giving $ZD = ZY$. Since $DP \cdot DQ = DA \cdot DX = 2DM \cdot DX = DM \cdot DY$, $PMQY$ is cyclic. It remains to show that $(PMQ)$ is tangent to $\omega$ at $Y$; to do that we will show that line $ZY$ is tangent to both circles at $Y$. It is clear from $ZD = ZY$ that $ZY$ is tangent to $\omega$. At the same time, $ZP \cdot ZQ = ZE \cdot ZF = ZD^2 = ZY^2$, giving $ZY$ tangent to $(PMQY)$ as well.

Let $AD$ meet the excircle again at $G$ and $(AEF)$ again at $N$(Which is obviously the midpoint of $DG$). We show $G$ is the desired tangency point. Let the tangent at $G$ to the excircle meet $BC$ again at $T$. It suffices to prove that $TG$ is tangent to $(MPQ)$. First we prove that $MPGQ$ is a cyclic quadrilateral. For this, note that since $D$ lies on the radical axis of $(MPQ)$ and $(AEF)$, $$PD.QD=AD.DN= \frac{AD}{2}.DG=MD.DG$$So the claim follows. Now note that since $TI_A$ is the perpendicular bisector of $DG$, $N$ lies on this line. Since $\Delta IDT$ is right angled at $D$ and $N$ is foot from $D$ to the hypotenuse, $TD^2=TN.TI_A$ So $TD^2=TN.TI_A=TP.TQ=TG^2$ so we are done by the converse of $PoP$.

I'm dumb

I'm dumb 2

Problem wrote: In triangle $ABC$, let $\omega$ be the excircle opposite to $A$. Let $D, E$ and $F$ be the points where $\omega$ is tangent to $BC, CA$, and $AB$, respectively. The circle $AEF$ intersects line $BC$ at $P$ and $Q$. Let $M$ be the midpoint of $AD$. Prove that the circle $MPQ$ is tangent to $\omega$. [asy][asy] import graph; size(9cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.5) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -11.88233553870968, xmax = 11.830618443060125, ymin = -14.609017933970518, ymax = 5.3283339819734525; /* image dimensions */ /* draw figures */ draw(circle((1.080363026950906,-9.048627177541055), 7.048627177541055), linewidth(0.5)); draw((-2,4)--(-5.606551884284705,-6.819655652854113), linewidth(0.5)); draw((-2,4)--(6.0644951022459805,-4.0644951022459805), linewidth(0.5)); draw(circle((-0.4598185286713953,-2.524313630917376), 6.703642839471553), linewidth(0.5)); draw((-7.142925780214831,-2)--(6.2232887228720415,-2), linewidth(0.5)); draw(circle((-0.45981852867139494,-7.943987089272077), 8.943987089272078), linewidth(0.5)); /* dots and labels */ dot((-2,4),dotstyle); label("$A$", (-1.8864969759476087,4.268992395158991), NE * labelscalefactor); dot((-4,-2),dotstyle); label("$B$", (-4.629920059749155,-2.6031466167399535), NE * labelscalefactor); dot((4,-2),dotstyle); label("$C$", (4.116438682667657,-1.7339432634562923), NE * labelscalefactor); dot((1.0803630269509057,-2),linewidth(4pt) + dotstyle); label("$D$", (1.18287736533531,-1.7882684730365213), NE * labelscalefactor); dot((1.080363026950906,-9.048627177541055),linewidth(4pt) + dotstyle); label("$I_A$", (1.4816660180265675,-9.556773443009241), NE * labelscalefactor); dot((6.0644951022459805,-4.0644951022459805),linewidth(4pt) + dotstyle); label("$E$", (6.262284461086689,-3.9612768562456737), NE * labelscalefactor); dot((-5.606551884284705,-6.819655652854113),linewidth(4pt) + dotstyle); label("$F$", (-6.1510259279955575,-7.08497640710883), NE * labelscalefactor); dot((-7.142925780214831,-2),linewidth(4pt) + dotstyle); label("$P$", (-7.8894326345628745,-1.9240814969870932), NE * labelscalefactor); dot((6.2232887228720415,-2),linewidth(4pt) + dotstyle); label("$Q$", (6.343772275457032,-1.7882684730365213), NE * labelscalefactor); dot((-0.45981848652454715,1),linewidth(4pt) + dotstyle); label("$M$", (-0.33822850291109224,1.2267806586661778), NE * labelscalefactor); dot((6.705912638138943,-13.296295179999717),dotstyle); label("$I$", (7.185813023950575,-13.332375508835144), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Consider a $\sqrt{bc}$ inversion that takes the $A-$Excircle to the $A-$Mixtillinear Incircle, so $D\mapsto T$, where $T$ is the $A-$Mixtillinear-touch point. Inverted problem wrote: The circumcircle of $\triangle ABC$ and the $A-$Mixtillinear Incircle are tangent at to each other $T$(map of $D$), the A-Mixtillinear Incircle is tangent with side $AC, AB$ at $E,F$ respectively. Points $P,Q$ are $EF\cap \odot(ABC)$. Let $M$ be the reflection of $A$ over(the new) $T$ ($M$ maps to the reflection of $A$ over $T$). Show that $\odot(MPQ)$ is tangent to the $A-$Mixtillinear incircle. [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(10cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.5) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -8.981474539737476, xmax = 5.275139359946031, ymin = -9.127837480782109, ymax = 2.969682290741172; /* image dimensions */ draw((-3,2)--(-4,-2)--(1,-2)--cycle, linewidth(0.5)); /* draw figures */ draw((-3,2)--(-4,-2), linewidth(0.5)); draw((-4,-2)--(1,-2), linewidth(0.5)); draw((1,-2)--(-3,2), linewidth(0.5)); draw(circle((-1.5,-0.5), 2.91547594742265), linewidth(0.5)); draw(circle((-2.031855077262087,-1.4953103618043528), 1.7869758191441576), linewidth(0.5)); draw((-4.325901454999577,-1.2171338554490905)--(1.2921528311248565,0.3389770960171954), linewidth(0.5)); draw(circle((-0.5318550772620875,-3.9953103618043526), 4.702451766566807), linewidth(0.5)); draw((-3,2)--(-2.7480912875706283,-8.14276134728984), linewidth(0.5)); draw((-5.655091427472534,-1.585298164194349)--(-2.9512465512633836,0.03700876153114119), linewidth(0.5)+dashed); draw((-5.655091427472534,-1.585298164194349)--(-4.325901454999577,-1.2171338554490905), linewidth(0.5)+dashed); draw((-5.655091427472534,-1.585298164194349)--(-2.874045643785314,-3.07138067364492), linewidth(0.5)+dashed); /* dots and labels */ dot((-3,2),linewidth(4pt) + dotstyle); label("$A$", (-2.932714653975803,2.129118927869664), NE * labelscalefactor); dot((-4,-2),linewidth(4pt) + dotstyle); label("$B$", (-4.467727053830559,-2.337404039545608), NE * labelscalefactor); dot((1,-2),linewidth(4pt) + dotstyle); label("$C$", (1.2041756221173296,-2.1725876938845277), NE * labelscalefactor); dot((-3.7654762661458525,-1.0619050645834114),linewidth(4pt) + dotstyle); label("$F$", (-3.674388209450668,-1.4144325038435592), NE * labelscalefactor); dot((-0.7682723577288666,-0.23172764227113343),linewidth(4pt) + dotstyle); label("$E$", (-0.46046946905958835,-0.4090527953109704), NE * labelscalefactor); dot((-2.874045643785314,-3.07138067364492),linewidth(4pt) + dotstyle); label("$T$", (-3.245865710731857,-3.3757470172104127), NE * labelscalefactor); dot((-4.325901454999577,-1.2171338554490905),linewidth(4pt) + dotstyle); label("$Q$", (-4.514951572322181,-1.71753877690714), NE * labelscalefactor); dot((1.2921528311248565,0.3389770960171954),linewidth(4pt) + dotstyle); label("$P$", (1.3525103332123027,0.4644738366927543), NE * labelscalefactor); dot((-2.7480912875706283,-8.14276134728984),linewidth(4pt) + dotstyle); label("$M$", (-2.685490135484182,-8.007086330286764), NE * labelscalefactor); dot((-2.9512465512633836,0.03700876153114119),linewidth(4pt) + dotstyle); label("$I$", (-2.883269750277479,0.16780441450281008), NE * labelscalefactor); dot((-2.9297030919645763,-0.830409297001144),linewidth(4pt) + dotstyle); label("$J$", (-2.8338248465791547,-1.1068324101042665), NE * labelscalefactor); /* end of picture */ [/asy][/asy] Let $I$ be the intersection of $AT$ and Mixillinear Incircle. The tangents from $I$ and $T$ concur with ${PQ}$ because of radax on Mixtillinear Incircle, $\odot(ABC)$ and $\odot(PQIM)$. For this we just need to show $\odot(PQIM)$. Let $IM\cap QP=J$. Because $(F,E;T,I)=-1$, we have $(A,J ; T,I) \stackrel{F}{=} (F,E ; T,I) = -1 $. So $JI\cdot JM=JA\cdot JT=JQ\cdot JP$. Now we invert back and we are done!

I fakesolved this problem before and then gave up. Here's my comeback! Let $AD\cap \omega=T$, and let the tangent to $\omega$ at $T$ meet $BC$ at $R$. Since $(EF;DT)=-1$, we get that $F,E,R$ are collinear. Now apply radax on $(PQT),(AEF),\omega$ to get that $RT$ is tangent to $(PQT)$, so now it just suffices to prove that $M,P,T,Q$ are actually cyclic. Orient $BC$ horizontally and let $A$ be above $BC$. If $N$ is the midpoint of $AI_A$, we get $NP=NQ$. Since $I_A$ is directly below $D$, $N$ is directly below $M$, so we also get $MP=MQ$. Now by the shooting lemma on $\omega$ and $(PQT)$, we get that $AD$ intersect the perpendicular bisector of $PQ$ lies on $(PQT)$, but of course, this is $M$.

Let $T$ be the second intersection of $AD$ with $\omega$. Then clearly $(E,F;D,T)=-1$. Let $X=EF \cap BC$, then we know that $XT$ is tangent to $\omega$. $XP \cdot XQ = XE \cdot XF = XT^2$, so $XT$ is tangent to $(TPQ)$. Let $(TPQ)$ intersect $AD$ again at $M'$. Then by taking a homothety centred at $T$, the tangent to $(TPQ)$ at $M'$ is parallel to $BC$, so $M'$ lies on the perpendicular bisector of $PQ$ (call this line $\perp_{PQ}$). Let $K$ be the $A$-excentre so $DK \perp BC$, and let $S$ be the centre of $(AEKF)$. By a homothety centred at $A$ with scale factor $2$, $M \rightarrow D$ and $S \rightarrow K$, so $MS \perp BC$. Also, $S$ lies on $\perp_{PQ}$, and thus so does $M$. Since $M$ and $M'$ both lie on $AD$ and $\perp_{PQ}$, they are the same point, so $MPQ$ and $\omega$ are both tangent to $TX$ at $T$. $\square$

We clearly have that $I_A$ lies on $(AEF)$ and that its the antipode of $A$. Thus we get that if $K=(AEF)\cap AD$ that $\angle I_AKA=90$ thus if we let $T$ be the second intersection of $AD$ with the $A$-excircle. We have that as $I_A$ is the centre of the $A$ excircle and $\angle I_AKA=90$, $K$ is the midpoint of $DT$ and we also have that $M$ is the midpoint of $AD$, as we have $2AM\cdot DK = PD \cdot DQ$, so we get that $2DK\cdot AM = 2AM \cdot DK = PD \cdot DQ$ which from POP implies that $MPQT$ is cyclic. Let the circumcentre of $(AEF)$ be $O$, clearly $O$ is the midpoint of $AI_A$, thus as $I_AD$ is perpendicular to $PQ$, $OM$ is perpendicular to $BC$ as $O$ is the circumcentre of $AEF$, $OM\cap PQ$ is the midpoint of $PQ$ so $M$ is the midarc of $PQ$, this means that $M$ is the midarc of $PQ$ for circle $MPQ$. Thus it suffices to prove that if we have a triangle $ABC$ and angle bisector $AD$ where $D$ is on $BC$ that the circle through $A$ and tangent to $BC$ at $D$ is tangent to $(ABC)$. To prove this take a root bc inversion at $A$, $D$ goes to the midarc of $BC$ that does not contain $A$, as the circle is tangent to $BC$ its tangent to $(ABC)$ after the inversion so we get that the line we get after inverting the circle is parallel to $BC$ and has one intersection at the point at infinity so the circle has one intersection with $ABC$ which proves the result.