[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(7.cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ real xmin = -1., xmax = 6., ymin = -1., ymax = 6.; /* image dimensions */ pen qqwuqq = rgb(0.,0.392156862745,0.); pen xdxdff = rgb(0.490196078431,0.490196078431,1.); pen qqffff = rgb(0.,1.,1.); pen ffdxqq = rgb(1.,0.843137254902,0.); pen bfffqq = rgb(0.749019607843,1.,0.); pair A = (2.52,2.42), C = (0.682887099703,5.21560658741), D = (-0.775182532809,1.84365628704), G = (-0.000564511004626,3.63504773574), H = (2.06529064264,4.18884817754), I = (1.17328212699,1.73139822855), J = (1.42845604051,2.43439277373), K = (1.95810238597,0.458640228594), L = (1.69327921324,1.44651650116), M = (2.69686346899,1.64361922649), O = (1.62386921699,1.70543828535), P = (1.25175936133,3.09352866053), Q = (1.81148283025,3.49817830179); draw(C--D--(5.7271885701,1.46903129608)--cycle, linewidth(1.2) + blue); /* draw figures */ draw(circle(A, 3.34520552433), qqwuqq); draw(C--D, linewidth(1.2) + blue); draw(D--(5.7271885701,1.46903129608), linewidth(1.2) + blue); draw((5.7271885701,1.46903129608)--C, linewidth(1.2) + blue); draw(circle(P, 1.364389237), linewidth(0.4) + qqffff); draw(circle(L, 1.02275649238), linewidth(0.4) + ffdxqq); draw(circle(Q, 2.05506404748), linewidth(0.4) + green); label("$A$",(0.68,5.46),SE*labelscalefactor,blue); label("$B$",(-0.78,2.08),SE*labelscalefactor,blue); label("$C$",(5.72,1.7),SE*labelscalefactor,blue); label("$I$",(1.32,3.46),SE*labelscalefactor,qqffff); label("$F$",(-0.22,3.96),SE*labelscalefactor,qqffff); label("$E$",(2.12,4.48),SE*labelscalefactor,qqffff); label("$Y$",(1.96,0.7),SE*labelscalefactor,ffdxqq); label("$X$",(1.42,2.68),SE*labelscalefactor,ffdxqq); label("$T$",(0.72,2.),SE*labelscalefactor,green); label("$S$",(2.7,1.88),SE*labelscalefactor,green); label("$V$",(1.78,1.48),SE*labelscalefactor,blue); label("$D$",(1.18,1.98),SE*labelscalefactor,qqffff); label("$K$",(1.62,1.94),SE*labelscalefactor,blue); draw(G--K, linewidth(0.4) + blue); draw(I--H, linewidth(0.4) + blue); draw(C--K, linewidth(0.4) + red); label("$L$",(1.82,3.74),SE*labelscalefactor,green); label("$O$",(2.62,2.72),SE*labelscalefactor,qqwuqq); draw(C--A, linewidth(0.4) + blue); draw(C--(0.718971820001,1.75757267968), linewidth(0.4) + linetype("2 2") + blue); draw(C--M, linewidth(0.4) + linetype("2 2") + blue); /* dots and labels */ dot(A,blue); dot(C,xdxdff); dot(D,blue); dot((5.7271885701,1.46903129608),blue); dot(G,qqffff); dot(H,qqffff); dot(I,qqffff); dot(J,ffdxqq); dot(K,ffdxqq); dot(L,blue); dot(M,green); dot((0.718971820001,1.75757267968),green); dot(O,blue); dot(P,qqffff); dot(Q,bfffqq); dot(A,qqwuqq); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] (a) WLOG $AB<AC$ and $T$ is between $B$ and $D$. Let $O$ be the circumcenter of triangle $ABC$. Let $I$ be the incenter of triangle $ABC$. $AI$ cuts $BC$ at $K$. We see that $\angle FAI= \angle IAC$ and $\angle AFY= \angle AIC(=90^{\circ}+ \tfrac 12 \angle ABC)$ so $\triangle AIC \sim \triangle AFY$ implies $\tfrac{AI}{AF}= \tfrac{AC}{AY}$ or $AI \cdot AY = AC \cdot AF= AC \cdot AE$. Hence, $I,Y,E,C,D$ are concyclic. Similarly, we have $\triangle ABI \sim \triangle AEX$ implies $\tfrac{AI}{AE}= \tfrac{AB}{AX}$, which follows that $B,F,I,D,X$ are concyclic. Also, note that $\tfrac{AI}{AE}= \tfrac{AI}{AF}$ so $\tfrac{AB}{AX}= \tfrac{AC}{AY}$ or $\tfrac{AX}{AY}= \tfrac{AB}{AC}$. Since $AB<AC$ so $X$ is inside triangle $ABC$. Since $BFIX$ is cyclic AND $IF \perp AB$ so $BX \perp AI$. Similarly, $CY \perp AI$. Hence, $BX \parallel CY$ implies $\tfrac{KX}{KY}= \tfrac{KB}{KC}= \tfrac{AB}{AC}$. Thus, we have $\tfrac{AX}{AY}= \tfrac{KX}{KY} \left( = \tfrac{AB}{AC} \right)$ implies $(AK,XY)=-1$. Note that $\angle XSY=\angle XTY=90^{\circ}$ so we follows that $TX,SX$ are bisectors of $\angle ATS, \angle AST$, repsectively. Therefore, $X$ is the incenter of triangle $ATS$. This means $\angle TAX= \angle SAX$ implies $\angle TAB= \angle SAC$. Similar we also obtain $Y$ is the $A$-excenter of $\triangle ATS$. Let $L$ be the circumcenter of triangle $AST$. If we let $H$ be the orthocenter of triangle $ABC$. It is well-known property that $\angle HAC= \angle OAB$. This property is also applied for $\triangle AT S$, which means $\angle HAS= \angle LAT$. We already have $\angle TAB= \angle SAC$ so $\angle HAS+\angle SAC= \angle LAT+ \angle TAB$ or $\angle HAC= \angle LAB$. Thus, $\angle LAB= \angle OAB(= \angle HAC)$. Since $L,O$ are on the same side wrt $AB$ so $A,L,O$ are collinear, which means $(L)$ and $(O)$ are tangent to each other. (b) $AI$ cuts $(L)$ the second time at $V$ Since $AI$ bisects $\angle TAS$ so $V$ is the midpoint of arc $TS$ of $(L)$. Since $X,Y$ is the incenter, $A$-excenter of $\triangle AST$ respectively so it is well-known that $V$ is the midpoint of $XY$ or $V$ is the center of $(XTYS)$. Note that $(AK,XY)=-1$ so from Newton's identity we have $VT^2=VS^2=VX^2=VK \cdot VA$. Consider an inversion about $V$ with radius $VX$ then since $V \in (L)$ so the inversion will send $(L)$ to a line $\ell$. Note that under this inversion, it sends $T$ to $T$ and $S$ to $S$ so $\ell$ is $BC$. We have $\angle IDE=\tfrac 12 \angle C = \angle AYD$ since $\triangle AFY \sim \triangle AIE$. This means $\triangle IDX \sim \triangle IYD$ implies $ID^2=IX \cdot IY$. Since $ID$ is the radius of $(I)$ so we obtain that $(I)$ and $(V, \tfrac{XY}{2})$ are orthogonal circles. This means that the image of $(I)$ under inversion about $V$ radius $VX$ is itself. Note that $BC$ is tangent to $(I)$ or under inversion, the image of $(I)$ is tangent to image of $(L)$. This follows that $(L)$ is tangent to $(I)$.

Another approach to part b) is to use that tangent point, $D$, and midpoint of $XY$ are collinear.

61plus wrote: Another approach to part b) is to use that tangent point, $D$, and midpoint of $XY$ are collinear. I tried to use that condition but failed. Can you post your solution using this idea? Thanks.

For $a)$ We know that $\angle BXA=\angle AYC=90 ^\circ$, so $CY$ and $BX$ are tangent to $(XYST)$, so $CY^2=CS\cdot CT$ and $BX^2=BT\cdot BS$. Thus $\frac{AC^2}{AB^2}=\frac{CT\cdot CS}{BT\cdot BS}$ , so $S$, $T$ are isogonal wrt $\angle BAC$. After inversion with center $A$ we have $S'T'||B'C'$ so we are done.

Part a) is easy, another proof using hamoric division! From $D(BA,FE)=-1\implies D(ZA,YX)=-1$ but $\angle XSY=\angle XTY=90^\circ$ so $SX$ is the bisector of $\angle ASZ,TX$ is the bisector of $\angle ATZ\implies AX$ is the bisector of $\angle STA$ i.e $\gamma $ tangent to $(ABC).\blacksquare$ Part b) is easy too, it is only the result of Sayawama theorem! Apply to the Sayawama theorem of triangle $ATS$ with $C$ lies on $ST$ and $DE$ passes through $X$ - the incenter of triangle $AST$ implies $\gamma $ tangents to $(I).\blacksquare$

shinichiman wrote: 61plus wrote: Another approach to part b) is to use that tangent point, $D$, and midpoint of $XY$ are collinear. I tried to use that condition but failed. Can you post your solution using this idea? Thanks. Though I'm not the one who proposed the idea, um I hope this helps. Let $P$ be the intersection between $VD$ and incircle of $ABC$. Prove that $A, Z, D, P$ are concyclic. Let $S$ be intersection between segment $AP$ and incircle of triangle $ABC$. Use the fact that incircle is tangent to $BC wrt D$. Then, you shoud be able to prove that $SD//AV$. Then $P$ is the homethenter of~

From the fact that $D(E,F;A,D)=-1$ and projecting this quadruple on the line $AI$ we get that if $AI \cap BC=Z$ then $(X,Y;Z,A)=-1$. Now, we see that $(SX,SY;SZ,SA)=-1$ and $\angle XSY=90^{\circ}$ so, $SX$ bisects angle $ASZ$ and so $\angle ASX=\angle XSZ=\angle XST=\angle AYT$. Therefore, $\triangle ASX \sim \triangle AYT$ and an inversion about $A$ of radius $\sqrt{AX.AY}$ followed by reflection in $XY$ sends $S$ to $T$. Thus, lines $AS,AT$ are isogonal in angle $BAC$ and result of part a.) follows. Now, we apply $\sqrt{bc}$ inversion (inversion about $A$ of radius $\sqrt{AB.AC}$ followed by reflection in the bisector of angle $BAC$). Denote by $X'$ the image of point $X$. We want to prove that the line $S'T'$ is tangent to the $A$ mixtilinear excircle and is closer to $BC$ than its parallel counterpart. Let $I_A$ be the excenter opposite $A$ and $T_A$ be the mixtilinear excenter opposite $A$ in triangle $ABC$. What we want to prove is equivalent to \begin{align*} \frac{AI_A}{AT_A}=\frac{d(A,BC)}{d(A,S'T')} \end{align*} Now, $\frac{AI}{AT_A}=\frac{d(A,\ell)}{d(A,\ell_a)}=\frac{(s-a)^2}{bc}$ where $\ell$ is the closer tangent to the incircle to $A$ parallel to $BC$ and $\ell_a$ is that for the mixtilinear excircle. We want $\ell_a=S'T'$. Note that $\frac{d(A,BC)}{d(A,S'T')}=\frac{AT}{AS'}=\frac{AS.AT}{AS.AS'}=\frac{AM^2-MD^2}{bc}$. Thus, we want to show this equal to $\frac{AI_A}{AT_A}$. Dividing the two, we want $\frac{AI}{AI_A}=\frac{s-a}{s}=\frac{b+c-a}{b+c+a}=\frac{(s-a)^2bc}{bc(AM^2-MD^2)}=\frac{(s-a)^2}{AM^2-MD^2}$ Now, clearly, $AM^2-MD^2=\frac{2b^2+2c^2-a^2-(b^2-2bc+c^2)}{4}=\frac{(b+c)^2-a^2}{4}=\frac{(b+c-a)(b+c+a)}{4}$. Putting this back, we conclude.

shinichiman wrote: 61plus wrote: Another approach to part b) is to use that tangent point, $D$, and midpoint of $XY$ are collinear. I tried to use that condition but failed. Can you post your solution using this idea? Thanks. Let $VD$ intersect $(AST)$ at $G$. First we show that $G$ lie on incircle. Call the midpoint of $DG$ as $M$. It suffice to show that $IM\perp VD \Leftrightarrow VI^2-DI^2=VM^2-DM^2=VD\cdot VG=VX^2\Leftrightarrow DI^2=VI^2-VX^2=IX\cdot IY$, which can be shown by an angle chase. Now consider the homothety from $G$ sending $D$ to $V$ and $I$ to $L'$. Thus $L'$ lie on perpendicular bisector of $GV$ and $L'V$ perpendicular to $BC$. But $LV$ is also perpendicular to $BC$ and also lie on p.b. of $GV$, thus they must be the same point $\Rightarrow G$ is the centre of homothety of two circles, and thus tangent point.

Proof of part (a) Lemma: Let $R\equiv XY\cap BC$. Then $(A,R;X,Y)=-1$. Proof: Let $\Gamma$ be the incircle; projecting from $D$ onto $\Gamma$, we get that $$(A,R;X,Y)\stackrel{D}{=}(AD\cap \Gamma, D; E, F)=-1$$as $AE, AF$ are tangent to $\Gamma$. Let $M$ be the midpoint of $XY$; it is well-known that $MX\cdot MY=MA\cdot MR$. Then it follows that $\{A,R\}$ swap under an inversion about the circle with diameter $XY$; therefore, $\odot(AST)$ maps to line $ST$. This implies $M\in \odot(AST)$, and since $M$ lies on the perpendicular bisector of segment $ST$, by Fact 5, $AM$ is an angle bisector of $\angle SAT$, so $AS$ and $AT$ are isogonal in $\angle BAC$. Extending $AS$ and $AT$ to intersect the circumcircle of $\odot(ABC)$ at $S'$ and $T'$ respectively, since the midpoint of minor arc $BC$ is also the midpoint of minor arc $S'T'$, it follows that $S'T'\parallel BC$, so $\triangle AST$ and $\triangle AS'T'$ are homothetic, as desired. Proof of part (b) Lemma: The circle with diameter $\overline{XY}$ is orthogonal to $\Gamma$. Proof: Let $I$ be the incenter; it suffices to show that $IX\cdot IY=r^2$, where $r$ is the inradius. Equivalently, we can show that $\{X,Y\}$ swap under an inversion about the incircle. Of course, it is well-known that $X$ and $Y$ are the projections of $B$ and $C$, respectively, onto the $A$-angle bisector. An inversion about the incircle sends line $DE$ to $\odot(CDEI)$ (as $CD, CE$ are tangent to the incircle), which is the circle with diameter $\overline{CI}$. As line $XI$ maps to itself under this inversion, it follows that the image of $X$ is the second intersection of the circle with diameter $\overline{CI}$ with line $XI$, which is simply $Y$ by our earlier observation. Hence the lemma is proved. Finally, an inversion about the circle with diameter $\overline{XY}$ fixes the incircle and maps $\odot(AST)$ to line $ST$, which is clearly tangent to the incircle, as desired.

It's quite easy if you make use of barycentric coordinates..

LukeMac wrote: It's quite easy if you make use of barycentric coordinates.. Is it though? I tried and $S, T$ are ugly; you can $\odot(AST)$ with some synthetic observation, I believe, but that's really the same as just doing the problem.

EulerMacaroni wrote: LukeMac wrote: It's quite easy if you make use of barycentric coordinates.. Is it though? I tried and $S, T$ are ugly; you can $\odot(AST)$ with some synthetic observation, I believe, but that's really the same as just doing the problem. Well, you can find $\odot(AST)$ proving (with barycentrics) that $BX \perp XY$ and keep in this way, but maybe yes, it makes use of many synthetic observations

6(a) can be done with complex bash. Invert about the incircle and let $P'$ denote the inverse of an arbitrary point $P$. Set $D=1, E=e, F=f$ with $|e|=|f|=1$. It suffices to show that $B'C', S'T'$, and the tangent to $(A'B'C')$ at $A'$ concur by considering the radical center of $(A'B'C'), (A'S'T'), (B'C'DS'T')$. $S'T'$ can be computed by subtracting the equations of $(XYS'T')$ and $(B'C'DS'T')$, and the other two lines are not hard to directly compute. $X$ and $Y$ also have manageable equations: $X= \frac{f(e+1)}{f+1}$ and $Y=\frac{e(f+1)}{e+1}$.

My solution copy and pasted: Edit: ok so this is too long to read, I'll get rid of some parts. Problem 6 Proof: a) Let $BC=a, CA=b, AB=c$ and let the capital letters denote the angles. Let the $I$ be the incenter and $I_A$ be excenter opposite $A$. Note that $S$ and $T$ are on the circle with diameter $XY$. Let this circle be $\omega$. Lemma (well-known ish): $BX \perp AX$ and $CY\perp AY$ Thus, $BX$ is tangent to $\omega$. By Power of a Point, $BX^2=BS\cdot BT$ and $CY^2=CS\cdot CT$. However, $\triangle ABX\sim \triangle ACY$ $$\dfrac{c^2}{b^2}=\dfrac{BX^2}{CY^2}=\dfrac{BS\cdot BT}{CS \cdot CT}$$However, by Law of Sines, we have $$\dfrac{BS}{CS}=\dfrac{\frac{c\sin BAS}{\sin ASB}}{\frac{b\sin CAS}{\sin ASC}}=\dfrac{c\sin BAS}{b\sin CAS}$$Similarly, $\dfrac{BT}{CT}=\dfrac{c\sin BAT}{b\sin CAT}$. Thus, $$\dfrac{c^2}{b^2}=\dfrac{BS}{CS}\cdot \dfrac{BT}{CT}=\dfrac{c\sin BAS}{b\sin CAS}\cdot \dfrac{c\sin BAT}{b\sin CAT}$$$$\dfrac{\sin BAS}{\sin CAS}=\dfrac{\sin CAT}{\sin BAT}$$and clearly $\angle BAS =\angle CAT$ and by homothety we're done. b) After applying $\sqrt{bc}$ inversion, it suffices to prove $S_1T_1$ is tangent to the mixtillinear excircle (oops im not sure about this name). Let $F_1$ be the where the excircle opposite $A$ meets $AB$. Clearly, $E'$ is where the mixtillinear excircle meets $AC$. Note that, by the fact that the inversion is radius is $\sqrt{bc}$, $AF\cdot AE'=bc$ and $$AF\cdot AF_1=AI\cos \frac{A}{2} \cdot AI_A \cos \frac{A}{2}=bc\cos^2\frac{A}{2}$$Thus $\dfrac{AF_1}{AE}=\cos^2 \dfrac{A}{2}$ Lemma: $AX\cdot AY=AS\cdot AT$ Proof of Lemma Just reflect across AI However, $AX\cdot AY=c\cos \dfrac{A}{2}\cdot b\cos \dfrac{A}{2}$. Thus, $AS\cdot AT=bc\cos^2 \dfrac{A}{2}$. However, by $\sqrt{bc}$ inversion, $AS\cdot AT_1=bc$, and $\dfrac{AT}{AT_1}=\cos^2 \dfrac A2$ Thus the homothety at center $A$ and ratio $\dfrac{1}{\cos^2 \frac A2}$ sends the mixtillinear excircle opposite $A$ to the excircle opposite $A$ and $S'T'$ to $ST$. Since the excircle opposite $A$ is tangent to $ST$ (which is $BC$), the other two are tangent and we are done.

First, we claim that $X$ and $Y$ are the incenter and excenter of $\triangle AST$. (This is Sharygin 2013, Problem 18, also problem 11.12 of my book). To see this, recall that $\angle AXB = \angle AYC$ are right angles (Iran). Now let $K = \overline{AXY} \cap \overline{BC}$ and let $L$ be the foot of the external $\angle A$-bisector. Then $(KL;BC)=-1$, so projection onto $\overline{AI}$ gives $(AK;XY)=-1$. Now, since $\angle YSX = 90^{\circ}$, we see that $\overline{SX}$ and $\overline{SY}$ are bisectors of $\angle AST$. The same statement holds for $\angle ATS$, which proves the claim. In particular, this implies that $\overline{AS}$ and $\overline{AT}$ are isogonal to each other, and therefore part (a) is solved. As for part (b), denote $(XSTY)$ by $\omega$, centered at a point $M$, which is midpoint of arc $ST$ of $\gamma$. Now, it's easy to see $\triangle IXD \sim \triangle IDY$, therefore $ID^2 = IX \cdot IY$ and thus the incircle is orthogonal to $\omega$. Therefore an inversion around $\omega$ fixes the incircle. Now $\gamma$ is mapped to line $BC$, which is obviously tangent to incircle. Therefore $\gamma$ was tangent too.

My solution: a Let $AI$ cut $BC$ at $R$. Then $\frac {RX}{RY}=\frac {BX}{CY}=\frac {AX}{AY}$$\implies $ $(AR,XY)=-1$$\implies $SX is the bisector of $\angle AST$. Similarly, we have $X$ is the incenter of $\triangle AST$$\implies $ $AS,AT$ are isogonal to each other, $\implies $ $(AST)$ is tangent to $(O)$. b Well-know that $BX$ and $CY$ are perpendicular to $AI$ and $\angle IDX=\angle IBX=\angle ICB=\angle IYD$$\implies $$ID^2=IX.IY$$\implies $$(I)$ and $(XY)$ are othogonal. Let $K$ be the midpoint of arc $TS$ of $(AST)$. Inversion with center $K$, radius $KT^2$, then $ST\mapsto (AST)$, $(I)\mapsto (I)$ $\implies $ $(AST)$ is tangent to $(I).$

My solution to part (b) Let circle $(S)$ be the A-mixtilinear circle and the circle $(T)$ which is tangent to the circumcircle$(O)$ internally at A and tangent to the incircle $(I)$ internally now according to part (a) it is sufficent to show that the middle point $M$ of $XY$ lies on circle $(T)$ applying homothetic transformation we get $\frac{r}{R_s}=\frac{R_T}{R}=\cos^2(\frac{A}{2})$ means $R_T=R\cos^2(\frac{A}{2})$ suppose circle $(T)$ meets the bisector of anlge $BAC$ again at point $Z$ now applying sine law we can get $AZ=2R\cos^2(\frac{A}{2})\cos(\frac{B-C}{2})=R\cos(\frac{A}{2})(\cos(\frac{A+B-C}{2})+\cos(\frac{A+C-B}{2}))=R\cos(\frac{A}{2})(\sin B+\sin C)=\frac{1}{2}(b+c)\cos(\frac{A}{2}))$ and applying sine law on $\triangle AEX$and $\triangle AFY$we can get $AX=c\cos(\frac{A}{2})$and $AY=b\cos(\frac{A}{2})$ so $AM=\frac{1}{2}(b+c)\cos(\frac{A}{2})=AZ$ which means $M=Z$ as disired!

(a) It is not difficult to get that $\angle AXB=\angle AYC=90^{\circ}$. Now we consider a inversion with power $\sqrt{AB\cdot AC}$ combined with a reflection about the angle bisector of $\angle A$. If $X,Y\leftrightarrow X',Y'$ then $\angle ACX'=\angle ABY'=90^{\circ}$. Then note if $AA_0$ is the diameter, then $A_0X'=A_0Y'$, and subsequently midpoint of $X'Y'$ is also midpoint of arc $BC$. Hence $S'T'//BC$ and thus $\gamma$ is tangent to the circumcircle. (b) By checking ratios, if we invert with power $\sqrt{AX\cdot AY}$ and reflect about the bisector of $\angle A$, then the incircle $\leftrightarrow$ $A$-excircle. From part (a), since $\angle SAI=\angle TAI$ and $(XSTY)$ is invariant, then $S\leftrightarrow T$ and $BC\leftrightarrow (AST)$. Hence problem follows from $BC$ tangent to $A$-excircle.

First, angle chase that X and Y are the projections of B and C onto the angle bisector. You use this to see that if circle AST meets AB,AC at P,Q then BX^2=BS.BT=BP.BA, where BX=AB.sin(A/2). Plug this in to find that AP/AB=w:=cos^2(A/2) So triangle APQ and ABC are homothetic from A with ratio w. Done with a) Now let m be the A-mixtilinear incircle. From the incenter on touch chord lemma we can compute that m and the incircle are homothetic from A with ratio w. Hence, since m touches circle ABC, we can homothety this to get that the incircle touches circle APQ, so done with b).

Let incircle is $\omega$ and $AI \cap \omega = U,W$ and $AI \cap BC=Z$ $K=SX \cap YT$ $L=TX \cap SY$. $M$= midpoint of $XY$ $-1=(E,F;D,DA\cap \omega)=(X,Y;Z;A)$ so polar line of $Z$ for $\gamma$ (which is $KL$ from brocard) passes through $A$ and $MZ \perp KL$ from isogonal line lemma for $(K,L)(X,Y)$from $A$ $\rightarrow $ $AI$ is angle bisector of $\angle SAT$ so $(SAT)$ is tangent to $(ABC)$ . since $MS=MT$ and $AM $ is angle bisector of $\angle SAT $ then $(AMST)$ is cyclic. $-1=(E,F;U,W)=(X,Y;U,W)$ then $MX^2=MU.MW$ since we know that $ST$ is tangent to $\omega$ after inversion centered $M$ with radius $MX$ we get that $(MXY)$ is tangent to $\omega$

Let $M = AI\cap BC$. Let $S'$ and $T'$ be the reflections of $S$ and $T$ over $AI$, respectively. By symmetry, $S'$ and $T'$ lie on $(XY)$, and $M$ lies on $S'T'$. If $EF$ intersects $AD$ and $BC$ at $P$ and $Q$, then we have $$-1 = (BC;DQ) \overset{A}{=}(FE;PQ) \overset{D}{=}(YX;AM).$$ Hence $M$ and $A$ are inverses about $XY$ by a well known lemma. But since $SS'TT'$ is an isosceles trapezoid, it follows that $A = ST'\cap TS'\cap XY$. In particular, $AS$ and $AS'$ are isogonal, which solves part (a). Now, since $M$ lies on $ST$ and $A$ is the inverse of $M$ about $(XY)$, $(AST)$ passes through the circumcenter $O$ of $(XY)$. Moreover, since $X$ and $Y$ are the intersections of the perpendicular bisector with opposite sides of $\triangle DEF$, $X$ and $Y$ are inverses about $(DEF)$, so $(DEF)$ and $(XY)$ are isogonal. Then inversion about $(XY)$ maps $(AST)$ to line $ST$ and fixes $(DEF)$, which solves part (b).

Don't worry Elmo. Proof for part (a): It is well-known that $AI$, $DE$, and the $B$-median are concurrent, hence $X$ is the projection from $B$ to $AI$. Analogously $Y$ is the projection from $C$ to $AI$. Set $K=AI\cap BC$. From $$\frac{AX}{AY}=\frac{AB}{AC}=\frac{BK}{CK}=\frac{KX}{KY}$$we know that $-1=(A,K;X,Y)$. Note that $\angle XSY=\angle XTY=90^\circ$, therefore $SX$ is the angle bisector of $\angle AST$, and $TX$ is the angle bisector of $\angle ATS$. This shows that $X$ is the incenter of $\triangle AST$, thus $AI$ is the angle bisector of $\angle SAT$, then $\angle BAS=\angle CAT$. Easy angle chasing shows that $\gamma$ is tangent to $(ABC)$ at $A$. Proof for part (b): Let $O'$ be the center of $\gamma$, and $L$ to be the midpoint of $XY$. From part (a) we know that $A$,$O'$ and $O$ are collinear. Because $LS=XY/2=LT$ and $O'S=O'T$, $O'L$ is perpendicular to $BC$, indicating that $AO'=O'L$. Because $$AL=\frac{AX+AY}{2}=\frac{c\cos \frac{A}{2}+b\cos \frac{A}{2}}{2}=2R\cos^2 \frac{A}{2}\cos \frac{B-C}{2}$$We know that $AO'=\frac{AL}{2\cos \frac{B-C}{2}}=R\cos^2\frac{A}{2}$. It is also known that $r=ID=4R\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}$, $AI=4R\sin\frac{B}{2}\sin\frac{C}{2}$. Now all we have to show is that $$AI^2+AO'^2-2AI\cdot AO'\cos \frac{B-C}{2}=(AO'-r)^2$$This is not so hard to check. Hence $IO'=AO'-r$, which means that $\gamma$ is tangent to the incircle of $\triangle ABC$.

Cute little projective/inversive problem, although easy for 3/6 if you understand configurations well. By Iran lemma $\overline{BX} \perp \overline{XY}$ and $\overline{CY} \perp \overline{XY}$. As $$-1 = (EF; \overline{AD} \cap (DEF), D) \stackrel D= (XY; A, \overline{BC} \cap \overline{XY}),$$by Apollonian incircle lemma $\overline{SX}$ bisects $\angle AST$, hence $X$ is the incenter of triangle $AST$. Part (a) now follows by noting that $\overline{AS}$ and $\overline{AT}$ are isogonal, so simple angle chasing shows that the tangents to $(AST)$ and $(ABC)$ at $A$ must coincide. For part (b), the key is that the incircle $\omega$ is orthogonal to $\gamma$. It suffices to show $IX \cdot IY = IF^2$. This follows by angle chasing and similar triangles because $$\angle IXF = \angle IXE = \angle DEC - \frac 12 A = \angle IFD.$$Then under inversion at $(XY)$, centered at the arc midpoint $M$ of $\widehat{ST}$ by incenter-excenter lemma, $\omega$ is fixed and $\gamma$ goes to $\overline{BC}$. The result follows.

Alternative solution first solving $b).$ First we claim the following: Claim 1. $X\in (BDFI)$ and $Y\in (CDEI).$ Proof This is easy, as $EF=EX$ implies that $\angle EXF=180^\circ-2\angle FED=\angle B,$ and similarly for $Y.$ $\hfill \blacksquare$ Notice that now inverting around the incircle we easily get that $X$ and $Y$ are inverses, which show that the incircle and $(XY)$ are orthogonal. Now let $N$ be the center of $(XY).$ If we invert around $(XY),$ then $(I)\to (I), S\to S, T\to T,$ so if we had that $A\to AI\cap BC,$ then we get that $(ASY)\to BC,$ showing that $(ASY)$ and the incircle are tangent. To do this, we let $K=AI\cap BC.$ Now the following claim shows that $A\to K:$ Claim 2. $(A,K;X,Y=-1)$ Proof. Notice that $AX\perp BX, AY\perp CY,$ so $\frac{AX}{AY}=\frac{AB}{AC},$ so we are left to show that $\frac{XK}{KY}=\frac{AB}{AC}.$ But this is trivially true as $\triangle BXK\sim \triangle BYK.$ $\hfill \blacksquare$ We're now left to show that $(ASY)$ is tangent $(ABC).$ For this apply Monge d'Alembert's Theorem to $(ASY),$ the incircle and the circumcenter. Notice how we wish to show that the tangency point of $(ASY)$ and the incircle lies on the line that goes through $A$ and the excimillicenter of $(ABC)$ and the incircle. Let $A'$ be the contact point of the $A-Mixtilinear$ with $(ABC).$ By Monge d'Alembert, it is well-known that $AA'$ goes through the excimillicenter of $(ABC)$ and the incircle. Now let $L=AA'\cap (I).$ We wish to show $L$ is the excimillicenter of $(ASY)$ and the incircle. By the previous inversion, we get that this is equivalent to showing $LDKA$ ciclic. In order to show the desired cyclic quadrilateral apply inversion at $A$ with radius $\sqrt{AB\cdot AC}$ followed by a reflection across the bisector of $\angle BAC.$ Now, if we let $\Omega_A$ be the external $A-Mixtilinear,$ we get that $D$ goes to the tangency point of $\Omega_A$ and $(ABC),$ which we will call $D',$ and $K$ goes to some $K'$ which is the midpoint of arc $BC.$ In order to treat the inverse of $L,$ we notice that $A'$ is well-known to be sent to the contact point of the excircle to $BC.$ So $L'$ will be a point on $\Omega_A$ homothetic to this contact point, which we will call $J.$ We now wish to show $L'-K'-D'.$ But this is easy as considering the homothety centered at $D'$ which sends $(ABC)$ to $\Omega_A,$ by the fact that $L'$ is the point homothetic to $J,$ then $L'$ and $K'$ are inverses, which ends the problem.

Take a $\sqrt{bc}$ inversion. Since $X,Y$ are the Iran Lemma points, they lie on the circles with diameters $AB,AC$ so under the inversion we find $X',Y'$ are the points on the angle bisector with $\angle ABY'=\angle ACX'=90.$ Then, the midpoint of $X'Y'$ is the point $M$ on the angle bisector with $AM=\frac{AB+AC}{2 \cos{\tfrac{A}{2}}},$ which is the arc midpoint of $BC$ on $(ABC).$ Then, the line through the intersections of the circle with diameter $X'Y'$ and the circumcircle of $ABC$ is perpendicular to the line through their centers, which is perpendicular to $BC.$ Thus, inverting back we get that $(AST)$ is tangent to $(ABC).$ Next, let $J,K$ be the intersections of the circle with diameter $X'Y'$ and the circumcircle. Since $JM=KM$ we find that $AJ,AK$ are isogonal with respect to $\angle A.$ Since $S,T$ invert to $J,K,$ we may assume that $A,S,J$ are collinear and $A,T,K$ are collinear. Then since the circles with diameters $XY$ and $X'Y'$ invert to each other we get that there is a homothety at $A$ sending these circles to each other. Since $JK \parallel ST$ we may check that this homothety also sends $ST$ and $JK$ to each other. Then, we may check that this homothety has scale factor $\frac{AX}{AY'}.$ Since $AX \perp XB$ and $AB \perp BY'$ we get $\frac{AX}{AY'}=\cos^2{\tfrac{A}{2}}.$ However, we can also check that this is the scale factor of the homothety sending the incenter $I$ to the midpoint of $EF.$ However, $I$ is the midpoint of the $A$-mixtilinear touch chord, so this homothety sends the incircle and $A$-mixtilinear incircle to each other. Therefore, we get that the incircle is the $A$-mixtilinear incircle of the triangle with vertices $A$ and the intersections of $(AST)$ with lines $AB,AC,$ so $(AST)$ is tangent to the incircle as well.

Time to actually solve this problem and help Elmo instead of providing a solution... Hi Elmo! Hope you're doing well. For part a), observe that the result is the same as showing that $AS$ and $AT$ are isogonal. Take a look at $(SXTY)$. Does this remind you of a configuration you've seen before? Do $X$ and $Y$ look familiar, viewing from $\triangle AST$? Incenter configs tend to have some harmonic interpretation... hopefully you can find it! In part b), take a look again at $(XSTY)$ and the incircle. Notice something about the two circles? If you did part a) in the way I hinted towards, maybe you'll be able to motivate something too... something that would invert your expectations perhaps? From there it shouldn't be too difficult to finish. Oh well, in any case, if you need to read a solution, here you go:

im sorry elmo WLOG let $AB<AC$. Let $M$ be the midpoint of $\overline{XY}$, so $XYST$ is cyclic with center $M$, and let $\overline{AXY} \cap \overline{BC}=K$. By the Iran Lemma, we have $\angle AXB=\angle AYC=90^\circ$, so $X$ and $Y$ can be redefined as the feet of the altitudes from $B$ and $C$ to the $\angle BAC$-bisector. We then have $AX=AB\cos\tfrac{\angle A}{2}$ and $AY=AC\cos\tfrac{\angle A}{2}$, so $AM=\tfrac{AB+AC}{2}\cos\tfrac{\angle A}{2}$. Now construct circles centered at $X$ and $Y$ tangent to $\overline{BC}$. These have radii $(AK-AX)\sin(\tfrac{\angle A}{2}+\angle B)$ and $(AY-AK)\sin(\tfrac{\angle A}{2}+\angle C)$ respectively. I claim that their exsimilicenter is $A$; indeed, since $(\tfrac{\angle A}{2}+\angle B)+(\tfrac{\angle A}{2}+\angle C)=180^\circ$ it suffices to show that $$\frac{AX}{AY}=\frac{AK-AX}{AY-AK} \iff AX\cdot AY=AK\cdot AM \iff AB\cdot AC\cos \frac{\angle A}{2}=\frac{AB+AC}{2}\cdot AK.$$We thus want to show by $\sqrt{bc}$ inversion that the length of the chord joining $A$ to the minor arc midpoint $Z$ of $BC$ is $\tfrac{AB+AC}{2\cos \frac{\angle A}{2}}$, which is evident since the feet of the perpendiculars from $Z$ to $\overline{AB}$, $\overline{AC}$ are a distance $\tfrac{AB+AC}{2}$ away from $A$ by symmetry. Thus consider the triangle $\triangle AS'T'$ with incenter $X$ and excenter $Y$, so $S',T'$ lie on $\overline{BC}$. Then by fact 5 $XYS'T'$ is cyclic, so $\{S',T'\}=\{S,T\}$. Then $M$ is the midpoint of minor arc $\overline{ST}$, so the conclusion follows by homothety at $A$, since $\gamma$ and $(ABC)$ have the same "lowest point". For part $B$, by a well-known lemma it suffices to show that $MX^2=\mathrm{Pow}_{\omega}(M)$, where $\omega$ is the incircle of $\triangle ABC$. Define $a,b,c,s$ as usual. By considering $\triangle AEI$ we find that $AI=\tfrac{s-a}{\cos \frac{\angle A}{2}}$ and the inradius is $(s-a)\tan \tfrac{\angle A}{2}$. Hence it suffices to show \begin{align*} \frac{1}{4}(b-c)^2\cos^2\frac{\angle A}{2}&=\left(\frac{1}{2}(b+c)\cos\frac{\angle A}{2}-\frac{s-a}{\cos \frac{\angle A}{2}}\right)^2-(s-a)^2\tan^2\frac{\angle A}{2}\\ (b-c)^2\cos^4\frac{\angle A}{2}&=(b+c)^2\cos^4\frac{\angle A}{2}-2(b+c)(b+c-a)\cos^2\frac{\angle A}{2}+(b+c-a)^2-(b+c-a)^2\left(1-\cos^2\frac{\angle A}{2}\right)\\ 4bc\cos^2\frac{\angle A}{2}&=2(b+c)(b+c-a)-(b+c-a)^2=(b+c-a)(b+c+a)\\ \frac{1+\cos \angle A}{2}&=\frac{b^2+2bc+c^2-a^2}{4bc}\\ \cos \angle A&=\frac{b^2+c^2-a^2}{2bc}, \end{align*}which is true. $\blacksquare$ Remark: If we want to actually be smart part a is equivalent to $(A,K;X,Y)=-1$ and this follows by projecting from the point at infinity along the line perpendicular to $\overline{AK}$

By the Iran lemma, $X$ and $Y$ are just the feet of the perpendiculars from $B$ and $C$ to $AI$. Claim 1: $AS$ and $AT$ are isogonal. We wish to show that $$\frac{BS\cdot BT}{CS\cdot CT}=\frac{AB^2}{AC^2}.$$However, since $BX$ is tangent to $(XY)$, we know that $$BS\cdot BT=BX^2$$and similarly $$CS\cdot CT=CY^2.$$Thus, it suffices to show that $$\frac{AB}{AC}=\frac{BX}{CY}.$$However, this follows easily as $$\triangle ABX\sim\triangle ACY.$$ Now, suppose that $(AST)$ intersects $AB$ at $C_2$ and $AC$ at $B_2$. We have $$BS\cdot BT=BC_2\cdot AB,$$and similarly for $C$, and due to $$\frac{BS\cdot BT}{CS\cdot CT}=\frac{AB^2}{AC^2}$$by the above claim, we have $$\frac{BC_2\cdot AB}{CB_2\cdot AC}=\frac{AB^2}{AC^2}$$which reduces to $$\frac{BC_2}{CB_2}=\frac{AB}{AC}.$$This means that $B_2C_2\parallel BC,$ so $(AB_2C_2)$ and $(ABC)$ are homothetic with center $A$, hence they are tangent at $A$, solving part (a). Claim 2: $(XY)$ and the incircle are orthogonal. We will show that $(XY)$ maps to itself under inversion around the incircle. Note that $$\angle IDX=\angle ICE=\gamma/2$$and $$\angle IYD=90-\angle EFD=90-(90-\gamma/2)=\gamma/2.$$This means that $$\triangle AXD\sim\triangle ADY,$$so $$IX\cdot IY=ID^2.$$This means that $X$ and $Y$ are inverses of each other. Furthermore, the image of $(XY)$ clearly also has its center on line $AI$. The only circle with center on $(AI)$ that passes through both $X$ and $Y$ is $(XY)$ itself, hence shown. Go back to the original diagram. Now, for the rest of the problem, we actually invert around $(XY).$ By the above lemma, the incircle inverts to itself. Claim 3: The image of $A$ is the foot of the angle bisector from $A$ to $BC$. Let the foot of the angle bisector be $L$. It suffices to show that $$(AL;XY)=-1.$$Let $AD$ intersect the incircle again at $D_2$. Projecting through $D$ onto the incircle, it suffices to show that $$(D_2E;DF)=-1,$$however this is clearly true since $EF$ is the polar of $A$ with respect to the incircle. This means that the image of $(AST)$ is $(LST)$ (aka line $BC$). This is clearly tangent to the incircle, solving part (b).

Assume WLOG $AB < AC$. The problem rests in the following two claims: Claim 1: $AS$ and $AT$ are isogonals. The angle condition on $S$ and $T$ and \[(A, BC \cap AI; X, Y) \overset{D}{=} (AD \cap (DEF), D; E, F) = -1\] tell us $SX$ and $TX$ are angle bisectors. Thus $X$ is the incenter of $\triangle AST$ (which also implies $Y$ is the excenter), giving the desired. ${\color{blue} \Box}$ Claim 2: The incircle of $\triangle ABC$ and $(XY)$ are orthogonal. This boils down to proving $\triangle IXD \sim \triangle IYD$, or \[\angle IDX = \frac C2 = 90 - \frac{A+B}{2} = \angle YFB - \angle YAF = \angle DYI. \quad {\color{blue} \Box}\] We now begin with part (a). If we extend $AS$ and $AT$ to meet $(ABC)$ at $S'$ and $T'$, our isogonality tells us $ST \parallel S'T'$. Thus there exists $h_A: (AST) \mapsto (ABC)$, which finishes. We use inversion at $(XY)$ for part (b). The incircle is sent to itself, and incenter-excenter lemma says $(AST)$ passes through the midpoint of $XY$, so it maps to $ST$, which finishes. $\blacksquare$

Here's a complex bash of part (b).

ELMO 2016 P6 [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(7.939146298858004cm); 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 = -5.24409, xmax = 2.695056298858004, ymin = -3.892578562921142, ymax = 0.7209030419910142; /* image dimensions */ pen qqqqcc = rgb(0.,0.,0.8); pen xdxdff = rgb(0.49019607843137253,0.49019607843137253,1.); pen zzccff = rgb(0.6,0.8,1.); pen qqzzcc = rgb(0.,0.6,0.8); pen qqffff = rgb(0.,1.,1.); pen afeeee = rgb(0.6862745098039216,0.9333333333333333,0.9333333333333333); pen eqeqeq = rgb(0.8784313725490196,0.8784313725490196,0.8784313725490196); /* draw figures */ draw(circle((-0.6309699530964192,-1.5856357285807214), 2.092545121445029), linewidth(1.) + qqqqcc); draw((-2.3335572186921834,-2.802164268449199)--(1.1340236495870901,-2.7097098666026214), linewidth(1.) + xdxdff); draw((-2.094030111572516,-0.0895742131465167)--(1.1340236495870901,-2.7097098666026214), linewidth(1.) + xdxdff); draw((-2.3335572186921834,-2.802164268449199)--(-2.094030111572516,-0.0895742131465167), linewidth(1.) + xdxdff); draw(circle((-1.3408942283283616,-1.8686696400668208), 0.9067055173100018), linewidth(1.) + zzccff); draw((-1.3167277737383822,-2.775053045471481)--(-0.7694816560384322,-1.1646791803662608), linewidth(1.) + qqzzcc); draw(circle((-0.9097426169026007,-2.8871576410537854), 0.6333203179032343), linewidth(1.)); draw(circle((-0.9532305411655596,-1.2561061046904733), 1.631631181924953), linewidth(1.) + dotted + qqffff); draw((-0.6628528961604037,-3.470372945089005)--(-2.094030111572516,-0.0895742131465167), linewidth(1.) + afeeee); draw((-2.2440853829604293,-1.788916061370997)--(-0.6628528961604037,-3.470372945089005), linewidth(1.) + qqzzcc); draw((-2.2440853829604293,-1.788916061370997)--(-0.7694816560384322,-1.1646791803662608), linewidth(1.) + qqzzcc); draw((-2.3335572186921834,-2.802164268449199)--(-1.1566323376447978,-2.303942337018565), linewidth(1.)); draw((1.1340236495870901,-2.7097098666026214)--(-0.6628528961604037,-3.470372945089005), linewidth(1.)); draw((-0.9097426169026007,-2.8871576410537854)--(-1.8257671210517157,-2.6348374891850703), linewidth(1.) + gray); draw((-1.3167277737383822,-2.775053045471481)--(-1.3408942283283616,-1.8686696400668208), linewidth(1.) + zzccff); draw(circle((-1.837225723510272,-2.3354169542580103), 0.6813207823149091), linewidth(1.) + eqeqeq); draw(circle((-0.10343528937063598,-2.2891897533347216), 1.3069589860700406), linewidth(1.) + eqeqeq); /* dots and labels */ dot((-2.3335572186921834,-2.802164268449199),dotstyle); label("$B$", (-2.3075964937379037,-2.732132269661029), NE * labelscalefactor); dot((1.1340236495870901,-2.7097098666026214),dotstyle); label("$C$", (1.1595886100414388,-2.640145673244069), NE * labelscalefactor); dot((-2.094030111572516,-0.0895742131465167),dotstyle); label("$A$", (-2.0670163028634185,-0.022065621376741097), NE * labelscalefactor); dot((-1.3408942283283616,-1.8686696400668208),linewidth(4.pt) + dotstyle); label("$I$", (-1.3098962904054807,-1.8122663054914276), NE * labelscalefactor); dot((-1.3167277737383822,-2.775053045471481),linewidth(4.pt) + dotstyle); label("$D$", (-1.2886686265047909,-2.7179804855968817), NE * labelscalefactor); dot((-2.2440853829604293,-1.788916061370997),linewidth(4.pt) + dotstyle); label("$F$", (-2.215609950168248,-1.734431493138615), NE * labelscalefactor); dot((-0.7694816560384322,-1.1646791803662608),linewidth(4.pt) + dotstyle); label("$E$", (-0.7438252530537512,-1.1046771022840416), NE * labelscalefactor); dot((-1.1566323376447978,-2.303942337018565),linewidth(4.pt) + dotstyle); label("$X$", (-1.1259232032661686,-2.243895719447933), NE * labelscalefactor); dot((-0.6628528961604037,-3.470372945089005),linewidth(4.pt) + dotstyle); label("$Y$", (-0.637686933550302,-3.41141790474012), NE * labelscalefactor); dot((-0.9097426169026007,-2.8871576410537854),linewidth(4.pt) + dotstyle); label("$K$", (-0.878267124424787,-2.831194758110063), NE * labelscalefactor); dot((-0.29196060922888234,-2.7477301829844247),linewidth(4.pt) + dotstyle); label("$T$", (-0.26266487130478133,-2.689676917468586), NE * labelscalefactor); dot((-1.5340765889598342,-2.7808481099463096),linewidth(4.pt) + dotstyle); label("$S$", (-1.508021153478586,-2.7250563776289556), NE * labelscalefactor); dot((-1.8257671210517157,-2.6348374891850703),linewidth(4.pt) + dotstyle); label("$Z$", (-1.7981325601213471,-2.5764626449554044), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Here is a sketch of the solution. Let $K$ be the center of circle $XY$. (a) We claim that $AS$ and $AT$ are isogonal, proving that will give us the desired tangency. (b) We show that the incircle and $(XY)$ are orthogonal, which follows by showing $\triangle IXD \sim \triangle IYD$ . To prove that we show that $BFIXD$ and $CEIDY$ are cyclic. Now as the incircle and $(XY)$ are orthogonal, consider an inversion around $(K)$, we get that the incircle is fixed under that inversion and that $(AST)$ is sent to line $BC$. Extend $KD$ until it meets the incircle again at $Z$, we show that its the desired tangency point. we get that $KD \cdot KZ= R^2$ where $R$ is the radius of circle $(K)$ this means that $Z$ is the inverse of $D$ wrt $(K)$, which means it lies on $(AST)$. Now as $BC$ is tangent to the incircle, hence its inverse $(AST)$ is tangent to it as well hence we are done

Solution for Part $a$: Let $AS,AT$ intersect $\gamma$ at $K,L$ respectively. $AI\cap BC=D$. By Iran Lemma, we have that $\measuredangle BXA=90=\measuredangle AYC$. Since $AXB\sim AYC$ and $BXD\sim CYD$, \[\frac{AX}{AY}=\frac{BX}{CY}=\frac{DX}{DY}\implies (A,D;X,Y)=-1\]\[-1=(A,D;X,Y)\overset{S}{=}(K,T;X,Y)\]$XY$ is diameter, subsequently $KT\perp XY$. $XY$ is the perpendicular bisector of $KT$ and $A$ lies on $XY,$ thus $\measuredangle XAS=\measuredangle TAX$. $AS,AT$ are isogonal lines in $\measuredangle CAB$ hence $(AST)$ and $(ABC)$ are tangent to each other.$\blacksquare$ Solution for Part $b$: Invert the diagram from $I$ with radius $ID$. New Problem Statement: $ABC$ is a triangle whose circumcenter is $O$. $M$ is the midpoint of $BC$ and $OM$ intersects $AB,AC$ at $X,Y$ respectively. If circles with diameter $(XY)$ and $(AO)$ meet at $S,T,$ then prove that $(MST)$ is tangent to $(ABC)$. Proof: Let $A'$ be the antipode of $A$. $A'M\cap (ABC)=Q$ which is $A-$queue point. Let $R$ be the altitude from $A$ to $OM$. $H$ is the orthocenter of $ABC$. Let $N$ be the midpoint of $XY$. Claim: $Q,A,N$ are collinear. Proof: Let $AD,BE,CF$ be the altitudes. \[(N,AD_{\infty};X,Y)=-1=(EF\cap BC,D;B,C)=(AQ,AD;AB,AC)=(AQ\cap XY,AD_{\infty};X,Y)\]Hence $QA$ passes through $N$.$\square$ Claim: $QHBMC\sim QAYNX$. Proof: \[\measuredangle BMQ=90-\measuredangle QNB=\measuredangle MNQ\]Set $B',C'$ as the antipodes of $B,C$. Pascal at $QA'ACBB'$ yields $M,O,AC\cap B'Q$ are collinear thus, $Q,Y,B'$ are collinear. Similarily $Q,X,C'$ are collinear. We get that $\measuredangle YQB=90=\measuredangle XQC$. These give that $Q,B,M,Y$ and $Q,C,X,M$ are concyclic. \[\measuredangle QBM=180-\measuredangle MYQ=\measuredangle QYX\]Hence $QBM\sim QYN$, similarily $QCM\sim QXN$. Combining these with $QHM\sim QAN$ implies $QHBMC\sim QAYNX$.$\square$ Lemma: $\frac{\sin A}{\sin B.\sin C}=\cot B+\cot C$. Proof: Let $D$ be the altitude from $A$. \[\cot B+\cot C=\frac{DB}{DA}+\frac{DC}{DA}=\frac{BC}{DA}=\frac{BC}{AC.\sin C}=\frac{\sin A}{\sin B.\sin C}\]Which completes the proof of lemma.$\square$ Claim: $Q,M,S,T$ are concyclic. Proof: We will show this by coaxiality lemma on $(XYST),(AOST)$. Since $QAYNX\sim QHBMC,$ if $G,J$ are the orthocenter of $QXY$ and $QG\cap XY,$ then $\measuredangle XAY=180-\measuredangle A=\measuredangle XGY$ gives that $A$ is $Q-$humpty point of $\triangle QXY$. $QA.QN=QG.QJ$ hence $Pow(Q,(XY))=QA.QN$. Let $I$ be the altitude from $O$ to $QA$. $Pow(Q,(AO))=QI.QA=\frac{QA^2}{2}$ and $Pow(M,(XY))=MX.MY, \ Pow(M,(AO))=MO.MR$. \[\frac{MX.MY}{MO.MR}\overset{?}{=}\frac{QA.QN}{\frac{QA^2}{2}}=\frac{2QN}{QA}=2.\frac{MN}{MO}.\frac{A'O}{A'A}=\frac{MN}{MO}\]\[MX.MY\overset{?}{=}MN.MR=(\frac{MX+MY}{2})MR\iff \frac{2}{MR}\overset{?}{=}\frac{1}{MX}+\frac{1}{MY}\]Multiply both sides by $MB$ to get \[\frac{BC}{MR}\overset{?}{=}\cot B+\cot C=\frac{\sin A}{\sin B.\sin C}\]Since $MYR\sim CYA,$ \[\frac{BC}{MR}=\frac{BC}{AC.\frac{MY}{CY}}=\frac{\sin A}{\sin B.\sin C}\]Which gives that $Q\in (MST)$. Claim: $(QMST)$ is tangent to $(ABCQ)$. Proof: Let $U$ be the midpoint of $AO$ and $V=UN\cap OQ$. \[\frac{VQ}{VO}=\frac{NQ}{NA}.\frac{UO}{UA}=\frac{NQ}{NA}=\frac{MQ}{MH}=\frac{MQ}{MA'}\]Hence $VM\parallel OA'\iff VM=VQ$. Circumcenter of $(QMST)$ is on $NU$ because of coaxiality lemma and it lies on the perpendicular bisector of $QM$. Intersection of those is $V$ hence $V$ is the circumcenter of $(QMST)$. $O,V,Q$ are collinear subsequently $(QMST)$ and $(QABC)$ are tangent to each other as desired.$\blacksquare$

(a) Let $I$ be the incenter of $\triangle{}ABC$. It suffices that $\overline{AS}$ and $\overline{AT}$ are isogonal in $\angle{}CAB$, or that $S$ and $T$ are inverses in the $A$-Apollonius circle $\omega$ of $\triangle{}ABC$, or that $(XYST)$ is orthogonal to $\omega$, or that $X$ lies on the polar of $Y$ and $Y$ lies on the polar of $X$ with respect to $\omega$, or that $\left(A,\overline{AI}\cap\overline{BC};X,Y\right)=-1$, which is true since $\left(A,\overline{AI}\cap\overline{BC};X,Y\right)\stackrel{D}{=}\left(A,B;F,\overline{AB}\cap\overline{DE}\right)=-1$. (b) First, since $X$ and $Y$ are the Iran Lemma points, they are inverses with respect to the incircle of $\triangle{}ABC$, so $(XYST)$ is orthogonal to the incircle of $\triangle{}ABC$. Let $\overline{AI}$ intersect $(ABC)$ at $A$ and $M_A$. Define $\Omega$ as the $A$-mixtilinear incircle touching $(ABC)$ at $T_1$ and let $\overline{M_AT_1}$ intersect $\Omega$ at $T_1$ and $T_2$, so that the tangent to $\Omega$ at $T_2$ is parallel to $\overline{BC}$. Taking a force overlaid inversion at $A$ swapping $B$ and $C$, which swaps $(XYST)$ and the circle $\gamma_1$ centered at $M_A$ orthogonal to $\Omega$ intersecting $(ABC)$ at points $S'$ and $T'$, gives that it suffices that $\overline{S'T'}$ is the tangent to $\Omega$ at $T_2$, for which it suffices that $\overline{S'T'}$ goes through $T_2$ since $\overline{S'T'}\parallel\overline{BC}$. This is true by inverting the fact that $M_A,T_A,S',$ and $T'$ lie on $(ABC)$ around $\gamma_1$.

For $(a)$, we prove that $G$ lies on $(AST)$ where $G$ is center of $(XY)$. That is sufficient because then $AG$ bisects $\angle SAT$ giving us that $(AST)$ and $(ABC)$ are homothetic and afterwards we are done by shooting lemma. To prove this, we need to prove $\angle AGT = \angle AST$. Simplifying, we need to prove $SX$ bisects $\angle AST$. Using the fact that $\angle XSY=90$, we just need to prove $(AL;XY)=-1$ where $L=AI \cap BC$. This is sufficient because of the famous lemma about right angles and angle bisectors from $EGMO$ chapter $9$. Now $$(AL;XY) \stackrel{D}{=} (AD \cap \omega,D ;EF)=-1$$where $\omega =(DEF)$. Hence we are done. For $(b)$, notice that $X$ and $Y$ are perpendiculars from $B$ and $C$ onto $AI$ due to the right angle intouch chord lemma from $EGMO$ chapter $1$. Hence $ X \longleftrightarrow Y$ under inversion about $\omega$. Since $XY$ is diameter in $(XSYT)$, it must be the diameter in the inverted circle, hence $(XY)$ is preserved under this inversion, giving us that the incircle and $(XY)$ are orthogonal. Now we invert about $(XY)$. The incircle is preserved and $(AST)$ is sent to $BC$, which is obviously tangent to $\omega$, hence we are done.