Here is the solution I had in mind (alas, I missed the inversion solution). Let triangle $ABC$ have circumcircle $\Gamma$. Let $\triangle XYZ$ be the tangential triangle of $\triangle ABC$ (hence $\Gamma$ is the incircle of $\triangle XYZ$), and denote by $\Omega$ its circumcircle. Suppose the symmedian $\overline{AX}$ meets $\Gamma$ again at $D$, and let $M$ be the midpoint of $\overline{AD}$. Finally, let $K$ be the Miquel point of quadrilateral $ZBCY$, meaning it is the intersection of $\Omega$ and the circumcircle of $\triangle BOC$ (other than $X$). [asy][asy] size(12cm); pair A = dir(115); pair B = dir(200); pair C = dir(340); draw(A--B--C--cycle, blue); draw(circumcircle(A, B, C), blue); pair O = circumcenter(A, B, C); pair X = 2*B*C/(B+C); pair Y = 2*C*A/(C+A); pair Z = 2*A*B/(A+B); draw(circumcircle(X, Y, Z), cyan); draw(X--Y--Z--cycle, cyan); draw(circumcircle(B, O, C), heavycyan); pair T = extension(Y, Z, B, C); draw(Z--T, lightcyan); draw(T--B, blue+dashed); pair D = -A+2*foot(O, A, X); pair U = foot(A, B, C); draw(circumcircle(T, A, U), heavygreen); pair V = circumcenter(X, Y, Z); pair M = midpoint(A--D); pair L = IP(V--(Y+Z-V), circumcircle(X, Y, Z)); pair K = extension(L, A, O, U); draw(A--X, heavycyan); draw(K--L, heavygreen+dashed); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$O$", O, dir(O)); dot("$X$", X, dir(X)); dot("$Y$", Y, dir(Y)); dot("$Z$", Z, dir(Z)); dot("$T$", T, dir(T)); dot("$D$", D, dir(D)); dot("$M$", M, dir(M)); dot("$L$", L, dir(L)); dot("$K$", K, dir(K)); [/asy][/asy] We first claim that $M$ and $K$ are $A_1$ and $A_2$. In that case $OM < OA < OK$, so $M = A_1$, $K = A_2$. To see that $M = A_1$, note that $\angle OMX = 90^{\circ}$, and moreover that $\overline{TA}$, $\overline{TD}$ are tangents to $\Gamma$, whence we also have $M = \overline{TO} \cap \overline{AD}$. Thus $M$ lies on both $(BOC)$ and $(AT)$. This solves part (a) of the problem: the concurrency point is the symmedian point of $\triangle ABC$. Now, note that since $K$ is the Miquel point, \[ \frac{ZK}{YK} = \frac{ZB}{YC} = \frac{ZA}{YA} \]and hence $\overline{KA}$ is an angle bisector of $\angle ZKY$. Thus from $(TA;YZ)=-1$ we obtain $\angle TKA = 90^{\circ}$. It remains to show $\overline{AK}$ passes through a fixed point on the Euler line. We claim it is the exsimilicenter of $\Gamma$ and $\Omega$. Let $L$ be the midpoint of the arc $YZ$ of $\triangle XYZ$ not containing $X$. Then we know that $K$, $A$, $L$ are collinear. Now the positive homothety sending $\Gamma$ to $\Omega$ maps $A$ to $L$; this proves the claim. Finally, it is well-known that the line through $O$ and the circumcenter of $\triangle XYZ$ coincides with the Euler line of $\triangle ABC$; hence done.

Let $DEF$ and $H$ be the orthic triangle and orthocenter of $ABC$, respectively. Let $\Phi$ denote the inversion about $(ABC)$. Note that $(AT)$ is orthogonal to $(ABC)$ since $\angle OAT=90^\circ$, so $\Phi((AT))=(AT)$. Since $\Phi((BOC))=BC$, $\Phi(\{A_1,A_2\})=\Phi((AT)\cap(BOC))=(AT)\cap BC=\{T,D\}$. Because $ABC$ is acute and scalene, $D$ lies on segment $BC$ while $T$ does not, so $OT>OD$. Since $OA_1<OA_2$, this means that $\Phi(A_1)=T,\Phi(A_2)=D$. (a) Note that $A$ and $A_1$ lie on the polar of $T$, which is the $A$-symmedian of $ABC$, so $AA_1,BB_1,CC_1$ concur at the symmedian point of $ABC$. (b) It suffices to show that $\Phi(AA_2)=(AOD),\Phi(BB_2)=(BOE),\Phi(CC_2)=(COF)$ are coaxial along $\Phi(OH)=OH$. But we clearly have that $\text p(O,(AOD))=\text p(O,(BOE))=\text p(O,(COF))=0$ and $\text p(H,(AOD))=\text p(H,(BOE))=\text p(H,(COF))=2\text p(H,(DEF))$, so we are done.

An even stronger characterization of $A_2$ holds: $AO$ and $TA_2$ intersect on the circumcircle of $BOC$. By defining $A_2'$ according to this characterization, you can prove $A_2' = A_2$ and also that $A_2$ inverts to the foot of the altitude from $A$ to $BC$ upon inversion about the circumcenter of $ABC$ (that preserves $A, B, C$).

Nice configuration. a) First notice that $TA$ is tangent to $\odot (ABC)$. Now consider an inversion $\Psi: X\mapsto X^*$ with pole $T$ and radius $TA$. We show that this inversion swaps $A_1$ and $O$. Notice that $$A_1 = \odot(AT)\cap \odot(BOC) \implies A_1^*=AO\cap \odot (BOC) = O.$$Therefore, $T, A_1, O$ are collinear. Choose $A'\neq A$ on $\odot (ABC)$ such that $TA'$ is tangent to $\odot (ABC)$. Then $\angle AA_1T = 90^{\circ} \implies A, A_1, A'$ are collinear. Now it is well known ($AA'BC$ is harmonic) that $AA'$ is a symmedian. It follows that the point of concurrency is the symmedian point of $\triangle ABC$. $\square$ b) Let $D, E, F$ be the feet of the altitudes from $A, B, C$. The central claim is that $A_2$ and $D$ are inverse points w.r.t. $\odot (ABC)$. Consider the inversion $\Psi: X\mapsto X^*$ with pole $O$ and radius $OA$. Notice that $\odot (AT)$ and $\odot (ABC)$ are orthogonal. Then $$A_2 = \odot (AT)\cap \odot (BOC) \implies A_2^* = \odot (AT)\cap BC = D.$$Then the concurrency reduces to showing that $\odot (AOD), \odot (BOE), \odot (COF)$ are coaxial, with the Euler line being the common axis. Now it is easy to check that $H$ has equal power w.r.t. each circle. Hence, $AA_2, BB_2, CC_2$ meet on line $OH$. $\square$ [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(12cm); 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 = -12.191595682657429, xmax = 14.606489302914257, ymin = -10.557497663135912, ymax = 5.976415148716769; /* image dimensions */ pen sqsqsq = rgb(0.12549019607843137,0.12549019607843137,0.12549019607843137); draw((-0.5369568351588,2.740419980647743)--(-2.230798055944935,-1.9113230734514821)--(4.468723190447987,-1.7849170121987856)--cycle, sqsqsq); /* draw figures */ draw((-0.5369568351588,2.740419980647743)--(-2.230798055944935,-1.9113230734514821)); draw((-0.5369568351588,2.740419980647743)--(4.468723190447987,-1.7849170121987856)); draw(circle((1.0932898029268947,-0.4874635336196818), 3.6162046516574713)); draw(circle((-5.288028542708942,0.34088522534988375), 5.322635570106767)); draw(circle((1.1839250350658481,-5.291130836984215), 4.804522276638789)); draw((-10.039100250259086,-2.0586495299479757)--(-0.5369568351588,2.740419980647743), dotted); draw((-10.039100250259086,-2.0586495299479757)--(4.468723190447987,-1.7849170121987856)); draw((-10.039100250259086,-2.0586495299479757)--(0.420071067976684,-4.04044997006736), dotted); draw((-0.5369568351588,2.740419980647743)--(-2.230798055944935,-1.9113230734514821), sqsqsq); draw((-2.230798055944935,-1.9113230734514821)--(4.468723190447987,-1.7849170121987856), sqsqsq); draw((4.468723190447987,-1.7849170121987856)--(-0.5369568351588,2.740419980647743), sqsqsq); draw((-10.039100250259086,-2.0586495299479757)--(1.093289802926895,-0.48746353361968187), blue); draw((-3.584305082742902,-4.701710909988256)--(1.093289802926895,-0.48746353361968187), red); draw((-0.5369568351588,2.740419980647743)--(0.420071067976684,-4.04044997006736), blue); draw((-0.5369568351588,2.740419980647743)--(-2.230798055944935,-1.9113230734514821)); draw((-2.230798055944935,-1.9113230734514821)--(4.468723190447987,-1.7849170121987856)); draw((4.468723190447987,-1.7849170121987856)--(-0.5369568351588,2.740419980647743)); /* dots and labels */ dot((-0.5369568351588,2.740419980647743),linewidth(3.pt) + dotstyle); label("$A$", (-0.63583198615664265,2.9921072541509782), NE * labelscalefactor); dot((-2.230798055944935,-1.9113230734514821),linewidth(3.pt) + dotstyle); label("$B$", (-2.761703513206261,-2.315822469460111), NE * labelscalefactor); dot((4.468723190447987,-1.7849170121987856),linewidth(3.pt) + dotstyle); label("$C$", (4.569848039450144,-1.9332297386955508), NE * labelscalefactor); dot((1.093289802926895,-0.48746353361968187),linewidth(3.pt) + dotstyle); label("$O$", (0.9821655978778742,-0.3438879139180482), NE * labelscalefactor); dot((-10.039100250259086,-2.0586495299479757),linewidth(3.pt) + dotstyle); label("$T$", (-10.641567792359429,-2.5113230734514828), NE * labelscalefactor); dot((-0.058442883591057315,-0.6500149947098082),linewidth(3.pt) + dotstyle); label("$A_1$", (-0.812800471910418,-0.6702939751707445), NE * labelscalefactor); dot((-3.584305082742902,-4.701710909988256),linewidth(3.pt) + dotstyle); label("$A_2$", (-4.276326550234306,-5.397880666021588), NE * labelscalefactor); dot((0.420071067976684,-4.04044997006736),linewidth(3.pt) + dotstyle); label("$A'$", (0.0973231691089978,-4.690006723006488), NE * labelscalefactor); dot((-0.44982212318496545,-1.8777197539654449),linewidth(3.pt) + dotstyle); label("$D$", (-0.585269561655564,-2.415822469460111), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy]

Sorry Evan (a) Notice that $A_1$ is simply the midpoint of the $A$-symmedian chord, so the three cevians concur at the symmedian point. (b) Invert about $\odot(ABC)$; evidently, the circle with diameter $\overline{AT}$ maps to itself and $\odot(BOC)$ maps to line $BC$. So the image of $A_2$ is the just the foot of the altitude from $A$, and we want to show that if the feet of the altitudes from $A, B$ and $C$ are $D, E$ and $F$, respectively, then $\odot(AOD), \odot(BOE), \odot(COF)$ are coaxial on the Euler line. But the orthocenter clearly has equal power to all three circles, so we're done.

Anybody bash this? This seems bary bashable

I know of at least one person who complex bashed.

This isn't pretty to complex bash considering there are multiple circles IMO.

First wrote: This isn't pretty to complex bash considering there are multiple circles IMO. Maybe if you're an inexperienced or fail to make any synthetic observations; however, after figuring out part a), part b) is not so hard. Define $X$ as the intersection of the tangents from $B,C$ to $(ABC)$. From a) we know that $A_2$ is the center of spiral similarity mapping $TA$ to $OX$. $X$ is well-known to be $\frac{2bc}{b+c}$ while $T$ can be computed as $AA\cap BC = \frac{a(2bc-ab-ac)}{bc-a^2}$ without too much effort. From there, you find $A_2$ using the spiral similarity formula (NOT the circle formula as you suggested; circles are much nastier) and it amounts to $\frac{2abc}{ab+ac+bc-a^2}$ which is not bad at all. After this, you compute $AA_2\cap OH$, both of which are not terribly nasty expressions, and solving gives that their intersection is $\frac{2abc(a+b+c)}{abc +(a+b+c)(ab+bc+ca)}$ which is symmetric in $a,b,c$, yielding the desired. I omitted the calculations but I have done them already; you can try it yourself to see that the bash isn't so bad. In general, you should try longer before commenting on how not-feasible a solution approach is.

For a) I proved (basically) that $O, A_1, T$ are colinear, and then the coordinate bash after that is rather straightforward, so I did it. (Oops maybe I should learn to complex bash better)

a.) Note that $A_1$ is the midpoint of the $A$ symmedian chord in triangle $ABC$, so lines $AA_1, BB_1, CC_1$ are concurrent at the symmedian point. b.) Note that inversion about the circle $(ABC)$ fixes the circle $(AT)$ and sends the circle $(BOC)$ to the line $BC$. Let $AD, BE, CF$ be the altitudes and $H$ be the orthocenter of triangle $ABC$. The previous discussion shows that line $AA_2$ maps to circle $(OAD)$. Points $O$ and $H$ have equal powers in the circles $(OAD), (OBE), (OFC)$. Thus, line $OH$ is their radical axis and their second common point lies on it. The result follows.

Let $X = BB\cap CC$. Note that $T=AA\cap BC$. Since the polar of $T$ is $AX$, $OT\perp AX$. Now it is clear that $OT\cap AX, B,C$ all lie on the circle with diameter $OX$ which is the circumcircle of $BOC$, so $A_1 = OT\cap AX$. Thus $AA_1, BB_1, CC_1$ concur at the symmedian point of $ABC$. Now since we know one of the intersections of the circles, we can complex bash part (b) by reflecting $A_1$ about the line joining the centres of the circles, or using spiral symmetry (the second option is clearly easier but in case you don't see it, the first option of reflecting would still be very doable).

Nice problem, though easy (I used inversion).

Denote with $\omega$ the circle of diameter $AT$.Let $\mathbb{I}$ be the inversion w.r.t. $\triangle{ABC}$. Denote with $X$ the intersection of the second tangent from $T$ to $(ABC)$. Denote with $Z=BB\cap CC$ $\text{(a)}$ $TA$ is tangent to $(ABC)$, because $\angle{OAT}=90$. Because $\mathbb{P}_{\omega}(O)=OA^2$, $\mathbb{I}$ fixes $\omega \implies \mathbb{I}(\omega)=\omega$. $\mathbb{I}(A_1)$ is on $\mathbb{I}((OBC))=BC$ and on $\mathbb{I}(\omega)=\omega$ $\implies \mathbb{I}(A_1)=T \implies O-A_1-T$. Because $(X,A;B,C)=-1 \implies Z-X-A$. But because $A$ is on the polar of $T$ w.r.t $(ABC)$ and that $AA_1$ is perpendicular on ${OT} \implies AA_1$ is the polar of $T$, so $A-A_1-X-Z$, so $AA_1$ is the $A$-symmedian of $\triangle{ABC}$. So $AA_1, BB_1, CC_1$ are concurrent in Lemoine Point. $\text{(b)}$Denote with $A_3$ the intersection of $\omega$ with $BC$. Notice that $\mathbb{I}(A_2)=A_3$,because $\mathbb{I}(A_2)=\mathbb{I}((OBC))\cap \mathbb{I}(\omega$) , But $A_3$ is the feet of the perpendicular from $A$ to $BC$, but since $ABA_3B_3$ is cyclic and isn't passing through $O$(in case it is passing the case is trivial) $\implies ABA_2B_2$ is cyclic. The intersection of $AA_3$, $BB_3$, $CC_3$ is on Euler line of $\triangle{ABC}$ $\implies$ the intersection of $AA_2$, $BB_2$, $CC_2$ (it exists because those lines are the radical axis of $CBC_2B_2$, $ABA_2B_2$, $ACA_2C_2$) is also on the Euler line.(the last one is because $HC_1\times{HC}=HB_1\times{HB}=HA_1\times{HA}$, so it must lie on the radical axis of $(AOA_1),(BOB_1),(COC_1)$)

Slightly different solution for (b) (assuming we have worked out the $A_2$ inversion): Let $H_A$ etc. be the feet of the $A$- etc. altitudes. Through inversion around $(ABC)$, $\overline{A_2H_A} \cap \overline{B_2H_B} = O$. It suffices to prove $\overline{AA_2} \cap \overline{BB_2}$, $\overline{A_2H_A} \cap \overline{B_2H_B}$, $\overline{AH_A} \cap \overline{BH_B}$ are collinear. By Desargues this is equivalent to proving that $\overline{AB}$, $\overline{H_AH_B}$, $\overline{A_2B_2}$ are concurrent, which is what the next four lines prove. By inversion $A_2B_2H_AH_B$ is cyclic. We also know that $ABH_AH_B$ is cyclic. Inverting this around $(ABC)$ yields $ABA_2B_2$ cyclic. Radical axes yields the desired result.

Here's a funny solution the TST group found: Let $O'$ and $H'$ be the isotomic conjugates of $O$ and $H$. Lemma: $H$ lies on $O'H'$. Proof: This is straightforward with bary: $H=(\tan A:\tan B:\tan C)$, $H'=(\cot A: \cot B: \cot C)$, $O'=(\csc 2A: \csc 2B: \csc 2C)$, and $\tan x+\cot x = 2\csc 2x$. A synthetic proof of this would be appreciated. Now, let $A_0$ be on $(ABC)$ with $AA_0\parallel BC$ and $A'$ be the antipode of $A$ on $(ABC)$. Take a $\sqrt{bc}$ inversion: $(BOC)$ becomes $(BHC)$ and $T$ goes to $A_0$ The circle with diameter $AT$ goes to the perpendicular to $AA_0$ through $A_0$, or $AH$ reflected over the perpendicular bisector of $BC$. The image of the two circles intersect at two points: the point $D$ such that $ABDC$ is a parallelogram and the reflection of $A'$ over $BC$, call it $A_3$. Define $B_3$ and $C_3$ similarly. $AA_1$ is isogonal to the $A$-median, which combined with the other vertices proves part (a). For part (b), let $\mathcal H$ be the Jerabek hyperbola, or the isogonal conjugate of the Euler line. It suffices to show that $AA_3$, $BB_3$, and $CC_3$ concur on $\mathcal H$. Since it passes through $O$ and $H$, it is also the isotomic conjugate of line $O'H'$, so $O,H,H'\in\mathcal H$. We claim that $AA_3$ passes through the harmonic conjugate of $O$ with respect to $H$ and $H'$ on $\mathcal H$, which proves the problem. This is clear by projecting $A(H,H';O,A_3)$ onto $A'A_3$, so we are done.

Writing the solution at the same time as solving it. Since $A_1$ etc are the midpoints of the symmedian chords, we have the required concurrency pt in (a) as the symmedian point. Inverting around the circumcircle we need to show that the circles $(AOD)$ etc are coaxal where $D$ is the foot of the $A$ altitude etc. But since a negative inversion at the orthocenter sends $DEF$ to $ABC$, the common radical axis is the Euler line of $ABC$, as required for (b).

Note that $A_1$ and $A_2$ are the inverses of $T$ and the foot from $A$ to $BC$ about $(ABC)$ respectively. We have 1) $A_1$ is the $A-$ Dumpty point, so $AA_1$, $BB_1$, $CC_1$ concur at the symmedian point. 2) If $\triangle ABC$ is the intouch triangle of $\triangle XYZ$ then $A_2$ is the $X-$ Sharkydevil point, so $AA_2$, $BB_2$, $CC_2$ concur at the exsimilicenter of $(XYZ)$ and $(ABC)$ which lies on the line through $O$ and the circumcenter of $\triangle XYZ$, which is the Euler line of $\triangle ABC$.

Let $X = \overline{BB} \cap \overline{CC}$, $Y = \overline{CC} \cap \overline{AA}$, $Z = \overline{AA} \cap \overline{BB}$ (where $AA, BB, CC$ are the respective tangents to $(ABC)$) so that $(ABC)$ is the incircle of $\triangle XYZ$. (a) Claim: $AA_1$ is the $A$-symmedian in $\triangle ABC$ Proof: First, we show that $A_1$ lies on $TO$. Let $A_1'$ be the second intersection of $TO$ with $(AT)$. Since $\triangle TA_1'A \sim \triangle TAO$, we have $TA_1' \cdot TO = TA^2 = TB \cdot TC$, so $A_1'$ lies on $(BOC)$ and thus $A_1' = A_1$. Now, note that $AX$ is the $A$-symmedian in $\triangle ABC$, and $X$ is also the $O$-antipode on $(BOC)$. But $\overline{AA_1} \perp \overline{OA_1}$ so $AA_1$ passes through $X$, which proves the claim. Similarly, $BB_1$ and $CC_1$ are the $B$ and $C$ symmedians in $\triangle ABC$, so the three lines concur at the symmedian point. (b) Let $A_3$ be the foot of the $A$-altitude in $\triangle ABC$, and consider an inversion about $(ABC)$. Then $(AT)$ stays fixed since $OA^2 = OA_1 \cdot OT$, so $A_2$ and $A_3$ swap (as $A_3$ lies on $(AT)$ and $A_1$ swaps with $T$). Therefore, since the nine-point circle of $\triangle ABC$ inverts to $(XYZ)$, we see that $A_2$ is just the $X$-sharkydevil point of $\triangle XYZ$, and hence $AA_2$ passes through the exsimilicenter of $(ABC)$ and $(XYZ)$. Analogously, we can prove that $BB_2$ and $CC_2$ pass through the exsimilicenter of $(ABC)$ and $(XYZ)$, so it suffices to show that this point lies on the Euler line on $\triangle ABC$. But this is true since the center of $(XYZ)$ lies on the line through $O$ and the nine-point center of $\triangle ABC$ by inversion about $(ABC)$, which is just the Euler line, as desired.

YAYY THE POWER OF INVERSION AND I CAN FINALLY POST LATEX Invert about (ABC), s.t. $A_1\mapsto(AT)\cap BC=T,A_2\mapsto(AT)\cap BC=D$ due to orthogonality. Part a) follows immediately since $OA_1O_A=OMT=90=OMO_A$ where $O_A$ is the antipode of O wrt (BOC) and M is the midpoint of BC, so $A_1\in AO_A$ implies $AA_1$ is a symmedian in ABC (since $AO_A$ is a symmedian since $O_A$ is the intersection of tangents at B and C), whence the lines concur at the symmedian point. Part b) follows since $AD$ etc. concur at the orthocenter, so their preimage of concurrence is a point on line OH.

My solution was kinda funny lol Invert about the circumcircle of $(ABC)$. Then, $A_1$ goes to $T_A$ (we use $T_A$, $T_B$, $T_C$). So, $AA_1$ goes to circumcircle of $AOT_A$, i.e. circle with diameter $OT_A$. Claim: $T_A, T_B, T_C$ are collinear Proof. Apply Pascal's on $ABBCCAA$. Now, the circle with diameters $OT_A$, passes through the feet from $O$ onto $T_A$, $T_B$, $T_C$, hence done. Now, $A_2$ goes to $D$, the feet from $A$ to $BC$. Thus, $AA_2$ goes to circumcircle of $AOD$. Note that defining $BOE$ and $COF$ similarly, Now, $O$ has equal power with respect to all the circles, and similarly, $HA\cdot HD = HB \cdot HE = HC \cdot HF$, since $ABDE$, and analogous quadrilaterals for $BC$ and $CA$ are cyclic. So, $OH$ is the common radical axis of all the circles and done. Edit: $A_1$ and $A_2$, both lie on the circumcircle of $BOC$, so after inversion go to $BC$, and the circumcircle of $AOT$ is orthogonal to $ABC$, so it inverts to itself, thus, $A_1$ goes to $T_A$ and $A_2$ to $D$.

Let $T_A$ be point on $BC$ such that $\angle{T_AAO} = 90^{\circ}$.Define points $T_B, T_C$ analogously. Let $O_A$ be midpoint of $AT_A$. Define points $O_B, O_C$ analogously. Now consider inversion around $(ABC)$. Points $A, B, C$ are fixed under inversion. Since $\angle{OAO_A} = 90^{\circ}$, so circle with diameter $AT_A$ is fixed under inversion. Note that $(BOC)^* = BC$, therefore $A_1^* = BC \cap (AT_AA_1)$. So $A_1^*$ is either foot of altitude $A$ to $BC$ or $T_A$. Let $D$ be foot of altitude $A$ to $BC$, then since $OD < OT_A$ and $OA_1 < OA_2$, so $A_1^* = T_A$. Therefore $AA_1, BB_1, CC_1$ concurrent $\iff$ the circles $(OAT_A), (OBT_B), (OCT_C)$ are coaxial. By Pascal's theorem on cyclic hexagon $AABBCC$ yields $T_A, T_B, T_C$ are collinear. Let $P_{AB} = (OAT_A) \cap (OBT_B)$. Then it's not hard to see that $T_A, T_B, P_{AB}$ are collinear. Thus $P_{AB}, P_{BC}, P_{CA}$ are collinear, thus $P_{AB} = P_{BC} = P_{CA}$. Hence the circles $(OAT_A), (OBT_B), (OCT_C)$ are coaxial. Now since $A_2^* = D$, therefore line $AA_2$ becomes $OAD$ under the inversion. Let $E, F$ be feet of the perpendicular from $B, C$ to $AC, AB$, respectively. Then lines $BB_2, CC_2$ become $(OBE), (OCF)$, respectively. Let $H$ be orthocenter of $ABC$. Then since $AH \cdot HD = BH \cdot HE = CH \cdot HF$, the circles $(OAD), (OBE), (OCF)$ are coaxial and radical axis of these circles is $OH$. Thus $AA_2, BB_2, CC_2, OH$ are concurrent. $\blacksquare$

Invert about the circumcircle. The circle with diameter $AT$ is orthogonal to the circumcircle, so it is fixed, and circle $(BOC)$ is sent to $BC.$ Thus, $A_1$ is sent to $T$ and $A_2$ is sent to the foot $D$ from $A$ to $BC.$ a. We get that $AA_1$ inverts to the circle with diameter $OT.$ First, by Pascal's on $AABBCC$ we find that the three $T$s are collinear. Then, all three of these circles pass through the foot from $O$ to the three $T$ line, so inverting back we get our desired result. b. We get $AA_2$ inverts to $(AOD).$ First, let $E,F$ be the feet from $B,C$ to $AC,AB$ respectively. Letting $P$ be the point on the Euler line such that $H$ is between $P$ and $O$ and $OH \cdot PH=AH \cdot DH=BH \cdot EH=CH \cdot FH.$ Then $(AOD),(BOE),(COF)$ pass through $P$ and inverting back we get our desired result.

It is morally correct to view the problem with $\triangle ABC$ as the intouch triangle most of the time: Restated problem wrote: Let $\triangle ABC$ have incenter $I$ and intouch triangle $DEF$ (with $D$ on $\overline{BC}$, etc.). Let $T=\overline{BC} \cap \overline{EF}$, and let the circle with diameter $\overline{DT}$ intersect $(AEFI)$ at points $A_1,A_2$ with $IA_1<IA_2$, and define $B_1,B_2,C_1,C_2$ similarly. Then prove that $\overline{DA_1}$, etc. concur, and $\overline{DA_2}$, etc. concur on the Euler line of $\triangle DEF$. For the first part, I claim that $A_1$ is the foot of $I$ onto $\overline{AD}$. This point obviously lies on $(AEFI)$, which has diameter $\overline{AI}$. Furthermore, since $T$ lies on the polar of $A$ wrt the incircle, $A$ lies on the polar of $T$. Since $D$ does as well, we conclude $\overline{AD}$ is the polar of $T$ and thus $T,A_1,I$ are collinear (since $A_1,T$ are now inverses) and $\angle TA_1D=\angle AA_1I=90^\circ$ as desired. The conclusion then follows from the well-known fact (proof by Ceva) that $\overline{AD},\overline{BE},\overline{CF}$ concur. For the second part, I claim that $A_2$ is the inverse of the foot of the altitude from $D$ to $\overline{EF}$ wrt the incircle. Note that since $\triangle DEF$ is always acute, $A_2$ will always lie outside the incircle, but since $T$ lies outside the incircle its inverse $A_1$ lies inside and we indeed have $IA_1<IA_2$. To see this, let $H$ be the foot of the altitude. An inversion about the incircle fixes the circle with diameter $\overline{DT}$, since it's clearly orthogonal, but swaps $(AEFI)$ with $\overline{EF}$, so clearly $A_1$ gets sent to $H$ and the conclusion follows. Now return to the original problem statement. We want to prove that if $D,E,F$ are the feet of the altitudes of a triangle $ABC$ and $O$ is its circumcenter, then $(AOD),(BOE),(COF)$ concur again at some point on the Euler line. This follows because $O$ lies on all three points, and the orthocenter $H$ of $ABC$ has equal power to all of these circles, since $HB\cdot HE=HC\cdot HF$ from $(BCEF)$ cyclic, and cyclic equalities hold as well. This establishes coaxiality, and since the common radical center is $\overline{OH}$ their intersection lies on the Euler line as well. $\blacksquare$

For part a, we make the following claim. Claim: $A_1$ is the $A$-dumpty point. Proof. We prove that the Dumpty point is $(AT)\cap(BOC)$. To do so, let $K$ be the intersection of the tangent from $T$ to the circumcircle. Then, as $D_A$ is the midpoint of $AK$, and $TA=TK$, $D_A$ lies on $AT$. As it is well known that $D_A$ lies on $(BOC)$ we conclude. $\blacksquare$ Now, for part b, note that $(AT)$ and $(ABC)$ are orthogonal. Hence $A_2$ inverts to $(AT)\cap BC=D$, the foot from $A$ to $BC$. It thus suffices to prove that the euler line is the radical axis of coaxial circles $(AOD)$, $(BOE)$, $(COF)$ which is clear since $O$ lies on all circles and $AH\cdot HD=CH\cdot HF=BH\cdot HE$.

Amazing Problem.

Claim 2:$AA_2C_2C$ is cyclic

So by Claim 2 and radax, we have $AA_2,BB_2,CC_2$ concurr. Let $H$ be the orthocenter of $\triangle ABC$ Let $U = CC \cap AB$ Let $G$ be the foot of altitude from $C$ to $AB$ It suffices to prove $AA_2,CC_2,OH$ concurrent. Claim 3:$OH$ is the radical axis of $(CU),(AT)$

By Claim 3, radax on $(CU),(AT),(AA_2C_2C)$, we're done

Clearly $T = AA \cap BC$. We begin by assuming $A_1$ lies inside the circle and $A_2$ lies outside, which we will soon prove. Inverting about $(ABC)$ tells us \[A_1^*, A_2^* = (AT)^* \cap (BOC)^* = (AT) \cap BC,\] so our inverted points correspond to $T$ and the foot from $A$ to $BC$, essentially proving our earlier statement. Solving (b) first, we need the radical axis of $(AOD)$, $(BOE)$, and $(COF)$ to be the Euler Line, where $\triangle DEF$ is the orthic triangle. This is well known, with a proof being to consider the power of $H$. We tackle (a) by simply noting that $A_1$ is the $A$-Dumpty point as it lies on $(AT)$, $(BOC)$, and is inside the circle. Hence the concurrence is the symmedian point. $\blacksquare$

Let $A'$ be the foot of the altitude from $A$ to $BC$, and define $B',C'$ analogously. Inverting about $(ABC)$, $(AT)$ is unchanged and $(BOC)$ goes to line $BC$, so $\{ A_1,A_2 \}$ goes to $\{T,A'\}$. Since $OA_1<OA_2$ and $OT>OA'$, $T$ is the inverse of $A_1$ and $A'$ is the inverse of $A_2$. Note that $A_1$ is the $A$-Dumpty point of $\Delta ABC$. Thus, $AA_1, BB_1, CC_1$ all pass through the symmedian point of $\Delta ABC$. Since line $AA_2$ is inverted to $(AOA')$, it suffices to show that the circles $(AOA'),(BOB'),(COC')$ have radical axis $OH$, where $H$ is the orthocentre of $\Delta ABC$. Trivially $O$ has equal power to all three circles, and $HA \cdot HA' = HB \cdot HB' = HC \cdot HC'$, so $OH$ is the radical axis of these three circles. $\square$

makes no sense how this isnt the canonical solution imo. maybe im just not inversionpilled enough also i might be fakesolving bc phantom point stuff Pass to the tangential triangle - the problem rephrases as the following; Rephrased Problem wrote: Let $\triangle ABC$ be a triangle with intouch triangle $\triangle DEF$, circumcenter $O$, and incenter $I$. Let $T = \overline{EF} \cap \overline{BC}$, and suppose that $(AI) \cap (DT)$ = $D_1$ and $X$ with $ID_1 < ID_2$. Define $E_1$, $Y$, $F_1$, and $Z$ similarly. Prove that $\overline{DD_1}$, $\overline{EE_1}$, and $\overline{FF_1}$ concur. Prove that $\overline{DX}$, $\overline{EY}$, and $\overline{FZ}$ concur on line $\overline{OI}$, Now, we can begin solving the problem. a) I claim that $\overline{AD_1D}$ collinear, which finishes (as then the requested concurrence point is the Gergonne point of $\triangle ABC$). Indeed, note that $\overline{IT} \perp \overline{AD}$ for pole-polar reasons, so if $\overline{AD} \cap (AI) = D'_1$, then $I$, $D'_1$, and $T$ are collinear, so $D'_1$ lies on $(AT)$ and $D_1 = D'_1$. b) In fact, the following characterization of $X$ holds: Claim: $X$ is the $A$-Sharkydevil point in $\triangle ABC$. Proof. Let $X'$ be the $A$-Sharkydevil point of $\triangle ABC$ - we wish to show that $X = X'$. If we let $N$ be the midpoint of arc $\widehat{BAC}$ and $M$ be the midpoint of arc $\widehat{BC}$. It is well-known that $\overline{X'DM}$ collinear. Note that \[ (N, M; B, C) \overset{X'}{=} (\overline{NX'} \cap \overline{BC}, D; B, C) \]so $\overline{TNX'}$ collinear. Hence $\measuredangle NX'M = \measuredangle TX'M = \measuredangle TX'D = 90$, so $X'$ lies on $(TD)$. By the definition of the Sharykdevil Point, $X'$ lies on $(AI)$, so $X = X'$ and we're done. $\square$ Now this reduces to Canada 2007/5, so the three lines mentioned concur at the exsimilicenter of the incircle and the circumcircle. Clearly this exsimilicenter lies on $\overline{IO}$.

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what ok Note that $(AT)$ is fixed by power of a point (since it remains tangent to $AO$). Thus $A_1,A_2$ are the intersections of $(AT)$ with $BC$. Let the foot from $A$ to $BC$ be $D$; then $A_1$ maps to $T$ and $A_2$ maps to $D$. (a) We claim that the concurrency point is the symmedian point $K$ of triangle $ABC$. If we invert, we want to prove that $K'TOA$ is concyclic. But $(TOA)$ passes through $M$, the midpoint of $BC$, and the inverse of $(AOM)$ is the line passing through $A$ and the pole of $BC$, which is the $A$-symmedian. (b) The inverse of $AA_2$ is $(AOD)$. Similarly define $E$ and $F$, then we want to prove that $(AOD),(BOE),(COF)$ are coaxial with their radical axis at the Euler line (which is fixed under our inversion). But this is just a simple power of a point calculation at $H$. $\blacksquare$