Seriously ?

Yes, this problem made it on despite appearing in contests as recent as 2016... see here and here. Even worse, apparently it was close to making the test!

+) South Korean TST #4 Can be solved by complex numbers..

Here is my unique solution using similar quadrilaterals. Replace point $T$ with $P$. Let $O$ be the circumcenter and $Q$ be the isogonal conjugate of $P$ WRT $\triangle ABC$. Since $A$ is circumcenter of $\triangle B_1PC_1$ and analogous relations for $B,C$, we get $$\begin{cases} \angle BOC = 2\angle BAC = \angle B_1AC_1 &\implies \triangle BOC\stackrel{-}{\sim}\triangle B_1AC_1\\ \angle QBC = \angle PBA = \frac{\angle PBC_1}{2} = \angle PA_1C_1 = \angle A_2B_1C_1 &\implies \triangle BQC\stackrel{-}{\sim}\triangle B_1A_2C_1 \end{cases}$$hence $\triangle BOC\cup Q\stackrel{-}{\sim}\triangle B_1AC_1\cup A_2$. Analogously, we get $\triangle AOC\cup Q\stackrel{-}{\sim}\triangle A_1BC_1\cup B_2$ and $\triangle AOB\cup Q\stackrel{-}{\sim}\triangle A_1CB_1\cup C_2$. Finally, let $X=BB_2\cap CC_2$ hence \begin{align*} \measuredangle B_2XC_2 &= \measuredangle BB_2A_1 + \measuredangle B_2A_1C_2 + \measuredangle A_1C_2C \\ &= (-\measuredangle OQA) + \measuredangle B_2A_1C_2 + (-\measuredangle AQO) \\ &= \measuredangle B_2A_1C_2 \end{align*}hence $X\in\odot(B_2A_1C_2)\equiv\Omega$, implying the conclusion.

Inverse+complex=killer

Hey doesn't this fail for T=orthocenter?

Redacted LOL... solution of some other problem at wrong place

AlastorMoody wrote: ISL 2018 G4 wrote: A point $T$ is chosen inside a triangle $ABC$. Let $A_1$, $B_1$, and $C_1$ be the reflections of $T$ in $BC$, $CA$, and $AB$, respectively. Let $\Omega$ be the circumcircle of the triangle $A_1B_1C_1$. The lines $A_1T$, $B_1T$, and $C_1T$ meet $\Omega$ again at $A_2$, $B_2$, and $C_2$, respectively. Prove that the lines $AA_2$, $BB_2$, and $CC_2$ are concurrent on $\Omega$. Solution: Denote $\Delta A'B'C'$ in place of $\Delta A_1B_1C_1$ and $\Delta A_1B_1C_1$ in place of $\Delta A_2B_2C_2$ and let $M \equiv T$. Let $\odot (C_1BA_1)$ $\cap$ $\odot (ABC)$ $=$ $D$. WLOG, assume $D$ lies on arc $AC$ not containing $B$. $$\angle BC_1A_1= \angle BDA_1=\angle BAM=\angle BDA' \implies D \in A'A_1$$$$\angle C_1DB=\angle BA_1C_1=\angle C'CB=\angle C'DB \implies D \in C'C_1$$Similarly, $B'B_1$ also passes through $D$ $\qquad \blacksquare$ Are you sure about this?

Wait, nvm I'm confused, when did the Problem statement change? ISL 2018 G4 wrote: A point $T$ is chosen inside a triangle $ABC$. Let $A_1$, $B_1$, and $C_1$ be the reflections of $T$ in $BC$, $CA$, and $AB$, respectively. Let $\Omega$ be the circumcircle of the triangle $A_1B_1C_1$. The lines $A_1T$, $B_1T$, and $C_1T$ meet $\Omega$ again at $A_2$, $B_2$, and $C_2$, respectively. Prove that the lines $AA_2$, $BB_2$, and $CC_2$ are concurrent on $\Omega$. Solution: From the definition of $A_1,B_1,C_1$, it implies that $A$ is center of $\odot (TC_1B_1)$, $B$ is center of $\odot (TA_1C_1)$ and $C$ is center of $\odot (TB_1A_1)$. Let $\odot (BAC_1)$ $\cap$ $\odot (A_2B_2C_2)$ $=$ $D$ $$\angle C_1DA_2=\angle C_1A_1T=\angle C_1BA=\angle C_1DA \implies D \in AA_2$$Similarly, $D$ lies on $BB_2,CC_2$ $\qquad \blacksquare$

Nineteenth problem in this pdf

Can we really do the second part using complex numbers??

a1267ab wrote: A point $T$ is chosen inside a triangle $ABC$. Let $T_A$, $T_B$, and $T_C$ be the reflections of $T$ in $BC$, $CA$, and $AB$, respectively. Let $\Omega$ be the circumcircle of the triangle $A_1B_1C_1$. The lines $A_1T$, $B_1T$, and $C_1T$ meet $\Omega$ again at $A_2$, $B_2$, and $C_2$, respectively. Prove that the lines $AA_2$, $BB_2$, and $CC_2$ are concurrent on $\Omega$. Proposed by Mongolia [asy][asy] import olympiad; import cse5; import geometry; size(10cm); defaultpen(fontsize(11pt)); linewidth(2); pair A=(1.7,2); pair B=(0,0); pair C=(3,0); pair T=(1.1,0.5); pair M=foot(T,B,C); pair Q=foot(T,A,C); pair R=foot(T,A,B); dot(M^^Q^^R); draw("$Q_A$",M,S); draw("$Q_B$",Q,E); draw("$Q_C$",R,W); pair TA=2*M-T; pair TB=2*Q-T; pair TC=2*R-T; pair MA=(B+C)/2; pair MB=(A+C)/2; pair MC=(A+B)/2; draw(circumcircle(MA,MB,MC),orange); dot(MA^^MB^^MC); draw("$M_A$",MA,S); draw("$M_B$",MB,E); draw("$M_C$",MC,W); draw("$A$",A,N); draw("$B$",B,W); draw("$C$",C,E); draw("$T$",T,S); draw("$T_A$",TA,S); draw("$T_B$",TB,E); draw("$T_C$",TC,W); dot(A^^B^^C^^T^^TA^^TB^^TC); draw(A--B--C--cycle,springgreen+opacity(0.25)); path abc=circumcircle(A,B,C); path tatbtc=circumcircle(TA,TB,TC); filldraw(abc,invisible,purple); filldraw(tatbtc,invisible,blue); pair O=circumcenter(TA,TB,TC); pair OA=foot(O,T,TA); pair A2=2*OA-TA; pair OB=foot(O,T,TB); pair B2=2*OB-TB; pair OC=foot(O,T,TC); pair C2=2*OC-TC; dot(A2^^B2^^C2); draw("$A_2$",A2,N); draw("$B_2$",B2,S); draw("$C_2$",C2,SW); pair RA=(T+A)/2; pair RB=(T+B)/2; pair RC=(T+C)/2; dot(RA^^RB^^RC); draw("$R_A$",RA,E); draw("$R_B$",RB,NW); draw("$R_C$",RC,NE); pair P=extension(A,A2,C,C2); draw(segment(A,A2),dashed+black); draw(segment(B,B2),dashed+black); draw(segment(C,C2),dashed+black); draw(segment(B2,P),dashed+black); draw(segment(C,P),dashed+black); draw(segment(A,P),dashed+black); dot(P); draw("$P$",P,E); draw(segment(T,TA),purple); draw(segment(T,TB),purple); draw(segment(T,TC),purple); draw(segment(T,A2),purple); draw(segment(T,B2),purple); draw(segment(T,C2),purple); filldraw(circumcircle(M,Q,R),invisible,red); pair Ponc=(T+P)/2; draw("$Q$",Ponc,W); draw(circumcircle(Ponc,RC,MA),dashed); dot(Ponc); draw(line(Ponc,RC),dotted); [/asy][/asy] Had to take a sneek-peek oops Let $\odot{Q_AQ_BQ_C}$ be the pedal triangle of $T$ WRT $ABC$. Let $R_A,R_B,R_C$ be mid-points of $TA,TB,TC$ respectivey. Let $Q$ be the Poncelet point of $ABCT$. Let $P$ be reflection of $T$ about $Q$. Its well known that $Q\in \odot{Q_AQ_BQ_C} \implies P\in \odot{T_AT_BT_C}$. Next note that $R_BR_CM_A$ is the nine-point circle of $BTC$ so $Q\in \odot{R_BR_CM_A}$.Thus if $X=Q_CT\cap \odot{Q_AQ_BQ_C}$ then \[\angle{Q_AQX=\angle{Q_AQ_CT}=\angle{Q_ABR_B}=\angle{R_CR_BQ_A}=\angle{R_CQQ_A}=\angle{Q_AQR_C}}\].Hence $X\in QR_C$.Thus $Q_CT\cap QR_C\in \odot{Q_AQ_BQ_C}$.By a homothety at $T$ we get $C_2,C,P$ collinear.Similarly $A_2,A,P$ and $B_2,B,P$ also are hence $AA_2\cap BB_2\cap CC_2=P\in \odot{A_2B_2C_2}$ as desired.$\blacksquare$

We only need to show that $BB_2$ and $CC_2$ intersect on $\omega$ . We invert in $T$ and note that $A$ goes to reflection of $T$ in $B_1C_1$. Now taking $\Delta A_1B_1C_1$ as main triangle we can restate the problem as New problem- A point $T$ is chosen inside $\Delta ABC$. Let $BT$ and $CT$ intersect the circumcircle again at $E$ and $F$ ,respectively. Let reflection of $T$ in $AC$ and $AB$ be $M,N$ respectively. Then the circles $\odot (TFN),\odot(TME) ,\odot(ABC)$ are concurrent. (Note-All angles considered are directed angles) Claim-$\Delta BFN$ is directly similar to $\Delta CEM$ Proof-As $\Delta BTF$ is similar to $\Delta CTE$, we have $\displaystyle \frac{BF}{CF}=\frac{BT}{CT}=\frac{BN}{CM}$. And $\angle NBF=\angle NBA+\angle ABF =\angle ABT+\angle ACF=\angle ACE +\angle MCA=\angle MCE $ Therefore by SAS test of similarity claim is proved. Main Proof- Let $\odot (TFN),\odot(TME) $ intersect at point $K$ . Then $\angle FKE=\angle FKT+\angle TKE=\angle FNT +\angle TME=\angle FNB +\angle BNT +\angle TMC+\angle CME $ From claim we get $\angle FNB+\angle CME =0$ $\therefore \angle FKE= \angle BNT+\angle TMC=\angle NTB+\angle CTM=\angle (NT,AB)+ \angle ABT +\angle TCA+\angle (AC,TM)=\angle ABT+\angle TCA+90+90 =\angle ABE +\angle FBA =\angle FBE$ Hence $K\in \odot(ABC)$ Hence proved.

Let $FB $ intersects $(DFE) $ at point $K $. Now $\angle KGD=\angle BFD=90^{\circ}-\left (\frac {\angle PBD}{2}\right)=90^{\circ}-\angle ABC $ Hence, $KG||AB $. Applying Reim's theorem, note that $J $ is the intersection of $GA $ and $(DFE) $. $J $, $A $, $B $, $F $ are concyclic. Hence, $J $ is the intersection of $P $ wrt the Poncelet point of $\{A,B,C,P\} $. $J $ is the reflection of $T $ wrt Poncelet point of $\{A,B,C,T\} $. $\blacksquare $

Key observation: $A_1C_2C\sim A_1B_2B$. This spiral similarity directly shows the desired result because $\measuredangle A_1B_2B = \measuredangle A_1C_2C$.

Nothing.... Move ahead

I think the following complex solution is less clever but maybe a bit more straightforward than other complex solutions above.

Really Nice Property about Antigonal Conjugate! ISL 2018 G4 wrote: A point $T$ is chosen inside a triangle $ABC$. Let $A_1$, $B_1$, and $C_1$ be the reflections of $T$ in $BC$, $CA$, and $AB$, respectively. Let $\Omega$ be the circumcircle of the triangle $A_1B_1C_1$. The lines $A_1T$, $B_1T$, and $C_1T$ meet $\Omega$ again at $A_2$, $B_2$, and $C_2$, respectively. Prove that the lines $AA_2$, $BB_2$, and $CC_2$ are concurrent on $\Omega$. Proposed by Mongolia Take a Negative Homothety at $T$ with a scale factor of $-2$. So the problem transforms to this one. Proposition:- $ABC$ be a triangle with a point $T$ in it's plane and let $A_1B_1C_1$ be the $T-\text{Pedal Triangle}$ WRT $\triangle ABC$. Let $\{A_1,B_1,C_1\}\cap\odot(A_1B_1C_1)=\{X,Y,Z\}$ and let $\{A_2,B_2,C_2\}$ be the midpoints of $\{PA,PB,PC\}$ respectively. Then $XA_2,YB_2ZC_2$ intersect on $\odot(A_1B_1C_1)$ Proof:- Let $\odot(A_1B_1C_1)\cap BC=\{A_1,D\}$ and let $K$ be the Poncelet Point of $\{ABCT\}$. We claim that $XA_2,YB_2,ZC_2$ concur at the Poncelet Point of $\{ABCT\}$. Notice that $\angle DKX=90^\circ$ and similarly we get that $\angle DKA_2=\angle DKZ+\angle ZKA_2=\angle ATC+90^\circ-\angle ATC=90^\circ\implies\overline{A_2-X-K}$. Similarly we get that $\overline{B_2-Y-K}$ and $\overline{C_2-Z-K}$. So, $XA_2,YB_2,ZC_2$ concur at the Poncelet Point of $\{ABCT\}$. So now transform back. We get that $AA_2,BB_2,CC_2$ are concurrent at $\odot(ABC)$. The Point of Concurrency is the $\text{Antigonal Conjugate}$ of $\{ABCT\}$. $\blacksquare$

Let $X$ be the intersection of $(A_1BC)$ with $\Omega$. It is easy to see that $C$ is the circumcenter of $\triangle A_1TB$, so we have \[\angle A_1XB = \angle A_1CB \stackrel{\text{circumcenter}}{=}\angle A_1B_1T = \angle A_2B_1B_2 = \angle A_1XB_2\]so $X$, $B$, and $B_2$ are collinear. Similarly, $X$, $C$, and $C_2$ are collinear. Finally, a symmetric argument (say $B$-centered instead of $A$-centered) yields that $AA_2$ intersects $\Omega$ at the same point as $BB_2$, yielding that $X$, $A$, and $A_2$ are collinear, as desired.

ISL 2018 G4 wrote: A point $T$ is chosen inside a triangle $ABC$. Let $A_1$, $B_1$, and $C_1$ be the reflections of $T$ in $BC$, $CA$, and $AB$, respectively. Let $\Omega$ be the circumcircle of the triangle $A_1B_1C_1$. The lines $A_1T$, $B_1T$, and $C_1T$ meet $\Omega$ again at $A_2$, $B_2$, and $C_2$, respectively. Prove that the lines $AA_2$, $BB_2$, and $CC_2$ are concurrent on $\Omega$. Let $TA_1 \cap BC = D, TB_1 \cap AC = E, TC_1 \cap AB = F, AB_1C \cap \odot A_1B_1C_1 = P$. So by Midpoint Theorem $DE \parallel A_1B_1$ and $\measuredangle TDC = \measuredangle TEC = 90^{\circ} \implies TECD$ is cyclic. Now, $$\measuredangle B_1PA = \measuredangle B_1CA = \measuredangle ACT = \measuredangle ECT = \measuredangle EDT = \measuredangle B_1A_1T = \measuredangle BA_1A_2 = \measuredangle BPA_2 \implies \overline{A - A_2 - P} \text{ are collinear.}$$ Similarly $\overline{B - B_2 - \odot ABC_1 \cap \odot A_1B_1C_1}$ and $\overline{C - C_2 - \odot A_1BC \cap \odot A_1B_1C_1}$ are collinear. Now, $$\measuredangle APC_1 = \measuredangle A_2PC_1 = \measuredangle A_2A_1C_1 = \measuredangle TA_1C_1 = \measuredangle TDF = \measuredangle TBF = \measuredangle TBA = \measuredangle ABC_1 \implies P \text{ lies on } \odot ABC_1.$$ Similarly $P$ lies on $\odot A_1BC$. So, $P = \odot A_1B_1C_1 \cap \odot A_1BC \cap \odot AB_1C \cap \odot ABC_1$ which in turn implies $\{ AA_2, BB_2, CC_2 \}$ all concur at $P$.

Focused on the wrong idea (inversion with center $T$), and it took me 2 hours to see this :0 Notice $C$ is the circumcenter of $(TB_1A_1)$, thus $\angle BCA_1=1/2.\angle TCA_1=\angle TB_1A_1$. Now, $\angle TB_1A_1=\angle B_2B_1A_1=\angle B_2C_2A_1$, thus $\angle B_2C_2A_1=\angle BCA_1; \angle C_2B_2A_1=\angle CBA_1$ (analougsly), thus $A_1$ is the miquel of $BB_2CC_2$ (by spiral similarity), which implies that the sides of the quadrilateral($BB_2,CC_2$) meet at the circumcircle of $(B_2A_1C_2)$, which implies the conclusion.

Maybe the shortest complex solution... $S=AA_2 \cap \Omega.$ Let $(A_1B_1C_1)$ be unit circle. Since $A$ is circumcenter of $(B_1C_1T)$ and $A_1,T,A_2$ is collinear we get: $$a=\frac{\begin{vmatrix} t & t\bar{t} & 1 \\ b_1 & 1 & 1 \\ c_1 & 1 & 1 \end{vmatrix} }{\begin{vmatrix} t & \bar{t} & 1 \\ b_1 & \bar{b_1} & 1 \\ c_1 & \bar{c_1} & 1 \end{vmatrix}} \implies a=\frac{b_1c_1(t\bar{t} -1) }{\bar{t}b_1c_1 +t - b_1 -c_1} \wedge\ \ \bar{a}=\frac{t\bar{t} -1}{\bar{t}b_1c_1 +t - b_1 -c_1} \ \ \text{ and } a_2=\frac{a_1-t}{a_1\bar{t}-1}$$$$s=\frac{a_2-a}{\bar{a}a_2-1} \implies s=\frac{-(a_1b_1+b_1c_1+c_1a_1)+t(a_1+b_1+c_1)+2a_1b_1c_1\bar{t}-a_1b_1c_1\bar{t}^2t}{\bar{t}(a_1b_1+b_1c_1+c_1a_1)-(a_1+b_1+c_1)+2t-a_1b_1c_1\bar{t}^2-t^2\bar{t}}$$Since the expression of $S$ is symmetric wrt $a_1,b_1,c_1$ implies $S \in BB_2, S \in CC_2 \qquad \blacksquare$

All angles are measured. Note that $A$ is the center of $(C_1TB_1)$, so \begin{align*} \angle C_1AB &= \angle C_1B_1B_2 = \angle C_1A_2B_2 \\ \angle C_1BA &= \angle C_1A_1A_2 = \angle C_1B_2A_2 \end{align*}implying that $\triangle C_1AB\sim^+ \triangle C_1A_2B_2$. Thus, $C_1$ is the center of spiral similitary taking $AB$ to $A_2B_2$, so $\angle C_1B_2B=\angle C_1A_2A$. Thus, $B_2B$ and $A_2A$ concur on $\Omega$. The analogous gives the result.

I found this sort of difficult... I feel like with inversion and isogonal conjugates so close, it's hard to believe a simple cyclic quad construction + angle chase solution actually cuts it (you don't even need homothety!?!?! ridiculous). I actually read the first sentence of the official solution because I was close to giving up. Upon seeing that it was super mundane, I immediately scrolled back up and solved within 10 minutes. -_- On a side note, I find the cyclic quadrilateral impossible to find until you consider degrees of freedom and thereby entirely remove a lot of points from the diagram. Again, without actually diving into the angle chase, it's not at all something you'd expect to be true (I hate this problem) Let $A_2C_2\cap AC=K$. Then \[\angle B_1C_2K=\angle B_1A_1T=\frac{1}{2}\angle B_1CT=\angle B_1CK\]where I'm too lazy to handle configuration issues (yay!). So $B_1CC_2K$ is cyclic, similarly $B_1KAA_2$. Now let $L=CC_2\cap AA_2$, notice that \[\angle LC_2A_2+\angle LA_2C_2=\angle CB_1K+\angle AB_1K=\angle AB_1C_1\]however since $\triangle AB_1C\sim \triangle A_2B_1C_2$ this finishes. Wow! I hate this problem. I hate this problem. I hate this problem. I hate geometry wait.. what did I say?

This felt easy for a G4

Let $(B_1AC) \cap (BA_1C) = K$, We Claim that $K$ is the desired point of concurrency. Claim 1: $K$ lies on $(A_1B_1C_1)$

Claim 2:$K-A-A_2$

I was going to do a complete writeup, but I closed the tab accidentally. Then I decided to just send the below screenshot (which was originally in the remark) and explanation that contains everything anyways. Then I clicked submit and it lagged out so here I am. Third time's the charm? This complex bash was carried out entirely in pastebin (aside from a geogebra diagram) with no paper. We use $A_1B_1C_1$ as the reference triangle, renaming it to $ABC$ and renaming the original $ABC$ to $DEF$. The first line is $A_2$. To calculate $D$, we take a homothety at $T$ and express the image of the intersection of the line through $B$ and the $B_2$-antipode with the line through $C$ and the $C_2$-antipode. The final line has a typo: it should be $-abct\overline{t}^2$ instead of $+abct\overline{t}^2$. Regardless, the final expression for the second intersection of $\overline{A_2D}$ with $(ABC)$ is symmetric in $a,b,c$, so $\overline{B_2E}$ and $\overline{C_2F}$ pass through it as well.

\usepackage{amsmath} Let $\triangle DEF$ be the pedal triangle of $T$ with respect to $\triangle ABC$. \begin{claim} $B_1$ is the center of spiral similarity sending $AC$ to $A_2C_2$ . \end{claim} \begin{proof} \text{ } \begin{itemize} \item $\angle B_1AC= \angle EAT =\angle EFT = \angle B_1C_1T=\angle B_1A_2C_2$ \item $\angle B_1CA=\angle ECT =\angle EDT = \angle B_1A_1T= \angle B_1C_2A_2$ \end{itemize} \end{proof} Hence , $\overline{AA_2}$ and $\overline{CC_2}$ meet on $\Omega$ . The same goes for $\overline{AA_2} \;,\; \overline{BB_2}$ , which indeed completes the proof .

Let $\triangle DEF$ be the pedal triangle of $T$ with respect to $\triangle ABC$. The main claim is that $B_1$ is the center of spiral similarity sending $AC$ to $A_2C_2$ . Proof : $$\angle B_1AC= \angle EAT =\angle EFT = \angle B_1C_1T=\angle B_1A_2C_2$$$$\angle B_1CA=\angle ECT =\angle EDT = \angle B_1A_1T= \angle B_1C_2A_2$$ Hence , $\overline{AA_2}$ and $\overline{CC_2}$ meet on $\Omega$ . The same goes for $\overline{AA_2} \;,\; \overline{BB_2}$ , which indeed completes the proof .

Pretty doable with complex numbers wasn't much involved. Set $(A_1B_1C_1)$ be the unit circle. Then, by the second intersection formula, \[a_2=\frac{a_1-t}{a_1\overline{t}-1}\]\[b_2=\frac{b_1-t}{b_1\overline{t}-1}\]\[c_2=\frac{c_1-t}{a_1\overline{t}-1}\]Then, observe that $A$ is the center of $(C_1TB_1)$. This means, \begin{align*} a &= \begin{vmatrix} b_1 & b_1\overline{b_1} & 1\\ c_1 & c_1\overline{c_1} & 1\\ t & t\overline{t} & 1 \end{vmatrix} \div \begin{vmatrix} b_1 & \overline{b_1} & 1\\ c_1 & \overline{c_1} & 1\\ t & \overline{t} & 1 \end{vmatrix}\\ &= \begin{vmatrix} b_1 & 1 & 1\\ c_1 & 1 & 1\\ t & t\overline{t} & 1 \end{vmatrix} \div \begin{vmatrix} b_1 & \frac{1}{b_1} & 1\\ c_1 & \frac{1}{c_1} & 1\\ t & \overline{t} & 1 \end{vmatrix}\\ &= \frac{b_1+t+c_1t\overline{t}-t-b_1t\overline{t}-c_1}{\frac{b_1}{c_1}+\frac{t}{b_1}+c_1\overline{t}-\frac{t}{c_1}-b_1\overline{t}-\frac{c_1}{b_1}}\\ &= \frac{(c_1-b_1)(t\overline{t}-1)}{\frac{b_1^2+tc_1+b_1c_1^2\overline{t}-b_1t-b_1^2c_1\overline{t}-c_1^2}{b_1c_1}}\\ &= \frac{b_1c_1(c_1-b_1)(t\overline{t}-1)}{(c_1-b_1)(b_1c_1\overline{t}-b_1-c_1+t)}\\ &= \frac{b_1c_1(t\overline{t}-1)}{b_1c_1\overline{t}-b_1-c_1+t} \end{align*}Similarly, we can obtain that \[b=\frac{a_1c_1(t\overline{t}-1)}{a_1c_1\overline{t}-a_1-c_1+t}\]and \[c=\frac{a_1b_1(t\overline{t}-1)}{a_1b_1\overline{t}-a_1-b_1+t}\]We now compute $X=\overline{AA_2}\cap \Omega$ and show that $x$ is symmetric in $a,b$ and $c$. Thus, it follows that $AA_2,BB_2$ and $CC_2$ intersect at $X$ on $\Omega$. Simply note that, \begin{align*} x&= \frac{a_2-a}{a_2\overline{a}-1}\\ &= \frac{\frac{a_1-t}{a_1\overline{t}-1}-\frac{b_1c_1(t\overline{t}-1)}{b_1c_1\overline{t}-b_1-c_1+t}}{\left(\frac{a_1-t}{a_1\overline{t}-1}\right)\left(\frac{t\overline{t}-1}{b_1c_1\overline{t}-b_1-c_1+t}\right)-1}\\ &= \frac{\frac{a_1b_1c_1\overline{t}+a_1t-a_1b_1-a_1c_1-b_1c_1t\overline{t}-t^2+b_1t+c_1t-a_1b_1c_1t\overline{t}^2 + a_1b_1c_1\overline{t}+b_1c_1t\overline{t}-b_1c_1}{(a_1\overline{t}-1)(b_1c_1\overline{t}-b_1-c_1+t)}}{\frac{a_1t\overline{t}-a_1-t^2\overline{t}+t-a_1b_1c_1\overline{t}^2-a_1t\overline{t}+a_1b_1\overline{t}+a_1c_1\overline{t}+b_1c_1\overline{t}+t-b_1-c_1}{(a_1\overline{t}-1)(b_1c_1\overline{t}-b_1-c_1+t)}}\\ &=\frac{2a_1b_1c_1\overline{t}-a_1b_1c_1t\overline{t}^2+(a_1+b_1+c_1)t-(a_1b_1+b_1c_1+c_1a_1)}{2t-t^2\overline{t}-a_1b_1c_1\overline{t}^2+(a_1b_1+b_1c_1+c_1a_1)\overline{t}-(a_1+b_1+c_1)} \end{align*}which is clearly symmetric in $a,b$ and $c$. Thus, a similar computation will show that $Y=BB_2\cap \Gamma$ and $Z=CC_2 \cap \Gamma$ are the same point as $X$ which is what we need to show.

HOW is this G4??? The following claim kills the problem. Claim: $B_1$ is the miquel point of $A_2ACC_2$. Proof. Trivial since \[\measuredangle B_1A_2C_2=\measuredangle B_1C_1C_2 = \measuredangle TC_1B_1 = \measuredangle TFE=\measuredangle TAE=\measuredangle TC=\measuredangle B_1AC\]where $E,F$ are feet from $T$ to $AC$ and $AB$. $\blacksquare$ Thus the intersection of $AA_2$ and $CC_2$ lies on $(B_1A_2C_2)$ as desired.

Call $X$ the antigonal conjugate of $T,$ then $XAB_1C,XA_1B_1C_1$ are cyclic by homothety at $T$ with scale factor $\tfrac12.$ Then if $P$ is the reflection of $B_1$ over $C,$ we have $TP\parallel AC$ and $A_1TB_1P$ cyclic with center $C.$ Thus \[\measuredangle(XA,A_2A_1)=\measuredangle XAC+\measuredangle ACB+90^\circ=\measuredangle XB_1C+\measuredangle PTA_1=\measuredangle XB_1C+\measuredangle PB_1A_1=\measuredangle XB_1A_1=\measuredangle XA_2A_1,\]so $X,A,A_2$ are collinear. Similarly $X,B,B_2$ and $X,C,C_2$ are collinear as well, so $X$ is the desired concurrency point.

not my sol but it's cool Note the midpoint of $T$ and its antigonal conjugate lies on $T$'s pedal circle in $ABC$ (because it's the Poncelet point of ABCT), so the antigonal conjugate of $T$ lies on the circumcircle of $A_1B_1C_1$ by homothety. Obviously it lies on the circumcircle of $AB_1C$ as well. Note that $C$ is the circumcenter of $A_1B_1T$ so we have angle $B_1, \text{antigonal} T A = B_1 C A = B_1 A_1 T = B_1 \text{antigonal T} A_2$, so we're done.

compleck