the only pure geometry in the IMO 2011 is P6! Weird!

There is a generalition for this Let $ABC$ be a triangle and a point $P$. A line pass through $P$ intersect circumcircle $(PBC),(PCA),(PAB)$ again at $P_a,P_b,P_c$, resp. Let $\ell_a, \ell_b, \ell_c,$ be tangets of circumcircle $(PBC),(PCA),(PAB)$ at $P_a,P_b,P_c$, resp. Prove that the circumcircle of the triangle determined by the lines $\ell_a, \ell_b, \ell_c$ is tangent to the circumcircle $(ABC)$. When $P\equiv H$ orthocenter we have problem 6. I don't have solution, yet. I think we can use invension...

Some idea. App steiner line wrt P and Migel theorem.

Denote the vertices of triangle determined by the lines $\ell_a$, $\ell_b$ and $\ell_c$ by $A_1$, $B_1$, $C_1$ respectively. Denote the intersection points of $l$ with lines $BC$, $CA$, $AB$ by $A'$, $B'$, $C'$ respectively. Denote the incenter of $A_1B_1C_1$ by $I$. Denote the circumcircle of triangle $ABC$ by $w$. Denote by $P$ tangent point of $l$ and $w$. WLOG points $C'$ and $B'$ lies on the sides of $\triangle$ $A_1B_1C_1$ and point $A'$ lies on extension of $B_1C_1$. Point $A$ is the excenter of triangle $A_1B'C'$ $\Longrightarrow$ $A_1A$ is the bisector of $\angle B_1A_1C_1$, similarly $B_1B$ and $C_1C$ - are the bisectors of angles $\angle A_1B_1C_1$ and $\angle A_1C_1B_1$ respectively, then (1) $AA_1$, $BB_1$, $CC_1$ - passes though $I$. $\angle B_1IC_1 = 90^{\circ}+ \frac{1}{2}\angle B_1A_1C_1$; $\angle C'AB' = 90^{\circ}- \frac{1}{2}\angle B'A_1C'$ $\Longrightarrow$ $\angle BIC + \angle BAC = 180^{\circ}$ $\Longrightarrow$ (2) $A, B, C, I$ lie on a circle $w$. Let $P'$ - be a reflection of $P$ with respect to the line $BC$. Then $P'$ lies on $B_1C_1$. Let $Q$ - be the intersection point of circumcircles of triangles $B_1BP'$ and $C_1CP'$. Let $\angle IB_1C_1 = \alpha$, $\angle IC_1B_1 = \beta$, $\angle BCP' = x$, $\angle CBP' = y$. Then $\angle BQC = \angle BQP' + \angle P'QC = \angle BB_1P'+ \angle P'C_1C = 180^{\circ} - \angle B_1IC_1$ $\Longrightarrow$ (3) points $I, B, C, Q$ lie on a circle $w$. $\angle B_1QC_1 = \alpha + \beta + x + y = 2\alpha +2\beta$, because $\angle BP'C = \angle B_1IC_1$ $\Longrightarrow$ (4) points $A_1, B_1, C_1, Q$ lie on a circumcircle of triangle $A_1B_1C_1$. $\angle B_1QB + \angle QC_1B_1 = x + \angle QC_1B_1 = \angle QCP' + \angle P'CB = \angle QCB$ $\Longrightarrow$ (5) $\angle B_1QB + \angle QC_1B_1 = \angle QCB$. Let $t$ - tangent line from $Q$ to the circumcircle of triangle $A_1B_1C_1$. Then $\angle (g, QB) = \angle (g, QB_1) + \angle (QB_1, QB) = \angle (QC_1, C_1B_1) + x = \angle (QC, CB)$ $\Longrightarrow$ $t$ is the tangent line from $Q$ to $w$ $\Longrightarrow$ (6) $w$ tangents to circumcircle of triangle $A_1B_1C_1$ at $Q$.

Very good classic geometry problem for IMO, In fact there are two main parts here: First is to prove that $AA_1$, $BB_1$, $CC_1$ - passes though $I$ and that $A, B, C, I$ lie on a circle $w$, that is not hard. Second part is to find the way to prove that two particular circles tangent to each other while exactly point of tangency is not given from the beginning. I know only two ways here - inversion and a trick with two circles when the point of tangency is defined as an intersection point of two other circles. The second way works here. In particular case the following theorem holds: Given a triangle $ABC$ with circumcircle $w$. Points $X$ and $Y$ are on segments $AB, AC$ respectively, $\angle BAC = \alpha$. Then there exist an arc of measure $\beta$ constructed on $XY$ (in other half-plane then point $A$) that tangent to $w$ if and only if there exist an arc of measure $\alpha + \beta$ constructed on $BC$ (in the same half-plane with point $A$) that tangent to $XY$. This theorem can be applied in this problem for triangle $IBC$ and segment $B_1C_1$ It is not too hard to repeat the trick with two circles but it is hard to reinvent such trick during the contest.

Yes, of course it's a very nice problem if you do it synthetically. But taking this at home (not working that efficiently, either, e.g. thinking about the fact that Harry Potter is over ), I was able to solve this in less than two hours with very straightforward complex numbers calculations (i.e. showing a discriminant is zero ...), which was extremely disappointing given the relatively simple nature of problems 4 and 5. There's only one "grind" at the end which some people like Yi Sun would not even call a grind in the first place. In the future, I feel like we need more problems like IMO 2008 #6 which are both very hard to bash (in the context of the other two problems in that day) and very nice. (No, I did not bash it. It's one of my favorite geometry problems.) Also, on a random note, WHERE IS THE GOOD, CHALLENGING NUMBER THEORY IN COMPETITIONS NOW?!!! Sorry, I just felt like saying that.

WakeUp wrote: Let $ABC$ be an acute triangle with circumcircle $\Gamma$. Let $\ell$ be a tangent line to $\Gamma$, and let $\ell_a, \ell_b$ and $\ell_c$ be the lines obtained by reflecting $\ell$ in the lines $BC$, $CA$ and $AB$, respectively. Show that the circumcircle of the triangle determined by the lines $\ell_a, \ell_b$ and $\ell_c$ is tangent to the circle $\Gamma$. Denote $D, E, F$ the intersections of $l$ and $BC,CA,AB$, respectively; $X,Y,Z$ the intersections of 3 lines $l_a,l_b,l_c$. Let $L$ be the point of contact of $l$ and $\Gamma$; $R,S,T$ be the reflections of $L$ wrt $AB,AC,BC; M$ be the Miquel point of the completed quadrilateral $XSRYTZ$. $\angle ZXY=\angle FEX+\angle EFX=180^o-2\angle LEC+2\angle LFA=180^o-2\angle BAC.$ We have the distances from $A$ to $l, DE, FY$ are equal so $XA$ is the bisector of angle $ZXY.$ We get $\angle ZXA=\frac{1}{2}\angle ZXY=90^o-\angle BAC. (1)$ On the other side, let $U, V$ be the projections of $L$ onto $AC, AB$ then $UV$ passes through the midpoint of $LQ$. $\angle ARL=\angle ALR=90^o-\angle LAV=90^o-\angle LUV=\angle UQL$, which follows that $L, A, Q, R$ are concyclic. We get $\angle ARQ=\angle LRQ-\angle LRA=180^o-\angle LAQ-90^o+\angle LAB=90^o-\angle BAC. (2)$ From $(1)$ and $(2)$ we obtain $A\in (XRS)$. Similarly with $B, C.$ So $\angle AMB=\angle XMY-\angle XMA-\angle BMY=2\angle ACB-\angle XRA-\angle BRY$ $=2\angle ACB-180^o+\angle ARB=2\angle CAB-180^o+\angle ALB=\angle ACB.$ Therefore $M\in \Gamma.$ Construct a tangent $Mt$ of $(XYZ)$. We will show that $Mt$ is also a tangent of $\Gamma$ iff $\angle tMA=\angle ABM$. $\Leftrightarrow \angle AMX+\angle XMt=\angle ABR+\angle RBM (3)$ But $\angle XMt=\angle XYM=\angle RBM, \angle AMX=ARX=\angle ALE=\angle ABL=\angle ABR.$ Hence $(3)$ is true. We are done. Another generalization: Given a triangle $ABC$ with its circumcenter $O$. Let $P$ be an arbitrary point in the plane. The line through $P$ intersects $(BPC), (CPA), (APB)$ again at $A_1, B_1, C_1$. Let $l_a$ be the tangent line through $A_1$ of $(BPC), l'_a$ be the reflection of $l_a$ wrt $BC$. Similarly we define $l'_b, l'_c$. Then the circumcircle of the triangle formed by $l'_a,l'_b,l'_c$ is tangent to $(O)$. When $P$ lies on $(O)$ we have IMO Pro.6

buratinogigle wrote: There is a generalition for this Let $ABC$ be a triangle and a point $P$. A line pass through $P$ intersect circumcircle $(PBC),(PCA),(PAB)$ again at $P_a,P_b,P_c$, resp. Let $\ell_a, \ell_b, \ell_c,$ be tangets of circumcircle $(PBC),(PCA),(PAB)$ at $P_a,P_b,P_c$, resp. Prove that the circumcircle of the triangle determined by the lines $\ell_a, \ell_b, \ell_c$ is tangent to the circumcircle $(ABC)$. When $P\equiv H$ orthocenter we have problem 6. I don't have solution, yet. I think we can use invension... Yes, we really may do it, and after inversion the corresponding circle will touch the circle $ABC$ in the Miquel point of lines $AB$, $BC$, $AC$ and our line passing through $P$. It may be easily checked by some direct angle-chasing.

Fedor Petrov wrote: buratinogigle wrote: There is a generalition for this Let $ABC$ be a triangle and a point $P$. A line pass through $P$ intersect circumcircle $(PBC),(PCA),(PAB)$ again at $P_a,P_b,P_c$, resp. Let $\ell_a, \ell_b, \ell_c,$ be tangets of circumcircle $(PBC),(PCA),(PAB)$ at $P_a,P_b,P_c$, resp. Prove that the circumcircle of the triangle determined by the lines $\ell_a, \ell_b, \ell_c$ is tangent to the circumcircle $(ABC)$. When $P\equiv H$ orthocenter we have problem 6. I don't have solution, yet. I think we can use invension... Yes, we really may do it, and after inversion the corresponding circle will touch the circle $ABC$ in the Miquel point of lines $AB$, $BC$, $AC$ and our line passing through $P$. It may be easily checked by some direct angle-chasing. what you said seems really intresting, would you please give us a detailed solution?? Thanks

Interestingly, the fact that the incenter of the determined triangle lies on the circumcircle of triangle $ABC$ appeared on a 1995 Iran olympiad: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=379391 (And it works for any line.)

livetolove212's generalization is exactly the same as buratino's. Congratulations, buratino and Fedor, for the first correct solution using inversion. (Nobody gave that in the exam. Then again, apparently only inversions with center $T$ were attempted, where $T$ is the point of tangency of $\ell$ with $\Gamma$.) Also there is a nice solution by a contestant whose name and country I don't remember, which shows a bit more: If the altitudes of triangle $ABC$ from the vertices $A$, $B$, $C$ intersect $\Gamma$ a second time at $X$, $Y$, $Z$, respectively, and if $X'$, $Y'$, $Z'$ are the reflections of $T$ (the point where $\ell$ touches $\Gamma$) in $BC$, $CA$, $AB$, then the lines $XX'$, $YY'$, $ZZ'$ concur at the point where $\Gamma$ touches $\Gamma '$. Try to reconstruct the solution from this. Also there exists an approach which uses Casey's theorem (with three circles being degenerate to the vertices $A'$, $B'$, $C'$, and the fourth circle being $\Gamma$). I think IRN1 found it, and maybe others too.

yeah ... darij is right ... she found that solution

buratinogigle wrote: Let $ABC$ be a triangle and a point $P.$ A line pass through $P$ intersect circumcircle $(PBC),$ $(PCA),$ $(PAB)$ again at $P_a,P_b,P_c,$ resp. Let $\ell_a, \ell_b, \ell_c,$ be tangets of circumcircle $(PBC),$ $(PCA),$ $(PAB)$ at $P_a,P_b,P_c,$ resp. Prove that the circumcircle of the triangle determined by the lines $\ell_a, \ell_b, \ell_c$ is tangent to the circumcircle $(ABC).$ Perform an inversion with center $P$ and arbitrary power. Label inverse points with primes. The given line $\ell$ through $P$ is double and circles $\odot(PBC),$ $\odot(PCA)$ and $\odot(PAB)$ go to the sidelines $B'C',$ $C'A',$ $A'B'$ of $\triangle A'B'C'.$ $\ell$ cuts $B'C',$ $C'A',$ $A'B'$ at ${P_a}',$ ${P_b}',$ ${P_c}'.$ Thus, tangents $\ell_a,\ell_b,\ell_c$ of $\odot(PBC),$ $\odot(PCA),$ $\odot(PAB)$ through $P_a,P_b,P_c$ go to the circles $\omega_a,\omega_b,\omega_c$ passing through $P$ and tangent to $B'C',$ $C'A',$ $A'B'$ through ${P_a}',$ ${P_b}',$ ${P_c}'.$ Pairwise circles $\omega_a,\omega_b,\omega_c$ meet at $D',E',F',$ the inverses of the vertices of $\triangle (\ell_a,\ell_b,\ell_c) \equiv \triangle DEF.$ Hence, in order to show that $\odot(DEF)$ is tangent to $\odot(ABC),$ we shall show that $\odot(D'E'F')$ is tangent to $\odot(A'B'C').$ For convenience, drop the primes from the new figure. Let $M$ be the Miquel point of $\triangle ABC \cup \ell.$ Henceforth, we'll use oriented angles (mod 180). From the tangencies of $\omega_b, CA$ and $\omega_c,AB$ we get $\angle AP_bP_c=\angle PDP_b$ and $\angle AP_cP_b=\angle PDP_c$ $\Longrightarrow$ $D \in \odot(AP_bP_c).$ Likewise, we have $E \in \odot(BP_cP_a)$ and $F \in \odot(CP_aP_b).$ Now, the rest is just simple angle chase using inscribed angles spanning the same arcs. $\angle EFD=\angle PFD+\angle PFE=\angle P_cP_bD+\angle P_cP_aE=\angle DMP_c+\angle EMP_c$ $\Longrightarrow \angle EFD=\angle EMD \Longrightarrow M \in \odot(DEF) \ (\star).$ $\angle MFE=\angle EFP_a+\angle MFP_a=\angle EP_aB+\angle MCB=\angle EMB+\angle MAB.$ $\Longrightarrow 90^{\circ}-\angle MFE+\angle EMB=90^{\circ}-\angle MAB$ $\Longrightarrow$ Circumcenters of $\triangle MEF$ and $\triangle MBA$ are collinear with $M.$ Together with $(\star),$ we deduce that $\odot(ABC)$ and $\odot(DEF)$ are tangent through $M,$ as desired.

darij grinberg wrote: Also there exists an approach which uses Casey's theorem (with three circles being degenerate to the vertices $A'$, $B'$, $C'$, and the fourth circle being $\Gamma$). I think IRN1 found it, and maybe others too. Could you say more about it? I thought about it few seconds after I read this problem, but I couldn't compute $d_X, d_Y, d_Z, XY, XZ, YZ$, where X, Y, Z are vertices of triangle given in this problem.

We can use angle chasing to prove Mr.Hung's generalization. The idea is the same as my proof for problem 6. Let $X, Y, Z$ be the triangle formed by $l_a,l_b,l_c; Q$ be the Miquel point of the completed quadrilateral $XYZP_aP_bP_c; R$ be the intersection of $AP_c$ and $CP_a$. $\angle AP_cX=\angle P_cPA=\angle P_bCA=\angle XP_bA$ then $A\in (XP_bP_c)$. Similarly with $B,C$. We will show that $Q$ lies on $(ABC)$. We have $\angle P_cAB+\angle P_aCB=\angle P_cPB+\angle P_aPB=180^o$ then $A,B,C,R$ are concyclic. We conclude that $\angle AQC=\angle AQP_b+\angle CQP_b=\angle P_bP_cA+\angle P_bP_aC=180^o-\angle ARC=\angle ABC$. Therefore $Q\in (ABC)$. Construct the tangent $Qt$ of $(XYZ)$. The idea is to show that $Qt$ is also the tangent of $(ABC)$, iff $\angle tQA=\angle ACQ$. $\Leftrightarrow \angle tQX+\angle XQA=\angle ACP_b+\angle P_bCQ (*)$ But $\angle tQX=\angle QZX=\angle QCP_b, \angle XQA=\angle XP_bA=\angle QCP_b$. So $(*)$ is true. We are done. We have a new problem: Given a triangle $ABC$ with its circumcircle $(O)$. Let $(O')$ be a circle tangent to $(O)$. $P$ is an arbitrary point on $(O)$. Three rays $PA, PB, PC$ (or $AP, BP, CP$) intersect $(O')$ again at $A_1,B_1,C_1$. Let $A_2B_2C_2$ be the triangle formed by the reflections of $A_1B_1$ wrt $AB, B_1C_1$ wrt $BC, C_1A_1$ wrt $CA$. Then $(A_2B_2C_2)$ is also tangent to $(O)$.

Try something I did at home - prove that $ OO'=R'-R $ , where $ O' $ is the curcumcenter of $ A'B'C' $ and $ R' $ is the curcumradius of $ A'B'C' $ . I did it using triangle $ OO'I $ and applying the cosine Th, where $ I $ is the incenter of $ A'B'C' $ . It works but you have to spend a lot of time calculating it!

Dear MLs If we reflect not just the line l, but also the circle (ABC) in the sides, we get get a triangle with sides tangent to Karnot circles. By Bobillier theorem this triangle remains similar when tangency point of l changes. A*B*C* is triangle determined by the lines l_a, l_b, l_c The tangency points of triangle A*B*C* with Karnot circles are collinear. From this it should be possible to quickly show that (ABC) and (A*B*C*) are incircle and 9pc of some triangle, and are internally tangent at F point. M.T.

Among the 6 contestants who got 7 in this problem, two of them are bronze medalists (and one from HKG!) interesting!

Dear Mathlinkers, having the same beginning ideas as Armpist, a synthetic proof without any calculation is possible and will appears next on my site. Sincerely Jean-Louis

woohoo first 45 MOHS! Honestly way more starightforward IMO than 2008/6 but that might just be since that was very hard for me. Anyways, great problem, using ggb for the problem helped a lot Also formatting is bad since source code lines are short like evan says Define $\ell_b \cap \ell_c=A'$ and cyclic. Also, define $\ell \cap BC = A_1$ and cyclic. Claim: Line $A'A$ bisects $\angle B'A'C'$ and cyclic variations hold true as well. Specifically, $AA',BB',CC'$ concur at the incenter of $\triangle A'B'C'$. Proof: Notice that by the reflections line $C_1B$ bisects $\angle B'C_1A_1$. Also, $A_1B$ bisects $\angle B'A_1C_1$. Thus $B$ is the incenter or excenter of $\triangle B'A_1C_1$ because config issues, but no matter what $BB'$ bisects $\angle A'B'C'$. The same thing holds true for other vertices implying the claim. Let $I'$ be the incenter of $\triangle A'B'C'$ Claim: $I'$ lies on $(ABC)$. Proof: Angle chase!!! \[\measuredangle BAC=\measuredangle C_1AB_1\overset{\text{incenter/excenter}}{=} \measuredangle B'I'C' = \measuredangle BI'C\]as desired. Let $H$ be the orthocenter of $\triangle ABC$ and say $\ell$ touches $(ABC)$ at $T$. Claim: $(AHB)$ is tangent to line $B'A'$ at the reflection of $T$ across $AB$ and cyclic variants. Proof: Reflecting the orthocenter gives $(AHB)$ and $(ABC)$ are symmetric across line $AB$. But lines $\ell$ and $B'A'$ are symmetric across $AB$ as well. This implies the result as $\ell$ is tangent to $(ABC)$. Say $(AHB)$ hits $A'B'$ at $F$, $(AHC)$ hits $A'C'$ at $E$, and $(BHC)$ hits $B'C'$ at $D$. Claim: $D,E,F,H$ are collinear and $B'FBD$ and variants are cyclic. Proof: By symmetry and tangents, we angle chase \[\measuredangle FHB = \measuredangle C_1FB= \measuredangle BTC_1 = \measuredangle BCT = \measuredangle DCB = \measuredangle DHB\]Repeat for the other points to get the collinearity. Also, \[\measuredangle B'FB = \measuredangle C_1FB = \measuredangle DCB = \measuredangle B'DB\]to get the cyclic quads. As a result from this, we find the three circumcircles concur as if $(B'FBD)\cap (DCC'E)=G\neq D,$ then \[\measuredangle FGE=\measuredangle FGD + \measuredangle DGE =\measuredangle A'B'C'+\measuredangle BCA' = \measuredangle B'A'C'= \measuredangle FA'E.\]By the same angle chasing, $G$ lies on $(A'B'C').$ Also, \[\measuredangle BGC = \measuredangle BGD + \measuredangle DGC =\measuredangle I'B'D+\measuredangle DEI' = \measuredangle B'I'C' =\measuredangle BI'C\]so $G$ lies on $(ABC).$ Let $GB'$ intersect $AF$ at $L.$ Claim: $L$ lies on $(ABC).$ Proof: Angle chasing FTW! \[\measuredangle LGB = \measuredangle B'GB = \measuredangle C_1FB= \measuredangle FAB = \measuredangle LAB\]boom. Now, let $\ell_1$ be the tangent to $(ABC)$ at $G.$ and let $X$ be a point on $\ell_1.$ Then, \[\measuredangle B'GX = \measuredangle LGX = \measuredangle LAG = \measuredangle FA'G = \measuredangle B'A'G\]thus DONE.

Define $E=l_b\cap l_c$, and similarly with $F,G$, D' the reflection of D which is the point in question that l passes through on (ABC), and let $T=(EAD')\cap(FBD')$. We make a couple of observations. E lies on a reflection of l over a line through C, and so does F, whence the composition of reflections that reflects E to F is over a line through C, so CG is the angle bisector (which is the line of reflection); similarly for the others, we get that they're concurrent at the incenter of EFG I. $BIC=180-GIF=90-FEG=BAC\implies I\in(ABC)$. $ATB=BTD'+ATD'=BFD'+AED'=180-AIB=ACB\implies T\in(ABC)$ $ETF=ETD'+FTD'=180-EAD'+180-FBD'=IAD'+IBD'=360-AIB-AD'B=360-AIB-ADB=2(180-EIF)=180-EGF\implies T\in(EFG)$ It's sufficient to solve the problem now, because ABT-EFT=ABT-D'BT=ABD'=AD'E=ATE. (This is because if we let some point T' be on the tangent through T to (EFG), with $AT\cap(EFG)=P$, we have T'TA=T'TP=EFT+PTE=EFT+ATE=ABT, as desired.)

45 MOHS??? Xooks (solved with hints ) Let $L$ be the tangency point of $\Gamma$ and $\ell$ $\ell \cap \ell_A=D$, $\ell \cap \ell_B=E$, $\ell \cap \ell_A=F$ $\ell_B \cap \ell_C=A'$, $\ell_A \cap \ell_C=B'$, $\ell_A \cap \ell_B=C'$ and $\measuredangle$ note directed angles mod $180^{\circ}$ Claim: $BB'$, $AA'$, and $CC'$ concur at the incenter of $\triangle A'B'C'$ Proof: Note that $BF$ bisects $\angle B'FE$, and $BD$ bisects $\angle B'DE$, hence $B$ lies on the angle bisector of $\angle A'B'C'$, similarly $A$ and $C$ lie on the angle bisectors of $\angle B'A'C'$ and $\angle A'C'B'$ respectively. Hence they concur at the incenter of $\triangle ABC$. $\square$ Claim: The incenter $I'$ of $\triangle A'B'C'$ lies on $\Gamma$ Proof: angle chasing $$\measuredangle B'C'A'= \measuredangle DC'E= \measuredangle B'DL+ \measuredangle A'EL= 2(\measuredangle BDL+ \measuredangle AEL)=2\measuredangle DCE= 2\measuredangle BCA$$Since $$\measuredangle C'B'A'+ \measuredangle B'A'C= 2 \measuredangle ACB$$$$\measuredangle I'B'A'+ \measuredangle B'A'I'= \measuredangle ACB$$Implying $I'$ lies on $\Gamma$. $\square$ Let $H$ be the orthocenter of $\triangle ABC$ Claim: $(BHC)$ is tangent to $B'C'$ with similar statements for $(AHC)$ and $(AHB)$ Proof: Let $L_{A}$ denote the reflection of $L$ across $BC$, define $L_{B}$, and $L_{C}$ similarly. Note that by orthocenter reflections $(ABC)$ and $(BHC)$ are reflections across $BC$. Also $B'C'$ and $\ell$ are reflections across $BC$. Since $\ell$ is tangent to $(ABC)$ at $L$. $B'C'$ is tangent to $(BHC)$ at $L_{A}$. Similar statements hold. $\square$ Claim:$L_A$, $L_B$, and $L_C$ and $H$ are collinear Proof: To show this note that $L_{A}$, $L_{B}$, and $L_{C}$, are just a homothety with a scale factor of $2$ with respect to $L$ of the simson line of $\triangle ABC$ with respect to $L$. Now Since the $HL$ bisects the simson line. This homothety sends the simson line to $L_{A}L_{B}$. Hence proven $\square$ Claim:$A'L_{B}AL_{C}$, $B'L_{A}BL_{C}$, and $C'L_{A}CL_{B}$ are cyclic Proof: $$\measuredangle BL_{C}B' = \measuredangle BL_{C}F = \measuredangle BCL_{A} = \measuredangle BL_{A}B'$$So $B'L_{A}BL_{C}$ is cyclic. Similar reasons give us the other cyclic quadrilaterals. $\square$ Claim:$(A'L_{B}AL_{C})$, $(B'L_{A}BL_{C})$, and $(C'L_{A}CL_{B})$ all intersect at a point $T$ on $(A'B'C')$, and $(ABC)$ Proof: Let $(B'L_{A}BL_{C})$ intersect $(L_{A}CC'L_{B})$ at $S$ which isn't $L_{A}$. Now we angle chase $$\measuredangle BSC = \measuredangle BSL_{A} + \measuredangle L_{A}SC = \measuredangle L_{A}L_{B}I'+ \measuredangle I'B'L_{A} = \measuredangle B'I'C' =\measuredangle BI'C$$Which means $S$ lies on $(C'L_{A}CL_{B})$, and $(A'B'C')$. Now $$\measuredangle L_{C}SL_{B}=\measuredangle L_{C}SL_{A} + \measuredangle L_{A}SL_{B} =\measuredangle BCA' +\measuredangle A'B'C'= \measuredangle B'A'C'= \measuredangle L_{C}A'L_{B}$$Which means $S$ lies on $(ABC)$.$\square$ Switch $S$ with $T$ from now on Claim:Let $B'T \cap AL_{C}= B'' $, define $A''$, and $C''$ similarly. $A''$, $B''$, and $C''$ lie on $\Gamma$ Proof: We angle chase $$\measuredangle ATA'' = \measuredangle ATA' = \measuredangle AL_{B}E= \measuredangle ACL_{B} = \measuredangle ACA'' $$So $A''$ lies on $\Gamma$. Similarly, $B''$, and $C''$, also lie on $\Gamma$. $\square$ Claim: $(ABC)$ is tangent to $(A'B'C')$ at $T$ !!!! Proof: Consider the tangent line $\ell_{1434}$ to $\Gamma$ at $T$ now $$\measuredangle(TA', \ell_{1434})= \measuredangle(TA'', \ell_{1434})= \measuredangle A''CT= \measuredangle L_{B}C'T = \measuredangle A'C'T$$$\square$

Too lazy to write down a full solution, but my method basically went as follows: First, $\ell_B \cap \ell_C=A'$, $\ell_A \cap \ell_C=B'$, $\ell_A \cap \ell_B=C'$ 1: Prove the Incenter of $A'B'C'$ lies on $\Gamma$ 2: Construct $BX||A'I$, where X lies on $\Gamma$ 3: Prove $A'IB'X$ is a parallelogram 4: Use reflecting the incenter

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/* end of picture */ [/asy][/asy] Define $K_a = \ell_b \cap \ell_c$, and $D_a$ as the reflection of $D$ over $\overline{BC}$. Cyclically define variants. Let $M$ be the Miquel Point of $K_bK_cD_bD_c$. We claim $M$ is the Anti-Steiner point of $D$ with respect to $(ABC)$. To show this phantom $M'$ as the Anti-Steiner Point of $D$. Note that, \begin{align*} \measuredangle M'D_aK_c &= \measuredangle(\ell, \overline{DH})\\ \end{align*}after reflection about $\overline{BC}$. However this is symmetric so we may similarly observe, \begin{align*} \measuredangle K_cD_bM' = \measuredangle(\overline{DH}, \ell) \end{align*}Together these imply that, \begin{align*} \measuredangle M'D_aK_c + \measuredangle K_cD_bM' = 0 \end{align*}and hence $MD_aD_bK_c$ is cyclic. From cyclic statements we may conclude that $M$ is indeed the Anti-Steiner point of $D$. This then implies that we have also $M \in (ABC)$ from properties of Anti-Steiner. Finally it suffices to show that $(ABC)$ and $(K_aK_bK_c)$ are both tangent at $M$. We will show this through homothety. Let $X_a = K_aM \cap (ABC)$ and similarly define $X_b$ and $X_c$. We claim that $\widehat{X_cD}$ has arc midpoint $C$. Indeed note that, \begin{align*} \measuredangle CDX_c &= \measuredangle CMK_c\\ &= \measuredangle K_cD_aC\\ &= \measuredangle (\overline{CD}, \ell)\\ &= \measuredangle DMC \end{align*}as desired. Similar conclusions hold for $X_a$ and $X_b$. Next we will show that in fact $D_a$, $X_b$, and $C$ are all collinear. Use complex numbers. We can easily see that, \begin{align*} d_a = b + c- bc/d \end{align*}by reflection. Similarly we may find, \begin{align*} x_b &= \frac{b/d}{1/b}\\ &= b^2/d \end{align*}Now rotate so that without loss of generality $d = 1$. Then we need to check $c$, $b + c - bc$ and $b^2$ are all collinear. Shifting by $c$ we need to check $0$, $b - bc$ and $b^2 - c$ are collinear. Then it suffices to check, \begin{align*} \begin{vmatrix} 0 & 0 & 1\\ b - bc & 1/b - 1/bc & 1\\ b^2 - c & 1/b^2 - 1/c & 1\\ \end{vmatrix} = 0\\ \iff (b - bc)(1/b^2 - 1/c) - (b^2 - c)(1/b - 1/bc) = 0\\ \iff 1/b - b/c - c/b + b - b + b/c + c/b - 1/b = 0\\ \iff 0 = 0 \end{align*}thus proven. Finally we will show $\overline{K_bK_c} \parallel \overline{X_bX_c}$. Indeed note that, \begin{align*} \measuredangle CX_bX_c &= \measuredangle DMC\\ &= \measuredangle (\ell, \overline{CD})\\ &= \measuredangle (\overline{D_aC}, \ell_a)\\ &= \measuredangle (\overline{DX_b}, \ell_a) \end{align*}and so we are finally done by homothety at $M$ mapping $(X_aX_bX_c) \mapsto (K_aK_bK_c)$.

Let ABC be an oxygonal (non-convex) triangle and \(\Gamma\) its circumscribed circle. Let \(\ell\) be a tangent line of \(\Gamma\), and let \(\ell_a\), \(\ell_b\), \(\ell_c\) be its symmetrical lines with respect to the axes of symmetry of the lines BC, CA, and AB, respectively. We want to prove that the circumscribed circle of the triangle defined by the lines \(\ell_a\), \(\ell_b\), and \(\ell_c\) is tangent to \(\Gamma\). Let \(S\) be the point of contact of \(\ell\) with \(\Gamma\). Without loss of generality, we assume that \(S\) is located in the small arc BC. Let the lines \(\ell_a\), \(\ell_b\), and \(\ell_c\) intersect at points \(P\), \(Q\), and \(R\). Let \(\Omega\) be the circumscribed circle of the triangle \(PQR\). To show that circles \(\Gamma\) and \(\Omega\) are tangent, we first show that they intersect at (at least) one point at which they have a common tangent. Due to symmetry, the lines \(\ell\), \(\ell_a\), and the line containing BC (denoted as \(\ell_{BC}\)) coexist, at least in the point A'. If \(\ell\) is parallel to BC, then we can consider A' as a point at infinity. Similarly, \(\ell\), \(\ell_b\), and \(\ell_{CA}\) coexist in B', and \(\ell\), \(\ell_c\), and \(\ell_{AB}\) coexist in C'. Let \(S_1\), \(S_2\), and \(S_3\) be the projections of \(S\) onto \(\ell_{BC}\), \(\ell_{CA}\), and \(\ell_{AB}\), respectively. Since \(S\) belongs to the circumscribed circle ABC, these points are collinear (Simpson's line of \(S\)). Let \(T_1\), \(T_2\), and \(T_3\) be the symmetrical points of \(S_1\), \(S_2\), and \(S_3\) with respect to \(S\). Then, due to scaling with respect to \(S\), points \(T_1\), \(T_2\), and \(T_3\) are collinear. Also, due to symmetries, \(T_1\) lies on \(\ell_c\), \(T_2\) lies on \(\ell_b\), and \(T_3\) lies on \(\ell_a\). We have that \( \angle RPQ = \pi - \angle PB'C' - \angle PC'B' = \pi - (\pi - 2\angle SC'A) - (\pi - 2\angle SB'A ) = 2(\angle SC'A + \angle SB'A) - \pi = 2(\pi - \angle A ) - \pi = \pi - 2\angle A \). Also, \( \angle PRQ = \angle RA'B' + \angle RB'A' = \angle RA'B' + (\pi - \angle RB'S ) = 2\angle CA'B' - \pi + 2\angle CB'S = \pi - 2( \angle CB'S - \angle CA'S ) = \pi - 2 \angle C \). So, we have that \( \angle RQP = \pi - (\pi - 2\angle A) - (\pi - 2\angle C ) = \pi - 2\angle B \). We observe that in the quadrilateral \(PRT_1T_3,T_2Q\) the circle \(\Omega\) coexists with the circumscribed circles of the triangles \(PT_2T_3\), \(RT_2T_1\), and \(QT_1T_3\), and this point, let it be \(M\), is the Miquel point of the quadrilateral. Consequently, \( \angle RMQ = \pi - \angle RPQ = 2 \angle A \). It is clear from the recordable that \( \angle T_1MT_2 = \angle T_1RP = \pi - 2 \angle B \) and \( \angle T_1RT_2 = 2 \angle C \). It is true that \(CS = CT_1\) because of symmetry with respect to \(\ell_{BC}\), and that \(CS = CT_2\) because of symmetry with respect to \(\ell_{CA}\). So, \(C\) is the circumcenter of \(T_1ST_2\). Thus, \( \angle T_1CT_2 = 2\angle T_1ST_2\) (registered - circumcenter) \(= 2\angle S_ 1SS_2\) (from the fact that \(S_1SS_2\) is isosceles) \(= 2\angle C\). Therefore, \(C\) is the circumcenter of \(T_1MT_2R\), and so \( \angle CMR = \angle CT_1R = \angle CT_1A' = \angle CSA' = \angle CMR = \angle SAC\). Similarly, we prove that \(B\) is the circumcenter of \(ST_1T_3\), homocyclic with \(T_1\), \(T_3\), \(Q\), \(M\), finally, that \( \angle BMQ = \angle BAS\). From the last relationships, we get that \(C\) is also on \(\Gamma\). Therefore, the requested circles intersect at least one point, and it remains to be shown that they touch at their intersection. Now let \(x\) be the tangent to \(\Omega\) at \(M\). Then \( \angle RMx = \angle RQM\) (from the quadrilateral \(RPQM\)) \(= \angle T_1QM = \angle T_1BM\) (from the quadrilateral \(T_1BT_3QM\)). In order for \(x\) to be tangent to \(\Gamma\), it must be \( \angle CMx = \angle CBM \Leftrightarrow \angle CMR + \angle RMx = \angle CBT_1 + \angle T_1BM \Leftrightarrow \angle CMR = \angle CBT_1 \Leftrightarrow \angle CT_1R = \angle CBS \Leftrightarrow \angle CT_1A' = \angle CSA'\) as follows from previous relationships!

present in most other bashes in this thread. Let $\Gamma$ be the unit circle and $\ell$ be tangent to $\Gamma$ at $1$. By Simson Line, it suffices to show that there is a unique point $z\in\Gamma$ such that the reflections of $z$ in $\ell_a$, $\ell_b$, and $\ell_c$. are collinear. To obtain the reflection, $z_a$, of $z$ in $\ell_a$, reflect it in $BC$, then $\ell$, then $BC$ again: \begin{align*} z&\mapsto b+c-bc/z\\ &\mapsto 2-1/b-1/c+z/bc\\ &\mapsto b+c-bc(2-b-c+bc/z)=b^2c+bc^2-2bc+b+c-b^2c^2/z=z_a \end{align*} We now prove that exactly one unit complex number $z$ satisfies \begin{align*} \left|\begin{matrix}z_a&\overline{z_a}&1\\z_b&\overline{z_b}&1\\z_c&\overline{z_c}&1\end{matrix}\right|&=0\\ \left|\begin{matrix}x_a-b^2c^2/z&a^2(x_a-z)&1\\x_b-c^2a^2/z&b^2(x_b-z)&1\\x_c-a^2b^2/z&c^2(x_c-z)&1\end{matrix}\right|&=0 \end{align*}where we multiply the second column by $a^2b^2c^2$ and substitute $x_a=b^2c+bc^2-2bc+b+c$ and similar expressions for $x_b$ and $x_c$. We now let the above determinant be $t_2z+t_1+t_0/z$. We compute each coefficient individually. Introduce the notation of the antisymmetric sum, $\sum_{\text{sym}'}f(a,b,c)=\sum_{\pi\in S_3}\text{sgn}(\pi)f(\pi(a,b,c))$. We have $t_2=-\sum_{\text{sym}'}x_ab^2=-\sum_{\text{sym}'}b^4c+b^3c^2-2b^3c+b^3+b^2c=(b-c)(c-a)(a-b)(a+b+c-1)^2$. (In complex bashing, I have occasionally found it useful to have memorized quantities of the form $\frac{\sum_{\text{sym}'}b^mc^n}{(b-c)(c-a)(a-b)}$). Since we multiplied an otherwise pure imaginary determinant by $a^2b^2c^2$, it follows that $t_0=-a^4b^4c^4\overline{t_2}=(b-c)(c-a)(a-b)(bc+ca+ab-abc)^2$. It now suffices to show that $\sum_{\text{sym}'}b^4c^2+b^2x_ax_b=t_1\stackrel?=2(b-c)(c-a)(a-b)(a+b+c-1)(bc+ca+ab-abc)$ \[\sum_{\text{sym}'}b^4c^2+b^2(b^2c+bc^2-2bc+b+c)(a^2c+ac^2-2ac+a+c)\stackrel?=2(b-c)(c-a)(a-b)(a+b+c-1)(bc+ca+ab-abc)\]\[\sum_{\text{sym}'}b^4c^2+a^2b^4c^2+ab^4c^3-2ab^4c^2+ab^4c+b^4c^2+a^2b^3c^3+ab^3c^4-2ab^3c^3+ab^3c^2+b^3c^3-2a^2b^3c^2-2ab^3c^3+4ab^3c^2-2ab^3c-2b^3c^2+a^2b^3c+ab^3c^2-2ab^3c+ab^3+cb^3+a^2b^2c^2+ab^2c^3-2ab^2c^2+ab^2c+b^2c^2\stackrel?=2(b-c)(c-a)(a-b)(a+b+c-1)(bc+ca+ab-abc)\]\[\sum_{\text{sym}'}2b^4c^2-2ab^4c^2+4ab^3c^2-2b^3c^2\stackrel?=2(b-c)(c-a)(a-b)(-abc(a+b+c)+\sum_{\text{sym}}a^2b+4abc-bc-ca-ab)\] which is true by inspection, if you have memorized sufficiently many of the aforementioned quantities $\frac{\sum_{\text{sym}'}b^mc^n}{(b-c)(c-a)(a-b)}$. The large expression above isn't that scary: you expand $5\cdot5$ terms, but most of them disappear under the antisymmetric sum (those of the form $a^lb^mc^n$ with not all of $l,m,n$ distinct), and some that remain cancel with each other.

Instead of considering a line $\ell$ tangent to $\Gamma$ consider a variable line $\ell$ passing through a fixed point $P$ on $\Gamma.$ Let $Q$ be the anti-stiner point of $PH$ where $H$ is the orthocenter. Let $P_A,P_B,P_C$ be the reflections of $P$ over $BC,CA,AB$ respectively. Let $A'$ be the intersection of the reflections of $\ell$ over $AB$ and $AC,$ and define $B',C'$ similarly. First we have \begin{align*} \measuredangle P_BAP_C &= \measuredangle P_BAC+\measuredangle CAB+\measuredangle BAP_C \\ &= \measuredangle CAP+\measuredangle CAB+\measuredangle PAB \\ &= 2\measuredangle CAB \\ &= 2\measuredangle(CA,\ell)+2\measuredangle(\ell,AB) \\ &= \measuredangle(P_BA',\ell)+\measuredangle(\ell,A'P_C) \\ &= \measuredangle P_BA'P_C \end{align*}so $A,A',P_B,P_C$ are concyclic, and similarly for the other two circles. But now notice that taking $\ell$ to be line $PH$ makes $A'=B'=C'=Q,$ so these three circles all pass through $Q.$ Next notice that $P_A,P_B,P_C$ are collinear along the stiner line of $P,$ so $P_BP_C$ intersects $B'C'$ at $P_A,$ and we have that $Q$ lies on both $(P_AP_BC')$ and $(P_AP_CB')$ so it is the center of spiral similarity taking $P_CP_B$ to $B'C'.$ Thus $\measuredangle B'QC'=\measuredangle P_CQP_B=\measuredangle P_CA'P_B=\measuredangle B'A'C',$ so $Q$ lies on the circumcircle of $A'B'C'.$ Now let $\ell'$ be the tangent at $Q$ to the circumcircle of $A'B'C'.$ We have \begin{align*} \measuredangle(\ell',QP) &= \measuredangle(\ell',QA')+\measuredangle A'QP \\ &= \measuredangle QB'A'+\measuredangle A'QA+\measuredangle AQP \\ &= \measuredangle QB'P_C+\measuredangle A'P_CA+\measuredangle ABP \\ &= \measuredangle QBP_C+\measuredangle(PA,\ell)+\measuredangle ABP \\ &= \measuredangle QBA+\measuredangle ABP_C+\measuredangle(PA,\ell)+\measuredangle ABP \\ &= \measuredangle QPA+\measuredangle PBA+\measuredangle(PA,\ell)+\measuredangle ABP \\ &= \measuredangle QPA+\measuredangle(PA,\ell) \\ &= \measuredangle(QP,\ell) \end{align*}which implies $\ell'$ is the reflection of $\ell$ over the perpendicular bisector of $PQ.$ But now taking $\ell$ to be the tangent to $\Gamma$ at $P$ makes $\ell$ the tangent to $\Gamma$ at $Q,$ so we are done.

Wow, I've just done the biggest angle chasing of my life. Basicly you just add the reflections of $F$ (point of tangency) in $AB$, $BC$, $CA$ and your point of tangency is Miquel for three lines in the problem and line formed by added reflections. At this point you just start calculating angles untill you get the desired. I guess you just won't have enough time on the contest to do such a big calculation

My solution https://nttuan.org/2023/07/28/imo2011p6/.

One-liner solution?! Let $\omega$ be the circumcircle of the triangle determined by $\ell_A$, $\ell_B$ and $\ell_C$. Let $H$ be the orthocenter of $\triangle ABC$. We prove a generalization: for any line $\ell$ passing through a point $J$ on $\Gamma$, $\omega$ passes through the anti-Steiner point $J'$ of line $HJ$. Applying this on line $JJ$ gives us the result of the problem. Let $D$, $E$ and $F$ be the reflections of $J$ across the sides of the triangle. (So, $D$, $E$ and $F$ are collinear.) Since $J'$ is the concurrency point of the three reflections of $HJ$ across the sides of the triangle, it follows from reflections that \[\angle (\overline{HJ}, \ell) = \angle (\overline{J' D}, \ell_A) = \angle (\overline{J' E}, \ell_B) = \angle (\overline{J' F}, \ell_C).\]Let $D'$, $E'$ and $F'$ be the feet from $J'$ to $\ell_A$, $\ell_B$ and $\ell_C$. From the above angles, we have that $J'DEF \sim J'D'E'F'$. Since $D$, $E$ and $F$ are collinear, $D'$, $E'$ and $F'$ are collinear, so by the converse of Simson's theorem, $J'$ lies on $\omega$ as desired.

Let $A'$ be the $A$ antipode and define $B'$ and $C'$ similarly. Suppose that $\ell$ is tangent to $\Gamma$ at $P$ and denote the reflections of $P$ across $AA'$, $BB'$ and $CC'$ by $D$, $E$ and $F$. It's not hard to see that $EF$ is parallel to $\ell_a$ and the similar ones. Let $\ell_b$ meet $\ell_c$ and $\ell_a$ at $M$ and $P$ and let $\ell_c$ meet $\ell_a$ at $N$. By homothety, it suffices to prove that $DM$, $EN$ and $FP$ concur on $\Gamma$. We proceed with complex numbers. Let $\Gamma$ be the unit circle and let $p=1$. Clearly, $d=a^2$, $e=b^2$ and $f=c^2$. Now we compute $N=\ell_a\cap\ell_c$ using the intersection formula. Notice that $1\pm i$ lie on $\ell$ so we will reflect only these points. For completeness, the reflections of $1+i$ and $1-i$ across $BC$ are $b+c-bc-bci$ and $b+c-bc+bci$ and the other follows from symmetry. We get \begin{align*} n&=\frac{2i(\frac{1}{c}+\frac{1}{b}-2+c+b)(-2abi)-2i(\frac{1}{a}+\frac{1}{b}-2+a+b)(-2bci)}{\frac{2i}{bc}(-2abi)-\frac{2i}{ab}(-2bci)}\\ &=\frac{(\frac{1}{c}+\frac{1}{b}-2+c+b)ab-(\frac{1}{a}+\frac{1}{b}-2+a+b)bc}{\frac{a}{c}-\frac{c}{a}}\\ &=\frac{b(a-c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-2+b)}{\frac{(a-c)(a+c)}{ac}}\\ &=\frac{b(a+c+\frac{ac}{b}-2ac+abc)}{a+c}\\ &=b+\frac{ac}{a+c}(b-1)^2 \end{align*} It remains to compute $NE\cap\Gamma=X\neq N$ \begin{align*} x&=\frac{e-n}{e\overline{n}-1}=\frac{b^2-b-\frac{ac}{a+c}(b-1)^2}{b^2\left(\frac{1}{b}+\frac{(b-1)^2}{a+c}\right)-1}\\ &=\frac{b-\frac{ac}{a+c}(b-1)}{1+\frac{b-1}{a+c}}\\ &=\frac{ab+bc+ca-abc}{a+b+c-1} \end{align*} This is symmetric in $a$, $b$, $c$ so we are done. $\blacksquare$

First time using argument of lines (learned it from puffypundo), hope I did that right.

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Here is my outline to prove this classic problem. you can find prove of why $I$ (incenter of $A_{1}B_{1}C_{1}$) lies on $(ABC)$. after this you have a new problem: In a complete quadrilateral $ABCD$ and $D:=AB \cap CD$ and $E:=AD \cap BC$ name a quadrilateral using angle bisectors of $ABC$ and $BCD$ and $CDA$ and $DAB$ as $I_{1}I_{2}I_{3}I_{4}$. prove that $(I_{1}I_{2}I_{3}I_{4})$(you can find they're cyclic by angle chasing) touches $AB$ iff touches $(EDC)$!! It's ain't that hard; let $M$ be the Miquel point of the quadrilateral $ABCD$. it's an classic geometry theorem that if you invert from $M$ whose takes $A$ to $C$ and $B$ to $D$; it wont change $(I_{1}I_{2}I_{3}I_{4})$(the inverted of $I_{1}$ will be $I_{1}' := (I_{1}I_{2}I_{3}) \cap (I_{4}DC)$ that lies on the circle!. you can prove it by adding $K$ on $(I_{4}CD)$ so that $\triangle AI_{2}B \sim \triangle DKC$. let $O$ be the circumcenter of $(I_{1}I_{2}I_{3})$ then you can find $MI_{1}OI_{1}'K$ are cyclic. with that and using angles you can prove $I_{1}$ is inverted of $I_{1}'$ through $M$!!) and if we invert through $M$ it takes $AB$ to $(EDC)$. so problem is sloved!!

Let's complex bash this! i love [redacted] btw. oops this is like very similar to #86 sorry Denote complex numbers with lowercase. Fix $(ABC)$ as the unit circle, and rotate it such that $p=1$. Analogous to most other solutions, to save ourselves computation, we'll show that if $\triangle A_1B_1C_1$ is the triangle formed by $\ell_a$, $\ell_b$, and $\ell_c$ (in the obvious way), and $A_2$, $B_2$, $C_2$, points not $P$ on $(ABC)$ such $AP=AA_2$, etc. Then $\triangle A_1B_1C_1$ is homothetic to $\triangle A_2B_2C_2$ with centre of homothety on $(ABC)$. Say $Q$. Clearly, $A_2=a^2$, $B_2=b^2$, $C_2=c^2$. Now, $\ell$ is just $\operatorname{Re}(z)=1$. By reflection formula, the reflection of $P$ across $AB$ is \[\frac{(a-b)ab+b^2-a^2}{b-a}=a+b-ab\]Further, $1-i$ lies on $\ell$, hence its reflection over $P$ is \[\frac{(a^2b-ab^2)(1+i)+b^2-a^2}{b-a}=a+b-(1+i)ab\]We use the intersection formula: Let's try and intersect $\ell_b$ with $\ell_c$. Our points are $a+b-ab$, $a+b-ab-iab$, $a+c-ac$ and $a+c-ac-iac$. Use the general intersection formula. We first evaluate $(a+b-ab)\left(\frac{1}{a}+\frac{1}{b}-\frac{1}{ab}+\frac{i}{ab}\right)-\left(\frac{1}{a}+\frac{1}{b}-\frac{1}{ab}\right)(a+b-ab-iab)$. Noting the obvious cancellation, this simplifies to \[\frac{i}{ab}(a+b-ab)+iab\left(\frac{1}{a}+\frac{1}{b}-\frac{1}{ab}\right)=i\left(\frac{1}{b}+\frac{1}{a}+a+b-2\right)\]Hence, \begin{align*} \frac{-ac\left(\frac{1}{b}+\frac{1}{a}+a+b-2\right)+ab\left(\frac{1}{c}+\frac{1}{a}+a+c-2\right)}{\frac{i}{ab}(iac)+\frac{-i}{ac}(iab)}&=\frac{a\left(-c\left(\frac{1}{b}+\frac{1}{a}+a+b-2\right)+b\left(\frac{1}{c}+\frac{1}{a}+a+c-2\right)\right)}{\frac{b}{c}-\frac{c}{b}}\\ &=\frac{a\left(\frac{b-c}{a}+a(b-c)-2(b-c)+\frac{b}{c}-\frac{c}{b}\right)}{\frac{b}{c}-\frac{c}{b}}\\ &=a+\frac{bc(1+a^2-2a)}{b+c}\\ &=a+\frac{bc(a-1)^2}{b+c} \end{align*}Now, we use the chord formula with $a^2$. We'll show that $A_2A_1$, $B_2B_1$, $C_2C_1$ concur at $Q$ on $(ABC)$. \begin{align*}\frac{a^2-a-\frac{bc(a-1)^2}{b+c}}{a^2\left(\frac{1}{a}+\frac{\left(\frac{1}{a}-1\right)^2}{bc\left(\frac{1}{b}+\frac{1}{c}\right)}\right)-1}&=\frac{a^2-a-\frac{bc(a-1)^2}{b+c}}{\frac{a^2\left(\frac{1}{a}-1\right)^2+(a-1)(b+c)}{b+c}}\\ &=\frac{a(a-1)(b+c)-bc(a-1)^2}{(a-1)^2+(a-1)(b+c)}\\ &=\frac{ab+ac+bc-ab}{a+b+c-1} \end{align*}But this is clearly symmetric in $a$, $b$ and $c$. Thus, the concurrence holds true. To finish, we'll show that $A_1B_1$ is parallel to $A_2B_2$. Indeed, this is equivalent to showing that \[\frac{a-b+\frac{bc(a-1)^2}{b+c}-\frac{ac(b-1)^2}{a+c}}{a^2-b^2}\]is self conjugate. This is the same as \begin{align*}\frac{1}{a+b}+\frac{c\left(\frac{b(a-1)^2}{b+c}-\frac{a(b-1)^2}{a+c}\right)}{a^2-b^2}&=\frac{ab}{a+b}+\frac{a^2b^2\left(\frac{c(a-1)^2}{a^2(b+c)}-\frac{c(b-1)^2}{b^2(a+c)}\right)}{c(b^2-a^2)}\\ &=\frac{ab}{a+b}+\frac{\frac{b^2(a-1)^2}{b+c}-\frac{a^2(b-1)^2}{a+c}}{b^2-a^2} \end{align*}Rearranging, we want \[ab-1=\frac{b(a-1)^2-a(b-1)^2}{a-b}\]Trivial as \[a^2b-ab^2+b-a=a^2b-2ab+b-ab^2+2ab-a\]Upon rearranging. Hence $A_1B_1\parallel A_2B_2$, same for other lines. Hence, the triangles are indeed homothetic, and we are done. This concludes the problem. (Other people said it was trivial to show parallelism but imagine if I have a skill issue in angle chasing)