Excellent problem! bgn wrote: Three circles $\omega_1,\omega_2,\omega_3$ are tangent to line $l$ at points $A,B,C$ ($B$ lies between $A,C$) and $\omega_2$ is externally tangent to the other two. Let $X,Y$ be the intersection points of $\omega_2$ with the other common external tangent of $\omega_1,\omega_3$. The perpendicular line through $B$ to $l$ meets $\omega_2$ again at $Z$. Prove that the circle with diameter $AC$ touches $ZX,ZY$. Proposed by Iman Maghsoudi - Siamak Ahmadpour Draw tangents $\overline{ZX'}, \overline{ZY'}$ to $\odot(AC)$ with $X',Y'$ on $\omega_2$. Let $V,W$ be contact points of $\{\omega_1, \omega_2\}, \{\omega_2, \omega_3\}$ respectively. Let $P, Q$ be contact points of $\overline{XY}$ with $\omega_1, \omega_3$, respectively. Let $M$ be the mid-point of $\overline{AC}$ and $T=\overline{PV} \cap \overline{QW}$. Lemma 1. $\overline{XY}, \overline{VW}, \overline{AC}$ are concurrent at a point $S$. (Proof) Note that $A,C$ lie on $\overline{ZV}, \overline{ZW}$, for homothety reasons. Thus, $\angle AVB=\angle CWB=90^{\circ}$, so we have $$ZV \cdot ZA=ZB^2=ZW \cdot ZC$$ hence $A,C,V,W$ lie on a circle. Consequently, $\overline{VW}$ passes through the exsimilicenter of $\omega_1, \omega_3$, proving the lemma. $\blacksquare$ Lemma 2. $T$ lies on line $\overline{ZM}$. (Proof) Notice that $T$ lies on $\omega_2$; in fact it is the mid-point of arc $\overline{XY}$ not containing $Z$, due to homothety reasons. Consequently, $T$ has equal powers in $\omega_1, \omega_3$. However, $\overline{ZM}$ is the radical axis of $\omega_2, \omega_3$; so the lemma must hold. $\blacksquare$ As an immediate corollary, we see that $\overline{ZM}$ bisects the angle $XZY$, so $\overline{XY} \parallel \overline{X'Y'}$. Now for the grand finale: Lemma 3. $S$ lies on $\overline{X'Y'}$. (Proof) By Dual of Desrague's Involution Theorem applied on $\odot(AC)$ and point $Z$, we see that $(\overline{ZA}, \overline{ZC}), (\overline{ZB}, \overline{ZB}), (\overline{ZX'}, \overline{ZY'})$ form pairs of an involution. Project on $\omega_2$, so $(V,W), (X', Y'), (B,B)$ are pairs of an involution on $\omega_2$. Consequently, lines $\overline{AC}, \overline{VW}, \overline{X'Y'}$ concur. $\blacksquare$ Now it is clear as day that $\overline{XY}, \overline{X'Y'}$ coincide; just as we desired. $\blacksquare$

Here is the solution I found during the contest ( unfortunately i wasn't able to fully write it down due to a lack of time ) . Let $J$ be the intersection point of $AC$ and $XY$ . Let $O_1$ and $O_3$ be the centers of $\omega_1$ and $\omega_3$ respectively.Then $J , O_1 , O_3$ are collinear .Let $D= \omega_1 \cap \omega_2$ and $G =\omega_3 \cap \omega_2$ .Consider the inversion centered at $J$ that sends $\omega_1$ to $\omega_3$ .This inversion also sends $\omega_3$ to $\omega_1$ and $AC$ to $AC$. Then the inversion must send $\omega_2$ to a circle which is tangent to $\omega_1 , AC$ and $\omega_3$ at three different points . But this type of circle must be unique and $\omega_2$ itself is such a circle .So $\omega_2$ is fixed under this inversion .Then this inversion takes $B$ to $B$ , $D$ to $G$. So $J ,D ,G$ are collinear and $JB^2=JA \cdot JC$ . Let the common tangent of $\omega_1$ and $\omega_2$ meet $AB$ at $N$ . Then $NA = ND= NB$ . So $D$ lies in the circle with diameter $AB$ (from now we will write the circle with diameter $AB$ as $(AB)$ ). So $\angle ADB =90^\circ$ .So , $\angle ADB + \angle BDZ= 90^\circ + 90^\circ=180^\circ$ . Thus $A , D ,Z$ are collinear , in the same way $C ,G ,Z$ are collinear . $X'Z$ and $Y'Z$ be the tangent from $Z$ to $(AC)$ such that $\{X' , Y'\} \in \omega_2$ .$X'Z$ and $Y'Z$ touches $(AC)$ at $R , P$ respectively .Let $K =RP \cap AC$. By la hires theorem the polar of $K$ wrt $(AC)$ goes though $Z$ .So $ZB$ is polar of $K$ wrt $(AC)$. So $K$ is the inverse of $B$ wrt $(AC)$ .Now its well known that $(K ,B ;A, C )=-1$ . Let $K'$ be the reflection of $B$ wrt $J$ . Now as $JB^2=JA \cdot JC$ so inversion wrt $(K'B)$ takes $A$ to $C \Longrightarrow (K' , B ;A ,C)=-1\Longrightarrow K=K'$ . So $J$ is the midpoint of $KB$ . Now let us prove a lemma , Lemma : Let $A ,B , C ,D$ be four points in circle $\omega$ . Tangents to $\omega$ at $A$ & $D$ meet at $M$ ,tangents to $\omega$ at $B$ & $C$ meet at $N$ . Let $J=AB \cap CD$ . Then $J ,M ,N$ are collinear . [asy][asy] import graph; size(7 cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.2) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -33., xmax = -6, ymin = -2., ymax = 18.5; /* image dimensions */ pen ffdxqq = rgb(1.,0.8431372549019608,0.); pen ffxfqq = rgb(1.,0.4980392156862745,0.); /* draw figures */ draw(circle((-15.591406596940846,4.170602969551047), 5.838881922231399), linewidth(0.4) + blue); draw((-32.09037581295504,0.7047951396101702)--(-10.059940647473814,2.3009969812082036), linewidth(0.4) + red); draw((-15.464881863893629,10.008113876923124)--(-32.09037581295504,0.7047951396101702), linewidth(0.4) + red); draw((-20.346084727918672,16.96841178922098)--(-20.795591909873973,1.5231518688861532), linewidth(0.4) + ffdxqq); draw((-10.059940647473814,2.3009969812082036)--(-20.346084727918672,16.96841178922098), linewidth(0.4) + ffdxqq); draw((-20.632815068113167,7.116234095227272)--(-22.24083871628236,4.364123020944482), linewidth(0.4) + red); draw((-20.795591909873973,1.5231518688861532)--(-22.24083871628236,4.364123020944482), linewidth(0.4) + red); draw((-15.464881863893629,10.008113876923124)--(-7.513228800524669,9.835766313536771), linewidth(0.4) + red); draw((-10.059940647473814,2.3009969812082036)--(-7.513228800524669,9.835766313536771), linewidth(0.4) + red); draw((-32.09037581295504,0.7047951396101702)--(-7.513228800524669,9.835766313536771), linewidth(0.4) + linetype("4 4")); draw((-15.464881863893629,10.008113876923124)--(-20.795591909873973,1.5231518688861532), linewidth(0.4) + yellow); draw((-20.632815068113167,7.116234095227272)--(-10.059940647473814,2.3009969812082036), linewidth(0.4) + ffdxqq); draw((-20.632815068113167,7.116234095227272)--(-20.795591909873973,1.5231518688861532), linewidth(0.4) + ffxfqq); draw((-15.464881863893629,10.008113876923124)--(-10.059940647473814,2.3009969812082036), linewidth(0.4) + ffxfqq); /* dots and labels */ dot((-20.795591909873973,1.5231518688861532),linewidth(3.pt) + dotstyle); label("$A$", (-21.161718685573057,0.5293791919886705), NE * labelscalefactor); dot((-10.059940647473814,2.3009969812082036),linewidth(3.pt) + dotstyle); label("$B$", (-9.759484813801835,1.5754556939860207), NE * labelscalefactor); dot((-20.632815068113167,7.116234095227272),linewidth(3.pt) + dotstyle); label("$D$", (-21.632453111471868,7.381180280071315), NE * labelscalefactor); dot((-15.464881863893629,10.008113876923124),linewidth(3.pt) + dotstyle); label("$C$", (-15.251386449287974,10.310194485663896), NE * labelscalefactor); dot((-20.346084727918672,16.96841178922098),linewidth(3.pt) + dotstyle); label("$E$", (-20.115642183575698,17.266603223946277), NE * labelscalefactor); dot((-32.09037581295504,0.7047951396101702),linewidth(3.pt) + dotstyle); label("$J$", (-32.35473725694481,-0.3074820096092098), NE * labelscalefactor); dot((-22.24083871628236,4.364123020944482),linewidth(3.pt) + dotstyle); label("$M$", (-22.8877449138687,3.4583933975812515), NE * labelscalefactor); label("$N$", (-7.405812684307776,8.95029503306734), NE * labelscalefactor); dot((-18.027242665298672,5.929570143860911),linewidth(3.pt) + dotstyle); label("$F$", (-18.180400654880582,4.818292850177807), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy][/asy] Proof: Let $E=AD \cap BC$ , $F=AC \cap BD$ . Now by brocards theorem $JF$ is the polar of $E$ . By la hires theorem the polar of $E$ goes though $M,N$ , So $J , M ,F, N$ are collinear . [asy][asy]import graph; size(18.8cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(8); defaultpen(dps); /* default pen style */ real xmin = -33., xmax = -4., ymin = -11., ymax = 9.; /* image dimensions */ pen xfqqff = rgb(0.4980392156862745,0.,1.); pen ffxfqq = rgb(1.,0.4980392156862745,0.); pair C_2 = (-23.001762946342133,-3.026280956652035), A_2 = (-9.038448471151908,-3.035942661766724), D_2 = (-18.963717497650077,-3.0290750213731505), Z = (-18.956311477197165,7.674273028079407), R_2 = (-21.25877248998085,1.584073070684142), T = (-15.48936252473688,3.9303444172584294), K = (-32.579197490828996,-3.0196539950886834), J = (-25.771457494239606,-3.024364508230916), A = (-23.00176294634213,-3.026280956652034), B = (-18.96371749765007,-3.0290750213731497), C = (-9.038448471151908,-3.035942661766723), O_3 = (-9.0352642722373,1.5659335359059465), O_1 = (-23.00123588881407,-2.2645655116171963), D = (-22.497708218366977,-1.693014479697557), O = (-18.962422022444255,-1.1568249910872843), F = (-17.72478176926395,0.24801252900858514), G = (-13.623825214074284,1.9157665830543051), I = (-19.918700912776877,0.4527875517520047), H = (-18.961126547238454,0.7154250391985535), P = (-11.946932705322933,2.639234106568399), R = (-22.41617265779375,-0.23220332486901085), M = (-16.02010570874702,-3.0311118092093796); /* draw figures */ draw((xmin, -6.919349364977759e-4*xmin-3.0421966800356484)--(xmax, -6.919349364977759e-4*xmax-3.0421966800356484), linewidth(0.4)); /* line */ draw(circle(M, 6.981658908912739), linewidth(0.4) + linetype("4 4") + red); draw(circle((-20.982740221996103,-3.0276779890125924), 2.019023207673717), linewidth(0.4) + linetype("4 4")); draw(circle((-14.001082984400991,-3.0325088415699373), 4.962635701239022), linewidth(0.4) + linetype("4 4")); draw((xmin, 0.2742737242123991*xmin + 5.915963832572232)--(xmax, 0.2742737242123991*xmax + 5.915963832572232), linewidth(0.4)); /* line */ draw(circle((-18.96001448742362,2.322599003353128), 5.3516753058471975), linewidth(0.4)); draw(O--(-19.918700912776877,0.45278755175200514), linewidth(0.4)); draw(D_2--(-20.82925480831267,-1.0144971871690105), linewidth(0.4)); draw((-17.724781769263945,0.2480125290085855)--D_2, linewidth(0.4)); draw(circle((-23.00123588881407,-2.264565511617198), 0.7617156273795733), linewidth(0.4)); draw(circle((-9.035264272237299,1.565933535905946), 4.6018772993017745), linewidth(0.4)); draw(G--(-9.035264272237299,1.565933535905946), linewidth(0.4)); draw((-9.035264272237299,1.565933535905946)--A_2, linewidth(0.4)); draw((-22.49770821836698,-1.693014479697558)--(-23.00123588881407,-2.264565511617198), linewidth(0.4)); draw((-23.00123588881407,-2.264565511617198)--C_2, linewidth(0.4)); draw((-25.77145749423961,-3.0243645082309167)--(-9.035264272237299,1.565933535905946), linewidth(0.4) + blue); draw((-25.77145749423961,-3.0243645082309167)--G, linewidth(0.4) + xfqqff); draw((-20.82925480831267,-1.0144971871690105)--O, linewidth(0.4)); draw((-17.724781769263945,0.2480125290085855)--O, linewidth(0.4)); draw(Z--(-11.946932705322935,2.639234106568399), linewidth(0.4)); draw(Z--(-22.888511240035744,-1.3115916691255134), linewidth(0.4)); draw(D_2--(-22.888511240035744,-1.3115916691255134), linewidth(0.4) + ffxfqq); draw(D_2--(-13.889598943492828,4.03470729171564), linewidth(0.4) + ffxfqq); draw(D_2--(-19.918700912776877,0.45278755175200514), linewidth(0.4) + ffxfqq); draw(Z--C_2, linewidth(0.4) + blue); draw(Z--A_2, linewidth(0.4) + blue); draw(T--C_2, linewidth(0.4) + blue); draw(R_2--A_2, linewidth(0.4) + blue); draw((-13.889598943492828,4.03470729171564)--J, linewidth(0.4) + green); draw(Z--M, linewidth(0.4) + yellow); draw(Z--D_2, linewidth(0.4)); /* dots and labels */ dot(C_2,linewidth(3.pt) + red); dot(A_2,linewidth(3.pt) + red); dot(D_2,linewidth(3.pt) + red); dot(Z,linewidth(3.pt) + red); label("$Z$", (-18.72764855770153,7.827165226927332), NE * labelscalefactor,red); label("$w_2$", (-22.318067521242696,7.350885160335141), NE * labelscalefactor); label("$w_1$", (-24.406372428608478,-1.6984361049165062), NE * labelscalefactor); label("$w_3$", (-9.75160114884862,6.654783524546552), NE * labelscalefactor); dot(R_2,linewidth(3.pt) + blue); dot(T,linewidth(3.pt) + blue); dot(K,linewidth(2.pt) + blue); label("$K$", (-32.7229551298722,-3.640193299484673), NE * labelscalefactor,blue); dot(J,linewidth(2.pt) + blue); label("$J$", (-25.90848648478386,-3.640193299484673), NE * labelscalefactor,blue); dot(A,linewidth(2.pt) + blue); label("$A$", (-23.453812295424086,-3.530282514886475), NE * labelscalefactor,blue); dot(B,linewidth(2.pt) + blue); label("$B$", (-18.83755934229973,-3.5669194430858746), NE * labelscalefactor,blue); dot(C,linewidth(2.pt) + blue); label("$C$", (-9.531779579652222,-3.6768302276840727), NE * labelscalefactor,blue); dot(O_3,linewidth(2.pt) + blue); label("$O_3$", (-8.79904101566423,1.0859704382378466), NE * labelscalefactor,blue); dot(O_1,linewidth(2.pt) + blue); label("$O_1$", (-23.673633864620484,-2.541085453502692), NE * labelscalefactor,blue); dot(D,linewidth(2.pt) + blue); label("$D$", (-23.01416915703129,-1.881620745913503), NE * labelscalefactor,blue); dot((-20.82925480831267,-1.01449718716901),linewidth(2.pt) + blue); label("$E$", (-20.99913810606431,-1.6617991767171068), NE * labelscalefactor,blue); dot(O,linewidth(2.pt) + blue); label("$O$", (-19.350476337091326,-1.7350730331159057), NE * labelscalefactor,blue); dot(F,linewidth(2.pt) + blue); label("$F$", (-17.665177639918944,-0.41614361793752797), NE * labelscalefactor,blue); dot(G,linewidth(2.pt) + blue); label("$G$", (-14.257943317374776,1.3057920074342428), NE * labelscalefactor,blue); dot((-22.88851124003574,-1.311591669125514),linewidth(2.pt) + blue); label("$X'$", (-23.527086151822886,-1.3687037511219118), NE * labelscalefactor,blue); dot(I,linewidth(2.pt) + blue); label("$I$", (-20.04657797287992,0.7562380844432521), NE * labelscalefactor,blue); dot(H,linewidth(2.pt) + blue); label("$H$", (-19,.35), NE * labelscalefactor,blue); dot(P,linewidth(2.pt) + blue); label("$P$", (-11.949816840812598,2.881179920008416), NE * labelscalefactor,blue); dot(R,linewidth(2.pt) + blue); label("$R$", (-22.97753222883189,-0.08641126414293357), NE * labelscalefactor,blue); dot((-13.889598943492828,4.03470729171564),linewidth(2.pt) + blue); label("$Y'$", (-13.74502632258318,4.053561622389196), NE * labelscalefactor,blue); dot(M,linewidth(2.pt) + blue); label("$M$", (-16.163063583743558,-3.6768302276840727), NE * labelscalefactor,blue); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy][/asy] Let $DG$ meet $(AB)$ & $(BC)$ again at $E$ & $F$ respectively .Let $ZB \cap O_1O_2 =O$ . Now $O_1A , O_1D , OB$ are tangent to $(AB)$ , so by our lemma $OE$ is tangent to $(AB)$ . In the same way $OF$ is tangent to $(BC) \Longrightarrow OB =OE=OF \Longrightarrow O$ is the circumcenter of $\triangle EBF$. Let $M$ be the midpoint of $AC$ .$\{R , B , P\} \in (ZM)$ . So $\angle EAB =\angle EDB= \angle BZC$. $BZ \perp AC\Longrightarrow AE \perp ZC$. In the same way $CF \perp ZA$ So $CF$ & $AZ$ meet at $(AC)$ . $AE ,CF ,ZB$ concur at the orthocenter $H$ of $\triangle ZAC$ .The inversion with center $Z$ and and radius $ZR$ takes $H$ to $B$ . $ZRBMP$ is cyclic $\Longrightarrow H \in RP$ .Let $I \in RP$ such that $BI \perp RP$ .By simson's theorem $I \in X'Y'$ . $\angle X'IB = \angle X'RB=\angle ZMB =\angle IBK$ . So $X'IY'$ goes though the midpoint $J$ of $KB$ . Now $\{ E , I , F \} \in (HB) \Longrightarrow OI =OB$ . Now as $JI=JB$ so the line $JX'IY'$ is the reflection of $AC$ wrt the line $O_1O_3$. So $X'Y'$ is the other common external tangent of $\omega_1$ and $\omega_3$ . $\blacksquare$

Here is a simpler solution. Let $O$ be the midpoint of $AC$, and $U,V$ the intersection of $ZX, ZY$ with $l$ respectively. We apply an inversion with center at $Z$ and radius $ZB$. The inversion maps $\omega_2, X,Y$ to $l, U, V$ respectively, and $\omega_1,\omega_3$ to themselves. Also the other common external tangent of $\omega_1,\omega_3$ maps onto a cirle that passes $Z, U, V$ and tangent to both $\omega_1$ and $\omega_3$. Now, by casey’s theorem for $U, V, Z, \omega_1$, we have $UV\cdot BZ+UA\cdot VZ=UZ\cdot VA$. Similarly, we have $UV\cdot BZ+UZ\cdot VC=UC\cdot VZ$ for $U, V, \omega_3, Z$. Then from these two equations we got $\frac{UZ}{VZ}=\frac{UO}{VO}$, which means $O$ lies on the bisector of $\angle UZV$, hence the distances from $O$ to $UZ$, $VZ$ are equal. Furthermore, we also got $UV\cdot BZ=AO\cdot (UZ+VZ)$. Finally, by the area relation we can easily deduce the distance from $O$ to $UZ$ is equal to $AO$, from which the claim will be obtained.

Brilliant problem! Here's my solution.

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anantmudgal09 wrote: Excellent problem! bgn wrote: Three circles $\omega_1,\omega_2,\omega_3$ are tangent to line $l$ at points $A,B,C$ ($B$ lies between $A,C$) and $\omega_2$ is externally tangent to the other two. Let $X,Y$ be the intersection points of $\omega_2$ with the other common external tangent of $\omega_1,\omega_3$. The perpendicular line through $B$ to $l$ meets $\omega_2$ again at $Z$. Prove that the circle with diameter $AC$ touches $ZX,ZY$. Proposed by Iman Maghsoudi - Siamak Ahmadpour Draw tangents $\overline{ZX'}, \overline{ZY'}$ to $\odot(AC)$ with $X',Y'$ on $\omega_2$. Let $V,W$ be contact points of $\{\omega_1, \omega_2\}, \{\omega_2, \omega_3\}$ respectively. Let $P, Q$ be contact points of $\overline{XY}$ with $\omega_1, \omega_3$, respectively. Let $M$ be the mid-point of $\overline{AC}$ and $T=\overline{PV} \cap \overline{QW}$. Lemma 1. $\overline{XY}, \overline{VW}, \overline{AC}$ are concurrent at a point $S$. (Proof) Note that $A,C$ lie on $\overline{ZV}, \overline{ZW}$, for homothety reasons. Thus, $\angle AVB=\angle CWB=90^{\circ}$, so we have $$ZV \cdot ZA=ZB^2=ZW \cdot ZC$$ hence $A,C,V,W$ lie on a circle. Consequently, $\overline{VW}$ passes through the exsimilicenter of $\omega_1, \omega_3$, proving the lemma. $\blacksquare$ Lemma 2. $T$ lies on line $\overline{ZM}$. (Proof) Notice that $T$ lies on $\omega_2$; in fact it is the mid-point of arc $\overline{XY}$ not containing $Z$, due to homothety reasons. Consequently, $T$ has equal powers in $\omega_1, \omega_3$. However, $\overline{ZM}$ is the radical axis of $\omega_2, \omega_3$; so the lemma must hold. $\blacksquare$ As an immediate corollary, we see that $\overline{ZM}$ bisects the angle $XZY$, so $\overline{XY} \parallel \overline{X'Y'}$. Now for the grand finale: Lemma 3. $S$ lies on $\overline{X'Y'}$. (Proof) By Dual of Desrague's Involution Theorem applied on $\odot(AC)$ and point $Z$, we see that $(\overline{ZA}, \overline{ZC}), (\overline{ZB}, \overline{ZB}), (\overline{ZX'}, \overline{ZY'})$ form pairs of an involution. Project on $\omega_2$, so $(V,W), (X', Y'), (B,B)$ are pairs of an involution on $\omega_2$. Consequently, lines $\overline{AC}, \overline{VW}, \overline{X'Y'}$ concur. $\blacksquare$ Now it is clear as day that $\overline{XY}, \overline{X'Y'}$ coincide; just as we desired. $\blacksquare$ I think we also have to consider the following two things in this proof: Prove that $Z$ does not lie inside $\odot(AC)$ (otherwise $X',Y'$ don't exist). Consider the special case when $S$ lies at infinity separately (otherwise Lemma 3 is not enough to imply $\{X,Y\} = \{X',Y'\}$). There might be ways to avoid considering these. Can someone please tell if there are ways to modify the main proof so that the above two things are also handled? Either way, I will put below a proof to handle these (in a different way though): To prove $Z$ doesn't lie inside $\odot(AC)$ : In fact $Z$ must strictly lie outside $\odot(AC)$. Let $\ell_2$ denote line $XY$. We will be invoking the fact that $\ell_2$ intersects $\omega_2$ at two distinct points. Let $O_i$ be center of $\omega_i$. Note that $$\text{dist}(O_1,\ell) = \text{dist}(O_1,\ell_2) ~,~ \text{dist}(O_3,\ell) = \text{dist}(O_3,\ell_2) ~~ \text{but} ~~ \text{dist}(O_2,\ell) > \text{dist}(O_2,\ell_2)$$ where in the end we use $\ell_2$ intersects $\omega_2$ at two distinct points. It follows that $O_2$ lies on the opposite side of line $O_1O_3$ as $B$. Since $U,V$ lie on segments $O_2O_1,O_2O_3$ respectively, so points $O_2,B$ lie on opposite of line $UV$. This means $$\angle UBV > 90^\circ$$ Hence it follows $$\angle AZC = \angle UZV = 180^\circ - \angle UBV < 90^\circ$$ thus $Z$ lies outside $\odot(AC)$, as desired. $\square$ Remark: Using a similar proof as above, one may not that if $\ell_2$ is tangent to $\omega_2$ (i.e. when $X \equiv Y$), then $Z \in \odot(AC)$. To resolve the case when $S$ is at infinity : Then $\ell \parallel \ell_2$. Since tangent to $\omega_2$ at $B$ is parallel to $\ell_2$, so there exists a (positive) inversion$\Phi$ at $B$ swapping $\omega_2$ and $\ell_2$. Note $\Phi(\omega_1),\Phi(\omega_3)$ are circles tangent to all three of $$\Phi(\ell) = \ell ~,~ \Phi(\omega_2) = \ell_2 ~,~ \Phi(\ell_2) = \omega_2$$ since $\Phi$ was positive, it follows $\omega_1,\omega_3$ must be fixed under $\Phi$. This means $A,C$ are fixed under $\Phi$, and thus radius of $\Phi$ must equal $BA,BC$. Next note that $X,Y$ are also fixed under $\Phi$ (since $X,Y$ are intersection points of $\omega_2,\ell_2$ which swap under $\Phi$). So radius of $\Phi$ must also equal $BX,BY$. We obtain $$BA = BC = BX = BY = \text{radius of } \Phi$$ It follows $X,Y \in \odot(AC)$ and $B$ is center of $\odot(AC)$. Since $\angle ZXB = \angle ZYB = 90^\circ$, it follows $ZX,ZY$ are tangents to $\odot(AC)$, as desired. $\square$