Let $O,O_1 $ and $O_2$ be the centers of $\omega , \omega_1$ and $\omega_2$. Obviously $O_1X_1O_2Y_2$ is a rhombus. In this figure there are two homotheties: $H_1$, which sends $\omega_1$ to $\omega$, and $H_2$, which sends $\omega_2$ to $\omega$. Let $Y_1$ and $Y_2$ denote $H_1(X_1)$ and $H_2(X_2)$. Obviously these two points are situated on the same side of $T_1T_2$. Due to the homothety properties we have $OY_1\parallel O_1X_1\parallel O_2X_2\parallel OY_2$. Now we may conclude that $Y_1\equiv Y_2$.

Apply an inversion $\Psi$ with centre $X_1$, with an arbitrary radius and denote by $\Psi(X)$ the image of $X$ about the inversion $\Psi$.The fact that the circles $\omega_1 $ and $\omega_2$ have the same radius implies that, after inversion, the images $\Psi(\omega_1),\Psi(\omega_2)$( which are lines, because these two circles pass through the pole) form an angle for which the image of the radical axis $\Psi(X_1X_2)=X_1\Psi(X_2)$ is an angle bisector. Now $\Psi(\omega)$ is a circle which is tangent to $\Psi(\omega_1),\Psi(\omega_2)$ and so from $\Psi(X_2)$ are drawn two equal tangents, which are $\Psi(X_2)\Psi(T_2),\Psi(X_2)\Psi(T_1)$, so the latter are equal. Proving the required is equivalent to proving that the intersection of $\Psi(X_1T_1)=X_1\Psi(T_1) $ and $\Psi(\omega)$ lies on the circle $\odot(X_1\Psi(X_2)\Psi(T_2))$.Denote the intersection mentioned above with $L$.We have $$m(\widehat{\Psi(\omega_1),\Psi(\omega_2)})=$$$=$ $$m( \widehat{\Psi(X_2) \Psi(T_2) \Psi(T_1)} )+m(\widehat{ \Psi(X_2) \Psi(T_1) \Psi(T_2)})$$, so$$ \frac{1}{2} \big( m(\widehat{\Psi(\omega_1),\Psi(\omega_2)}) \big )=$$$=$ $$ \frac{1}{2} \big ( m(\widehat{\Psi(X_2)\Psi(T_2)\Psi(T_1))}+m(\widehat{\Psi(X_2)\Psi(T_1)\Psi(T_2) } \big ) $$. From here we have $$m(\widehat{X_1 \Psi(X_2) \Psi(T_2)} )=m(\widehat{X_1 L \Psi(T_2)})$$The problem is solved.

Hmm, so basically by homothety centred about $T_2$ we only need to prove that lines $X_1T_1$ and $XX_2$ are parallel where $X$ is the second intersection of ray $T_2T_1$ with circle $\omega_1$. For this, we see that proving $T_2T_1$ bisects $X_1X_2$ is sufficient. (Since, then, rays $X_1T_2$ and $X_2X$ will be symmetric with respect to the midpoint of segment $X_1X_2$.) Now, we let $O$ denote the centre of $\omega_2$ and apply inversion about $T_2$ of radius $\sqrt{T_2X_1.T_2X_2}$ followed by reflection in the bisector of angle $X_1T_2X_2$. Let $T$ be the intersection of the tangents to $\omega_2$ at points $X_1,X_2$. Now, it is clear that circle $\omega_1$ is mapped onto the circumcircle of triangle $X_1OX_2$ and circle $\omega$ is mapped to a line tangent to this new circle parallel to $X_1X_2$. Thus, point $T_2$ being the tangency point, is mapped onto point $T$ and we see that $T_2T$ is a symmedian. Thus, line $T_2T_1$ is a median in triangle $X_1T_2X_2$. Our proof is complete now, by the proposition made in the second paragraph.

Consider the composition of homotheties \[ \omega_1 \xrightarrow{T_1} \omega \xrightarrow{T_2} \omega_2. \]This is a negative homothety, and since $\omega_1$ and $\omega_2$ have equal radius, it follows that the composition is just reflection about the midpoint of $\overline{X_1X_2}$. In particular, it sends $X_1$ to $X_2$, and thus $\overline{X_1T_1} \cap \overline{X_2T_2}$ is just the image of $X_1$ under the first homothety.

Call $O_1, O_2$ and $O$ the centers of $w_1, w_2$ and $w$ respectively. And $A \equiv X_1T_2 \cap w$ $B \equiv X_2T_2 \cap w$. Notice that $AB \parallel X_1X_2$. Hence $\triangle T_2AB \sim \triangle T_2X_1X_2$ and $AO \parallel X_1O_2$ which implies $\triangle X_2X_1O_2 \sim \triangle BAO$. Hence: $\frac{OB}{O_2X_2} = \frac{AB}{X_1X_2} = \frac{T_2B}{T_2X_2}$. Hence $OB \parallel O_2X_2$. Since $O_1X_1=O_1X_2=O_2X_1=O_2X_2$ we get $O_1X_1O_2X_2$ is a parallelogram. Hence $X_1O_1 \parallel X_2O_2$ Since $O_1, T_1, O$ are collinear and $O_1X_1 \parallel OB$ by Reim's theorem, we get $X_1, T_1$ and $B$ are collinear. Done!

Hello. My solution,without transformations. [asy][asy]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 = -5.3891428373224155, xmax = 10.896539640418274, ymin = -4.263644286871532, ymax = 4.950623430797529; /* image dimensions */ /* draw figures */ draw(circle((0.,0.), 2.)); draw(circle((3.024023033799519,0.), 2.)); draw((0.,0.)--(3.024023033799519,0.)); draw((1.5120115168997594,1.3091299296717986)--(1.5120115168997594,-1.3091299296717986)); draw((1.5120115168997594,0.)--(3.9000895026873623,1.7979175570894195), linetype("2 2")); draw((3.9000895026873623,1.7979175570894195)--(3.024023033799519,0.)); draw(circle((3.306000063645372,0.578690625266576), 1.3562656719328152)); draw((3.9000895026873623,1.7979175570894195)--(1.5120115168997594,-1.3091299296717986)); draw((1.1652346831652585,4.942650325192146)--(2.6490780130185403,-3.534394107098305)); label("y",(1.666965830194931,2.792464380645806),SE*labelscalefactor); label("y'",(2.416964365354042,-1.3860988866692336),SE*labelscalefactor); draw((0.,0.)--(3.306000063645372,0.578690625266576)); draw((1.970046704760118,0.3448419653461572)--(1.5120115168997594,1.3091299296717986)); draw((1.970046704760118,0.3448419653461572)--(1.5120115168997594,-1.3091299296717986)); /* dots and labels */ dot((0.,0.),linewidth(3.pt) + dotstyle); label("$O_1$", (-0.32282620185985217,0.2516530166373936), NE * labelscalefactor); label("$\omega_1$", (-1.0116003668018925,1.3996099582074593), NE * labelscalefactor); dot((3.024023033799519,0.),linewidth(3.pt) + dotstyle); label("$O_2$", (3.136350715404617,-0.17691757488209764), NE * labelscalefactor); label("$\omega_2$", (2.432270457908309,2.042465845486696), NE * labelscalefactor); dot((1.5120115168997594,1.3091299296717986),linewidth(3.pt) + dotstyle); label("$X_1$", (1.3579866269894526,1.5985891614129375), NE * labelscalefactor); dot((1.5120115168997594,-1.3091299296717986),linewidth(3.pt) + dotstyle); label("$X_2$", (1.42206834932665,-1.6616085526460493), NE * labelscalefactor); dot((3.9000895026873623,1.7979175570894195),linewidth(3.pt) + dotstyle); label("$T_2$", (3.9628797133350653,1.8894049199440208), NE * labelscalefactor); dot((1.5120115168997594,0.),linewidth(3.pt) + dotstyle); label("$M$", (1.207783053566904,-0.2381419450991678), NE * labelscalefactor); dot((1.970046704760118,0.3448419653461572),linewidth(3.pt) + dotstyle); label("$T_1$", (2.04961814405162,0.6802236081568848), NE * labelscalefactor); dot((3.306000063645372,0.578690625266576),linewidth(3.pt) + dotstyle); label("$O$", (3.549615214369841,0.5577748677227444), NE * labelscalefactor); dot((2.2806554056762685,-0.3090733665902651),linewidth(3.pt) + dotstyle); label("$S$", (2.2945156249199012,-0.6973247217271941), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy][/asy] Let $O_1,O_2,O$ be the centers of $(\omega_1),(\omega_2),(\omega )$ respectively,and $M$ the midpoint of $O_1O_2$.Since$(\omega _1),(\omega _2)$ are equal,$M$ lies on $X_1X_2$ and $X_1X_2$ is the perpendicular bisector of $O_1O_2$. Lemma:$T_1,T_2,M$ are collinear. Proof:Let $r$ be the radius of $(\omega)$ and $R$ that of $(\omega_1),(\omega_2)$.Obviously,$T_1,O_1,O$ are collinear,and $T_2,O_2,)$ are collinear,since $(\omega)$ is tangent to $(\omega_1),(\omega_2)$. We have $\frac{T_2O}{T_2O_2}\cdot \frac{O_2M}{O_1M}\cdot \frac{O_1T_1}{OT_1}=\frac{r}{R}\cdot \frac{R}{r}=1$,and the converse of Menelaus's theorem gives the desired result. The theorem of medians gives $OM^2=\frac{2O_2^2+2O_1O^2-O_1O_2^2}{4}=\frac{2(R+r)^2+2(R-r)^2-4O_1M^2}{4}=R^2+r^2-O_1M^2$. Hence, PoP gives $MT_1\cdot MT_2=OM^2-r^2=R^2-O_1M^2$.Also,$MX_1^2=O_1X^2-O_1M^2=R^2-O_1M^2$. Thus $MX_2^2=MT_1\cdot MT_2$,which gives that $MX_2$ is tangent to the circumcircle of $\triangle{T_1T_2X_2}$. It follows that $\angle{T_1T_2X_2}=\angle{T_1X_2X_1} \ (1)$. Let $\overline{yT_1y'}$ be the common tangent of $(\omega),(\omega _1)$ and $S\equiv T_2X_2\cap (\omega)$. We have $\angle{X_1X_2T_1}=\angle{X_1T_1y}$ and $\angle{T_1T_2S}=\angle{ST_1y'}$. Now $(1)$ gives $\angle{X_1T_1y}=\angle{ST_1y'}\Rightarrow X_1,T_1,S$ collinear.

Let $P$ be the radical center of $\omega,\omega_1,\omega_2$. Denote by $T$ the reflection of $T_1$ across line $X_1X_2$ (obviously $T \in \omega_2$). Let $X = X_1T_1 \cap \omega$, by simple angle chasing we get that $\angle T_1T_2X = \angle X_1T_2T$ so we need to show that $T_1T_2$ and $TT_2$ are isogonal lines in $\triangle X_2T_2X_1$. Using the fact that $PT$ is tangent to $\omega_2$, we get that $PT = PT_1 = PT_2$, so $P$ is the circumcenter of $\triangle TT_1T_2$, in particular $\angle X_1PT = \angle TT_2T_1$. Using some more angle chasing and since $\angle TT_2X_2 = \angle TX_1X_2$, we are done.

Consider the problem from the perspective of triangle $T_2X_1X_2$. By an inversion about $X_1$, it is easy to show that $T_2T_1$ passes through the midpoint of $X_1X_2$. Let $T_3$ be the second intersection of $T_1T_2$ and $\omega_1$. Since the circles are congruent, $X_1T_2X_2T_3$ is a parallelogram. Now, the homothety centered at $T_1$ sending $\omega_1$ to $\omega$ sends $T_3X_2$ to a line parallel to it. $T_3$ goes to $T_2$, and $X_2$ goes to $X_2T_2\cap \omega$, so $X_2T_2\cap\omega$ lies on $X_1T_1$ as desired.

Invert about $X_1$ with the radius of the circles. The problem becomes the following: Let $\omega \equiv \odot(X_1,R)$ be a circle and let $\ell_1, \ell_2$ be lines intersecting inside $\omega$ such that the internal angle bisector of the lines passes through $X_1$. Define $\gamma$ is a circle tangent to $\ell_1$ and $\ell_2$ such that the tangency point $T_1$ on $\ell_1$ is on the opposite side of $\ell_2$ as $X_1$ and the tangency point $T_2$ on $\ell_2$ is on the same side of $\ell_1$ as $X_1$. If $X_2\equiv\ell_1\cap\ell_2$, then $\odot(X_1X_2T_2)$ and line $X_1T_1$ intersect on $\gamma$. To prove this, let $K'\equiv X_1T_1\cap \gamma$. Then $$\angle X_2T_2K'=\angle T_2T_1K'=\angle T_2T_1X_1=\angle X_2X_1T_1=\angle X_2X_1K'$$as $X_1X_2 \parallel T_1T_2$ since they are clearly both perpendicular to the external angle bisector of $\ell_1,\ell_2$. Hence, $K\equiv K'$ and we're done.

By de monge's theorem,$T_1,T_2$ and midpoint of $X_1X_2,$(the negative centre of homothety mapping $\omega_1$ to $\omega_2$)say $M$ are collinear. Now consider the composition of homotheties \[ \omega_1 \xrightarrow{T_1} \omega \xrightarrow{T_2} \omega_2 \xrightarrow{M} \omega_1\]Since this composition has dilation factor $1$ and $\omega_1$ ie fixed, we have that the entire plane is fixed under this transformation. So if $X_1$ was mapped to a point say $X$ on $\omega$ by $\omega_1 \xrightarrow{T_1} \omega $ then $\omega \xrightarrow{T_2} \omega_2$ take it to $X_2$, so that after $\omega_2 \xrightarrow{M} \omega_1$ takes it to $X_1$. So $X$ is the required point lying on $\omega$. Done

Nice problem, I have seen general problem and extension as following Problem. Let two circles $(K),(L)$ intersect at $A,B$. $I$ is insimilicenter of $(K),(L)$. $BI$ cuts $(K),(L)$ again at $M,Q$. $AI$ cuts $(K),(L)$ again at $N,R$. A circle $(\omega)$ is tangent to $(K)$ externally at $S$ and is tangent to $(L)$ internally at $T$. a) Prove that $NS$ and $AT$ intersect at $E$ on $(\omega)$. $MS$ and $BT$ intersect at $F$ on $(\omega)$. $QT$ and $BS$ intersect at $G$ on $(\omega)$. $RT$ and $AS$ intersect at $H$ on $(\omega)$. b) Prove that $FH,EG$ and $AB$ are concurrent.

Wait, is my solution valid?

Let tangents to $\omega$ at $T_1$ and $T_2$ meet at $X$. Let $\angle X_1T_2X_2 = \theta, \angle X_1T_1X = \angle X_1X_2T_1 = \beta$, and $\angle X_1T_2X = \angle X_1X_2T_2 = \gamma$, where we used some tangency formulas. Hence, $\angle T_1X_2T_2 = \gamma - \beta$ Because of same radii, $\angle X_1T_1X_2 = 180 - \theta$. Thus, in quadrilateral $XT_1X_2T_2$, $180 - \theta + \beta = \angle T_1XT_2 + \gamma + \theta + \gamma - \beta$ i.e. $\angle XT_1T_2 = \angle T_1T_2X = \frac{180 - \angle T_1XT_2}{2} = \theta + \gamma - \beta$, so $\angle T_1T_2X_2 = \gamma + \theta - (\gamma + \theta - \beta) = \beta$ and $\angle X_1T_1T_2 = \beta + \theta + \gamma - \beta = \theta + \gamma$ i.e. if $P$ is the intersection of lines $X_1T_1$ and $X_2T_2$, then $\angle T_1PT_2 = \theta + \gamma - \beta = \angle T_1T_2X$, showing that $XT_2$ is tangent to the circumcircle of $T_1T_2P$ i.e. $P$ lies on $\omega$. Does this work? Thanks!

Problem authors: me + Charles Leytem

Cody is my idol, he is like god for me!!

Please, please. I'm just a man.

seems like homothety is an apparent part in elemgeo The homothety taking w_1->w->w_2 is the same as reflecting over the midpoint of X_1X_2 since the homothety is negative and equal radii; in particular, if X_0=X_1T_1\cap w, X_1->X_0 due to being collinear and on their respective circles, from which it follows that homothety at T_2 should take X_0->X_2 (because our w_1->w->w_2 takes X_1->X_0->X_2), whence T_2X_0X_2 is indeed a line. $\blacksquare$

First bash sol We complex bash. Let $\omega_1$ be the unit circle. Orient the circle such that $T_1T_2$ is parallel to the imaginary axis. Let $M$ be the midpoint of $T_1T_2$, let $A$ and $B$ be the respective centers of the circles. Now computing coordinates we have $a = 0$, $T_1 = t$ and $T_2 = \frac{1}{t}$. Also $m = \frac{t^2 + 1}{2t}$. Clearly $B$ is the reflection of $A$ over $M$ so $b = \frac{t^2+1}{t}$. Now note that it follows form simple angle chasing that $X_1X_2$ is a diameter of $\omega$ so it suffices to show $T_1X_1 \perp T_2X_2$. Note that $x_1 = 1$ and $x_2$ = $\frac{t^2+1}{t} + 1$. Now it suffices to show, \begin{align*} \frac{t - 1}{\frac{1}{t} - \frac{t^2+t+1}{t}} &= - \overline{\left( \frac{t - 1}{\frac{1}{t} - \frac{t^2+t+1}{t}} \right)}\\ &= -\overline{\left(\frac{t^2 - t}{-t^2 - t} \right)}\\ &= \frac{1 - t}{1+t} \end{align*}It is also easy to see the LHS simplifies to exactly this, so we are done.

Consider the following homotheties: \[h_1(\omega_1)\overset{T_1}=\omega,\]\[h_2(\omega)\overset{T_2}=\omega_2.\] The homothety of $h_2(h_1(\omega_1))=\omega_2,$ so this center must lie on the line containing the two centers of $\omega_1,\omega_2$. Note that this homothety has a factor of $-1$. Therefore, the center is the midpoint of $X_1X_2$. Let us say: \[h_1(X_2)=K,\]\[h_2(K)=P.\]The center of the homothety from $X_2$ to $P$ is the midpoint of $X_1,X_2$, with a factor of $-1$. Thus, $P=X_1$. As a result, $T_1X_2$ and $T_2X_1$ concur at $K$ on $\omega$.

Consider the composition of homotheties \[ \omega_1 \xrightarrow{T_1} \omega \xrightarrow{T_2} \omega_2. \] This is simply a negative homothety, and since $\omega_1$ and $\omega_2$ have equal radii, the composition is just a reflection about the midpoint of $\overline{X_1X_2}$. In particular, the intersection point happens to be the image of $X_1$ over the first homothety. $\square$

??? Define functions $A(X)$, and $B(X)$, as homotheties that transform, $\omega_{1}$ to $\omega$, and $\omega$ to $\omega_{2}$. Consider the composition $B(A(X))$, this just reflects $X$ over the midpoint of $X_{1}X_{2}$, now $A(X_{1})$ is the point where $X_{1}T_{1}$ intersects $\omega$. Now $B(A(X_{1}))=X_{2}$ so the point's $X_{2}$, $A(X_{1})$, and $T_{2}$ are collinear. So $T_{2}X_{2} \cap X_{1}T_{1}=A(X_{2})$ which lies on $\omega$ finishing.

Using phantom points, define $A_1 = X_1T_1 \cap \omega$ and $A_2 = X_2T_2 \cap \omega$, as well as $O$, $O_1$, and $O_2$ as the centers of $\omega$, $\omega_1$, and $\omega_2$ respectively. Homotheties tell us \[OA_1 \parallel O_1X_1 \parallel O_2X_2 \parallel OA_2 \implies A_1 = A_2. \quad \blacksquare\]

$\overline{X_1T_1} \cap \omega = Y$. Let $\Psi_1$ be the homothety sending $\omega_1 \to \omega$ with center $T_1$, and let $\Psi_2$ be the homothety sending $\omega \to \omega_2$ with center $T_2$. $\Psi_3 = \Psi_2(\Psi_1(P))$ is a reflection of $P$ over the midpoint of $\overline{X_1X_2}$. Notice that $\Psi_1(X_1) = Y$. Since $\Psi_3(X_1) = X_2$, it follows that $\Psi_3(X_1) = \Psi_2(Y) = X_2$ so $T_2$, $Y$, and $X_2$ are collinear.

Consider the negative homothety $\omega_1 \to \omega$ and the positive homothety $\omega \to \omega_2$. The composition of these two homotheties is just a negative homothety sending $\omega_1$ to $\omega_2$, or in other words, a reflection through the midpoint of segment $X_1 X_2$. Now suppose ray $X_1 T_1$ intersects $\omega$ at $P \ne T_1$, and ray $T_2 P$ intersects $\omega_2$ at $Q$. It suffices to show that $Q = X_2$. But $Q$ is just the composition of the two homotheties stated earlier on $X_1$, so $Q$ is the reflection of $X_1$ over the midpoint of segment $X_1 X_2$. But this is just $X_2$. We may conclude.

Suppose that $P = X_1T_1 \cap \omega$. Then, \[ \angle O_1X_1T_1 = \angle O_1T_1X_1 = \angle PT_1O = \angle T_1PO. \]Therefore, $O_1X_1 \parallel PO$. Now, suppose $Q = X_2T_2 \cap \omega$. Then, a homothety centered at $T_2$ sends $\omega \to \omega_2$, $O \to O_2$, and $Q \to X_2$. Thus, $OQ \parallel X_2O_2$. Since $O_1X_1 \parallel O_2X_2$, then $OP \parallel OQ$ which holds if $P = Q$. Thus, $X_1T_1$ and $X_2T_2$ intersect at a point lying on $\omega$.

umm i never used equal radii ? so where did i go wrong, collinearities are preserved under homothety right? Let the point of intersection from $X_1$ to $\omega$ be called $A$. We will prove $X_2, A, T_2$ are collinear. Consider the series of homotheties that will bring $\omega_1$ to $\omega$ and $\omega$ to $\omega_2$. The first homothety, centered at $T_1$ will bring $X_1$ to $A$, so $X_1, T_1, A$ are collinear. Then the second homothety, centered at $T_2$, will bring $A$ to $X_2$ so $T_2$, $A$, $X_2$ are collinear, so we are done.

I made a huge mistake by thinking the center of $\omega$ lies on the line connecting the centers of $\omega_{1}$ and $\omega_{2}$. I realized that mistake after looking at someone else's diagram. Plz anyone check if this is correct Let $O_{1}$ and $O_{2}$ be the centers of the circles $\omega_{1}$ and $\omega_{2}$ and let $O$ be the center of the rhombus $O_{1}X_{1}O_{2}X_{2}$. By Monge d'Alembert $O$, $T_{1}$ and $T_{2}$ are collinear. Let $N=\overline{X_{1}T_{1}} \cap \overline{X_{2}T_{2}}$, $M=\overline{X_{2}T_{1}} \cap \overline{X_{1}T_{2}}$ and let $\overline{T_{2}T_{1}O}$ intersect $\omega_{2}$ again at $Y$. Applying Ceva on $\triangle T_{2}X_{1}X_{2}$ gives us \[q=\frac{T_{2}M}{MX_{1}}=\frac{T_{2}N}{NX_{2}}\]Now applying Van Aubel's Theorem gives us \[\frac{T_{2}T_{1}}{T_{1}Y}=\frac{1}{2}\frac{T_{2}T_{1}}{T_{1}O}=\frac{1}{2}2q=q\]Hence the homothety that maps $T_{2}MN$ to $T_{2}X_{1}X_{2}$ is the same as the homothety that maps $\omega$ to $\omega_{2}$, therefore $\omega$ is the circumcircle of $T_{2}MN$ as desired.

Consider the homothety $\mathcal{H}_1$ centered at $T_1$ sending $\omega_1$ to $\omega$ and the homothety $\mathcal{H}_2$ centered at $T_2$ sending $\omega$ to $\omega_2$. Their composition $\mathcal{H}$ (taking $\omega_1$ to $\omega_2$) is a homothety centered at $M$ with ratio $- \frac{r_{\omega}}{\cancel{r_{\omega_1}}} \cdot \frac{\cancel{r_{\omega_2}}}{r_{\omega}} = -1$. Let $X$ be the image of $X_1$ under $\mathcal{H}_1$. Since $\mathcal{H}_1$ is the homothety taking $\omega_1$ to $\omega$, $X$ must be on $\omega$. Further, since $T_1$ is the center of $\mathcal{H}_1$, we have that $X_1$, $T_1$, and $X$ are collinear. Then the image of $X$ in $\mathcal{H}_2$ must be the image $X_2$ of $X_1$ in $\mathcal{H}$, implying that $T_2, X, X_2$ are collinear. Hence $X$ is the intersection of $X_1T_1$ and $X_2T_2$.

Let $O_1, O_2, O_3$ be the centers of $w_1, w_2, w$ respectively. Define $A_1=X_1T_1 \cap w, A_2= X_2T_2 \cap w.$ Notice that from the homothety $w_1 \rightarrow w,$ $$O_3A_1 \parallel X_1O_1.$$From the homothety $w_2 \rightarrow w,$ $$O_3A_2 \parallel O_2X_2.$$But by symmetry $O_1X_1 \parallel O_2X_2$ so it follows that $O_3A_1 \parallel O_3A_2 \implies A_1=A_2.$ QED