This is the best geo I've ever done! EDIT: the solution was slippery at end so I’ll try to fix and repost it later.

Official Solution. Suppose ray $BC$ contains $X$, ie ray $BC$ intersects $\omega_3$, then we claim that the desired excircle is the $B$-excircle of $\triangle ABC$. Let $AB$ and $AC$ meet $\omega_2$ for the second time at $D$ and $E$ respectively. We have $BC \parallel DE$ since $\omega_1$ and $\omega_2$ are tangent at $A$. Let $f$ be the composition of the following maps: $\bullet$ Inversion with center $A$ and radius $\sqrt{AB \times AE}=\sqrt{AC\times AD}$. $\bullet$ Reflection across the external angle bisector of $\measuredangle BAC$. It is clear that this map is an involution (i.e. $f(f(x)) = x$) Note that $f(E) = B, f(D) = C$, $f(DE) = (ABC) \equiv \omega_1$, $f(BC) = (ADE) \equiv \omega_2$. Since $\omega_3$ is tangent to $AB, AC, \omega_1, \omega_2$, therefore its image is tangent to $AB, AC, DE, BC$, which is the $E$-excircle of $\bigtriangleup ADE$ and $B$-excircle of $\bigtriangleup ABC$. Call this circle $\Omega\equiv f(\omega_3)$. Claim 1: Let $I$ be the center of $\Omega$, then $f(I)=I$. Proof: First, $AI$ is the angle bisector of $\angle CAD$, and moreover we can angle chase that $\angle ICA=90^{\circ}-\frac12\angle BCA=90^{\circ}-\frac12\angle DEA=\angle AID$. So $\bigtriangleup ADI \sim \bigtriangleup AIC$, which means that $AD \times AC = AI^2$. This gives $f(I) = I$. $\square$ Let $T\equiv f(X)$ be the image of $X$ under $f$. Since $X$ is one of the intersections of $BC$ and $\omega_3$, $T$ is one of the intersections of $\omega_2$ and $\Omega$. Now let $\Gamma$ be a circle passing through $A$ and $X$ which is tangent to $\omega_3$ at $X$. Let the lines $AB$ and $AC$ meet $\Gamma$ for the second time at $P$ and $Q$ respectively, and let them be tangent to $\Omega$ at $K$ and $L$ respectively. Claim 2: $\Omega$ is the incircle of $\bigtriangleup APQ$, in particular $PQ$ is tangent to $\Omega$. \textit{Proof:} As $\omega_3$ is the $A$-mixtilinear incircle of $\triangle APQ$, and $\Omega_3$ is a circle tangent to both $AP$ and $AQ$, it suffices to prove $K, I, L$ are colinear (via the special case of Sawayama Thebault theorem), which will prove that $\Omega$ is the incircle. To prove this, since $\Omega$ and $\omega_3$ maps to each other via $f$, if we let $L'$ as the point where $AC$ touches $\Omega$, we get: \[ AL' \times AL = AB \times AE = AI^2 \implies \bigtriangleup ALI \sim \bigtriangleup AIL' \implies \angle AIL = \angle AL'I = 90^{\circ}. \]Similarly $\angle AIK = 90^{\circ}$, so $K, I, L$ are indeed colinear. Thus $PQ$ is tangent to $\Omega$. $\blacksquare$ We will also prove a classical result on mixtilinear incircles. Claim 3: Given a triangle $ABC$ with incircle $\omega$ and $X$ is the tangency point of the $A$-mixtilinear incircle with $(ABC)$. Then if the two tangents of $X$ to $\omega$ meet $(ABC)$ again at $R$ and $S$, then $RS$ is tangent to $\omega$ and $RS\parallel BC$. Proof: By Poncelet Porism, since $XR, XS$ are tangent to $\omega$, then so as $RS$. Let $I$ be the incenter of $\triangle ABC$, and $N$ be the midpoint of major arc $BAC$. It is well known that the points $X, I, N$ are colinear. Moreover, $I$ is the incenter of the triangle $\triangle XRS$, hence $N$ must be the midpoint of minor arc $RS$. This directly implies $RS\parallel BC$. $\square$ By Claim $2$, let $U$ be the tangency point of $PQ$ with $\Omega$. Since $AT$ is the reflection of $AX$ across $AI$, $T$ must be the point diametrically opposite to $U$ in $\Omega$. Let $Y\neq X$ be the second intersection of $\Gamma$ and $BC$, and let $Z\neq A$ be the second intersection of $\Gamma$ and $\omega_2$. We claim that $XZ$ is the desired common tangent of $\omega_2$ and $\Omega$. Define the line $\ell$ to be the tangent line of $Y$ to $\Omega$. Since $T$ is diametrically opposite to $U$ in $\Omega$, from Claim $3$, $YT$ is tangent to $\Omega$, thus $\ell$ must be the line $YT$ and therefore parallel to $PQ$. Claim 4: We have $f(\ell)=\Gamma$, $f(Y)=Z$ and $Y, T, Z$ are colinear. Proof: Note that $f(\ell)$ is a circle passing through $A$ and tangent to $\omega_3$ at $X$, which is $\Gamma$. Since $Y$ is located on $\ell$, $\Gamma$, and $BC$, therefore $f(Y)$ is located on $\Gamma$, $\ell$, and $\omega_2$. Since $Z$ is the second intersection of $\Gamma$ and $\omega_2$, then $f(Y)=Z$ must hold. This also means that $Z \in \ell$, thus $Y, T, Z$ are colinear. $\square$ By Claim $3$ again, $XZ$ must be tangent to $\Omega$, because $YZ$ is parallel to $PQ$. Finally, $XZ$ is also tangent to $\omega_2$ because $$\angle XZT=\angle XAY=\angle TAZ$$where the last equality is due to the fact that $f(X)=T$ and $f(Y)=Z$. Therefore, $XZ$ is the desired line passing through $X$ that is tangent to $\omega_2$ and the $B$-excircle of $\triangle ABC$. $\blacksquare$ Remark. The problem remains true if $X$ is the closer intersection instead, but due to configuration issues the proposer decided to specify which choice of $X$ instead.

Amazing problem! My favorite problem from the TST! Here's my solution: WLG, let $B,C,X$ be collinear in that order. Let the required line tangent to $\omega_2$ passing through $X$ be $\ell$. We claim that $\ell$ touches the $B$-excircle of $ABC$. Denote the excircles of $ABC$ corresponding to $B,C$ as $\omega_b,\omega_c$. Let the other line parallel to $BC$ touching $\omega_b$ intersect $AB,AC$ at $B',C'$ respectively. Lemma 1: $B',C' \in \omega_2$ Proof: Let the $B_1,C_1$ be the intersections of the other line parallel to $BC$ touching $\omega_c$ with $AB,AC$. By a $\sqrt{bc}$ inversion, $\omega_3$ goes to $\omega_c$ and so $\omega_2$ goes to $B_1C_1$. So it suffices to $B_1,C_1$ goes to $C',B'$ under the inversion. This is equivalent to showing $AB\cdot AC = AB_1\cdot AC'$. By a homothety at $A$ sending $B_1C_1$ to $BC$, we get that $BC$ goes to $B'C'$ under homothety and so $AB^2 = AB_1\cdot AC' \implies AB\cdot AC = AB_1\cdot AC'.\square$ Let $I$ be the center of $\omega_b$. Let $\mathcal{T}$ denote the composition of inversion in circle centered at $A$ passing through $I$ followed by a reflection in $AI$. By mixti config, $\omega_b \xrightarrow{\mathcal{T}} \omega_3$. Also, angle chase gives $ACI\sim AIB'$ and so $AC\cdot AB' = AI^2 \implies C\xrightarrow{\mathcal{T}} B'$ and similarly $B\xrightarrow{\mathcal{T}} C'$. Hence $BC \xrightarrow{\mathcal{T}} \omega_2$. Now, comes the key idea of solution, we want to show the other tangent to $\omega_b$ through $X$ is tangent to $\omega_2$ which is the image of $BC$ in $\mathcal{T}$, which we now rephrase in the following lemma: Lemma 2: Let $P$ be a point and $\mathcal{C}$ be a circle with center $O$. Let $\mathcal{T}$ denote the composition of an inversion in the circle centered at $P$ with radius $PO$ and then a reflection in $PO$. Let $\mathcal{C}\xrightarrow{\mathcal{T}}\mathcal{C}'$ and let $Q$ be an arbitrary point on $\mathcal{C}'$. Let $\ell_1,\ell_2$ be the tangents from $Q$ to $\mathcal{C}$. Then, the image of $\ell_1$ in $\mathcal{T}$ touches $\ell_2$. Proof: Let $\ell_1,\ell_2$ touch $\mathcal{C}$ at $U,V$. Construct points $X,Y$ on $\ell_1,\ell_2$ so that $\measuredangle XPO = \measuredangle OPY = \measuredangle UOQ$. So, as $\measuredangle XPY = \measuredangle UOV = \measuredangle XQY\implies P,Q,X,Y$ are concyclic on a circle say $\omega$. Let $PO,QO$ meet $\omega$ again at $M,N$. Note that as $PO,QO$ bisect $\angle XPY,\angle XQY$ we have that $M,N$ are actually midpoints of opposite arcs of $XY$ on $\omega$. Let $Q\xrightarrow{\mathcal{T}}Q'$. We know that $Q'\in \mathcal{C}$ and so $\frac{PO}{OQ'} = \frac{PQ}{QO} \implies \frac{PO}{PQ} = \frac{OQ'}{OQ} = \frac{OV}{OQ}$ Hence as $\triangle OPQ \sim \triangle ONM$, $\frac{NO}{NM} = \frac{OV}{OQ} = \frac{NY}{NM} \implies NO = NY \implies$ $O$ is the incenter of $QXY$. Hence, as $P,O,M$ are collinear, $P$ is $Q$ mixtilinear incircle touch point in $QXY$ and $PO^2 = PX\cdot PY$.

$\measuredangle PXQ = \measuredangle PYQ\implies \ell_1'$ touches $\ell_2$ at $Y$ and we are done!

excellent solution!!

Solution by the user @Zhaom We will prove the following statement. The problem is analogous, where $\omega_3$ is instead the image of the incircle, the $B$-excircle, or the $C$-excircle of $\triangle{}ABC$ under force overlaid inversion at $A$ swapping $B$ and $C$. Let $\triangle{}ABC$ have circumcircle $\omega_1$, incircle $\omega$, and $A$-mixtilinear incircle $\omega_3$. Let $\omega_2$ be the circle tangent to $\omega_1$ at $A$ and tangent to $\omega_3$. Then, let $X$ and $Y$ be the intersections of $\omega_3$ with $\overline{BC}$. Prove that the tangents to $\omega$ at $X$ and $Y$ other than $\overline{BC}$ are tangent to $\omega_2$. We prove the following more general problem. Let $\triangle{}ABC$ have circumcircle $\omega_1$, incircle $\omega$, and $A$-mixtilinear intouch point $T$. Let $X$ and $Y$ be points on $\overline{BC}$ such that $(TXY)$ is tangent to $\omega_1$, and let $\omega_3$ be the circumcircle of $\triangle{}TXY$. Let $\omega_2$ be the circle tangent to $\omega_1$ at $A$ and tangent to $\omega_3$. Prove that the tangents $\ell_X$ and $\ell_Y$ to $\omega$ at $X$ and $Y$, respectively, other than $\overline{BC}$ are tangent to $\omega_2$. Let $T'=\overline{AT}\cap\overline{BC}$, let $I$ be the incenter of $\triangle{}ABC$, let $I_A$ be the $A$-excenter of $\triangle{}ABC$, and let $D$ be the foot of the altitude from $I$ to $\overline{BC}$. Claim $1$. We have that $\overline{TX}$ and $\overline{TY}$ are isogonal conjugates with respect to $\angle{}BTC$. Proof. Let $X'$ and $Y'$ be the images of $X$ and $Y$ under homothety centered at $T$ sending $\omega_3$ to $\omega_1$. Then, we see that $\overline{X'Y'}\parallel\overline{BC}$, so $B,C,X',$ and $Y'$ form an isosceles trapezoid with $\overline{X'Y'}\parallel\overline{BC}$ and circumcircle $\omega_1$. This means that the arcs $BX'$ not containing $T$ and $CY'$ not containing $T$ on $\omega_1$ have the same measure, so $\angle{}XTB=\angle{}X'TB=\angle{}Y'TC=\angle{}YTC$, proving the claim. Claim $2$. Either the insimilicenter or exsimilicenter $P$ of $\omega$ and $\omega_2$ is on $\overline{AT}$. Proof. Applying Monge's Theorem on $\omega,\omega_2,$ and $\omega_1$ and noting that the exsimilicenter of $\omega$ and $\omega_1$ is on $\overline{AT}$ by Monge's Theorem on $\omega$, the $A$-mixtilinear incircle, and $\omega_1$ gives the result. Claim $3$. There exists an involution swapping points $X$ and $Y$, points $B$ and $C$, and points $T'$ and $D$. Proof. For this it suffices that $\overline{TT'}$ and $\overline{TD}$ are isogonal conjugates with respect to $\angle{}BTC$, which is well known. Claim $4$. We have that $\ell_X$ and $\ell_Y$ intersect at a point $P'$ on $\overline{AT}$. Proof. Dual of Desargues's Involution Theorem on $ABDC$ with inconic $\omega$ and point $P'$ gives that there exists an involution swapping lines $\overline{P'A}$ and $\overline{P'D}$, swapping lines $\overline{P'B}$ and $\overline{P'C}$, and swapping lines $\ell_X$ and $\ell_Y$. Projecting onto $\overline{BC}$ gives that there exists an involution swapping points $\overline{P'A}\cap\overline{BC}$ and $D$, points $B$ and $C$, and points $X$ and $Y$. By claim $3$ this implies that $\overline{P'A}\cap\overline{BC}=T'$, proving the claim. Claim $2$ and claim $4$ together imply that if we can prove that $\overline{PI}=\overline{P'I}$, then we would be done. Note that $I$ is the incenter of $\triangle{}PXY$. Then, let $I'_A$ be the $P'$-excenter of $\triangle{}P'XY$, or the antipode of $I$ on $(IXY)$ by the incenter-excenter lemma, and let $O_1$ be the center of $\omega_2$. Since $P',I,$ and $I'_A$ are collinear and $P,O_1,$ and $I$ are also collinear, it suffices that $O_1,I,$ and $I'_A$ are collinear. Claim $5$. Let $E$ be the intersection of $(AIT)$ with $\omega_3$ other than $T$. Then, the circles $\omega_2$ and $\omega_3$ are tangent at $E$. Also, the circle $(AIT)$ is orthogonal to $\omega_1,\omega_2,$ and $\omega_3$. Proof. Let $E'$ be the tangency point of $\omega_2$ and $\omega_3$. It suffices that $E'$ is on $(AIT)$ to prove the first part of the claim. Let $O_2$ be the intersection of the tangents to $\omega_1$ at $A$ and $T$. It is well known that $O_2$ lies on the perpendicular bisector of $\overline{AI}$, so $O_2I=O_2A=O_2T$, implying that $O_2$ is the center of $(AIT)$. However, also note that $O_2$ is the radical center of $\omega_1,\omega_2,$ and $\omega_3$, so $\overline{O_2E'}$ is the radical axis of $\omega_2$ and $\omega_3$, implying that $O_2I=O_2A=O_2T=O_2E'$, meaning that $E'$ is on $(AIT)$. Then, note that the tangents to $\omega_k$ at the intersections of $\omega_k$ with $(AIT)$ intersect at $O_2$ the center of $(AIT)$ for $k=1,2,3$, so $(AIT)$ is orthogonal to $\omega_1,\omega_2,$ and $\omega_3$, proving the claim. Let $O'$ be the center of $\omega_3$, so $\overline{TO'}$ is tangent to $(AIT)$ since $\omega_3$ is orthogonal to $(AIT)$. Also, let $O'$ Claim $6$. We have that $O_1$ is the intersection of $\overline{AO}$ and the tangent to $(AIT)$ through $O'$ other than $\overline{TO'}$. Proof. First, note that the homothety centered at $A$ sending $\omega_2$ to $\omega_1$ sends $O_1$ to $O$, so $O_1$ is on $\overline{AO}$. Then, note that the tangent to $(AIT)$ through $O'$ other than $\overline{TO'}$ is $\overline{EO'}$ be claim $5$ since $\omega_3$ is orthogonal to $(AIT)$. By claim $5$, we see that one of the tangents to $(AIT)$ through $O_1$ is $\overline{O_1E}$ since $\omega_2$ is orthogonal to $(AIT)$. Therefore, we see that $\overline{EO'}=\overline{O_1E}$ is the tangent to $(AIT)$ at $E$, so $O_1$ lies on the tangent to $(AIT)$ through $O'$ other than $\overline{TO'}$. This proves the claim. Claim $7$. Point $I'_A$ lies on $\overline{I_AT'}$. Proof. Let $T_1$ be the intersection of $\overline{BC}$ with the tangent to $\omega_1$ at $T$ and let $\gamma$ be the circle centered at $T_1$ through $T$. Since $\overline{TX}$ and $\overline{TY}$ are isogonal conjugates with respect to $\angle{}BTC$, we see that $X$ and $Y$ swap under inversion at $\gamma$. This implies that $\omega_3$ is orthogonal to $\gamma$, so the inverse $I^*$ of $I$ with respect to $\gamma$ is on $\omega_3$. Then, note that $I'_A$ is on the altitude $\ell$ from $I^*$ to $\overline{I^*I}$. As $X$ and $Y$ vary, the line $\ell$ remains fixed. Taking $X=B$ gives that $Y=C$, so $I'_A=I_A$ lies on $\ell$. Taking $X=T'$ gives that $Y=D$ since $\overline{TT'}$ and $\overline{TD}$ are isogonal conjugates with respect to $\angle{}BTC$. Since $\angle{}IDT'=90^\circ$, we see that $I'_A=T'$ lies on $\ell$, so $\ell=\overline{I_AT'}$. Since $I'_A$ lies on $\ell$ in general, this proves the claim. Claim $8$. Let $M$ be the midpoint of $\overline{XY}$ and let $D''$ be the reflection of $D$ over $M$. Then, point $I'_A$ is the intersection of $\overline{I_AT'}$ and the altitude from $D''$ to $\overline{BC}$. Proof. We see that $I'_A$ is on $\overline{I_AT'}$ by claim $7$ and that $I'_A$ is on the altitude from $D''$ to $\overline{BC}$ since $\omega$ is the incircle of $\triangle{}P'XY$ and $I'_A$ is the $P'$-excenter of $\triangle{}P'XY$. Therefore, we can rephrase the problem in terms of $O'$. Let $\triangle{}ABC$ have circumcircle $\omega_1$ with center $O$, incircle $\omega$ with center $I$ touching $\overline{BC}$ at $D$, and $A$-mixtilinear intouch point $T$. Also, let the $A$-excenter of $\triangle{}ABC$ be $I_A$. Let $T'=\overline{AT}\cap\overline{BC}$. Then, let $O'$ be a point on $\overline{TO}$, let $M$ be the foot of the altitude from $O'$ to $\overline{BC}$, let $D''$ be the reflection of $D$ over $M$, let $I'_A$ be the intersection of $\overline{I_AT'}$ and the altitude from $D''$ to $\overline{BC}$, and let $O_1$ be the intersection of $\overline{AO}$ and the tangent to $(AIT)$ through $O'$ other than $\overline{TO}$. Prove that $I'_A,I,$ and $O_1$ are collinear. Let $\overline{O'O_1}$ touch $(AIT)$ at $E$ and let $\psi$ denote a point pole duality around $(AIT)$. Let $\infty_{\overline{ID}}$ denote the point at infinity on $\overline{ID}$ and let $I_N$ denote the point at infinity on $\overline{BC}$. Then, let $D$ be the foot of the altitude from $I$ to $\overline{BC}$, let $M_1$ be the foot of the altitude from $T$ to $\overline{BC}$, let $M_M$ be the midpoint of $\overline{BC}$, and let $D'$ be the foot of the altitude from $I_A$ to $\overline{BC}$. Also, let $D''_1$ denote the reflection of $D$ over $M_1$, and let $J$ and $D_1$ denote the intersections of the internal and external angle bisectors of $\angle{}CAB$ with $\overline{BC}$, respectively. Also, let $M_A$ and $M'_A$ be the intersections of internal and external angles bisectors of $\angle{}CAB$ with $\omega_1$ other than $A$, respectively. Claim $9$. The line $\overline{ID}$ is tangent to $(AIT)$. Proof. Well known. Claim $10$. The altitude from $\overline{D''_1}$ to $\overline{BC}$, the line $\overline{IO}$, and the line $\overline{I_AT'}$ concur at a point $O'_1$. Proof. We will prove that if $O''_1=\overline{I_AT'}\cap\overline{OI}$, then the foot of the altitude from $O''_1$ to $\overline{BC}$ is $D''_1$. Let $Z'$ be the foot of the altitude from $\overline{I_AT'}\cap\overline{IT}$ to $\overline{BC}$ and let $I'$ be the reflection of $I$ over $T$. By Dual of Desargues's Involution Theorem on $TOAO$ with inconic $(AIT)$ and point $I$ there exists an involution swapping $\overline{IT}$ and $\overline{IA}$ fixing $\overline{IO}$ and $\overline{ID}$ by claim $9$. Projecting onto $\overline{I_AT'}$ gives that there exists an involution swapping $\overline{I_AT'}\cap\overline{IT}$ and $I_A$ fixing $O''_1$ and $\overline{I_AT'}\cap\overline{ID}$. Projecting through $\infty_{\overline{ID}}$ onto $\overline{BC}$ gives that there exists an involution swapping $Z'$ and $D'$ fixing the foot of the altitude from $O''_1$ to $\overline{BC}$ and $D$, so it suffices that there exists an involution swapping $Z'$ and $D'$ fixing $D''_1$ and $D$. Projecting through $\infty_{\overline{ID}}$ onto $\overline{IT}$ gives that it suffices that there exists an involution fixing $I$ and $I'$ swapping $\overline{I_AT'}\cap\overline{IT}$ and $\overline{I_AD'}\cap\overline{IT}$. Projecting through $I_A$ onto $\overline{BC}$ gives that it suffices that there exists an involution fixing $J$ and $\overline{I_AI'}\cap\overline{BC}$ swapping $D'$ and $T'$. The involution fixing $J$ and swapping $D'$ and $T'$ is the involution taking a point $P$ on $\overline{BC}$ and giving the point $Q$ on $\overline{BC}$ such that $\overline{AP}$ and $\overline{AQ}$ are isogonal conjugates in $\angle{}CAB$. This involution fixes $J$ and $D_1$, so it suffices that $\overline{I_AI'}$ passes through $D_1$. Claim $10.1$. We have that $A,I,D,$ and $I'$ are concyclic. Proof. Note that $(AID)$ is preserved under force overlaid inversion at $T$ swapping $B$ and $C$ since $I$ is sent to itself and $A$ and $D$ swap. This implies that $(AID)$ is sent to the reflection of it over $\overline{TI}$ under inversion at $T$ fixing $I$. This implies that $(AID)$ is sent to the reflection of it over $T$ under inversion at $T$ fixing $I$. Therefore, we see that $I'$ is on $(AID)$, proving the claim. Now, note that $D_1$ is on $(AID)$ since $\angle{}IAD_1=\angle{}IDD_1=90^\circ$. This implies that $\angle{}II'D_1=\angle{}IDD_1=90^\circ$. However, note that $\angle{}II'I_A=\angle{}ITM_A=90^\circ$, so $\overline{I_AI'}$ passes through $D_1$, proving the claim. Then, note that \begin{align*} \left(O,\overline{ID}\cap\overline{AO};A,O_1\right)&=\left(\psi\left(\overline{AT}\right),\psi\left(\overline{AI}\right);\psi\left(\overline{AO}\right),\psi\left(\overline{AE}\right)\right)\\ &=\left(\overline{AT},\overline{AI};\overline{AO},\overline{AE}\right)\\ &=(T,I;A,E)\\ &=\left(\overline{TO},\overline{TI};\overline{TA},\overline{TE}\right)\\ &=\left(\psi\left(\overline{TO}\right),\psi\left(\overline{TI}\right);\psi\left(\overline{TA}\right),\psi\left(\overline{TE}\right)\right)\\ &=\left(T,\overline{TO}\cap\overline{ID};O,O'\right)\\ &\stackrel{\infty_{\overline{ID}}}{=}\left(M_1,D;M_M,M\right)\\ &=\left(D''_1,D;D',D''\right)\\ &\stackrel{\infty_{\overline{ID}}}{=}\left(O'_1,\overline{ID}\cap\overline{I_AT'};I_A,I'_A\right)\\ &\stackrel{I}{=}\left(O,\overline{ID}\cap\overline{AO};A,\overline{II'_A}\cap\overline{AO}\right), \end{align*}meaning that $\overline{II'_A}\cap\overline{AO}=O_1$, so $I,I'_A,$ and $O_1$ are collinear, so we are done.

Did anyone solve this problem during the test?