Let $AB$ cut $\omega_2$ at $X,Y.$ Since $\omega_1 \perp \omega_2$ $\Longrightarrow$ $OA^2=OB^2=OX \cdot OY$ $\Longrightarrow$ $(X,Y,A,B)=-1$ $\Longrightarrow$ $B$ is on the polar of $A$ WRT $\omega_2.$ Inversion WRT $\omega_1$ takes $\omega_2$ into itself and carries the circles passing through $O,A$ tangent to $\omega_2$ into two lines tangent to $\omega_2$ at the inverses $F^*,G^*$ of $F,G.$ These lines will meet at the double point $A$ $\Longrightarrow$ $F^*G^*$ is the polar of $A$ WRT $\omega_2$ $\Longrightarrow$ $B,F^*,G^*$ are collinear. Since $B$ is double, then $B,F,G,O$ lie on the inverse circle of $F^*G^*.$

Nice and easy.

any solution without inversion ?

Can someone explain me why from the relation $OB^2 = OM \times ON,$ wa can deduce that $B$ lies on $F_2G_2$ ,thank you

@above The condition implies the $M$ and $N$ are inverses under the inversion WRT $\omega_1$. Hence, $M$ is the intersection of the polar of $N$ WRT $\omega_1$ and $AB$. Denote $C$ and $D$ as the points of tangency from $N$ to $\omega_1$. It's well known $CBDA$ is harmonic and by perspective at $C$ we get $(AB;MN)=-1$. If we define $B'$ as the intersection of the polar of $A$ WRT $\omega_2$ and $AB$, then using a similar argument as before we get $(AB';MN)=-1$. This implies $B=B'$ as desired.

Invert about $\omega_1,$ and denote inverses with $'$. By inversion properties, $AF'.AG'$ are tangent to $\omega_2$ and it suffices to prove $B\in F'G'.$ If $M,N$ are the midpoints of $AF',AG'$ then it suffices to prove $O\in MN$ by homothety, but this follows as $O$ is on the radical axis of the point-circle centered at $A$ and $\omega_2$ (from orthogonality).

Probably same as others... ELMO Shorlist 2013 G5 wrote: Let $\omega_1$ and $\omega_2$ be two orthogonal circles, and let the center of $\omega_1$ be $O$. Diameter $AB$ of $\omega_1$ is selected so that $B$ lies strictly inside $\omega_2$. The two circles tangent to $\omega_2$, passing through $O$ and $A$, touch $\omega_2$ at $F$ and $G$. Prove that $FGOB$ is cyclic. Proposed by Eric Chen Solution:- Let the the two circles passing through $AO$ and tangent to $\omega_2$ are $\gamma_1$ and $\gamma_2$ respectively. Now by an Inversion $\Psi$ around $\omega_1$ we see that $A,B$ remains fixed under this inversion, $\Psi(\omega_2)\mapsto\omega_2$ as $\omega_1$ and $\omega_2$ are orthogonal. $\Psi(\gamma_1)\mapsto$ A line tangent to $\omega_2$ through $A$ as $\gamma_1$ passes through the center of $\omega_1$ and is tangent to $\omega_2$, similarly $\gamma_2\mapsto$ A line tangent to $\omega_2$ passing through $A$. So, if $F',G'$ are the point of tangencies from $A$ to $\omega_2$ then $\Psi(F)\mapsto F'$ and $\Psi(G)\mapsto G'$. Let $AB\cap\omega_2=P,Q$ and $\omega_1\cap\omega_2=X,Y$. Now notice that $OX^2=OP.OQ=OB^2\implies (A,B;P,Q)=-1\implies B\in$ Polar of $A$ WRT $\omega_2$. Hence, $F',B,G'$ are collinear, now inverting back we get that $FGOB$ is a cyclic quadrilateral. $\blacksquare$

A bit different approach. Invert around $A$. That gives you this figure. Let $\omega_2'$ be a circle with center $X$ and $O'$ point outside of this circle. Let $\omega_1'$ be a line passing through $X$. Denote $O'F'$ and $O'G'$ lines tangent to $\omega_2'$ ($F'$, $G'$ lies on $\omega_2'$). Denote $B'$ feet from $O'$ to $\omega_1'$. Prove, that $F'G'O'B'$ is concyclic. This is trivial, as $O'X$ will be diameter of this circle. We have actualy prooven a little generalization. It's not nessesary, that $O$ is the center of $\omega_1$. We just need that it lies on diameter $AB$ and that it is outside of $\omega_2$.

So first of all_let me define point $E$ as the center of $\omega_2$ We are first tempted to do an inversion $\psi_1(\omega_1)$, because under this inversion $\omega_2 \xrightarrow{\psi_1} \omega_2$,and the tangent circles are transformed into tangents of $\omega_2$,this way $F \xrightarrow{\psi_1} F'$ and $G \xrightarrow{\psi_1} G'$ Because of the inversion $AF'$ and $AG'$ are tangent to $\omega_2$. One more thing about this inversion,it does this to $B$, $B \xrightarrow{\psi_1} B$,so the point $B$ is fixed under this inversion. Now if the quad $FOGB$ were cyclic then the following must hold: $$ 180 = \angle OGB + \angle OFB = \angle OB'G' + \angle OB'F' = \angle OBG' + \angle OBF' $$Thus we need to show that the points $G',B$ and $F'$ are collinear. We now consider a second inversion,inversion $\psi_2(\omega_2)$,now under this inversion we have the following: We can easily construct $A'$ because of the tangents (here we define this as $A \xrightarrow{\psi_2} A'$), we also have $\omega_1 \xrightarrow{\psi_2} \omega_1$,thus we know that the point $B'$ stays on $\omega_1$ (here $B \xrightarrow{\psi_2} B'$) If the points $F',A',B,G'$ are all collinear,then we must show that the point $B'$ is on the circumcircle of the $\triangle AEF'$ or in other words is on the circle of the cyclic quad $AF'EG'$,but this is easily seen because we have the following: $$\angle EB'A + \angle EF'A = \angle BB'A + 90 = 90 + 90 = 180 $$Thus the points $F',A',B',G'$ are all collinear. Thus the quad $FOGB$ is cyclic.....

Let $O_1$ be the center of $\omega_2$. We first invert around $\omega_1$. Under this inversion, $F$ and $G$ invert to points of tangency from $A$ to $\omega_2$, and everything else remains fixed. We wish to prove that $F',B'G'$ are collinear. But consider the second inversion around $\omega_2$. This inversion sends point $B$ to the second intersection of ray $O_1B$ with $\omega_1$, and we wish to show that $F'O_1G'B'$ is cyclic. Notice that both $AF'O_1G'$ and $AB'O_1F'$ are cyclic, which proves the desired result. The ELMO 2013 Geometry Shortlist seems really easy for some reason? (Sorry if this isn't a great writeup, idrk how to write inversive solutions or how much detail to put in them)

Cute problem; almost as cute as audio-off Invert about $\omega_1.$ It then suffices to prove that $B$ lies on the polar of $A$ wrt $\omega_2.$ Letting $\angle AG'F'= \angle AF'G' = \theta,$ it is obvious since $AB$ is a diameter that $$\angle F'BG' = 2(90-\theta) + (180-(180-2\theta)) = 180,$$as desired. $\blacksquare$

v_Enhance wrote: Let $\omega_1$ and $\omega_2$ be two orthogonal circles, and let the center of $\omega_1$ be $O$. Diameter $AB$ of $\omega_1$ is selected so that $B$ lies strictly inside $\omega_2$. The two circles tangent to $\omega_2$, passing through $O$ and $A$, touch $\omega_2$ at $F$ and $G$. Prove that $FGOB$ is cyclic. Proposed by Eric Chen

Consider the inversion about $\omega_1$. Notice $(AOF)$ goes to $AF^*$ and $(AOG)$ goes to $AG^*$. Furthermore, since $\omega_1$ and $\omega_2$ are orthogonal, we know $F^*, G^* \in \omega_2$. But $(AOF), (AOG)$ are tangent to $\omega_2$ at $F$ and $G$ respectively, so we know $AF^*$ and $AG^*$ are also tangents to $\omega_2$. The desired conclusion is equivalent to showing $B, F^*, G^*$ are collinear. Because $F^*G^*$ is the polar of $A$ wrt $\omega_2$, it suffices to prove $A$ lies on the polar of $B$ by La Hire's. Let $B'$ denote the image of $B$ under the inversion about $\omega_2$. Since $B' \in \omega_1$ by orthogonality, we know $$\angle OB'A = \angle BB'A = 90^{\circ}$$where the last equality follows from Thales', as desired. $\blacksquare$ mathlogician wrote: The ELMO 2013 Geometry Shortlist seems really easy for some reason? (Sorry if this isn't a great writeup, idrk how to write inversive solutions or how much detail to put in them) I agree with this sentiment. We must remember, however, that this Geometry Shortlist had $14$ problems...

Let the inversion around $\omega_1$ be $\Psi_1,$ and notice that $\Psi_1(F)$ and $\Psi_1(G)$ are the tangents from $A$ to $\omega_2.$ Let the inversion around $\omega_2$ be $\Psi_2$ and notice that $\Psi_2(A)$ is the midpoint of $\overline{\Psi_1(F)\Psi_1(G)}$ and also on $\omega_1.$ Thus $$\angle B\Psi_2(A)A=\angle \Psi_1(G)\Psi_2(A)A=90$$and $\Psi_1(F),\Psi_2(G),$ and $B$ are collinear. $\square$

https://www.geogebra.org/m/drudhjrs WE CAN EASILY PROVE THAT D IS THE INVERSE OF F AND E IS THE INVERSE OF G.

Clearly $AF',AG'$ tangent to $\omega_2,$ where $F',G'$ are inverses of $F,G$ wrt $\omega_1.$ Furthermore $O$ lies on radical axis of $A,\omega_2$ (because of orthogonality), which is the $A-\text{midline}$ of $AF'G',$ so $B\in F'G'.$ Thus $FGOB$ is cyclic.

Let $X$ and $Y$ be the points on $\omega_2$ such that $\overline{AX}$ and $\overline{AY}$ are tangent to $\omega_2$, and let $T$ be the second intersection of $\overline{O_2B}$ with $\omega_1$, where $O_2$ is the center of $\omega_2$. Inverting about $\omega_1$, it suffices to show that $X,Y,B$ are collinear. Inverting about $\omega_2$, we find that it suffices to show $O_2XYT$ cyclic. In fact, $O_2XYTA$ is cyclic, since $\angle AXO_2=\angle AYO_2=90^\circ$ and $\angle ATO_2=\angle ATB=90^\circ$; done. $\blacksquare$

nukelauncher wrote:

good work

Let the center of $\omega_2$ be $P$. Invert about $\omega_1$: $F_1$ and $G_1$ are the touch points of the tangents at $A$ to $\omega_2$, and $B$ is fixed. Hence we need $B$ to lie on $F_1G_1$. Invert about $\omega_2$: $B_1 = PB \cap \omega_1$, $O_1 = P$, and $F_1$ and $G_1$ are fixed. Hence we need $F_1$, $O_1$, $G_1$, $B_1$ to be concyclic, which holds as each point lies on $(AP)$. $\blacksquare$

After inversion about $\omega_1$, it is equivalent to showing that $B$ lies on the polar of $A$ wrt $\omega_2$, which is a well-known result of orthogonal circles.