Can anyone provide a figure? I don't understand the statement. Thanks.

GGPiku wrote: Can anyone provide a figure? I don't understand the statement. Thanks. There is information missing when it says: YanYau wrote: and the line intersects the circle $\omega$ through $AO_1O_2$ at $F$. I'm quite sure this should be "the line AE intersects...". Then the problem makes sense to me and seems to be true.

Sorry about that, fixed the typo

Let $G=O_1C\cap O_2D$. As $\angle{CAD}=\angle{CAB}+\angle{BAD}=\angle{ECD}+\angle{EDC}=180-\angle{DEC}$, the quadrilateral $ACDE$ is cyclic. As $ \angle{ECG}=\angle{GDE}=90$, we have $G\in \odot{(ACDE)}$. Note that $\angle{DEA}=\angle{DCA}=\dfrac{1}{2}\angle{BO_1A}$. Since $AO_1BO_2$ is a kite we have $\dfrac{1}{2}\angle{BO_1A}=\angle{O_2O_1A}=\angle{O_2FA}$. Hence we have $\angle{DEA}=\angle{O_2FA}$ which results in $FO_2\parallel ED$. Similarly $FO_1\parallel EC$ which implies that $G\in \omega$ and $G$ is the antipode of $F$. Let $H$ be the feet of altitude of $F$ on $EG$. ( Note that $H$ lies on $\omega$.) Thus we have $\angle{GEA}=\angle{GCA}=\dfrac{1}{2}\angle{GO_1A}=\dfrac{1}{2}\angle{GHA}$. Hence $\angle{HEA}=\angle{EAH}=\angle{FGH}$. Hence $\triangle{GFE}$ is isosceles which gives us $EF=FG$. Since $FG$ is the diameter of $\omega$, we are done.

This is Romania BMO TST 2016,Azerbaijan BMO TST 2017.

Maybe from IMO shortlist ...

tchebytchev wrote: Maybe from IMO shortlist ... IMO shortlist? Which year?

It s easy to get $AO_1O_2\overset{+}{\sim} ACD$ then $A $ is the similicenter of similarity that sends $O_1\to C,O_2\to D$ thus if $K=CO_1\cap DO_2$ then $KAO_1C$and $KAO_2D$ are cyclic more over wiyh easy angle chase we get $ACDE$ cyclic thus Reim's implies $O_1F\parallel CE$ since $CE \perp CO_1$ then $CO_1 \perp FO_1$ which means that $ FK,EK$ are diameters of resp. $(AO_1O_2),(ACD)$ . $\angle AEK=\frac{\pi}{2}-\angle EKA =\frac{\pi}{2}-\angle ECA=\frac{\pi}{2}-\angle ABD=\frac{\pi}{2}-\frac{\angle AO_2D}{2}=\frac{\angle AFK}{2}$ therefore $EFK$ is isoceles . RH HAS