After Inverting at $B$, the problem becomes trivial. (B becomes ortho centre of A*C*D* )

$BE$ is a diameter of $T_2$. $BF$ is a diameter of $T_1$. And you obtain $FEA$ is a straight line parrallel to $O_1O_2$. And rest is angle chasing.

$$2\angle CFE = 2\angle BFC = \angle BO_1C = \angle O_1CO_2 - \angle O_1BO_2 = \angle O_1DO_2 - \angle O_1BO_2$$$$=\angle CO_2B = \angle CO_2B = 2\angle CFB = 2\angle CFD \ \blacksquare$$

Let $O_1B$ meet $\Gamma_2$ at $D'$, we have $$\angle BDO_2=\angle DBO_2=180-\angle O_1BO_2=180-\angle O_1AO_2$$. Hence $AO_1O_2D'$ is cyclic implying $D'=D$ giving $O_1,B,D$ are collinear. Similarly, $O_2,B,C$ are collinear. This gives $BE$ and $BF$ are diameters of $\Gamma_2$ and $\Gamma_1$ respectively, giving $\angle FCE=\angle EDF=90$. Hence, alternate interior angles are equal and we are done!

rd1452002 wrote: $$2\angle CFE = 2\angle BFC = \angle BO_1C = \angle O_1CO_2 - \angle O_1BO_2 = \angle O_1DO_2 - \angle O_1BO_2$$$$\angle CO_2B = \angle CO_2B = 2\angle CFB = 2\angle CFD \ \blacksquare$$ My solution

Drunken_Master wrote: Let $O_1B$ meet $\Gamma_2$ at $D'$, we have $$\angle BDO_2=\angle DBO_2=180-\angle O_1BO_2=180-\angle O_1AO_2$$. Hence $AO_1O_2D'$ is cyclic implying $D'=D$ giving $O_1,B,D$ are collinear. Similarly, $O_2,B,C$ are collinear. This gives $BE$ and $BF$ are diameters of $\Gamma_2$ and $\Gamma_1$ respectively, giving $\angle FCE=\angle EDF=90$. Hence, alternate interior angles are equal and we are done! Same solution. I wonder why did they give only one geo this year.

AnArtist wrote: Drunken_Master wrote: Let $O_1B$ meet $\Gamma_2$ at $D'$, we have $$\angle BDO_2=\angle DBO_2=180-\angle O_1BO_2=180-\angle O_1AO_2$$. Hence $AO_1O_2D'$ is cyclic implying $D'=D$ giving $O_1,B,D$ are collinear. Similarly, $O_2,B,C$ are collinear. This gives $BE$ and $BF$ are diameters of $\Gamma_2$ and $\Gamma_1$ respectively, giving $\angle FCE=\angle EDF=90$. Hence, alternate interior angles are equal and we are done! Same solution. I wonder why did they give only one geo this year. Usually they give 2 to 2.5 geo

I prepared for 2-3 geometry questions because I am weak in geometry. Only 1 came :/ I should spent that time doing combi or algebra.

Prove by angle chasing that $B$ is the incenter of $\triangle ACD$.

Found a nice solution in the exam using phantom points!

My solution (something like this): Let $O_1B \cap \Gamma_2 = D'$ and $O_2B \cap \Gamma_1 = C'$. As $\triangle O_1D'A$ is not isosceles, $\angle AO_1O_2 = \angle O_2O_1B = \angle O_2O_1D'$ and $\angle O_2AD'=\angle O_2D'A$ gives $O_1AO_2D'$ is cyclic. So, $D'\equiv D$. Similarly, $C'\equiv C$. Proved.

We can also use the concept of projective geometry to show O1,B,Dand F collinear.

Claim 1 : $C , B , O_2$ and $D , B , O_1$ are co-linear. Proof: From the cyclic $\square O_1 A O_2 C$ we get, $\angle O_1 C O_2 = 180^o - \angle O_1 A O_2$ $\triangle O_1 A O_2 \cong \triangle O_1 B O_2 \Rightarrow \angle O_1 A O_2 = \angle O_1 B O_2$ Let the line $O_2 B$ intersects $\tau_1$ at $K$ $\therefore \angle O_1 B K = 180^o - \angle O_1 B O_2 = 180^o - \angle O_1 A O_2$ $O_1 B = O_1 K$ $\therefore \angle O_1 K B = \angle O_1 B K = 180^o - \angle O_1 A O_2 = \angle O_1 C O_2$ Therefore, $K$ coincides with $C$ Thus , $C, B , O_2$ are co-linear. (Proved) Similarly, $D, B , O_1$ are co-linear. Claim 2 : $F , A , E$ are co-linear. Proof : Trivial. Now we complete our proof, Let $\angle A E O_2 = \angle F E C = \alpha$ $\therefore E A O_2 = \alpha \Rightarrow \angle A O_2 B = 2\alpha$ $\Rightarrow \angle O_1 O_2 B = \angle O_1 O_2 C = \angle O_1 D C = \angle F D C = \alpha$ $\therefore \angle F E C = \angle F D C$ So, $C , D , E , F$ are concyclic [Q.E.D.]

all the ques pls?

@chandramauli see here https://artofproblemsolving.com/community/c596374_2018_india_national_olympiad

For those who didn't get Kayak's slick solution, he says: Invert about $B$. Then $A \to A^*$, $B \to B^*$ and $C \to C^*$, while $\Gamma_1$ goes to the straight line $A^*C^*$ and $\Gamma_2$ goes to the straight line $A^*D^*$. (Because $\angle O_1AO_2$ is obtuse, $C$ and $D$, and hence $C^*$ and $D^*$, are different from B.) $\odot (O_{1}AO_{2}) \to \odot (O_{1}^*A^*O_{2}^*)$, which is also the circumcircle of $\Delta A^*C^*D^*$. Now note: 1. $O_1^*$ is the reflection of $B$ in $A^*C^*$, 2. $O_2^*$ is the reflection of $B$ in $A^*D^*$. This forces $B$ to be the orthocenter of $A^*C^*D^*$ because there is only one point other than $A^*$ whose reflections in $A^*C^*$ and $A^*D^*$ respectively lie on $\odot (A^*C^*D^*)$ (why?) $-$ and the orthocenter clearly satisfies this condition. Hence, $B^*$ is the orthocenter of $\Delta A^*C^*D^*$. Since $E^*$ and $F^*$ are the feet of altitudes through $D^*$ and $C^*$ in $\Delta A^*C^*D^*$, it's trivial to note that $C^*D^*E^*F^*$ is cyclic.

Extention: Let $AB$ meet the circle $\odot (O_1AO_2)$ again at $O$. Prove that $O$ is the circumcenter of $BCD$. (Note: This is easier than the original problem )

Synthetic_Potato wrote: Extention: Let $AB$ meet the circle $\odot (O_1AO_2)$ again at $O$. Prove that $O$ is the circumcenter of $BCD$. (Note: This is easier than the original problem ) Just note that $B$ is the incentre of $ACD$

A different approach: After some easy angle chasing we get that $C,B,O_2,E$ and $ D,B,O_1,F$ are collinar in that order . By Power of point $B$ on $(CDO_2O_1) \Longleftrightarrow BD \cdot BO_1=BC \cdot BO_2$ and since $BO_1$ and $BO_2$ are the radiuses of their respective circles , we get that $ BD \cdot BF=BC \cdot BE$ wich completes the proof.

TheDarkPrince wrote: My solution (something like this): Let $O_1B \cap \Gamma_2 = D'$ and $O_2B \cap \Gamma_1 = C'$. As $\triangle O_1D'A$ is not isosceles, $\angle AO_1O_2 = \angle O_2O_1B = \angle O_2O_1D'$ and $\angle O_2AD'=\angle O_2D'A$ gives $O_1AO_2D'$ is cyclic. So, $D'\equiv D$. Similarly, $C'\equiv C$. Proved. I think $\angle AO_1O_2=\angle BAO_2$ which can never be equal to $\angle BAO_2$ so your claim is wrong

hydrohelium wrote: TheDarkPrince wrote: My solution (something like this): Let $O_1B \cap \Gamma_2 = D'$ and $O_2B \cap \Gamma_1 = C'$. As $\triangle O_1D'A$ is not isosceles, $\angle AO_1O_2 = \angle O_2O_1B = \angle O_2O_1D'$ and $\angle O_2AD'=\angle O_2D'A$ gives $O_1AO_2D'$ is cyclic. So, $D'\equiv D$. Similarly, $C'\equiv C$. Proved. I think $\angle AO_1O_2=\angle BAO_2$ which can never be equal to $\angle BAO_2$ so your claim is wrong How is $\angle AO_1O_2 = \angle BAO_2$ I think my solution works fine.

TheDarkPrince wrote: hydrohelium wrote: TheDarkPrince wrote: My solution (something like this): Let $O_1B \cap \Gamma_2 = D'$ and $O_2B \cap \Gamma_1 = C'$. As $\triangle O_1D'A$ is not isosceles, $\angle AO_1O_2 = \angle O_2O_1B = \angle O_2O_1D'$ and $\angle O_2AD'=\angle O_2D'A$ gives $O_1AO_2D'$ is cyclic. So, $D'\equiv D$. Similarly, $C'\equiv C$. Proved. I think $\angle AO_1O_2=\angle BAO_2$ which can never be equal to $\angle BAO_2$ so your claim is wrong How is $\angle AO_1O_2 = \angle BAO_2$ I think my solution works fine. People actually look up posts that old?

Too easy for INMO, anyways here is an alternate proof. Angle chase to get $C,B,O_{2},E$ are collinear. Similarly $D,B,O_{1},F$ are collinear. Then notice $\angle FCE = \angle EDF = 90^{\circ}$ (Angle in a semicircle). By inscribed angles theorem, $C,D,E$ and $F$ are collinear. $\blacksquare$

vlohani wrote: For those who didn't get Kayak's slick solution, he says: Invert about $B$. Then $A \to A^*$, $B \to B^*$ and $C \to C^*$, while $\Gamma_1$ goes to the straight line $A^*C^*$ and $\Gamma_2$ goes to the straight line $A^*D^*$. (Because $\angle O_1AO_2$ is obtuse, $C$ and $D$, and hence $C^*$ and $D^*$, are different from B.) $\odot (O_{1}AO_{2}) \to \odot (O_{1}^*A^*O_{2}^*)$, which is also the circumcircle of $\Delta A^*C^*D^*$. Now note: 1. $O_1^*$ is the reflection of $B$ in $A^*C^*$, 2. $O_2^*$ is the reflection of $B$ in $A^*D^*$. This forces $B$ to be the orthocenter of $A^*C^*D^*$ because there is only one point other than $A^*$ whose reflections in $A^*C^*$ and $A^*D^*$ respectively lie on $\odot (A^*C^*D^*)$ (why?) $-$ and the orthocenter clearly satisfies this condition. Hence, $B^*$ is the orthocenter of $\Delta A^*C^*D^*$. Since $E^*$ and $F^*$ are the feet of altitudes through $D^*$ and $C^*$ in $\Delta A^*C^*D^*$, it's trivial to note that $C^*D^*E^*F^*$ is cyclic. So how does O1*B perpendicular to A*C* imply that O1* is reflection of B in A*C*??

$\angle ACO_2=\angle AO_1O_2=\frac{1}{2} \angle AO_1B=\angle ACB$ $\implies$ $BC$ passes through $O_2$, similarly, $BD$ passes through $O_1$, hence, $\angle FCE$ $=$ $90^{\circ}$ $=$ $\angle EDF$ which implies the conclusion

What about radical axis

$\textcolor{red}{\text{Claim:}}$ $C,B,O_2$ are coliner. Similarly $O_1,B,D$ are similar . Let's join $B,O_2$ and $B,0_1$ .Note that $AB \perp O_1O_2$ since it's radial axis of $\Gamma_1,\Gamma_2$. Note that $\triangle AO_2B$ is isosceles .And hence $\angle AO_2O_1=\angle BO_2O_1=a$ and by similar argument we have $\angle BO_1O_2=\angle AO_1O_2=b$ and hence $ \angle O_1AO_2=\angle O_1BO_2=180-(a+b)$. Suppose$O_2B$ met $\Gamma_1$ at $C'$ . So $\angle O_1O_2C'=a$. Again$\angle O_1C'B=\angle C'O_1B$ . Which means $A,O_1,C',D$ are concyclic . $\Rightarrow C'=C$. Similarly we can show that $O_1,B,D$ are coliner. $\textcolor{red}{\text{Claim:}}$ $E,A,F$ are Coliner and $FE $ parallel to $O_1O_2$. In $\Gamma_2$ we get $\angle AO_2E =180-2b$ which gives $\angle AEB=\angle O_1O_2B =b$. So $AE$ parallel to $O_1O_2$ by similar argument $ AF$ parallel to $O_1O_2$. Our claim is proved. Note that $C,O_1,O_2,D$ are concyclic so $CB.BO_2=DB.BO_1 \cdots (*)$. Since $ \triangle BO_1O_2 \sim BEF$ hence $ \frac{BO_1}{BF}=\frac{BO_2}{BE}$. From $(*)$ we get $ CB.BE =DB .BF$ so we are done .

XbenX wrote: A different approach: After some easy angle chasing we get that $C,B,O_2,E$ and $ D,B,O_1,F$ are collinar in that order . By Power of point $B$ on $(CDO_2O_1) \Longleftrightarrow BD \cdot BO_1=BC \cdot BO_2$ and since $BO_1$ and $BO_2$ are the radiuses of their respective circles , we get that $ BD \cdot BF=BC \cdot BE$ wich completes the proof. First you need to prove that $(CDO_2O_1) $ is cyclic..

Wait what? i just realised this problem is a part of https://artofproblemsolving.com/community/q1h1952359p21527833 $C-B-O_2$ are collinear, so is $D-B-O_1$ $FB$ and $BE$ is diameter so $\angle FCE = 90^\circ = \angle EDF$ and we are done. edit- arulxz wrote: Prove by angle chasing that $B$ is the incenter of $\triangle ACD$. yes indeed, this was the brazil problem lol

$$\angle CAB =\angle DAB,\angle FAB = \angle EAB \implies FB \cdot BD = CB \cdot BE$$

$$\measuredangle DO_1O_2 =\measuredangle DAO_2 =\measuredangle O_2DA =\measuredangle O_2O1A =\measuredangle BO_1O_2$$$\implies D,B,O_1$ are collinear, hece $BE$ is the diameter of $\Gamma_1$ similiarly we get $BF$ as the diameter of $\Gamma_2$, $\implies \measuredangle FDE = \measuredangle FCE = 90^{\circ}$, Hence $C,D,E,F$ are concyclic

Claim: $C,B,O_2$ and $D,B,O_1$ are collinear. Proof : $\angle ADO_1 = \angle AO_2O_1 = \angle AO_2B/2 = \angle ADB$ so $D,B,O_1$ are collinear. we prove the other one with same approach. Now we have $\angle ECF = \angle BCF = \angle 90 = \angle BDE = \angle FDE$ so $CDEF$ is cyclic.

Just invert about $B$, using the inversion distance formula we find out that $O_1'$ and $O_2'$ are the reflection of $B$ over $A'C'$ and $A'D'$ respectively, then note that since both $O_1'$ and $O_2'$ lie on $(A'C'D')$, $B'$ is the orthocenter of $\triangle{A'C'D'}$ and therefore, $C,B,O_2,E$ and $D,B,O_1,F$ are in fact collinear, the conclusion now follows immediately.

I hope you enjoyed reading my proof

Let $\angle O_1BC=\angle O_1CB=\alpha$, $\angle O_1AB=\angle O_1BA=\beta$, and $\angle O_2AB=\angle O_2BA=\gamma.$ Main Claim: $C,B,O_2$ are collinear. We wish to show that $\alpha+\beta+\gamma=180$. Note that $\angle AO_2O_1=90-\gamma$ since $AB\perp O_1O_2$. Thus, from cyclic $AO_1CO_2$, we have $\angle ACO_1=90-\gamma$. Since $O_1C=O_1A$, $\angle AO_1C=2\gamma$. However, we have $\angle AO_1B=180-2\beta$, and $\angle BO_1C=180-2\alpha$. Hence, $$2\gamma=180-2\beta+180-2\alpha\rightarrow \alpha+\beta+\gamma=180,$$as desired. Hence, $BE$ is a diameter of $\Gamma_2$, so $\angle BDE=90$. Similarly, $\angle BCF=90$. Hence, $C,D,E,F$ are concyclic and we are done.

Bro what? Why Invert this ? Anyways here is the sketch (typing full solns is too stressful due to my 50 WPM ). You get that $E$ and $F$ are the $B-$antipodes in $\Gamma_1$ and $\Gamma_2$ respectively from some angle chasing. Then a bit of homothety shows $EF\parallel O_1O_2$ and $\overline{A-E-F}$ and so you get $\measuredangle ECF=\measuredangle ECB=90^\circ=\measuredangle BDF=\measuredangle EDF$ which finishes.

To be fair, inversion (especially about $A$) doesn't help a ton, but I'll just stick with it In the inverted picture, $\omega_1^*$ and $\omega_2^*$ become two lines that intersect at $B$, and $(O_1AO_2)$ becomes the line $\overline{CD}$, on which the reflections of $A$ over $\omega_1^*$ and $\omega_2^*$ lie. Now, observe $O_1$ and $O_2$ lie on $(ABD)$ and $(ACD)$, the two images of the lines, respectively, by the converse of Fact 5; $A, E, F$ are collinear as $\angle BAE = \angle BAF = 90^\circ$; $CDFE$ is cyclic by a quick angle chase. Hence done!

We invert about $A$ with radius $1$. We can see that $F^*C^*B^*$ are collinear, $E^*D^*B^*$ are collinear, $O_1^*C^*D^*O_2^*$ are collinear, $F^*O_1^*B^*D^*$ are concyclic, and $E^*O_2^*B^*C^*$ are concyclic. \begin{lemma} $O_1^*$ is the reflection of $A$ over $F^*B^*$. \end{lemma} \begin{proof} EGMO lemma 8.10 \end{proof} This we have \[\angle AF^*C^* = \angle C^*F^*O_1^* = \angle C^*D^*B^*\]$\blacksquare$

Me when I wanted a medium geo and tried this only for it to turn out trivial. The following observation is key. Claim : The points $D$ , $B$ and $O_1$ are collinear. Similarly, the points $C$ , $B$ and $O_2$ are collinear. Proof : Simply note that \[\measuredangle AO_1D = \measuredangle AO_2D = 2\measuredangle ADO_2 = 2\measuredangle AO_1O_2 = \measuredangle AO_1B\]from which it is clear that $D-B-O_1$ as claimed. Similarly, we can also show $C-B-O_2$. Now, note that this means $BF$ and $BE$ are diameters of $\Gamma_1$ and $\Gamma_2$ respectively. This then gives us, \[\measuredangle FCE = \measuredangle FCB = 90^\circ = \measuredangle BDE = \measuredangle FDE\]from which it is clear that the points $C, D, E, F$ are concyclic as required.

Let $O_2B$ intersect $\Gamma_1$ at $C'$ (phantom point), then, $$\angle O_1C'B = \angle O_1BC' = 180^o - \angle O_1BO_2$$As $O_1$ and $O_2$ are the centers of $\Gamma_1$ and $\Gamma_2$, we have $O_1B=O_1A$, $O_2B=O_2A$, which implies that, $$ \angle O_1BA = \angle O_1AB , \angle O_2BA = \angle O_2AB$$$\therefore \angle O_1BO_2 = \angle O_1AO_2$ Hence, $\angle O_1C'B = \angle O_1AO_2 = \angle O_1CB$, $\Rightarrow C \equiv C'$, $\therefore C,B,O_2,E $ are collinear, Similarly, $ D,B,O_1,F$ are collinear, $\Rightarrow BF$ and $BE$ are diameters of $\Gamma_1$ and $\Gamma_2$ respectively. $\therefore \angle BDE= \angle FDE = \angle FCE = 90^o$, $\Rightarrow CDEF$ is a cyclic quadrilateral.

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