2006 China Western Math Olympiad

[FahimFerdous]

AB is a diametre of a circle with centre O. C is a point on AB extended. A line through C cuts the circle with centre O at D, E. OF is a diametre of the circumcircle of triangle BOD with centre O'. Line CF intersect the circumcirle again at G. Prove that, O, A, E, G are concyclic.

[sourav das]

As, $FB\perp OB$ and $FD\perp OD$ it follows that $F$ is the pole of $BD$. Let $AE\cap BD= H$, $AD\cap BE=M$. As $H\in BD$ Polar of C, By La-hire's theorem, $F\in $Polar of H. and as we know that $HMC$ is self polar wrt. circle $ABE$ it follows that $C\in $Polar of H. That means $CF$ is the pole of $H$. But as $OG\perp CF$ , $OG$ meet $H$.
Now by power of point theorem : $HD.HB=HG.HO=HE.HA$ implies $O,G,E,A$ cyclic.

[*Mahi*]

My solution is quite ugly I had to use inversion and complex

[FahimFerdous]

I used pole polar too. But the last part of my solution is quite different from Sourav. And Mahi, sometimes Euclid rocks.

[*Mahi*]

I know that (though I'm short of time now and can't spend my time finding nicer proofs )

[zadid xcalibured]

euclid always rocks.algebraic methods have their styles.but ugly.

[zadid xcalibured]

when i was working with xcaliburs,i faced this problem.as i knew it had to be solved by pole polar,i found no difficulties solving this problem but how do u figure out how to prove it?ur solution is marvellous.sourav da.

[Phlembac Adib Hasan]

Sourav Vaia, your proof is really beautiful.I like.
My Proof :
Let $Q$ be the intersection of arc $BD$ and $OF$.It is easy to prove $F$ and $Q$ are the midpoints of arc $BD$ of two circles (Notice that $\triangle BOD$ isosceles).\[\angle BGF=\angle BOF=\frac {1}{2}\angle BOD=\angle BED\]So $B,C,E,G$ con-cyclic.As $BE\perp AE$ and $GC\perp OG$,\[\angle AEG=90^0-\angle BEG=90^0-\angle BCG=\angle COG.\]And we are done.
Figure :

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