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A Very Nice Problem

Posted: Sun Feb 24, 2013 11:10 am
by FahimFerdous
Let $O$ be the circumcircle of triangle $ABC$. Let $AS$ and $AM$ be the symmedian and median respectively. Let $AS \cap O=S$ and $AM \cap BC=M$. Let $SM \cap O=A_1$ and let $A_2$ be the diametrically opposite point of $A$. Let $A_1A_2 \cap BC=A_3$. Similarly define $B_3$ and $C_3$. Prove that, $AA_3, BB_3, CC_3$ are concurrent.

Re: A Very Nice Problem

Posted: Sun Feb 24, 2013 7:07 pm
by Tahmid Hasan
Similar to APMO-2012-4.
Here's the sketch of my solution.
Lemma: In $\triangle ABC, P \in BC$. Then $\frac{MB}{MC}=\frac{AB}{AC}.\frac{\sin BAP}{\sin CAP}$.
Use this lemma and sine law to prove $ABSC$ is a harmonic quadrilateral i.e. $\frac{AB}{AC}=\frac{BS}{SC}$.
Use them again to prove $\frac{BS}{SC}=\frac{A_1C}{A_1B}$ which will imply $A,A_1$ are symmetric wrt the perpendicular bisector of $BC$.
Let $AH_a \perp BC,H_a \in O$. It is well-known that $H_a,A_2$ are symmetric wrt the perpendicular bisector of $BC$. So we conclude $A_1A_2 \perp BC$. Now use Ceva to finish things off.

Re: A Very Nice Problem

Posted: Sun Feb 24, 2013 9:58 pm
by FahimFerdous
Hmm, Sourav gave the same one. This is the solution most people would give. I was hoping for another way. Can anyone try it in a different way?

Re: A Very Nice Problem

Posted: Mon Feb 25, 2013 12:43 am
by *Mahi*
FahimFerdous wrote:Hmm, Sourav gave the same one. This is the solution most people would give. I was hoping for another way. Can anyone try it in a different way?
Same solution here, but Zadid told me you used something like butterfly's theorem, if nobody posts that, can you post it?

Re: A Very Nice Problem

Posted: Mon Feb 25, 2013 6:57 pm
by FahimFerdous
Yeah, yeah, this is a problem that Zadid made himself thinking of using Butterfly. But as now it's exposed, anyone can give the proof using butterfly. If anyone doesn't, then I'd give Zadid's solution. Or Zadid could give it himself.

Re: A Very Nice Problem

Posted: Fri Mar 01, 2013 5:43 pm
by FahimFerdous
The solution with butterfly seems long but uses many beautiful ideas. Let $H$ be the orthocentre and $AH \cap BC=D$. By homothety, we can show that, $A_2, M, H$ are collinear. Let $AH \cap O=P$. By angle chasing we can show that, $P, D, S$ are collinear. Now, using butterfly on chords $A_2P$ and $SA_1$ we get $BD=A_3C$. Now apply ceva and kill it as we know that the perpendiculars in a triangle are concuurent at $H$. :mrgreen:

Re: A Very Nice Problem

Posted: Sat Mar 02, 2013 3:22 pm
by zadid xcalibured
Mahi and Fahim vai, cool avatars. :mrgreen:
This is my 200th post. :mrgreen:

Re: A Very Nice Problem

Posted: Sun Mar 03, 2013 12:24 am
by FahimFerdous
My avatar is the sign of an anime character. He's awesome! :mrgreen:

Re: A Very Nice Problem

Posted: Sun Mar 03, 2013 4:07 pm
by Tahmid Hasan
FahimFerdous wrote:My avatar is the sign of an anime character. He's awesome! :mrgreen:
Actually Mahi's avatar also symbolizes another popular anime character. Do you remember Gaara especially his forehead seal ;) ?

Re: A Very Nice Problem

Posted: Sun Mar 03, 2013 10:29 pm
by FahimFerdous
Oops, missed that one!