A Very Nice Problem

For discussing Olympiad level Geometry Problems
FahimFerdous
Posts: 176
Joined: Thu Dec 09, 2010 12:50 am

A Very Nice Problem

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.

Tahmid Hasan
Posts: 665
Joined: Thu Dec 09, 2010 5:34 pm

Re: A Very Nice Problem

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.
বড় ভালবাসি তোমায়,মা

FahimFerdous
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Joined: Thu Dec 09, 2010 12:50 am

Re: A Very Nice Problem

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?

*Mahi*
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Re: A Very Nice Problem

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?

Use $L^AT_EX$, It makes our work a lot easier!

FahimFerdous
Posts: 176
Joined: Thu Dec 09, 2010 12:50 am

Re: A Very Nice Problem

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.

FahimFerdous
Posts: 176
Joined: Thu Dec 09, 2010 12:50 am

Re: A Very Nice Problem

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 $H$.

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Re: A Very Nice Problem

Mahi and Fahim vai, cool avatars.
This is my 200th post.

FahimFerdous
Posts: 176
Joined: Thu Dec 09, 2010 12:50 am

Re: A Very Nice Problem

My avatar is the sign of an anime character. He's awesome!

Tahmid Hasan
Posts: 665
Joined: Thu Dec 09, 2010 5:34 pm

Re: A Very Nice Problem

FahimFerdous wrote:My avatar is the sign of an anime character. He's awesome!
Actually Mahi's avatar also symbolizes another popular anime character. Do you remember Gaara especially his forehead seal ?
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FahimFerdous
Posts: 176
Joined: Thu Dec 09, 2010 12:50 am