## IMO Marathon

Discussion on International Mathematical Olympiad (IMO)
sourav das
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### Re: IMO Marathon

Come on guys, IMO standard problems won't be easy to crack (Doesn't mean ugly). They'll be smart and hidden. So please post a new one.
You spin my head right round right round,
When you go down, when you go down down......
(-$from$ "$THE$ $UGLY$ $TRUTH$" )

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### Re: IMO Marathon

sourav das wrote:Come on guys, IMO standard problems won't be easy to crack (Doesn't mean ugly). They'll be smart and hidden. So please post a new one.
Why don't you post one A samrt one ?
$\frac{1}{0}$

*Mahi*
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### Re: IMO Marathon

Problem $\boxed 2$ is a easy one if you just remember this:
If $a+x \mid b+x$ for infinitely many $x$, then $a=b$.
(And also, guys, sorry I wasn't here for day 1 as today there was ACM preliminary round , I hope I'll be regular here from now on)

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

sourav das
Posts: 461
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### Re: IMO Marathon

sourav das wrote:Come on guys, IMO standard problems won't be easy to crack (Doesn't mean ugly). They'll be smart and hidden. So please post a new one.
Why don't you post one A samrt one ?
A good one (I assume):
IberoAmerican 2012:
Problem 3:
Let $ABC$ be a triangle, $P$ and $Q$ the intersections of the parallel line to $BC$ that passes through $A$ with the external angle bisectors of angles $B$ and $C$, respectively. The perpendicular to $BP$ at $P$ and the perpendicular to $CQ$ at $Q$ meet at $R$. Let $I$ be the incenter of $ABC$. Show that $AI = AR$.
You spin my head right round right round,
When you go down, when you go down down......
(-$from$ "$THE$ $UGLY$ $TRUTH$" )

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### Re: IMO Marathon

Solution: (problem 3)
Let $D$ be the ex-center opposite to $A$ of $\triangle ABC$.
We know $BCDI$ con-cyclic and $A,I,D$ collinear. Next notice that $\triangle IBC$ and $\triangle RPQ$ are homothetic. So $R,I,D,A$ collinear. Now notice that $RFIE$ is a parallelogram. So the result follows immediately.
Attachments geo.png (25.58 KiB) Viewed 2825 times

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

### Re: IMO Marathon

sourav das wrote:Problem 3:
Let $ABC$ be a triangle, $P$ and $Q$ the intersections of the parallel line to $BC$ that passes through $A$ with the external angle bisectors of angles $B$ and $C$, respectively. The perpendicular to $BP$ at $P$ and the perpendicular to $CQ$ at $Q$ meet at $R$. Let $I$ be the incenter of $ABC$. Show that $AI = AR$.
Let $I_a$ be the $A$ excircle of $\triangle ABC$.
So $\angle RPI_a+\angle RQI_a=180^{\circ} \Rightarrow RPI_aQ$ is concyclic.
$\angle AQC=90^{\circ}-\frac{1}{2}\angle C \Rightarrow \angle RQP=\frac{1}{2}\angle C$
So $\angle BI_aR=\frac{1}{2}\angle C.$
Again $\angle BI_aA=\frac{1}{2}\angle C$.
Hence $\angle BI_aA=\angle BI_aR \Rightarrow R,A,I_a$ are collinear.
Note that $AB=AP,AC=AQ$.
By power of point $AR.AI_a=bc$.
So it suffices to prove that $AI.AI_a=bc$
$\leftrightarrow \frac{s.AI^2}{s-a}=bc$
$\leftrightarrow \frac{s.(s-a)^2}{(s-a).\cos^2\frac{A}{2}}=bc$
$\leftrightarrow \cos^2\frac{A}{2}=\frac{s.(s-a)}{bc}$
which is actually true.
বড় ভালবাসি তোমায়,মা

*Mahi*
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### Re: IMO Marathon

Solution to problem $\boxed 3$:
[Sorry beforehands for a little confusion in the names of the points]
Attachments IMOmarathon-3.png (63.68 KiB) Viewed 2821 times

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Tahmid Hasan
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### Re: IMO Marathon

Problem 4:A circle $K$ passes through the vertices $B;C$ of $\triangle ABC$ and another circle $\omega$ touches
$AB;AC;K$ at $P;Q;T;$ respectively. If $M$ is the midpoint of the arc $BTC$ of $K$; show that $BC;PQ;MT$ concur.
Source:Luis González
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FahimFerdous
Posts: 176
Joined: Thu Dec 09, 2010 12:50 am

### Re: IMO Marathon

At first I'm very sorry for not posting till now. I've been checking and trying to solve the problems but as I'm using a CLASSIC Nokia 6101, (if you know what I mean) I can't post very comfortably. I have a solution for problem 3, but as I dnt have latex, it'd be hard to read for you, if anyone wants to read it in the first case. I'm sorry for that. [Edit: don't worry, it's LaTeXed now ]

Extend $PB$ and $QC$, let them meet at $T$. Then $T$ is the excentre opposite $A$. Now, as it's well known that $A, I, T$ are collinear and $\triangle RPQ$ and $\triangle IBC$ are homothetic, it implies that $R$ lies on line $AI$. Now, $\triangle RPA$ and $\triangle IBD$ are homothetic too where $D$ is the intersection point of $AI$ and $BC$. Now from sine law we can show that $\frac{ID}{AR}=\frac{BD}{AP}=\frac{BD}{AB}=\frac{DI}{AI}$. Which implies $AR=AI$. And $AB=AP$ comes from simple angle chasing.
Last edited by *Mahi* on Sun Nov 11, 2012 1:44 pm, edited 1 time in total.
Reason: LaTeXed $\color{blue}{\textit{To}} \color{red}{\textit{ problems }} \color{blue}{\textit{I am encountering with-}} \color{green}{\textit{AVADA KEDAVRA!}}$