IMO Marathon

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

Unread post by sourav das » Sat Nov 10, 2012 9:21 pm

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.
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Nadim Ul Abrar
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Re: IMO Marathon

Unread post by Nadim Ul Abrar » Sat Nov 10, 2012 9:30 pm

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 ?
$\frac{1}{0}$

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*Mahi*
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Re: IMO Marathon

Unread post by *Mahi* » Sat Nov 10, 2012 9:35 pm

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)
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sourav das
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Re: IMO Marathon

Unread post by sourav das » Sat Nov 10, 2012 9:40 pm

Nadim Ul Abrar wrote:
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 $AI = AR$.
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When you go down, when you go down down......
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Phlembac Adib Hasan
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Re: IMO Marathon

Unread post by Phlembac Adib Hasan » Sat Nov 10, 2012 10:19 pm

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.
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Tahmid Hasan
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Re: IMO Marathon

Unread post by Tahmid Hasan » Sat Nov 10, 2012 10:29 pm

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.
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*Mahi*
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Re: IMO Marathon

Unread post by *Mahi* » Sat Nov 10, 2012 10:30 pm

Solution to problem $\boxed 3$:
[Sorry beforehands for a little confusion in the names of the points]
$D$ is actually $P$, and $E$ is $Q$ of the statement.
$IBRC$ and $DJER$ are homothetic as $BI || RD$ and $CI || ER$.
Again, $\angle ADB = \angle CBJ = \angle ABD$, so $\triangle ABD$ is isosceles.
So perpendicular from $A$ on $BD$(i.e. $K$) is the midpoint of $BD$.
And similarly, $F$ is the midpoint of $CE$, with $AF \perp CE$.
As $AK || RD$, $AF || RE$ so $DJER$ and $AKJF$ are also homothetic.
So $\frac {RA}{AI}=\frac {DK}{KB} = 1$, which proves the statement.
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Tahmid Hasan
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Re: IMO Marathon

Unread post by Tahmid Hasan » Sat Nov 10, 2012 10:35 pm

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
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Re: IMO Marathon

Unread post by FahimFerdous » Sun Nov 11, 2012 12:15 pm

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
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SANZEED
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Re: IMO Marathon

Unread post by SANZEED » Sun Nov 11, 2012 11:59 pm

Can anyone confirm if this is the correct figure?
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$\color{blue}{\textit{To}} \color{red}{\textit{ problems }} \color{blue}{\textit{I am encountering with-}} \color{green}{\textit{AVADA KEDAVRA!}}$

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