IMO Marathon

 Posts: 461
 Joined: Wed Dec 15, 2010 10:05 am
 Location: Dhaka
 Contact:
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......($from$ "$THE$ $UGLY$ $TRUTH$" )
When you go down, when you go down down......($from$ "$THE$ $UGLY$ $TRUTH$" )
 Nadim Ul Abrar
 Posts: 244
 Joined: Sat May 07, 2011 12:36 pm
 Location: B.A.R.D , kotbari , Comilla
Re: IMO Marathon
Why don't you post one A samrt one ?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.
$\frac{1}{0}$
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)
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)
Please read Forum Guide and Rules before you post.
Use $L^AT_EX$, It makes our work a lot easier!
Nur Muhammad Shafiullah  Mahi
Use $L^AT_EX$, It makes our work a lot easier!
Nur Muhammad Shafiullah  Mahi

 Posts: 461
 Joined: Wed Dec 15, 2010 10:05 am
 Location: Dhaka
 Contact:
Re: IMO Marathon
A good one (I assume):Nadim Ul Abrar wrote:Why don't you post one A samrt one ?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.
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......($from$ "$THE$ $UGLY$ $TRUTH$" )
When you go down, when you go down down......($from$ "$THE$ $UGLY$ $TRUTH$" )
 Phlembac Adib Hasan
 Posts: 1016
 Joined: Tue Nov 22, 2011 7:49 pm
 Location: 127.0.0.1
 Contact:
Re: IMO Marathon
Solution: (problem 3)
Let $D$ be the excenter opposite to $A$ of $\triangle ABC$.
We know $BCDI$ concyclic 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.
Let $D$ be the excenter opposite to $A$ of $\triangle ABC$.
We know $BCDI$ concyclic 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 2698 times
 Tahmid Hasan
 Posts: 665
 Joined: Thu Dec 09, 2010 5:34 pm
 Location: Khulna,Bangladesh.
Re: IMO Marathon
Let $I_a$ be the $A$ excircle of $\triangle ABC$.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 $AI = AR$.
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 $
$\leftrightarrow \frac{s.AI^2}{sa}=bc$
$\leftrightarrow \frac{s.(sa)^2}{(sa).\cos^2\frac{A}{2}}=bc$
$\leftrightarrow \cos^2\frac{A}{2}=\frac{s.(sa)}{bc}$
which is actually true.
বড় ভালবাসি তোমায়,মা
Re: IMO Marathon
Solution to problem $\boxed 3$:
[Sorry beforehands for a little confusion in the names of the points]
[Sorry beforehands for a little confusion in the names of the points]
 Attachments

 IMOmarathon3.png (63.68 KiB) Viewed 2694 times
Please read Forum Guide and Rules before you post.
Use $L^AT_EX$, It makes our work a lot easier!
Nur Muhammad Shafiullah  Mahi
Use $L^AT_EX$, It makes our work a lot easier!
Nur Muhammad Shafiullah  Mahi
 Tahmid Hasan
 Posts: 665
 Joined: Thu Dec 09, 2010 5:34 pm
 Location: Khulna,Bangladesh.
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
$AB;AC;K$ 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
বড় ভালবাসি তোমায়,মা
 FahimFerdous
 Posts: 176
 Joined: Thu Dec 09, 2010 12:50 am
 Location: Mymensingh, Bangladesh
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.
[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
Reason: LaTeXed
Your hot head might dominate your good heart!
Re: IMO Marathon
Can anyone confirm if this is the correct figure?
 Attachments

 Marathon4.ggb.png (174.41 KiB) Viewed 2592 times
$\color{blue}{\textit{To}} \color{red}{\textit{ problems }} \color{blue}{\textit{I am encountering with}} \color{green}{\textit{AVADA KEDAVRA!}}$