BdMO Online Forum

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.

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 ?

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)

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 $AI = AR$.

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.

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.

Solution to problem $\boxed 3$:
[Sorry beforehands for a little confusion in the names of the points]

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

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.

Can anyone confirm if this is the correct figure?