## IMO Marathon

### Re: IMO Marathon

Problem $\boxed {23}$:

Let $a,b,c$ be positive real numbers such that $a+b+c=1$. Prove that

$\displaystyle\frac{ab}{\sqrt {ab+bc}}+\displaystyle\frac{bc}{\sqrt {bc+ca}}+\displaystyle\frac{ca}{\sqrt {ca+ab}} \leq \displaystyle\frac{1}{\sqrt {2}}$

Source: Chinese MO 2006

Let $a,b,c$ be positive real numbers such that $a+b+c=1$. Prove that

$\displaystyle\frac{ab}{\sqrt {ab+bc}}+\displaystyle\frac{bc}{\sqrt {bc+ca}}+\displaystyle\frac{ca}{\sqrt {ca+ab}} \leq \displaystyle\frac{1}{\sqrt {2}}$

Source: Chinese MO 2006

$\color{blue}{\textit{To}} \color{red}{\textit{ problems }} \color{blue}{\textit{I am encountering with-}} \color{green}{\textit{AVADA KEDAVRA!}}$

- Nadim Ul Abrar
**Posts:**244**Joined:**Sat May 07, 2011 12:36 pm**Location:**B.A.R.D , kotbari , Comilla

### Re: IMO Marathon

Another proof for $\boxed {22}$

Edit [ solution removed due to some bug . ( I dunno why i did consider CL ,CK as bisector ) ]

I'll fix it ,when my laptop be fixed

Edit [ solution removed due to some bug . ( I dunno why i did consider CL ,CK as bisector ) ]

I'll fix it ,when my laptop be fixed

Last edited by Nadim Ul Abrar on Thu Jan 24, 2013 10:19 pm, edited 2 times in total.

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- Tahmid Hasan
**Posts:**665**Joined:**Thu Dec 09, 2010 5:34 pm**Location:**Khulna,Bangladesh.

### Re: IMO Marathon

I'm a little confused.Nadim Ul Abrar wrote: Now $\displaystyle \angle DCL=180-(\frac{\angle D}{2}+\angle CLD)=180-(90-\angle ALD)=90+\angle ALD$

Here you concluded $\frac{\angle D}{2}+\angle CLD=90^{\circ}-\angle ALD$.

But according to my figure $90^{\circ}-\angle ALD=\angle CYL,\frac{\angle D}{2}+\angle CLD=\angle YCL$.

So for your arguement to be true $CL$ must be equal to $LY$ which isn't true according to my figure and neither can I find any reasoning behind that.

Please clarify and provide a decent figure.

বড় ভালবাসি তোমায়,মা

### Re: IMO Marathon

Important tip: use \Leftrightarrow or \Longleftrightarrow instead of \leftrightarrow for $\Leftrightarrow$ or $\Longleftrightarrow$.Tahmid Hasan wrote:

Been out of home for a few days, hope I'll be regular in this marathon for at least next few days.

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

Reviving the marathon...

Problem $24$: Circles $W_1,W_2$ intersect at $P,K$. $XY$ is common tangent of two circles which is nearer to $P$ and $X$ is on $W_1$ and $Y$ is on $W_2$. $XP$ intersects $W_2$ for the second time in $C$ and $YP$ intersects $W_1$ in $B$. Let $A$ be intersection point of $BX$ and $CY$. Prove that if $Q$ is the second intersection point of circumcircles of $ABC$ and $AXY$,$\angle QXA=\angle QKP$.

Source: Iran TST 2010-5.

Problem $24$: Circles $W_1,W_2$ intersect at $P,K$. $XY$ is common tangent of two circles which is nearer to $P$ and $X$ is on $W_1$ and $Y$ is on $W_2$. $XP$ intersects $W_2$ for the second time in $C$ and $YP$ intersects $W_1$ in $B$. Let $A$ be intersection point of $BX$ and $CY$. Prove that if $Q$ is the second intersection point of circumcircles of $ABC$ and $AXY$,$\angle QXA=\angle QKP$.

Source: Iran TST 2010-5.

বড় ভালবাসি তোমায়,মা

- Nadim Ul Abrar
**Posts:**244**Joined:**Sat May 07, 2011 12:36 pm**Location:**B.A.R.D , kotbari , Comilla

### Re: IMO Marathon

Let $(ABC)\cap W_1=D, DC \cap W_1=E$

Observe some facts (Can be proved easily )

1.$XY||BE$

2.$XE||AC$

$\angle KEX=\angle KPC=\angle KYC$ imply $Y,K,E$ are colinear .

Let $QY \cap BE =G$ . Now $\angle QGB =\angle QYX=180- \angle QAB$ imply $G$ lie on $(ABC)$ .

So $\angle GQD=\angle YQD=\angle GBD=\angle DKE$ imply $D,K,Y,Q$ cyclic .

Note that $AX=AY$ .

Now $\angle AYX=\angle AXY=\angle XBE =\angle XDE=\angle XDC$ imply $X,Y,C,D$ cyclic .(1)

So Due to some radical center stuff $DX,CY,KP$ concur at some point $F$

Now $\angle CYK=\angle CPK=\angle KDX $ imply $D,K,Y,F$ cyclic .(2)

(1),(2) imply ,$F,Q,Y,K$ cyclic .

So $\angle QXA=\angle QYA=\angle QYF=\angle QKF=\angle QKP$

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- Nadim Ul Abrar
**Posts:**244**Joined:**Sat May 07, 2011 12:36 pm**Location:**B.A.R.D , kotbari , Comilla

### Re: IMO Marathon

Problem $\boxed {25}$

Let $m$ be a positive odd integer. Prove that there exist infinitely many positive integer $n$ such that $\displaystyle \frac{2^n−1}{mn+1}$ is an integer.

source : Mongolia TST 2011

Let $m$ be a positive odd integer. Prove that there exist infinitely many positive integer $n$ such that $\displaystyle \frac{2^n−1}{mn+1}$ is an integer.

source : Mongolia TST 2011

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- Tahmid Hasan
**Posts:**665**Joined:**Thu Dec 09, 2010 5:34 pm**Location:**Khulna,Bangladesh.

### Re: IMO Marathon

Nice Solution, Nadim vai , I would like to post mine as well.

Since there are two major spiral similarities in the figure, I tried to exploit them as much as possible.

From alternate segment theorem, $\angle YXK=\angle XBK,\angle XYK=\angle YCK$.

Since a spiral similarity centered at $K$ sends $BX$ to $YC$, we get $\triangle KBX \sim \triangle KXY \sim \triangle KYC$.

Let $M,Z,N$ be the midpoints of $BX,XY,YC$ respectively. From radical axis theorem we get $K,P,Z$ are collinear.

So $\triangle KBM \sim \triangle KXZ \sim \triangle KYN$ and $\triangle KMX \sim KZY \sim KNC$.

Again a spiral similarity centered at $Q$ sends $XY$ to $BC$. So $\triangle QXY \sim \triangle QBC$.

Since $M,N$ are corresponding points, we conclude $\triangle QXY \sim \triangle QMN \sim \triangle QBC$.

$\triangle KBM \sim \triangle KYN \Rightarrow \angle KMB=\angle KNY \Rightarrow AMKN$ is cyclic.

$\triangle QXY \sim \triangle QMN \Rightarrow \angle QMN=\angle QXY=180^{\circ}-\angle QAZ \Rightarrow AQMN$ is cyclic.

So $A,Q,M,K,N$ are concyclic. Now $\angle QXA=\angle QXY-\angle AXY=\angle QMN-\angle BKX=\angle QKN-\angle ZKN=\angle QKP$.

Done! Awesome use of spiral similarity, quite as awesome as IMO-2006-G9.

Since there are two major spiral similarities in the figure, I tried to exploit them as much as possible.

From alternate segment theorem, $\angle YXK=\angle XBK,\angle XYK=\angle YCK$.

Since a spiral similarity centered at $K$ sends $BX$ to $YC$, we get $\triangle KBX \sim \triangle KXY \sim \triangle KYC$.

Let $M,Z,N$ be the midpoints of $BX,XY,YC$ respectively. From radical axis theorem we get $K,P,Z$ are collinear.

So $\triangle KBM \sim \triangle KXZ \sim \triangle KYN$ and $\triangle KMX \sim KZY \sim KNC$.

Again a spiral similarity centered at $Q$ sends $XY$ to $BC$. So $\triangle QXY \sim \triangle QBC$.

Since $M,N$ are corresponding points, we conclude $\triangle QXY \sim \triangle QMN \sim \triangle QBC$.

$\triangle KBM \sim \triangle KYN \Rightarrow \angle KMB=\angle KNY \Rightarrow AMKN$ is cyclic.

$\triangle QXY \sim \triangle QMN \Rightarrow \angle QMN=\angle QXY=180^{\circ}-\angle QAZ \Rightarrow AQMN$ is cyclic.

So $A,Q,M,K,N$ are concyclic. Now $\angle QXA=\angle QXY-\angle AXY=\angle QMN-\angle BKX=\angle QKN-\angle ZKN=\angle QKP$.

Done! Awesome use of spiral similarity, quite as awesome as IMO-2006-G9.

বড় ভালবাসি তোমায়,মা

### Re: IMO Marathon

My বিশ্রী solution:Nadim Ul Abrar wrote:Problem $\boxed {25}$

Let $m$ be a positive odd integer. Prove that there exist infinitely many positive integer $n$ such that $\displaystyle \frac{2^n−1}{mn+1}$ is an integer.

source : Mongolia TST 2011

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

- Nadim Ul Abrar
**Posts:**244**Joined:**Sat May 07, 2011 12:36 pm**Location:**B.A.R.D , kotbari , Comilla

### Re: IMO Marathon

I don't see why this is ugly . Is it for dirichlet ?*Mahi* wrote: My বিশ্রী solution:

problem $\boxed {26}$

Let $I$ be the incenter of acute $\triangle ABC$. Let $\omega$ be a circle with center $I$ that lies inside $\triangle ABC$ . $D, E, F$ are the intersection points of circle $\omega$ with the perpendicular rays from $I$ to

sides $BC, CA, AB$ respectively. Prove that lines $AD, BE, CF$ are concurrent.

Source : Excalibur .

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