IMO Marathon

Discussion on International Mathematical Olympiad (IMO)
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SANZEED
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Re: IMO Marathon

Unread post by SANZEED » Wed Jan 23, 2013 10:36 pm

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
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Nadim Ul Abrar
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Re: IMO Marathon

Unread post by Nadim Ul Abrar » Thu Jan 24, 2013 12:15 am

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
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
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Re: IMO Marathon

Unread post by Tahmid Hasan » Thu Jan 24, 2013 11:11 am

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

Unread post by *Mahi* » Fri Jan 25, 2013 9:33 pm

Tahmid Hasan wrote:
$MK=ML \leftrightarrow \angle MLK=\angle MKL$
$\leftrightarrow \angle MLB+\angle BLK=\angle MCL$
$\leftrightarrow \angle MCB+\angle SPL=\angle MCS-\angle LCS$
$\leftrightarrow \angle MCB+\angle SPL=\angle SPC-\angle LBC$
$\leftrightarrow \angle MCB+\angle LBC=\angle SPC-\angle SPL$
$\leftrightarrow \angle LPC=\angle LPC$
Important tip: use \Leftrightarrow or \Longleftrightarrow instead of \leftrightarrow for $\Leftrightarrow$ or $\Longleftrightarrow$.
Been out of home for a few days, hope I'll be regular in this marathon for at least next few days.
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Tahmid Hasan
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Re: IMO Marathon

Unread post by Tahmid Hasan » Sat Jan 26, 2013 9:02 pm

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.
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Nadim Ul Abrar
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Re: IMO Marathon

Unread post by Nadim Ul Abrar » Mon Jan 28, 2013 10:47 pm

Iran 10-5.PNG
Iran 10-5.PNG (57.32 KiB) Viewed 2754 times
Solution $\boxed {24}$
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
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Re: IMO Marathon

Unread post by Nadim Ul Abrar » Mon Jan 28, 2013 11:42 pm

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
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Tahmid Hasan
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Re: IMO Marathon

Unread post by Tahmid Hasan » Tue Jan 29, 2013 12:52 pm

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. :)
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*Mahi*
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Re: IMO Marathon

Unread post by *Mahi* » Tue Jan 29, 2013 4:05 pm

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
My বিশ্রী solution:
Let $k$ be any prime of the form $x \phi (m)+ 1$, as $\gcd (\phi(m) , 1) = 1$, there are infinitely many of them by Dirichlet Theorem.
As $k$ is a prime, $k \mid 2^{k-1} - 1$, also $m \mid 2^{\phi(m)} - 1 \mid 2^{x \phi(m) +1 -1} - 1 = 2^{k-1}-1$.
So, now, let's set $n = \frac {2^k - 2}{ m}$. So, $mn+1 = 2^k - 1$.
So, $\displaystyle \frac{2^n−1}{mn+1} = \frac {2^n - 1} {2^k - 1}$ which will be an integer if $k \mid n = \frac {2^k - 2}{ m}$.
As $k$ is a prime, $(k,m) = 1$ and thus $k \mid \frac{2^k - 2}{ m}$, so we have proved the claim.
Sorry again for my ugly solution
Anyone can take my turn to post a problem.
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Nadim Ul Abrar
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Re: IMO Marathon

Unread post by Nadim Ul Abrar » Tue Jan 29, 2013 6:59 pm

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

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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