IMO Marathon
 Phlembac Adib Hasan
 Posts: 1016
 Joined: Tue Nov 22, 2011 7:49 pm
 Location: 127.0.0.1
 Contact:
IMO Marathon
Here we'll talk about IMO level problems.
Rules:
1. You can post any 'mathproblem' from anywhere if you are sure it has a solution. Also you should give the source.(Like book name, link or selfmade)
2. If a problem remains unsolved for two days, the proposer must post the solution (for selfmade problems) or the official solution will be posted. (for contest problems)
3. Anyone can post a new problem iff the previous problem has been solved already.
4. Don't forget to type the problem number.
Rules:
1. You can post any 'mathproblem' from anywhere if you are sure it has a solution. Also you should give the source.(Like book name, link or selfmade)
2. If a problem remains unsolved for two days, the proposer must post the solution (for selfmade problems) or the official solution will be posted. (for contest problems)
3. Anyone can post a new problem iff the previous problem has been solved already.
4. Don't forget to type the problem number.
 Phlembac Adib Hasan
 Posts: 1016
 Joined: Tue Nov 22, 2011 7:49 pm
 Location: 127.0.0.1
 Contact:
Re: IMO Marathon
Problem 1:
Circles $\omega _1$ with center $O$, $\omega _2$ with center $O'$, intersect at points $P,Q$. line $\ell$ passes through $P$ and intersects $\omega _1$, $\omega _2$ at $K,L$ ,respectively. points $A,B$ are on arcs $KQ,LQ$ (arcs do not contain $P$ ) respectively. $\angle KPA=\angle LPB,\angle KAP=90\angle LBP$. Prove that $OO'$ is parallel to $KL$.
source: http://www.artofproblemsolving.com/Foru ... 6&t=506365
Circles $\omega _1$ with center $O$, $\omega _2$ with center $O'$, intersect at points $P,Q$. line $\ell$ passes through $P$ and intersects $\omega _1$, $\omega _2$ at $K,L$ ,respectively. points $A,B$ are on arcs $KQ,LQ$ (arcs do not contain $P$ ) respectively. $\angle KPA=\angle LPB,\angle KAP=90\angle LBP$. Prove that $OO'$ is parallel to $KL$.
source: http://www.artofproblemsolving.com/Foru ... 6&t=506365
Welcome to BdMO Online Forum. Check out Forum Guides & Rules
 Nadim Ul Abrar
 Posts: 244
 Joined: Sat May 07, 2011 12:36 pm
 Location: B.A.R.D , kotbari , Comilla
Re: IMO Marathon
Waiting for confirmation
 Attachments

 Capture.PNG (35.4 KiB) Viewed 4472 times
Last edited by Nadim Ul Abrar on Sat Nov 10, 2012 8:05 pm, edited 1 time in total.
$\frac{1}{0}$
Re: IMO Marathon
Well,I am confused about your last conclusion, Nadim vai. I have the following proof in favor of the statement.And I think there is an error in your proof.
Adib,please confirm which one of us is correct.
$\color{blue}{\textit{To}} \color{red}{\textit{ problems }} \color{blue}{\textit{I am encountering with}} \color{green}{\textit{AVADA KEDAVRA!}}$
 Nadim Ul Abrar
 Posts: 244
 Joined: Sat May 07, 2011 12:36 pm
 Location: B.A.R.D , kotbari , Comilla
 Tahmid Hasan
 Posts: 665
 Joined: Thu Dec 09, 2010 5:34 pm
 Location: Khulna,Bangladesh.
Re: IMO Marathon
Sorry I couldn't check the proofs of Sanzeed and Nadim vai since I'm a little busy with my texts.
I found another constructive disproof.
Choose any point $K$ on $\omega_1$ such that $KQ$ is not perpendicular to $PQ$.[We still haven't defined $\omega_2$].
Now draw $KQ \bot QL$ suth that $L \in KP$. Draw the circumcircle of $\triangle PQL$. Let this be $\omega_2$.
Now take any point $A$ on arc $KQ$. Take point $B$ on arc $QL$ such that $\angle KPA=\angle LPB$.
Notice that the construction satisfies all the properties of the problem yet doesn't satisfies the requirement.
Note: This is a confusing problem. So I'm posting another problem.
Problem 2: Determine all functions $f:\mathbb{N}\to\mathbb{N}$ such that for every pair $(m,n)\in\mathbb{N}^2$ we have that:
\[f(m)+f(n)m+n\]
Source: Iran NMO2004P4.
I found another constructive disproof.
Choose any point $K$ on $\omega_1$ such that $KQ$ is not perpendicular to $PQ$.[We still haven't defined $\omega_2$].
Now draw $KQ \bot QL$ suth that $L \in KP$. Draw the circumcircle of $\triangle PQL$. Let this be $\omega_2$.
Now take any point $A$ on arc $KQ$. Take point $B$ on arc $QL$ such that $\angle KPA=\angle LPB$.
Notice that the construction satisfies all the properties of the problem yet doesn't satisfies the requirement.
Note: This is a confusing problem. So I'm posting another problem.
Problem 2: Determine all functions $f:\mathbb{N}\to\mathbb{N}$ such that for every pair $(m,n)\in\mathbb{N}^2$ we have that:
\[f(m)+f(n)m+n\]
Source: Iran NMO2004P4.
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
What if Q' lie on LQ ?SANZEED wrote: $LQ\parallel LQ'$, which is impossible.
$\frac{1}{0}$
Re: IMO Marathon
For Problem 2
Let us denote the statement with $P(m,n)$. Now,
$P(1,1)\Rightarrow 2f(1)2\Rightarrow f(1)=1$ since $f(m)\in \mathbb N$
Let $p$ be a prime.
$P(p1,1)\Rightarrow f(p1)+f(1)p1+1=p$. Since $f(p1)1>1$ we must have $f(p1)=p1$.
Now, $f(p1)+f(n)p1+n\Rightarrow (p1+f(n))(p1+f(n))+(nf(n))$.
So, $(p1+f(n))(nf(n))$.
If we fix $n$ now then we can take arbitrarily large value of $p$, such that $p1+f(n)>nf(n)$. Still $(p1+f(n))$ will divide $(nf(n))$
so we must have $nf(n)=0$ i.e. $f(n)=n\forall n\in \mathbb N$ which is indeed a solution.
Let us denote the statement with $P(m,n)$. Now,
$P(1,1)\Rightarrow 2f(1)2\Rightarrow f(1)=1$ since $f(m)\in \mathbb N$
Let $p$ be a prime.
$P(p1,1)\Rightarrow f(p1)+f(1)p1+1=p$. Since $f(p1)1>1$ we must have $f(p1)=p1$.
Now, $f(p1)+f(n)p1+n\Rightarrow (p1+f(n))(p1+f(n))+(nf(n))$.
So, $(p1+f(n))(nf(n))$.
If we fix $n$ now then we can take arbitrarily large value of $p$, such that $p1+f(n)>nf(n)$. Still $(p1+f(n))$ will divide $(nf(n))$
so we must have $nf(n)=0$ i.e. $f(n)=n\forall n\in \mathbb N$ which is indeed a solution.
Last edited by SANZEED on Sat Nov 10, 2012 9:00 pm, edited 1 time in total.
$\color{blue}{\textit{To}} \color{red}{\textit{ problems }} \color{blue}{\textit{I am encountering with}} \color{green}{\textit{AVADA KEDAVRA!}}$
Re: IMO Marathon
Then connect $C$ and the midpoint of $KQ$ say $N$. Then $CN\parallel LQ$ and again $CO\parallel LQ$ which will bring the necessary contradiction,I think.Nadim Ul Abrar wrote:What if Q' lie on LQ ?SANZEED wrote: $LQ\parallel LQ'$, which is impossible.
$\color{blue}{\textit{To}} \color{red}{\textit{ problems }} \color{blue}{\textit{I am encountering with}} \color{green}{\textit{AVADA KEDAVRA!}}$

 Posts: 461
 Joined: Wed Dec 15, 2010 10:05 am
 Location: Dhaka
 Contact:
Re: IMO Marathon
For problem 1; Note that condition 2 implies those two circles are orthogonal. And 1st condition can be found for any $l$ for such two circles. So problem is certainly not true.
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......($from$ "$THE$ $UGLY$ $TRUTH$" )