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### IMO Marathon

Posted: Sat Nov 10, 2012 10:29 am
Here we'll talk about IMO level problems.
Rules:
1. You can post any 'math-problem' from anywhere if you are sure it has a solution. Also you should give the source.(Like book name, link or self-made)
2. If a problem remains unsolved for two days, the proposer must post the solution (for self-made 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.

### Re: IMO Marathon

Posted: Sat Nov 10, 2012 10:36 am
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

### Re: IMO Marathon

Posted: Sat Nov 10, 2012 6:48 pm
Waiting for confirmation

### Re: IMO Marathon

Posted: Sat Nov 10, 2012 7:40 pm
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.

### Re: IMO Marathon

Posted: Sat Nov 10, 2012 8:07 pm
Edited some comma ,.. :S

### Re: IMO Marathon

Posted: Sat Nov 10, 2012 8:15 pm
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 NMO-2004-P4.

### Re: IMO Marathon

Posted: Sat Nov 10, 2012 8:28 pm
SANZEED wrote: $LQ\parallel LQ'$, which is impossible.
What if Q' lie on LQ ?

### Re: IMO Marathon

Posted: Sat Nov 10, 2012 8:51 pm
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.
$P(p-1,1)\Rightarrow f(p-1)+f(1)|p-1+1=p$. Since $f(p-1)1>1$ we must have $f(p-1)=p-1$.
Now, $f(p-1)+f(n)|p-1+n\Rightarrow (p-1+f(n))|(p-1+f(n))+(n-f(n))$.
So, $(p-1+f(n))|(n-f(n))$.
If we fix $n$ now then we can take arbitrarily large value of $p$, such that $p-1+f(n)>n-f(n)$. Still $(p-1+f(n))$ will divide $(n-f(n))$
so we must have $n-f(n)=0$ i.e. $f(n)=n\forall n\in \mathbb N$ which is indeed a solution.

### Re: IMO Marathon

Posted: Sat Nov 10, 2012 9:00 pm
SANZEED wrote: $LQ\parallel LQ'$, which is impossible.
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.
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.