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

by Phlembac Adib Hasan

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

by Nadim Ul Abrar

Waiting for confirmation

It seems the problem statement is not true .We may disprove the statement .

The condition $\angle KAP =90- \angle LBP $ says $\triangle KQP$ is right angled triangle with right angle $Q$ .
Now draw a perpendicular $QX$ from $Q$ to $KL$ so that $KL\cap QX=X$ .

Then the line through the circumcenters $O,O'$ of circles $(KXQ)$ and $(LQX)$ will be parellal to $KL$.

But for no other position of $X$ , So that $OO'$ be parellal to $KL$ .

So $PQ$ must be perpendicular to $KL$ .

But for some other position of $X$ on $KL$ say $X'$, We can find infinity pairs of point $A,B$ on circles $(KX'Q) , (LX'Q)$ so that $\angle KX'A=\angle LX'B$ though $OO'$ is not parellal to $KL$

Re: IMO Marathon

Posted: Sat Nov 10, 2012 7:40 pm

by SANZEED

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.

$\angle KQL=\angle KQP+\angle LQP=\angle KAP+\angle LBP=90^{\circ}$. So the midpoint of $KL$, say $C$ is the center of $\bigodot KQL$. Also, $KQ$ is the radical axis of $\bigodot KQL,\bigodot AKPQ$. Thus by a well-known fact, $CO\perp KQ$. Similarly $CO'\perp LQ$. Now let $O\notin KQ$. Then draw the diameter of $\bigodot AKPQ$ through $K$. Let this diameter intersect the circle again at $Q'$. Then, $\frac{KO}{OQ'}=\frac{KC}{CL}=1$. Thus $CO\parallel LQ'$. But since both $CO,LQ$ are perpendicular to $KQ$, we have that $CO\parallel LQ$. Thus , $LQ\parallel LQ'$, which is impossible. So,$O\in KQ$. Similarly $O'\in LQ$. Now, $\frac{KO}{OQ}=\frac{LO'}{O'Q}=1$ and as a result $KL\parallel OO'$

Adib,please confirm which one of us is correct.

Re: IMO Marathon

Posted: Sat Nov 10, 2012 8:07 pm

by Nadim Ul Abrar

Edited some comma ,.. :S

Re: IMO Marathon

Posted: Sat Nov 10, 2012 8:15 pm

by Tahmid Hasan

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

by Nadim Ul Abrar

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

by SANZEED

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

by SANZEED

Nadim Ul Abrar wrote:

SANZEED wrote: $LQ\parallel LQ'$, which is impossible.

What if Q' lie on LQ ?

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.

Re: IMO Marathon

Posted: Sat Nov 10, 2012 9:10 pm

by sourav das

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.