Is this a JOKE? http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2669997&sid=a30c270e0ea868f1b6b6b3a82bf9aa9d#p2669997

It seems that it is not (or maybe very well prepared joke ). You can also find problems (in Bosnian but as I can see it is same problem) at http://www.infima.ba/images/pdf/BiHBosnian.pdf

Also on the AoPS Wiki https://www.artofproblemsolving.com/Wiki/index.php/2012_USAMO_Problems/Problem_4. If this is really the case, the shame and stain on Balkan MO will be hard to wipe out, and an investigation into the whereabouts of how this problem got on the shortlist, and then has been selected for the actual paper, is in high order.

The simplest explanation is that the author of the problem proposed it for both olympiads...

Seems the Indian Olympiad setters are not the only ones who take problems from other contests.

Igor wrote: The simplest explanation is that the author of the problem proposed it for both olympiads... I have only one thing to say here. The (well-known) problems 2-4 proposed by Saudi-Arabia. You know about the observers of Saudi-Arabia these years...

silouan wrote: I have only one thing to say here. The (well-known) problems 2-4 proposed by Saudi-Arabia. You know about the observers of Saudi-Arabia these years... I don't know about the observers of Saudi Arabia these years, and I don't think it is very polite to make insinuations like that. If problems 2-4 are well known, could you cite where they have previously appeared? Moreover, I doubt whether Problem 4 is well known; as was pointed out, it appeared in the USAMO this year. And I can't see how a Saudi Arabian proposer could get a problem on the USA Olympiad.

what happened with problem 3? I saw it some hours ago and now I can't find the topic.

anonymouslonely wrote: what happened with problem 3? I saw it some hours ago and now I can't find the topic. See here.

mymath7 wrote: If problems 2-4 are well known, could you cite where they have previously appeared? As far as I know problem 4 was proposed for IMO 2010 and was not selected for Shortlist and then appeared in USAMO 2012. mymath7 wrote: And I can't see how a Saudi Arabian proposer could get a problem on the USA Olympiad.

cefer wrote: As far as I know problem 3 was proposed for IMO 2010 and was not selected for Shortlist and then appeared in USAMO 2012. I suppose you meant Problem 4, since Problem 3 did not appear in USAMO 2012. cefer wrote:

Thank you for your helpful hint. I would rather not say more at this point until more is known.

How do you know problems 2 and 4 were proposed by Saudi Arabia? First of all, is Saudi Arabia, a guest country to the Balkan Mathematical Olympiad, even allowed to submit problems? (I thought guest countries cannot even vote at motions of the International Jury); secondly, usually the authors are only disclosed later on. If however, this is the case, please take note that Saudi Arabia trainers are not only foreign, but traditionally during the past years there are prominent USA specialists working there - which may explain how a same problem will find itself on both USAMO and BMO problem lists ...

mavropnevma wrote: First of all, is Saudi Arabia, a guest country to the Balkan Mathematical Olympiad, even allowed to submit problems? (I thought guest countries cannot even vote at motions of the International Jury) Well, problem 1 of 2010 BMO was proposed by Saudi Arabia. Check the site http://www.math.md/bmo2010/Files/Official_Results_BMO_2010_By_Countries.pdf to see how well they performed, as opposed to the other three problems.

Quote: Well, problem 1 of 2010 BMO was proposed by Saudi Arabia. Check the site http://www.math.md/bmo2010/Files/Official_Results_BMO_2010_By_Countries.pdf to see how well they performed, as opposed to the other three problems. It was very difficult to perform well on such a difficult problem If you look at other BMO 2010 problems, there is a huge difference in difficulty. Also SA team "invests" a lot in inequalities (a lot of problems involving them in their camps and trainings).

Quote: How do you know problems 2 and 4 were proposed by Saudi Arabia? First of all, is Saudi Arabia, a guest country to the Balkan Mathematical Olympiad, even allowed to submit problems? (I thought guest countries cannot even vote at motions of the International Jury); secondly, usually the authors are only disclosed later on. And also Problem 1 of 2011 BMO was proposed by United Kingdom. I do not know if guest countries can vote during problem selection???

delegat wrote: Quote: I do not know if guest countries can vote during problem selection??? No, guest countries cannot vote for the problem selection, but they can submit problems to the problem selection committee.

They can participate in the discussions, but not in the votes.

@mymath7: you are right, I made a typo mistake. mavropnevma wrote: How do you know problems 2 and 4 were proposed by Saudi Arabia? Almost everyone staying in BMO host hotel knows it.

So two of the four problems in the 2012 Balkan Mathematical Olympiad were proposed by the Saudi Arabia (trainers), one of which has been previously used in the USAMO. I suggest we move this competition in the Persian Gulf (and the Romanian Master of Mathematics in the Russian republic, and ...) As an unrelated topic, as of now, April 28, 23:30 Turkey hour, the problems have not been posted on the 2012 BMO site http://bmo2012.tubitak.gov.tr/problems, in the approved manner of the Cyprus 2011 JBMO and Kazakhstan 2010 IMO, where the sites are just nicely designed places to set pictures, speeches and other meta-competition tidbits of information, and not the very meat of it. Oh, well ...

mavropnevma wrote: hasan4444 wrote: This year, the second inequality was kinda easy as well. Most of the team got 10s. For this particular problem, only one student managed to get a 10 and the rest got nothing at all. Well, not many people got a 10 or close on Problem 4, so I only wish we will find Al Yazeed PASUNI's name on the list of medalists at the incoming IMO also ... In fact he is already a medalist in IMO 2011 and even he got almost problem 3 at that IMO and you can check that ..!!

mavropnevma wrote: hasan4444 wrote: This year, the second inequality was kinda easy as well. Most of the team got 10s. For this particular problem, only one student managed to get a 10 and the rest got nothing at all. Well, not many people got a 10 or close on Problem 4, so I only wish we will find Al Yazeed PASUNI's name on the list of medalists at the incoming IMO also ... mavropnevma , excuse me I have some kind of retardation . but what do you mean saying this ? do you mean that SAUDI ARABIA team's members are cheaters ? it's will be great if you had proof ,

mavropnevma wrote: hasan4444 wrote: This year, the second inequality was kinda easy as well. Most of the team got 10s. For this particular problem, only one student managed to get a 10 and the rest got nothing at all. Well, not many people got a 10 or close on Problem 4, so I only wish we will find Al Yazeed PASUNI's name on the list of medalists at the incoming IMO also ... Well, in proper understanding of the English language, my remark just says: - in reply to "only one student managed to get a 10 (on problem 4) and the rest got nothing at all", I said "Well, not many people got a 10 or close on Problem 4", meaning it was not expected from any country to get many scores of $10$ on problem 4, so that was to be expected; - then, I extended my best wishes to that student, by saying "I only wish we will find Al Yazeed PASUNI's name on the list of medalists at the incoming IMO also". Please, Abu3gab, do not put words in my mouth, where you may have a different impression.

mavropnevma wrote: ... so I only wish we will find Al Yazeed PASUNI's name on the list of medalists at the incoming IMO also ... It so happens that I do know Al Yazeed personally, and mathematically too, and I can guarantee you that he really is a student with a great potential/capacity in competitive math. Rest assured, he will certainly be on the medalists' list at the IMO, and hopefully even higher than last year's (if I am not wrong, he got the only bronze medal for Saudi Arabia).

Any news as to who proposed this problem?

I will offer a wild guess: Titu Andreescu

Wrong, and I guess I need to present quite a lot of apologies for this very unfortunate situation. I gave this problem to Titu in 2010, as a proposal for IMO 2010, where it was rejected (which was not a surprise for me, since I considered the problem almost trivial). The problem was proposed by Saudi Arabia, Titu being one of the coaches. Unfortunately, being occupied with my research I did not keep track of the problem, so I sent it again to Titu for USAMO 2012 (up to here nothing unfortunate, since the problem was not selected on the IMO Shortlist, so it was still unknown), as a very easy problem. Unfortunately, it was also sent to BMO by Saudi Arabia (being on their list of submitted problems for IMO) and neither of us seems to have taken time to ask the other one what he's doing with the problem. This terrible mistake actually taught me one thing: never submit again problems for mathematical competitions. So, my apologies for this very nasty situation, but I would like to clarify that there is absolutely nothing between me and the team of Saudi Arabia. Also, there has never been such a relation and there will never be (this is just to avoid some useless discussion).

harazi wrote: Wrong, and I guess I need to present quite a lot of apologies for this very unfortunate situation. Thank you, harazi, for your willingness to come forward, accept your mistake, and clarify what happened. This unfortunate situation occurred mainly because the two contests took place a mere few days apart; hence the coincidence of the problem escaped notice. However, the extreme proximity between the contests also meant that the damage was limited. Indeed, as mavropnevma pointed out, not many people got a 10 or close on this particular problem. I would like to extend my congratulations to you for this extremely nice (and, in my humble opinion, non-trivial) problem. Lastly, I do hope that you will continue to propose problems for competitions in the future.

harazi wrote: This terrible mistake actually taught me one thing: never submit again problems for mathematical competitions. Why so ? I agree this is a very infortunate situation, but it is so rare that I doubt that could happen more than once... And it would be a very great loss if you didn't submit problems to competitions anymore ...

harazi wrote: This terrible mistake actually taught me one thing: never submit again problems for mathematical competitions. So, my apologies for this very nasty situation... I recommend to only submit as many problems as your time allows to keep track of, Gabriel needs to prevail as great problem proposer!

As I see no one can catch solution from this link, so it is better to look up for solutions given the link below http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2669997&sid=a30c270e0ea868f1b6b6b3a82bf9aa9d#p2669997

The only solutions are $f\equiv1$, $f(x)=x$, and $f\equiv2$, which clearly work. Now I will show that no other function works. Note that $f(1)=\{1,2\}$ and $f(2)=\{1,2\}$. Let $P(m,n)$ denote the assertion that $m-n|f(m)-f(n)$. Case 1: $f(1)=f(2)=1$. If $k$ is a factorial, $P(k,1): k-1|f(k)-1 \implies f(k)\equiv1\pmod{k-1}$. $P(k,2):k-2|f(k)-1 \implies f(k)\equiv1\pmod{k-2}$. Since one of $\{k-1,k-2\}$ is even, we have $k$ is odd, so $k=1$. Since $f(n!)=1$ for all positive integers $n$, we have $\boxed{f\equiv1}$ is the only solution for this case. Case 2: $f(1)=1, f(2)=2$. $P(6,1): f(6)\equiv1\pmod5$ $P(6,2): f(6)\equiv2\pmod4$. Since $f(6)$ is a factorial, we have $f(6)=6$. So $f(3)=3$. Claim: We have infinitely many fixed points: Proof: Note that if $n$ is a fixed point, then $n!$ is also a fixed point. So $3!, 3!!, 3!!!. \ldots$ are all fixed points. Set $m$ as an arbitrarily large fixed point. $P(m,x): m-n|m-f(n) \implies m-n|n-f(n)$, so $n-f(n)=0\implies \boxed{f(x)=x}$ is the only solution for this case. ($n-f(n)=0$ because there are infinitely many values of $m$ we can choose, so $n-f(n)$ must have infinitely many divisors) Case 3: $f(1)=2, f(2)=1$. Let $k$ be an even factorial. $P(k,2): f(k)\equiv1\pmod{k-2}$. Since $k-2$ is even, $f(k)$ is odd, so $f(k)=1$. But from $P(k,1)$, we also have $k-1|f(k)-2\implies f(k)\equiv2\pmod{k-1}$, so $1\equiv2\pmod{k-1}$, which is not possible for $k>2$. So there are no solutions for this case. Case 4: $f(1)=f(2)=2$. Let $k$ be a factorial. $P(6,1): f(6)\equiv2\pmod5$. $P(6,2): f(6)\equiv2\pmod4$. Since $f(6)$ is a factorial and $f(6)\equiv2\pmod{20}$, $f(6)=2$. Let $k$ be a factorial that is divisible by $4$. $P(6,k): 6-k|2-f(k)$, so $3|2-f(k)$, so $f(k)\equiv2\pmod3$. Thus, $f(k)=2$. If $f(n!)=2$, then $f(n)=2$, so the only solution for this case is $\boxed{f\equiv2}$.

Let $P(m,n)$ be the assertion $m-n\mid f(m)-f(n)$. If $x=x!$ for $x\in\mathbb N$ then $x\in\{1,2\}$. We can prove this by noting that if $x\ge3$ then $x!\ge x(x-1)>x$. This easily gives that $\{f(1),f(2)\}\subseteq\{1,2\}$. $\textbf{Case 1: }f(2)=1$ For $x>2$, we have: $P(x!,2)\Rightarrow x!-2\mid f(x)!-1\Rightarrow2\mid f(x)!-1\Rightarrow2\nmid f(x)!$ since $x!-2$ is even, so $f(x)=1\forall x\ge2$. Also, for large enough $x$: $P(x,1)\Rightarrow x-1\mid1-f(1)\Rightarrow f(1)=1$, so $\boxed{f(x)=1}$, which works. $\textbf{Case 2: }f(2)=2,f(1)=1$ $P(6,1)\Rightarrow5\mid f(6)-1\Rightarrow f(3)!\equiv1\pmod5\Rightarrow f(3)\in\{1,3\}$ since $f(3)<5$ (we can just test each factorial residue). Assume FTSOC that $f(3)=1$. Then $f(m)=1$ for arbitrarily large $m$ ($3!,(3!)!,\ldots$), so: $P(m,x)\Rightarrow m-x\mid1-f(x)\Rightarrow f(x)=1$ This is absurd, so we must have $f(3)=3$. By induction, we can show that $f$ has arbitrarily large fixed points at $3,3!,(3!)!$, etc. If $f$ has a sufficiently large fixed point $m$, then: $P(m,x)\Rightarrow m-x\mid m-f(x)\Rightarrow m-x\mid x-f(x)\Rightarrow\boxed{f(x)=x}$, which works. $\textbf{Case 3: }f(2)=2,f(1)=2$ $P(6,1)\Rightarrow5\mid f(6)-2\Rightarrow f(3)!\equiv2\pmod5\Rightarrow f(3)=2$ since $f(3)<5$ here. Then $f(3!)=2,f((3!)!)=2$, etc. Let $m$ be sufficiently large such that $f(m)=2$. $P(m,x)\Rightarrow m-x\Rightarrow2-f(x)\Rightarrow\boxed{f(x)=2}$, this also works.

Posted here as well as the USAMO thread for completeness... We claim the only working functions are $f(x) = 1 \forall x$, $f(x) = 2 \forall x$ and $f(x) = x \forall x$, all of which work. Now we prove they are the only working functions. Note that $f(1), f(2) \in \{1, 2\}$. We now take four cases, in order of how interesting they are. Case 1: $f(1) = 2, f(2) = 1$. Note that $4 = 3!-2 \mid f(3)! - f(2) = f(3)! - 1 \implies f(3) = 1$. But then we have $2 = 3-1 \mid f(3) - f(1) = -1$, which isn't true. Therefore this case yields no solution. Case 2a: $f(1) = f(2) = 1$. Note that we have $4 \mid f(3)!-1 \implies f(3) = 1 \implies f(720) = f((3!)!) = (f(3)!)! = 1$. Now for $n \ge 4$, $4 \mid 720 -- n! \mid f(720) - f(n)! = 1-f(n)! \implies f(n) = 1$. Thus $f\equiv 1$. Case 2b: $f(1) = f(2) = 2$. Similarly, $4 \mid f(3)! - 2 \implies f(3) \in \{ 2, 3\}$. Now assume $f(3) = 3$. Then we have $2 = 3-1 \mid f(3) - f(1) = 1$, contradiction. So $f(3) = 2$ $\implies f(720) = f((3!)!) = 2$. Now for $n \ge 4$, we have $4 \mid 720 - n! \mid f(720) - f(n)! = 2-f(n)! \implies f(n) = 2$. Thus $f \equiv 2$. Case 3: $f(1) = 1, f(2) = 2$. Note that once again $4 \mid f(3)! - 2 \implies f(3) \in \{ 2, 3\}$. Now if $f(3) = 2$, then $2 = 3-1 \mid f(3)-f(1) = 1$, contradiction, so $f(3) = 3$. This is where the boring part ends, and the fun part begins. We let $g(n) = n!, g^k(n) = g(g^{k-1}(n))$. (So for example $g^2(3) = 720$.) Note that the sequence $\{g(3), g^2(3), g^3(3), \dots\}$ has infinitely many terms in it. Further, $f(g^k(3)) = g^k(3)$ as $f(3) = 3$. Then we have $g^k(3) - n \mid g^k(3) - f(n) \implies g^k(3) - n \mid n-f(n)$. But then $n- f(n)$ has infinitely many divisors, so $f(n) = n \forall n$. These three cases overall must give us all the solutions, so we are done.