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### Re: IMO Marathon

Posted: Mon Nov 12, 2012 12:21 am
SANZEED wrote:Can anyone confirm if this is the correct figure?
Yes but there are four possible constructions- two when externally tangent two when internally tangent.[Maybe!]
For each of which there are cases when $\omega$ touches $AB,AC$ on segments or on extensions.
So to avoid complications I would recommend using directed lenghs.
Since no one posted a solution, here's a little hint to shed some light:
Let $A$ be a point on $\odot \alpha$. A circle $\beta$ is tangent (internally or externally) to $\alpha$ with tangency point $B$. Let $\ell$ be the length of the tangent from $A$ to $\beta$. Let the radius of $\alpha,\beta$ be $R,r$ respectively. Prove that $AB=\ell.\sqrt{\frac{R}{R \pm r}}$[The sign is negative when internally tangent and positive when externally tangent.]
Use the hint, the fact whether $K,\omega$ are externally or internally tangent or whether $\omega$ touches $AB,AC$ on segments or extension won't matter.

### Re: IMO Marathon

Posted: Mon Nov 12, 2012 12:46 am
SANZEED wrote: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.
Good job.
It was a problem from Iran.

### Re: IMO Marathon

Posted: Mon Nov 12, 2012 2:10 am
I think it's time for a new problem.
Problem 5:
If $a^p-b^p$ is an integer for every prime $p$ and rational $a,b$ then $a,b$ is integer.
This is a modified version of a recent AoPS problem.

### Re: IMO Marathon

Posted: Mon Nov 12, 2012 10:44 am
@Tahmid vai, by "source", I meant where you have seen the problem. It can be a book with page number or a link. It can be useful when someone is personally interested in a problem and look for further information. So please give a link to problem 4. Moderators, please include this in the rules.
Masum wrote:I think it's time for a new problem.
Problem 5:
If $a^p-b^p$ is an integer for every prime $p$ and rational $a,b$ then $a,b$ is integer.
This is a modified version of a recent AoPS problem.
I had solved the AoPS version in the IMO camp, anyway.
Solution:(You should have mentioned $a\neq b$)
Let $a=m/n, b=x/y$ where $m,x\in \mathbb Z$, $n,y\in \mathbb N$ and $(m,n)=(x,y)=1$
So $a^p-b^p=\displaystyle \frac {(my)^p-(nx)^p}{(ny)^p}\Rightarrow n|my\Rightarrow n|y$ and similarly $y|n$. So $n=y$.
Now re-write $a^p-b^p=\displaystyle \frac {m^p-x^p}{n^p}$
Let $q$ be any prime such that $q|n$. So Fermat's little theorem asserts $q|m-x.....(i)$
Let $n^r||m-x$
Take a very large prime $w$ such that $w>n^{101r}$.
$m^w-x^w=(m-x)(m^{w-1}+m^{w-2}x+...+x^{w-1})\equiv (m-x)(m^{w-1}+m^{w-2}\times m$ $+...+m^{w-1}) =(m-x)wm^{w-1}\equiv m-x(\bmod \; q)$
So $v_q(m^w-x^w)=v_q(m-x)<v_q(n^w)$
A contradiction. So $n|m$ and $n|x$.

### Re: IMO Marathon

Posted: Mon Nov 12, 2012 2:20 pm
Tahmid, I proved your hint but still can't use it. Give a link please.

### Re: IMO Marathon

Posted: Mon Nov 12, 2012 2:27 pm
I am taking the liberty and posting a problem. :/
Problem 6:
Point $D$ lies inside triangle $ABC$ such that $\angle DAC =\angle DCA = 30^{\circ}$ and $\angle DBA = 60^{\circ}$. Point $E$ is the midpoint of segment $BC$. Point $F$ lies on segment $AC$ with $AF = 2FC$. Prove that $DE\perp EF$.

Source: http://www.artofproblemsolving.com/Foru ... d#p1358815

### Re: IMO Marathon

Posted: Mon Nov 12, 2012 2:38 pm
Phlembac Adib Hasan wrote: Let $q$ be any prime such that $q|n$. So Fermat's little theorem asserts $q|m-x.....(i)$
Not really. The exponent is not $q$, it's $p$. So it does not follow from Fermat's theorem, at least not directly. Also you didn't say anything about $\gcd(m,x)$. A lot depends on this gcd. You can't conclude anything about the exponent of $q$ in $m^w-x^w$ without saying $\gcd(m,x)=1$. This point is not much important. But the first one is important. Fix that.

### Re: IMO Marathon

Posted: Mon Nov 12, 2012 4:27 pm
Masum wrote:
Phlembac Adib Hasan wrote: Let $q$ be any prime such that $q|n$. So Fermat's little theorem asserts $q|m-x.....(i)$
Not really. The exponent is not $q$, it's $p$. So it does not follow from Fermat's theorem, at least not directly. Also you didn't say anything about $\gcd(m,x)$. A lot depends on this gcd. You can't conclude anything about the exponent of $q$ in $m^w-x^w$ without saying $\gcd(m,x)=1$. This point is not much important. But the first one is important. Fix that.
Why? Can't I skip that part in such a forum? Since $q|n$, we have $q|m^q-x^q$ and $(m,n)=(x,n)=1$, So $m^q\equiv m(\bmod \; q)$ and $x^q\equiv x(\bmod \; q)$. So $0\equiv m^q-x^q\equiv m-x(\bmod \; q)$.

### Re: IMO Marathon

Posted: Mon Nov 12, 2012 6:53 pm
OW, not so fast. The expression is $q|m^p-x^p$, not $q|m^q-x^q$. So I actually find that a bit confusing. I think you need to deal with this in this way:
Let $e$ be the smallest positive integer such that $m^e\equiv x^e\pmod q$. Then we have $e|p$ for all prime, giving $e=1$. And the fact is it is not actually obvious that $\gcd(m,x)=1$ if you understand. So it would be rather considered as a common mistake if you didn't prove it. The fact if even if $\gcd(m,x)=g$, we can't have $q|g$ for the sake of $\gcd(m,n)=1$. So that $g$ won't matter.
What I meant is, these two facts weren't obvious, rather you might make a mistake like $a|bc,\gcd(b,c)=1\Longrightarrow a|b$ or $a|c$. So I had to make sure. The part wasn't about skipping in this forum.

### Re: IMO Marathon

Posted: Mon Nov 12, 2012 7:08 pm
Phlembac Adib Hasan wrote:$(m-x)wm^{w-1}\equiv m-x(\bmod \; q)$
I don't find this correct. This only means $\dfrac{m^w-x^w}{m-x}$ is not divisible by $q$, not $(m-x)wm^{w-1}\equiv m-x(\bmod \; q)$
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