IMO Marathon

Discussion on International Mathematical Olympiad (IMO)
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Tahmid Hasan
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Re: IMO Marathon

Unread post by Tahmid Hasan » Mon Nov 12, 2012 7:09 pm

FahimFerdous wrote:Tahmid, I proved your hint but still can't use it. Give a link please.
Since no one posted any solution, I'm giving it.
Using the hint we get $\frac{BT}{TC}=\frac{BP}{CQ}$.
Let $PQ$ intersect $BC$ at $U$.
Then applying Manelaus' theorem on $\triangle ABC$ with transversal $PQ$ we get
$\frac{UB}{UC}=\frac{PB}{AP},\frac{AQ}{CQ}=\frac{BT}{CT}$
which implies $MT$ bisects $\angle BTC$ externally.
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Re: IMO Marathon

Unread post by Phlembac Adib Hasan » Mon Nov 12, 2012 8:03 pm

Masum wrote:
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 $(m-x)wm^{w-1}\equiv m-x(\bmod \; q)$
This is what I needed to say $v_q(m^w-x^w)=v_q(m-x)<v_q(n^w)$.
And can't I write it? Because $w, m^{w-1}$ are co-primes with $q$.
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Re: IMO Marathon

Unread post by Phlembac Adib Hasan » Mon Nov 12, 2012 8:19 pm

FahimFerdous wrote: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
Solution:
Assume that $CD$ meets $AB$ at $W$. Let $F'$ be the mid-point of $AF$ and $M$ is that of $AC$. Now apply cosine rules to show $DFF'$ is an equilateral triangle. So $\angle DBA=\angle DF'F=60^{\circ}$ and $ABDF'$ is cyclic.
Since $\angle A+\angle B+\angle C=180^{\circ}$, we must have $\angle DAB+\angle DBC+\angle DCB=60^{\circ}$
In other words $\angle WDB+\angle DAB=60^{\circ}$
So $\angle EFD=\angle EFA-60^{\circ}=\angle BF'A-60^{\circ} =\angle BDA-60^{\circ}=\angle BDW=60^{\circ}-\angle DAB$
At the same time $\angle EMD=\angle DXA\; (X=MD\cap AB)\; =90^{\circ}-\angle A=60^{\circ}-\angle DAB$
Therefore $\angle EFD=\angle EMD$ and $MDEF$ is cyclic. So $\angle DEF=180^{\circ}-\angle DMF=90^{\circ}$
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Re: IMO Marathon

Unread post by Phlembac Adib Hasan » Mon Nov 12, 2012 8:33 pm

Tahmid Hasan wrote:
FahimFerdous wrote:Tahmid, I proved your hint but still can't use it. Give a link please.
Since no one posted any solution, I'm giving it.
Using the hint we get $\frac{BT}{TC}=\frac{BP}{CQ}$.
Let $PQ$ intersect $BC$ at $U$.
Then applying Manelaus' theorem on $\triangle ABC$ with transversal $PQ$ we get
$\frac{UB}{UC}=\frac{PB}{AP},\frac{AQ}{CQ}=\frac{BT}{CT}$
which implies $MT$ bisects $\angle BTC$ externally.
Hello, bro, can't you share the link? Or is there any taboo on it? We all want to know this!
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Re: IMO Marathon

Unread post by sourav das » Mon Nov 12, 2012 9:39 pm

Sorry for being late. Today was chemistry 2nd part exam.So I was busy.(তারপরও *** খাইসি)

Finally, I've found my own solution of problem 4:

Let $BT,CT$ intersect $\omega$ at $F,G$. Since the homothety from $T$
sends $F,G$ to $B,C$; So, $FG||BC$ and so $\frac{TF}{TB}=\frac{TG}{TC}$
Note that $MT$ is the external angle bisector of $\angle BTC$. Let $MT\cap BC=E$
Then $\frac{BE}{CE}=\frac{BT}{CT}$..........(i).

Now using power of point $CQ^2=CG.CT$ and $BP^2=BF.BT$ . It means that
$\frac{CQ^2}{BP^2}=\frac{CG.CT}{BF.BT}=\frac{CT^2}{BT^2}$
Thus, $\frac{CQ}{BP}=\frac{CT}{BT}$.........(ii)

Now using Manulus theorem in $\triangle ABC$ using (i),(ii),
$\frac{AP}{PB} \frac{BE}{EC} \frac{CQ}{QA}=\frac{AP.BE.CQ}{QA.EC.PB}=1 $
$P,Q,E$ are collinear. Hence $MT,BC,PQ$ are concurrent . (Proved)

I thought the solution would be ugly, but It's Pankha (পাঙ্খা)
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SANZEED
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Re: IMO Marathon

Unread post by SANZEED » Mon Nov 12, 2012 10:29 pm

New problem!!
Problem $7$
Find all positive integers $n$ such that there exists $2n$ different positive integers $x_{1} , ... , x_{n} , y_{1} , ... , y_{n}$, such that $(11x_{1}^{2} + 12y_{1}^{2})\cdot ...\cdot (11x_{n}^{2} + 12y_{n}^{2})$ is a perfect square.

Source: Argentina National Round $2012$
Last edited by SANZEED on Mon Nov 12, 2012 10:36 pm, edited 1 time in total.
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Tahmid Hasan
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Re: IMO Marathon

Unread post by Tahmid Hasan » Mon Nov 12, 2012 10:32 pm

Phlembac Adib Hasan wrote: Hello, bro, can't you share the link? Or is there any taboo on it? We all want to know this!
Absolutely. Here's the link http://www.artofproblemsolving.com/Foru ... 6&t=399496
The proof provided by Mr. González only holds when internally tangent but the requirement is also satisfied when externally tangent. So I edited the problem a little. :)
Actually I learned the hint in this excalibur PDF
http://www.math.ust.hk/excalibur/v16_n5.pdf
By the way, cool solution Sourav vai, good use of the homothety. If you notice closely, the general Casey's chord theorem can be proved using homothety. So thanks for using the basic method.
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Re: IMO Marathon

Unread post by Masum » Tue Nov 13, 2012 1:39 am

Phlembac Adib Hasan wrote:
Masum wrote:
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 $(m-x)wm^{w-1}\equiv m-x(\bmod \; q)$
This is what I needed to say $v_q(m^w-x^w)=v_q(m-x)<v_q(n^w)$.
And can't I write it? Because $w, m^{w-1}$ are co-primes with $q$.
Yeah, you can write this, just not $(m-x)wm^{w-1}\equiv m-x(\bmod \; q)$ this. Because you think what would that mean?
One one thing is neutral in the universe, that is $0$.

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SANZEED
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Re: IMO Marathon

Unread post by SANZEED » Wed Nov 14, 2012 11:44 pm

I think it's time for the solution of Problem $7$.I am really grateful to Ignacio Darago from Argentina for the help from him about this problem.
I claim that for all even $n$ there exist numbers, and for odd $n$ there don't.
Indeed, let's explore the case $n = 2$. Due to Cauchy-Schwarz inequality
$(11x_{1}^{2}+12y_{1}^{2})(11x_{2}^{2}+12y_{2}^{2})\geq (11x_{1}x_{2} + 12y_{1}y_{2})^{2}$ with equality if and only if $\frac{x_{1}}{x_{2}}=\frac{y_{1}}{y_{2}}= k$ with constant $k$. We take the equality case as it's an integer
and a perfect square. Then we have infinite quadruples $(x_{1},x_{2},y_{1},y_{2})$ for which it's a perfect square.
For all even $n$, we can divide it in many products of two numbers, each of which is a perfect square.
Thus, for even $n$ we can accomplish our goal.
It only remains to prove that it's impossible for odd $n$. Suppose there exists tuples $(x_{1},.....,x_{n},y_{1},.....,y_{n})$, and take the one with the least sum. Let's look at the product modulo $3$. We have, $2x_{1}^{2}\cdot .....\cdot 2x_{n}^{2}\equiv 2^{n}\cdot x_{1}^{2}\cdot ...\cdot x_{n}^{2}$
$\equiv 2\cdot x_{1}^{2}\cdot ...\cdot x_{n}^{2}\equiv c^{2}(mod 3)$
It's trivial that $x^{2}\equiv 0,1(mod 3)$. Then, if $3\not |x_{i}$ for all $i$, we have that ,
$2\cdot x_{1}^{2}\cdot ...\cdot x_{n}^{2}\equiv 2 (mod 3)$, which is not a quadratic residue. Then $3|x_{i}$ for some $i$, and so $x_{i} = 3a_{i}$.
Now, let's analyze the factor $(11x_{i}^{2}+12y_{i}^{2})$ . It's clear that $3\not |y_{i}$, because if it divided it we could have the same tuple with $\frac{x_{i}}{3}$ and $\frac{y_{i}}{3}$,, which would still be a perfect square but would have less sum. Then $3\parallel (11x_{i}^{2} +12y_{i}^{2})$.
As the product must be a perfect square, we must have that $3$ divides the product of all other factors except $11x_{i}^{2}+12y_{i}^{2}$ . Looking modulo $3$ we have $3|x_{k}$ for some $k$, and so $x_{k} = 3a_{k}$. We have again that $3\parallel (11x_{k}^{2}+12y_{k}^{2})$. Consider the whole product, but divide the factors $(11x_{i}^{2}+12y_{i}^{2})$ and $(11x_{k}^{2}+12y_{k}^{2})$ by $3$. We have:
$(11x_{1}^{2}+12y_{1}^{2})...(11\cdot 3a_{i}^{2} + 4y_{i}^{2})...(11\cdot 3a_{k}^{2}+4y_{k}^{2})...(11x_{n}^{2}+12y_{n}^{2}=c'^{2}$.
Looking modulo $3$ again, we will see $3$ divides some $x_{\ell}$, with $\ell \neq i,k$. If we continue with this procedure,we will always find pairs of numbers $(x_{a},x_{b})$ such that $3$ divides both of them and we discard them.But as $n$ is odd, there will always remain a number in the end. We've reached a contradiction and we're
done.
P.S. Someone post a new problem here please!!
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Tahmid Hasan
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Re: IMO Marathon

Unread post by Tahmid Hasan » Thu Nov 15, 2012 10:13 pm

Problem $8$:Let $ABC$ be a triangle such that $AB \neq AC$. The internal bisector lines of the angles $ABC$ and $ACB$ meet the opposite sides of the triangle at points $B_0$ and $C_0$, respectively, and the circumcircle $ABC$ at points $B_1$ and $C_1$, respectively. Further, let $I$ be the incentre of the triangle $ABC$. Prove that the lines $B_0C_0$ and $B_1C_1$ meet at some point lying on the parallel through $I$ to the line $BC$.
Source:Romania TST-2010-Exam-3-Problem-2.
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