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Re: IMO Marathon

Posted: Mon Nov 12, 2012 7:09 pm
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.

Re: IMO Marathon

Posted: 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 $q$.

Re: IMO Marathon

Posted: 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 $\angle DEF=180^{\circ}-\angle DMF=90^{\circ}$

Re: IMO Marathon

Posted: Mon Nov 12, 2012 8:33 pm
Tahmid Hasan wrote:
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!

Re: IMO Marathon

Posted: 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)

Re: IMO Marathon

Posted: 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 $2012$

Re: IMO Marathon

Posted: 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.

Re: IMO Marathon

Posted: Tue Nov 13, 2012 1:39 am
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?

Re: IMO Marathon

Posted: 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.
P.S. Someone post a new problem here please!!

Re: IMO Marathon

Posted: 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.